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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth
+%                                                                         %
+% This eBook is for the use of anyone anywhere at no cost and with        %
+% almost no restrictions whatsoever.  You may copy it, give it away or    %
+% re-use it under the terms of the Project Gutenberg License included     %
+% with this eBook or online at www.gutenberg.org                          %
+%                                                                         %
+%                                                                         %
+% Title: Plane Geometry                                                   %
+%                                                                         %
+% Author: George Albert Wentworth                                         %
+%                                                                         %
+% Release Date: July 3, 2010 [EBook #33063]                               %
+%                                                                         %
+% Language: English                                                       %
+%                                                                         %
+% Character set encoding: ISO-8859-1                                      %
+%                                                                         %
+% *** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***            %
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+
+%%%% PG BOILERPLATE %%%%
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+
+\begin{center}
+\begin{minipage}{\textwidth}
+\small
+\begin{PGtext}
+The Project Gutenberg EBook of Plane Geometry, by George Albert Wentworth
+
+This eBook is for the use of anyone anywhere at no cost and with
+almost no restrictions whatsoever.  You may copy it, give it away or
+re-use it under the terms of the Project Gutenberg License included
+with this eBook or online at www.gutenberg.org
+
+
+Title: Plane Geometry
+
+Author: George Albert Wentworth
+
+Release Date: July 3, 2010 [EBook #33063]
+
+Language: English
+
+Character set encoding: ISO-8859-1
+
+*** START OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***
+\end{PGtext}
+\end{minipage}
+\end{center}
+
+\clearpage
+
+
+%%%% Credits and transcriber's note %%%%
+\begin{center}
+\begin{minipage}{\textwidth}
+\begin{PGtext}
+Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy
+and the Online Distributed Proofreading Team at
+http://www.pgdp.net
+\end{PGtext}
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+\end{center}
+\vfill
+
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+
+%%%%%%%%%%%%%%%%%%%%%%%%%%% FRONT MATTER %%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\frontmatter
+\pagestyle{fancy}
+\thispagestyle{empty}
+
+\begin{titlepage}
+\null\vfil
+\begin{center}
+{\LARGE PLANE GEOMETRY \par}
+\vspace{3em}
+{BY \par}
+{\large G.A. WENTWORTH \par}
+\textsc{Author of a Series of Text-Books in Mathematics}
+\vspace{3em}
+
+\emph{REVISED EDITION}
+
+\vspace{3em}
+
+\vfil\null
+
+{\large GINN \& COMPANY}
+
+BOSTON · NEW YORK · CHICAGO · LONDON
+
+\end{center}
+\end{titlepage}
+
+\newpage
+\thispagestyle{empty}
+
+\begin{center}
+\null\vfil
+Entered, according to Act of Congress, in the year 1888, by
+
+G.A. WENTWORTH
+
+in the Office Of the Librarian of Congress, at Washington
+
+\smallrule
+
+\textsc{Copyright, 1899}
+
+\textsc{By G.A. WENTWORTH}
+
+\smallrule
+
+{\small ALL RIGHTS RESERVED}
+
+{\small 67 10}
+
+\null\vfil
+
+The Athenæum Press
+\smallrule
+
+\settowidth{\TmpLen}{PRIETORS · BOSTON · U.S.A.}%
+\parbox{\TmpLen}{GINN \& COMPANY · PRO\-PRIETORS ·
+BOSTON · U.S.A.}
+\end{center}
+
+\cleardoublepage
+\SetRunningHeads
+
+\section*{PREFACE.}
+\pdfbookmark[0]{PREFACE.}{PREFACE}
+
+Most persons do not possess, and do not easily acquire, the power of
+abstraction requisite for apprehending geometrical conceptions, and for
+keeping in mind the successive steps of a continuous argument. Hence,
+with a very large proportion of beginners in Geometry, it depends mainly
+upon the \emph{form} in which the subject is presented whether they pursue the
+study with indifference, not to say aversion, or with increasing interest
+and pleasure.
+
+Great care, therefore, has been taken to make the pages attractive.
+The figures have been carefully drawn and placed in the middle of
+the page, so that they fall directly under the eye in immediate connection
+with the text; and in no case is it necessary to turn the page in
+reading a demonstration. Full, long-dashed, and short-dashed lines of
+the figures indicate given, resulting, and auxiliary lines, respectively.
+Bold-faced, italic, and roman type has been skilfully used to distinguish
+the hypothesis, the conclusion to be proved, and the proof.
+
+As a further concession to the beginner, the reason for each statement
+in the early proofs is printed in small italics, immediately following the
+statement. This prevents the necessity of interrupting the logical train
+of thought by turning to a previous section, and compels the learner to
+become familiar with a large number of geometrical truths by constantly
+seeing and repeating them. This help is gradually discarded, and the
+pupil is left to depend upon the knowledge already acquired, or to find
+the reason for a step by turning to the given reference.
+
+It must not be inferred, because this is not a geometry of interrogation
+points, that the author has lost sight of the real object of the study.
+The training to be obtained from carefully following the logical steps
+of a complete proof has been provided for by the Propositions of the
+\scanpage{005.png}%
+Geometry, and the development of the power to grasp and prove new
+truths has been provided for by original exercises. The chief value of
+any Geometry consists in the happy combination of these two kinds
+of training. The exercises have been arranged according to the test
+of experience, and are so abundant that it is not expected that any
+one class will work them all out. The methods of attacking and proving
+original theorems are fully explained in the first Book, and illustrated
+by sufficient examples; and the methods of attacking and solving
+original problems are explained in the second Book, and illustrated by
+examples worked out in full. None but the very simplest exercises are
+inserted until the student has become familiar with geometrical methods,
+and is furnished with elementary but much needed instruction in the
+art of handling original propositions; and he is assisted by diagrams
+and hints as long as these helps are necessary to develop his mental
+powers sufficiently to enable him to carry on the work by himself.
+
+The law of converse theorems, the distinction between positive and
+negative quantities, and the principles of reciprocity and continuity
+have been briefly explained; but the application of these principles is
+left mainly to the discretion of teachers.
+
+The author desires to express his appreciation of the valuable suggestions
+and assistance which he has received from distinguished educators
+in all parts of the country. He also desires to acknowledge his obligation
+to Mr.~Charles Hamilton, the Superintendent of the composition
+room of the Athen\ae{}um Press, and to Mr.~I.~F. White, the compositor,
+for the excellent typography of the book.
+
+Criticisms and corrections will be thankfully received.
+
+\hfill G.~A. WENTWORTH.
+
+\textsc{Exeter}, N.H., June, 1899.
+\scanpage{006.png}%
+
+
+\clearpage
+\section*{NOTE TO TEACHERS.}
+\pdfbookmark[0]{NOTE TO TEACHERS.}{NOTE TO TEACHERS}
+
+It is intended to have the first \PageName\ pages of this book simply read
+in the class, with such running comment and discussion as may be useful
+to help the beginner catch the spirit of the subject-matter, and not
+leave him to the mere letter of dry definitions. In like manner, the
+definitions at the beginning of each Book should be read and discussed
+in the recitation room. There is a decided advantage in having the
+definitions for each Book in a single group so that they can be included
+in one survey and discussion.
+
+For a similar reason the theorems of limits are considered together.
+The subject of limits is exceedingly interesting in itself, and it was
+thought best to include in the theory of limits in the second Book every
+principle required for Plane and Solid Geometry.
+
+When the pupil is reading each Book for the first time, it will be
+well to let him write his proofs on the blackboard in his own language,
+care being taken that his language be the simplest possible,
+that the arrangement of work be vertical, and that the figures be
+accurately constructed.
+
+This method will furnish a valuable exercise as a language lesson,
+will cultivate the habit of neat and orderly arrangement of work, and
+will allow a brief interval for deliberating on each step.
+
+After a Book has been read in this way, the pupil should review
+the Book, and should be required to draw the figures free-hand. He
+should state and prove the propositions orally, using a pointer to indicate
+on the figure every line and angle named. He should be encouraged
+in reviewing each Book, to do the original exercises; to state
+the converse propositions, and determine whether they are true or
+false; and also to give well-considered answers to questions which
+may be asked him on many propositions.
+\scanpage{007.png}%
+
+The Teacher is strongly advised to illustrate, geometrically and
+arithmetically, the principles of limits. Thus, a rectangle with a constant
+base $b$, and a variable altitude $x$, will afford an obvious illustration
+of the truth that the product of a constant and a variable is
+also a variable; and that the limit of the product of a constant and a
+variable is the product of the constant by the limit of the variable.
+If $x$ increases and approaches the altitude $a$ as a limit, the area of the
+rectangle increases and approaches the area of the rectangle $ab$ as a
+limit; if, however, $x$ decreases and approaches zero as a limit, the area
+of the rectangle decreases and approaches zero as a limit.
+
+An arithmetical illustration of this truth may be given by multiplying
+the approximate values of any repetend by a constant. If, for example,
+we take the repetend $0.3333$ etc., the approximate values of the repetend
+will be $\frac{3}{10}$, $\frac{33}{100}$, $\frac{333}{1000}$,
+$\frac{3333}{10000}$, etc., and these values multiplied by $60$
+give the series $18$, $19.8$, $19.98$, $19.998$, etc., which evidently approaches
+$20$ as a limit; but the product of $60$ into $\frac{1}{3}$ (the limit of the repetend
+$0.333$ etc.) is also $20$.
+
+Again, if we multiply $60$ into the different values of the decreasing
+series  $\frac{1}{30}$, $\frac{1}{300}$, $\frac{1}{3000}$, $\frac{1}{30000}$,
+etc., which approaches zero as a limit, we
+shall get the decreasing series $2$, $\frac{1}{5}$, $\frac{1}{50}$,
+$\frac{1}{500}$, etc.; and this series evidently
+approaches zero as a limit.
+
+The Teacher is likewise advised to give frequent written examinations.
+These should not be too difficult, and sufficient time should be
+allowed for accurately constructing the figures, for choosing the best
+language, and for determining the best arrangement.
+
+The time necessary for the reading of examination books will be
+diminished by more than one half, if the use of symbols is allowed.
+
+\textsc{Exeter}, N.H., 1899.
+\scanpage{008.png}%
+
+
+\clearpage
+\phantomsection\pdfbookmark[0]{TABLE OF CONTENTS.}{CONTENTS}
+\tableofcontents
+
+\mainmatter
+\scanpage{009.png}%
+\scanpage{010.png}%
+
+
+\phantomsection%
+\part{GEOMETRY.}
+\markboth{GEOMETRY.}{} % fake a chapter heading, no chapters yet
+
+\section{INTRODUCTION.}
+
+\begin{point}%
+If a block of wood or stone is cut in the shape represented
+in Fig.~1, it will have \emph{six flat
+faces}.
+
+Each face of the block is called a
+surface\label{surface}; and if the faces are made
+smooth by polishing, so that, when a
+straight edge is applied to any one
+of them, the straight edge in every
+part will touch the surface, the faces
+are called \textbf{plane surfaces}, or \indexbf{planes}.
+
+% [Illustration: Fig. 1.]
+\centering{\includegraphics{./images/woodcutsmall.pdf}}
+
+\centering{\small\textsc{Fig. 1.}}
+\end{point}
+
+\pp{The intersection\label{intersection} of any two of these surfaces is called
+a \indexbf{line}.}
+
+\pp{The intersection of any three of these lines is called
+a \indexbf{point}.}
+
+\begin{point}%
+The block extends in three principal directions:
+
+\begin{list}{}{\setlength{\itemsep}{0pt}}
+\item From left to right, $A$ to $B$.
+
+\item From front to back, $A$ to $C$.
+
+\item From top to bottom, $A$ to $D$.
+\end{list}
+
+These are called the \textbf{dimensions}\label{dimensions} of the block, and are named
+in the order given, \textbf{length}, \textbf{breadth} (or \textit{width}), and \textbf{thickness}
+(\textit{height} or \textit{depth}).
+\end{point}
+\scanpage{011.png}%
+
+\begin{point}%
+A \textbf{solid}, in common language, is a limited portion of
+space \textit{filled with matter}; but in Geometry we have nothing
+to do with \textit{the matter} of which a body is composed; we study
+simply its \textit{shape} and \textit{size}; that is, we regard a solid as a
+limited portion of space which may be occupied by a physical
+body, or marked out in some other way. Hence,
+
+\textit{A geometrical solid\label{geometrical1} is a limited portion of space.}
+\end{point}
+
+\begin{point}%
+The surface\label{surface2} of a solid is simply the boundary of the
+solid, that which separates it from surrounding space. The
+surface is no part of a solid and has no thickness. Hence,
+
+\textit{A surface has only two dimensions, length and breadth.}
+\end{point}
+
+\begin{point}%
+A line\label{line2} is simply a boundary of a surface, or the intersection
+of two surfaces. Since the surfaces have no thickness,
+a line has no thickness. Moreover, a line is no part of a
+surface and has no width. Hence,
+
+\textit{A line has only one dimension, length.}
+\end{point}
+
+\begin{point}%
+A point\label{point2} is simply the extremity of a line, or the intersection
+of two lines. A point, therefore, has no thickness,
+width, or length; therefore, no magnitude. Hence,
+
+\textit{A point has no dimension, but denotes position simply.}
+\end{point}
+
+\begin{point}%
+It must be distinctly understood at the outset that the
+points, lines, surfaces, and solids of Geometry are \textit{purely ideal},
+though they are represented to the eye in a material way.
+Lines, for example, drawn on paper or on the blackboard, will
+have some width and some thickness, and will so far fail of
+being \textit{true lines}; yet, when they are used to help the mind
+in reasoning, it is assumed that they represent true lines,
+without breadth and without thickness.
+\end{point}
+\scanpage{012.png}%
+
+\figc{012aaZ10}{}
+\begin{point}%
+A point is \textit{represented} to the eye by a fine dot, and
+named by a letter, as $A$ (Fig.~2). A line is named by two
+letters, placed one at each end, as $BF$.
+A surface\label{surface3} is represented and named by
+the lines which bound it, as $BCDF$. A
+solid\label{geometrical2} is represented by the faces which
+bound it.
+\end{point}
+
+\pp{A point in space may be considered by itself, without
+reference to a line\label{line3}.}
+
+\pp{If a point moves in space, its path is a line. This line
+may be considered apart from the idea of a surface.}
+
+\pp{If a line moves in space, it generates, in general, a surface.
+A surface can then be considered apart from the idea of a solid.}
+
+\begin{point}%
+If a surface moves in space, it generates, in general, a
+solid.
+
+\filbreak
+\figc{012bbZ14}{}
+Thus, let the upright surface $ABCD$
+(Fig.~3) move to the right to the position
+$EFGH$, the points $A$, $B$, $C$, and $D$ generating
+the lines $AE$, $BF$, $CG$, and $DH$,
+respectively. The lines $AB$, $BC$, $CD$,
+and $DA$ will generate the surfaces $AF$.
+$BG$, $CH$, and $DE$, respectively. The surface $ABCD$ will generate the
+solid $AG$.
+\end{point}
+
+\pp{\indexbf{Geometry} is the science which treats of \textit{position, form},
+and \textit{magnitude}.}
+
+\pp{A \indexbf{geometrical figure} is a combination of points, lines,
+surfaces, or solids.}
+
+\begin{point}%
+{\indexbf{Plane Geometry} treats of figures all points of which are
+in the same plane.}
+
+\indexbf{Solid Geometry} treats of figures all points of which are not
+in the same plane.
+\end{point}
+\scanpage{013.png}%
+
+
+\section{GENERAL TERMS.}
+
+\begin{point}%
+A \indexbf{proof} is a course of reasoning by which the truth or
+falsity of any statement is logically established.
+\end{point}
+
+\begin{point}%
+An \textbf{axiom}\label{axiom} is a statement admitted to be true without
+proof.
+\end{point}
+
+\begin{point}%
+A \indexbf{theorem} is a statement to be proved.
+\end{point}
+
+\begin{point}%
+A \textbf{construction}\label{construction} is the representation of a required figure
+by means of points and lines.
+\end{point}
+
+\begin{point}%
+A \indexbf{postulate} is a construction admitted to be possible.
+\end{point}
+
+\begin{point}%
+A \textbf{problem} is a construction to be made so that it shall
+satisfy certain given conditions.
+\end{point}
+
+\begin{point}%
+A \indexbf{proposition} is an axiom, a theorem, a postulate, or a
+problem.
+\end{point}
+
+\begin{point}%
+A \textbf{corollary} is a truth that is easily deduced from known
+truths.
+\end{point}
+
+\begin{point}%
+A \indexbf{scholium} is a remark upon some particular feature of
+a proposition.
+\end{point}
+
+\begin{point}%
+The \textbf{solution of a problem} consists of four parts:
+
+1. The \emph{analysis}\label{analysis}, or course of thought by which the construction
+of the required figure is discovered.
+
+2. The \emph{construction} of the figure with the aid of ruler and
+compasses.
+
+3. The \emph{proof} that the figure satisfies all the conditions.
+
+4. The \emph{discussion} of the limitations, if any, within which
+the solution is possible.
+\end{point}
+\scanpage{014.png}%
+
+\begin{point}%
+A theorem consists of two parts: the \indexbf{hypothesis}, or
+that which is assumed; and the \indexbf{conclusion}, or that which is
+asserted to follow from the hypothesis.
+\end{point}
+
+\settowidth{\TmpLen}{\qquad\qquad Its contradictory:\quad}%
+\begin{point}%
+The \textbf{contradictory} of a theorem\label{contradictory} is a theorem which must
+be true if the given theorem is false, and must be false if the
+given theorem is true. Thus, \\[0.5\baselineskip]
+\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
+\parbox{\TmpLen}{\qquad\qquad Its contradictory:} If $A$ is $B$, then $C$ is not $D$.
+\end{point}
+
+\begin{point}%
+The \indexbf{opposite} of a theorem is obtained by making \emph{both
+the hypothesis and the conclusion negative}. Thus, \\[0.5\baselineskip]
+\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
+\parbox{\TmpLen}{\qquad\qquad Its opposite:} If $A$ is not $B$, then $C$ is not $D$.
+\end{point}
+
+\begin{point}%
+The \textbf{converse}\label{converse1} of a theorem is obtained by \emph{interchanging
+the hypothesis and conclusion}. Thus, \\[0.5\baselineskip]
+\parbox{\TmpLen}{\qquad\qquad A theorem:} If $A$ is $B$, then $C$ is $D$.\\
+\parbox{\TmpLen}{\qquad\qquad Its converse:} If $C$ is $D$, then $A$ is $B$.
+\end{point}
+
+\begin{point}%
+The converse of a truth is not \emph{necessarily} true.
+
+Thus, Every horse is a quadruped is true, but the converse, Every
+quad\-ru\-ped is a horse, is not true.
+\end{point}
+
+\begin{point}%
+\textit{If a direct proposition and its opposite are true, the
+converse proposition is true; and if a direct proposition and
+its converse are true, the opposite proposition is true}.
+
+Thus, if it were true that
+
+1. If an animal is a horse, the animal is a quadruped;
+
+2. If an animal is not a horse, the animal is not a quadruped;\\
+it would follow that
+
+3. If an animal is a quadruped, the animal is a horse.
+
+Moreover, if 1 and 3 were true, then 2 would be true.
+\end{point}
+\scanpage{015.png}%
+
+
+\section[GENERAL AXIOMS.]{\pointno\hsp GENERAL AXIOMS.\hsp\phantom{XX.}}
+\label{generalaxioms}
+
+1. Magnitudes which are equal to the same magnitude, or
+equal magnitudes, are equal to each other.
+
+2. If equals are added to equals, the sums are equal.
+
+3. If equals are taken from equals, the remainders are equal.
+
+4. If equals are added to unequals, the sums are unequal
+in the same order; if unequals are added to unequals in the
+same order, the sums are unequal in that order.
+
+5. If equals are taken from unequals, the remainders are
+unequal in the same order; if unequals are taken from equals,
+the remainders are unequal in the reverse order.
+
+6. The doubles of the same magnitude, or of equal magnitudes
+are equal; and the doubles of unequals are unequal.
+
+7. The halves of the same magnitude, or of equal magnitudes
+are equal; and the halves of unequals are unequal.
+
+8. The whole is greater than any of its parts.
+
+9. The whole is equal to the sum of all its parts.
+
+
+\section[SYMBOLS AND ABBREVIATIONS.]
+  {\pointno\hsp SYMBOLS AND ABBREVIATIONS. \hsp\phantom{XX.}}
+
+\label{abbr}\label{symbols}%
+\noindent\small\enlargethispage{8pt}%
+\begin{tabular*}{\dentwidth}{rl@{\extracolsep{\fill}}l@{\extracolsep{0pt}}@{\dots}l}
+
+$>$ & is (or are) greater than. & Def. & definition. \\
+$<$ & is (or are) less than. & Ax. & axiom. \\
+$\Bumpeq$ & is (or are) equivalent to. & Hyp. & hypothesis. \\
+$\therefore$ & therefore. & Cor. & corollary. \\
+$\perp$ & perpendicular. & Scho. & scholium. \\
+$\perp_s$ & perpendiculars. & Ex. & exercise. \\
+$\parallel$ & parallel.\qquad $\parallel_s$ parallels. & Adj. & adjacent. \\
+$\angle$ & angle.\qquad $\angle_s$ angles. & Iden. & identical. \\
+$\triangle$ & triangle.\qquad $\triangle_s$ triangles. & Const. & construction. \\
+$\Par$ & parallelogram. & Sup. & supplementary. \\
+$\Par_s$ & parallelograms. & Ext. & exterior. \\
+$\odot$ & circle.\qquad $\odot_s$ circles. & Int. & interior. \\
+rt. & right.\qquad  st.\ straight. & Alt. & alternate. \\
+\end{tabular*}
+
+\qed\ stands for quod erat demonstrandum, \emph{which was to be proved}.
+
+\qef\ stands for quod erat faciendum, \emph{which was to be done.}
+
+The signs $+$, $-$, $×$, $\div$, $=$, have the same meaning as in Algebra.
+\scanpage{016.png}%
+
+\normalsize
+
+\phantomsection%
+\part{PLANE GEOMETRY.}
+
+\chapter{BOOK I\@. RECTILINEAR FIGURES.}
+
+\section{DEFINITIONS.}
+
+\vspace{2ex}
+\figc{016aaZ37}{}
+\begin{point}%
+A \indexbf{straight line} is a line such that any part of it, however
+placed on any other part, will lie wholly in that part if
+its extremities lie in that part, as~$AB$.
+\end{point}
+
+\begin{point}%
+A \indexbf{curved line} is a line no part of
+which is straight, as~$CD$.
+\end{point}
+
+\begin{point}%
+A \textbf{broken line} is made up of different
+straight lines, as~$EF$.
+\end{point}
+
+\note{A straight line is often called simply a \emph{line}.}
+
+\begin{point}%
+A \textbf{plane surface}, or a \indexbf{plane}, is a surface in which, if any
+two points are taken, the straight line joining these points
+lies wholly in the surface.
+\end{point}
+
+\begin{point}%
+A \textbf{curved surface}\label{curvedsurf} is a surface no part of which is plane.
+\end{point}
+
+\begin{point}%
+A \indexbf{plane figure} is a figure all points of which are in the
+same plane.
+\end{point}
+
+\begin{point}%
+Plane figures which are bounded by straight lines are
+called \indexbf{rectilinear} figures; by curved lines, \indexbf{curvilinear} figures.
+\end{point}
+
+\begin{point}%
+Figures that have the \emph{same shape} are called \indexbf{similar}.
+Figures that have the \textit{same size but not the same shape} are
+called \textbf{equivalent}\label{equivalent1}. Figures that have the \textit{same shape and the
+same size} are called \textbf{equal}\label{equal} or \textbf{congruent}\label{congruent}.
+\end{point}
+\scanpage{017.png}%
+
+
+\section{THE STRAIGHT LINE.}
+
+\pp{\textbf{Postulate.} A straight line can be drawn from one point
+to another.}
+
+\pp{\textbf{Postulate.} A straight line can be produced indefinitely.}
+
+% footnote keeps this from being a normal \ax
+\pp{\textbf{Axiom.}\footnote{The general axioms on page \pageref{generalaxioms} apply to all magnitudes. Special
+geometrical axioms will be given when required.}
+\textit{Only one straight line can be drawn from one
+point to another.} Hence, two points \textit{determine} a straight line.}
+
+\pp{\cor[1]{Two straight lines which have two points in
+common coincide and form but one line.}}
+
+\pp{\cor[2]{Two straight lines can intersect in only one
+point.}}
+
+For if they had two points common, they would coincide
+and not intersect.
+
+Hence, two intersecting lines \textit{determine} a point.
+
+\pp{\ax{A straight line is the shortest line that can be
+drawn from one point to another.}\label{axiomstraight}}
+
+\pp{\defn{The \textbf{distance}\label{distance1} between two points is the length of
+the straight line that joins them.}}
+
+\pp{A straight line determined by two points may be considered
+as prolonged indefinitely.}
+
+\pp{If only the part of the line between two fixed points is
+considered, this part is called a \textbf{segment} of the line\label{lineseg}.}
+
+\pp{For brevity, we say ``the line $AB$,'' to designate a segment
+of a line limited by the points $A$ and $B$.}
+
+\pp{If a line is considered as extending from a fixed point,
+this point is called the \indexbf{origin} of the line.}
+\scanpage{018.png}%
+
+\figc{018aaZ55}{}
+\begin{point}%
+If any point, $C$, is taken in a given straight line, $AB$,
+the two parts $CA$ and $CB$ are
+said to have \emph{opposite directions}
+from the point $C$ (Fig.~5).
+
+Every straight line, as $AB$, may be considered as extending
+in either of two opposite directions, namely, from $A$ towards
+$B$, which is expressed by $AB$, and read segment $AB$; and
+from $B$ towards $A$, which is expressed by $BA$, and read segment
+$BA$.
+\end{point}
+
+\begin{point}%
+If the magnitude of a given line is changed, it becomes
+longer or shorter.
+
+Thus (Fig.~5), by prolonging $AC$ to $B$ we add $CB$ to $AC$, and $AB = AC+CB$.
+By diminishing $AB$ to $C$, we subtract $CB$ from $AB$, and
+$AC = AB-CB$.
+
+If a given line increases so that it is prolonged by its own
+magnitude several times in succession, the line is \emph{multiplied},
+and the resulting line is called a \emph{multiple} of the given line.
+
+\figc{018bbZ56}{}
+Thus (Fig.~6), if $AB = BC = CD = DE$,
+then $AC = 2AB$, $AD = 3AB$,
+and $AE = 4AB$. Hence,
+
+\textit{Lines of given length may be added and subtracted; they
+may also be multiplied by a number.}
+\end{point}
+
+
+\section{THE PLANE ANGLE.}
+
+\label{angle}
+\figc{018ccZ57}{}
+\begin{point}%
+The \emph{opening} between two straight lines drawn from the
+same point is called a \textbf{plane angle}. The two
+lines, $ED$ and $EF$, are called the \textbf{sides}\label{anglesides}, and $E$,
+the point of meeting, is called the \indexbf{vertex} of
+the angle.
+
+The size of an angle depends upon the \emph{extent
+of opening} of its sides, and not upon the length of its sides.
+\end{point}
+\scanpage{019.png}%
+
+\begin{point}%
+If there is but one angle at a given vertex, the angle is
+designated by a capital letter placed at the vertex, and is read
+by simply naming the letter.
+
+\figcc{019aaZ58}{019bbZ58}
+
+If two or more angles have the same vertex, each
+angle is designated by three letters, and is read by
+naming the three letters, the one at the vertex
+between the others. Thus, $DAC$ (Fig.~8) is the
+angle formed by the sides $AD$ and $AC$.
+
+An angle is often designated by placing a
+small \emph{italic} letter between the sides and near
+the vertex, as in Fig.~9.
+\end{point}
+
+\begin{point}%
+\textbf{Postulate of Superposition.}\label{superposition} Any figure may be moved
+from one place to another without altering its size or
+shape.
+\end{point}
+
+\begin{point}%
+The \textbf{test of equality} of two geometrical magnitudes is
+that they may be made to coincide throughout their whole
+extent. Thus,
+
+Two straight lines are equal, if they can be placed one upon
+the other so that the points at their extremities coincide.
+
+Two angles are equal, if they can be placed one upon the
+other so that their vertices coincide and their sides coincide,
+each with each.
+\end{point}
+
+\begin{point}%
+A line or plane that divides a geometric magnitude into
+\emph{two equal parts} is called the \textbf{bisector}\label{bisector} of the magnitude.
+
+If the angles $BAD$ and $CAD$ (Fig.~8) are equal, $AD$ \emph{bisects}
+the angle $BAC$.
+\end{point}
+
+\begin{point}%
+Two angles are called \textbf{adjacent angles}\label{adjacent1}
+when they have the same vertex and a common
+side between them; as the angles $BOD$
+and $AOD$ (Fig.~10).
+\end{point}
+\scanpage{020.png}%
+
+\figcc{019ccZ62}{020aaZ63}
+\begin{point}%
+When one straight line meets another
+straight line and makes the \emph{adjacent angles
+equal}, each of these angles is called a \textbf{right
+angle}\label{right}; as angles $DCA$ and $DCB$ (Fig.~11).
+\end{point}
+
+\begin{point}%
+A \indexbf{perpendicular} to a straight line is a straight line that
+makes a right angle with it.
+
+Thus, if the angle $DCA$ (Fig.~11) is a right angle, $DC$ is perpendicular
+to $AB$, and $AB$ is perpendicular to $DC$.
+\end{point}
+
+\begin{point}%
+The point (as $C$, Fig.~11) where a perpendicular meets
+another line is called the \indexbf{foot} of the perpendicular.
+\end{point}
+
+\begin{point}%
+When the sides of an angle extend in opposite directions,
+so as to be in the same straight line, the angle is called a
+\textbf{straight angle}.\label{straight}
+
+\figc{020bbZ66}{}
+
+Thus, the angle formed at $C$ (Fig.~12) with its sides $CA$ and $CB$ extending
+in opposite directions from $C$ is a straight angle.
+\end{point}
+
+\pp{\cor{A right angle is half a straight angle.}}
+
+\figcc{020ccZ68}{020ddZ69}
+\begin{point}%
+An angle less than a right angle is called
+an \textbf{acute angle}\label{acute}; as, angle $A$ (Fig.~13).
+\end{point}
+
+\begin{point}%
+An angle greater than a right angle and
+less than a straight angle is called an \textbf{obtuse
+angle}\label{obtuse}; as, angle $AOD$ (Fig.~14).
+\end{point}
+
+\begin{point}%
+An angle greater than a straight angle and less than
+two straight angles is called a \textbf{reflex angle}\label{reflex}; as, angle $DOA$,
+indicated by the dotted line (Fig.~14).
+\end{point}
+\scanpage{021.png}%
+
+\begin{point}%
+Angles that are neither right nor straight angles are
+called \textbf{oblique angles}\label{oblique}; and intersecting lines that are not perpendicular
+to each other are called \indexbf{oblique lines}.
+\end{point}
+
+\subsection{EXTENSION OF THE MEANING OF ANGLES.}
+
+\figc{021aaZ72}{}
+\begin{point}%
+Suppose the straight line $OC$ (Fig.~15) to move in
+the plane of the paper from coincidence with $OA$, about the
+point $O$ as a pivot, to the position $OC$; then the line $OC$
+describes or generates \emph{the angle $AOC$}, and
+the magnitude of the angle $AOC$ depends
+upon the \emph{amount of rotation} of the line
+from the position $OA$ to the position $OC$.
+
+If the rotating line moves from the position
+$OA$ to the position $OB$, \emph{perpendicular}
+to $OA$, it generates the right angle $AOB$;
+if it moves to the position $OD$, it generates
+the obtuse angle $AOD$; if it moves to
+the position $OA'$, it generates the straight angle $AOA'$; if it
+moves to the position $OB'$ it generates the reflex angle $AOB'$,
+indicated by the dotted line; and if it moves to the position
+$OA$ again, it generates two straight angles. Hence,
+\end{point}
+
+\begin{point}%
+\textit{The angular magnitude about a point in a plane is equal
+to two straight angles, or four right angles; and the angular
+magnitude about a point on one side of a straight line drawn
+through the point is equal to a straight angle, or two right
+angles.}
+\end{point}
+
+\begin{point}%
+The whole angular magnitude about a point in a plane
+is called a \indexbf{perigon}; and two angles whose sum is a perigon are
+called \indexbf{conjugate angles}.
+\end{point}
+
+\note{This \emph{extension of the meaning of angles} is necessary in the
+applications of Geometry, as in Trigonometry, Mechanics, etc.}
+\scanpage{022.png}%
+
+\figccc{022aaZ75}{022bbZ76}{022ccZ77}
+\begin{point}%
+When two angles have the same vertex, and
+the sides of the one are prolongations of the sides of
+the other, they are called \indexbf{vertical angles}; as, angles
+$a$ and $b$, $c$ and $d$ (Fig.~16).
+\end{point}
+
+\begin{point}%
+Two angles are called \textbf{complementary}\label{complementary} when
+their sum is equal to a right angle; and each is
+called the \emph{complement}\label{complement} of the other; as, angles $DOB$
+and $DOC$ (Fig.~17).
+\end{point}
+
+\begin{point}%
+Two angles are called \textbf{supplementary}\label{supplementary} when
+their sum is equal to a straight angle; and each
+is called the \indexemph{supplement} of the other; as, angles
+$DOB$ and $DOA$ (Fig.~18).
+\end{point}
+
+
+\subsection{UNIT OF ANGLES.}
+
+\begin{point}%
+By adopting a suitable unit for measuring angles we
+are able to express the magnitudes of angles by numbers.
+
+If we suppose $OC$ (Fig.~15) to turn about $O$ from coincidence
+with $OA$ until it makes \emph{one three hundred sixtieth} of a
+revolution, it generates an angle at $O$, which is taken as the
+unit for measuring angles. This unit is called a \emph{degree}.
+
+The degree is subdivided into sixty equal parts, called
+\emph{minutes}, and the minute into sixty equal parts, called \emph{seconds}.
+
+Degrees, minutes, and seconds are denoted by symbols.
+Thus, $5$~degrees $13$~minutes $12$~seconds is written $5°\ 13'\ 12''$.
+
+A right angle is generated when $OC$ has made \emph{one fourth} of
+a revolution and contains~$90°$; a straight angle, when $OC$ has
+made \emph{half} of a revolution and contains~$180°$; and a perigon,
+when $OC$ has made a complete revolution and contains~$360°$.
+\end{point}
+
+\note{The natural angular unit is one complete revolution. But
+this unit would require us to express the values of most angles by fractions.
+The advantage of using the degree as the unit consists in its convenient
+size, and in the fact that $360$~is divisible by so many different
+integral numbers.}
+\scanpage{023.png}%
+
+\figc{023aaZ79}{}
+
+\begin{point}%
+By the method of superposition we are able to compare
+magnitudes of the same kind. Suppose we have two angles,
+$ABC$ and $DEF$ (Fig.~19). Let
+the side $ED$ be placed on the
+side $BA$, so that the vertex $E$
+shall fall on $B$; then, if the
+side $EF$ falls on $BC$, the angle
+$DEF$ equals the angle $ABC$;
+if the side $EF$ falls between
+$BC$ and $BA$ in the position shown by the dotted line $BG$, the
+angle $DEF$ is less than the angle $ABC$; but if the side $EF$
+falls in the position shown by the dotted line $BH$, the angle
+$DEF$ is greater than the angle $ABC$.
+\end{point}
+
+\figc{023bcZ80}{}
+
+\begin{point}%
+If we have the angles $ABC$ and $DEF$ (Fig.~20), and
+place the vertex $E$ on $B$ and the side $ED$ on $BC$, so that the
+angle $DEF$ takes the position $CBH$, the angles $DEF$ and $ABC$
+will together be equal to the angle $ABH$.
+
+If the vertex $E$ is placed on $B$, and the side $ED$ on $BA$, so
+that the angle $DEF$ takes the position $ABF$, the angle $FBC$
+will be the difference between the angles $ABC$ and $DEF$.
+
+If an angle is increased by its own magnitude two or more
+times in succession, the angle is \emph{multiplied} by a number.
+
+Thus, if the angles $ABM$, $MBC$, $CBP$, $PBD$ (Fig.~21) are all equal,
+the angle $ABD$ is $4$~times the angle $ABM$. Therefore,
+
+\textit{Angles may be added and subtracted; they may also be multiplied
+by a number.}
+\label{page:PageName}% [** TN: For verbal ref. in Note to Teachers]
+\end{point}
+\scanpage{024.png}%
+
+
+\pagebreak
+\section{PERPENDICULAR AND OBLIQUE LINES.}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{All straight angles are equal.}
+
+\figc{024aaZ81}{Let the angles $ACB$ and $DEF$ be any two straight angles.}
+
+\prove{$\angle ACB = \angle DEF$}.
+
+\textbf{Proof.} Place the $\angle ACB$ on the $\angle DEF$, so that
+the vertex $C$ shall fall on the vertex $E$, and the side $CB$ on the
+side $EF$.
+
+\step{Then $CA$ will fall on $ED$,}{§~47}
+
+\pnote{(because $ACB$ and $DEF$ are straight lines).}
+
+\step{$\therefore \angle ACB = \angle DEF$.}{§~60}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{All right angles are equal.}\hfill~Ax.~7}
+
+\begin{point}%
+\cor[2]{At a given point in a given line
+  there can be but one perpendicular to the line.}
+
+\figc{024bbZ83}{}
+For, if there could be two $\perp_s$, we should have rt.~$\angle_s$ of
+different magnitudes; but this is impossible, §~82.
+\end{point}
+
+\pp{\cor[3]{The complements of the same angle
+  or of equal angles are equal.}\hfill~Ax.~3}
+
+\pp{\cor[4]{The supplements of the same angle
+  or of equal angles are equal.}\hfill~Ax.~3}
+
+\note{The beginner must not forget that in Plane Geometry all
+the points of a figure are in the same plane.  Without this
+restriction in Cor.~2, an indefinite number of perpendiculars can be
+erected at a given point in a given line.}
+\scanpage{025.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two adjacent angles have their exterior sides
+in a straight line, these angles are supplementary.}
+
+\figc{025aaZ86}{Let the exterior sides $OA$ and $OB$ of the adjacent angles $AOD$ and
+$BOD$ be in the straight line $AB$.}
+
+\prove{$\angle_s AOD$ and $BOD$ are supplementary.}
+
+\step[\indent\textbf{Proof.}]{$AOB$ is a straight line.}{Hyp.}
+
+\step{$\therefore \angle AOB$ is a st.~$\angle$.}{§~66}
+
+\step[\indent But]{$\angle AOD + \angle BOD =$ the st.~$\angle AOB$.}{Ax.~9}
+
+\step{$\therefore$ the $\angle_s AOD$ and $BOD$ are supplementary.}{§~77}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\defn{Adjacent angles that are supplements of each
+other are called \emph{supplementary-adjacent angles}\label{suppladj}.}
+
+Since the angular magnitude about a point is neither increased
+nor diminished by the number of lines which radiate
+from the point, it follows that,
+\end{point}
+
+\pp{\cor[1]{The sum of all the angles about a point in a
+plane is equal to a perigon, or two straight angles.}}
+
+\pp{\cor[2]{The sum of all the angles about a point in a
+plane, on the same side of a straight line passing through the
+point, is equal to a straight angle, or two right angles.}}
+\scanpage{026.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} If two adjacent angles are supplementary,
+their exterior sides are in the same straight
+line.}
+
+\figc{026aaZ90}{Let the adjacent angles $OCA$ and $OCB$ be supplementary.}
+
+\prove{$AC$ and $CB$ are in the same straight line.}
+
+\textbf{Proof.} Suppose $CF$ to be in the same line with $AC$.
+
+\step[\indent Then]{$\angle_s OCA$ and $OCF$ are supplementary,}{§~86}
+
+\pnote{(if two adjacent angles have their exterior sides in a straight line, these
+angles are supplementary).}
+
+\step[\indent But]{$\angle_s OCA$ and $OCB$ are supplementary.}{Hyp.}
+
+\step{$\therefore \angle_s OCF$ and $OCB$ have the same supplement.}{}
+
+\eq{$\therefore \angle OCF$}{$= \angle OCB$.}{§~85}
+
+\step{$\therefore CB$ and $CF$ coincide.}{§~60}
+
+\step{$\therefore AC$ and $CB$ are in the same straight line.}{\qed}
+
+Since Propositions II.\ and III.\ are true, their opposites are
+true. Hence,~\hfill§~33
+
+\end{proof}
+
+\pp{\cor[1]{If the exterior sides of two adjacent angles are
+not in a straight line, these angles are not supplementary.}}
+
+\pp{\cor[2]{If two adjacent angles are not supplementary,
+their exterior sides are not in the same straight line.}}
+\scanpage{027.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If one straight line intersects another straight
+line, the vertical angles are equal.}
+
+\figc{027aaZ93}{Let the lines $OP$ and $AB$ intersect at $C$.}
+
+\proveq{$\angle OCB$}{$= \angle ACP$.}
+
+\step[\indent\textbf{Proof.}]{$\angle OCA$ and $\angle OCB$ are supplementary.}{§~86}
+
+\step{$\angle OCA$ and $\angle ACP$ are supplementary,}{§~86}
+
+\pnote{(if two adjacent angles have their exterior sides in a straight line, these
+angles are supplementary).}
+
+\step{$\therefore \angle_s OCB$ and $ACP$ have the same supplement.}{}
+
+\eq{$\therefore \angle OCB$}{$= \angle ACP$.}{§~85}
+
+\eq[\indent In like manner,]{$\angle ACO$}{$= \angle PCB$.}{\qed}
+
+\end{proof}
+
+\pp{\cor{If one of the four angles formed by the intersection
+of two straight lines is a right angle, the other three
+angles are right angles.}}
+
+\ex{Find the complement and the supplement of an angle of~$49°$.}
+
+\ex{Find the number of degrees in an angle if it is double its complement;
+if it is one fourth of its complement.}
+
+\ex{Find the number of degrees in an angle if it is double its supplement;
+if it is one third of its supplement.}
+\scanpage{028.png}%
+
+\proposition{Theorem}
+
+\begin{proof}%
+\obs{Two straight lines drawn from a point in a perpendicular
+to a given line, cutting off on the given line
+equal segments from the foot of the perpendicular, are
+equal and make equal angles with the perpendicular.}
+
+\figc{028aaZ95}{Let $CF$ be a perpendicular to the line $AB$, and $CE$ and $CK$ two
+straight lines cutting off on $AB$ equal segments $FE$ and $FK$ from $F$.}
+
+\prove{$CE = CK$; and $\angle FCE = \angle FCK$.}
+
+\textbf{Proof.} Fold over $CFA$, on $CF$ as an axis, until it falls on the
+plane at the right of $CF$.
+
+\step{$FA$ will fall along $FB$,}{}
+
+\pnote{(since $\angle CFA = \angle CFB$, each being a rt.~$\angle$, by hyp.).}
+
+\step{Point $E$ will fall on point $K$,}{}
+
+\pnote{(since $FE = FK$, by hyp.).}
+
+\eq{$\therefore CE$}{$= CK$,}{§~60}
+
+\pnote{(their extremities being the same points);}
+
+\eq{and $\angle FCE$}{$= \angle FCK$,}{§~60}
+
+\pnote{(since their vertices coincide, and their sides coincide, each with each).}
+
+\hfill\qed
+
+\end{proof}
+
+\ex{Find the number of degrees in the angle included by the hands
+of a clock at $1$~o'clock. $3$~o'clock. $4$~o'clock. $6$~o'clock.}
+\scanpage{029.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Only one perpendicular can be drawn to a given
+line from a given external point.}
+
+\figc{029aaZ96}{Let $AB$ be the given line, $P$ the given external point, $PC$ a perpendicular
+to $AB$ from $P$, and $PD$ any other line from $P$ to $AB$.}
+
+\proveq{$PD$ is not}{$\perp$ to $AB$.}
+
+\textbf{Proof.} Produce $PC$ to $P'$, making $CP'$ equal to $PC$.
+
+\step{Draw $DP'$.}{}
+
+\step{By construction, $PCP'$ is a straight line.}{}
+
+\step{$\therefore PDP'$ is not a straight line,}{§~46}
+
+\pnote{(only one straight line can be drawn from one point to another).}
+
+\step{Hence, $\angle PDP'$ is not a straight angle.}{}
+
+\step{Since $PC$ is $\perp$ to $DC$, and $PC = CP'$,}{}
+
+\step{$AC$ is $\perp$ to $PP'$ at its middle point.}{}
+
+\step{$\therefore \angle PDC = \angle P'DC$,}{§~95}
+
+\pnote{(two straight lines from a point in a $\perp$ to a line, cutting off on the line equal
+segments from the foot of the $\perp$, make equal $\angle_s$ with the $\perp$)}
+
+\step{Since $\angle PDP'$ is not a straight angle,}{}
+
+\step{$\angle PDC$, the half of $\angle PDP'$, is not a right angle.}{}
+
+\step{$\therefore PD$ is not $\perp$ to $AB$.}{\qed}
+
+\end{proof}
+\scanpage{030.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The perpendicular is the shortest line that can be
+drawn to a straight line from an external point.}
+
+\figc{030aaZ97}{Let $AB$ be the given straight line, $P$ the given point, $PC$ the perpendicular,
+and $PD$ any other line drawn from $P$ to $AB$.}
+
+\proveq{$PC$}{$< PD$.}
+
+\textbf{Proof.} Produce $PC$ to $P'$, making $CP' = PC$.
+
+\step{Draw $DP'$.}{}
+
+\eq[\indent Then]{$PD$}{$= DP'$,}{§~95}
+
+\pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the
+line equal segments from the foot of the $\perp$, are equal).}
+
+\eq{$\therefore PD + DP'$}{$= 2PD$,}{}
+
+\eq[and]{$PC + CP'$}{$= 2PC$.}{Const.}
+
+\eq[\indent But]{$PC + CP'$}{$< PD + DP'$.}{§~49}
+
+\eq{$\therefore 2 PC$}{$< 2 PD$.}{}
+
+\eq{$\therefore PC$}{$< PD$.}{Ax.~7}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{The shortest line that can be drawn from a
+point to a given line is perpendicular to the given line.}}
+
+\pp{\defn{The \textbf{distance}\label{distance2} of a point from a line is the length
+of the perpendicular from the point to the line.}}
+\scanpage{031.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The sum of two lines drawn from a point to the
+extremities of a straight line is greater than the sum of
+two other lines similarly drawn, but included by them.}
+
+\figc{031aa100}{Let $CA$ and $CB$ be two lines drawn from the point $C$ to the extremities
+of the straight line $AB$. Let $OA$ and $OB$ be two lines similarly
+drawn, but included by $CA$ and $CB$.}
+
+\proveq{$CA + CB$}{$> OA + OB$.}
+
+\textbf{Proof.} Produce $AO$ to meet the line $CB$ at $E$.
+
+\eq[\indent Then]{$CA + CE$}{$> OA + OE$,}{}
+
+\eq[and]{$BE + OE$}{$> OB$,}{§~49}
+
+\pnote{(a straight line is the shortest line from one point to another).}
+
+\step{Add these inequalities, and we have}{}
+
+\eq{$CA + CE + BE + OE$}{$> OA + OE + OB$.}{Ax.~4}
+
+\step{Substitute for $CE + BE$ its equal $CB$, then}{}
+
+\eq{$CA + CB + OE$}{$> OA + OE + OB$.}{}
+
+\step{Take away $OE$ from each side of the inequality.}{}
+
+\eq{$CA + CB$}{$> OA + OB$.}{Ax.~5}
+
+\hfill\qed
+\end{proof}
+\scanpage{032.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of two straight lines drawn from the same point
+in a perpendicular to a given line, cutting off on the
+line unequal segments from the foot of the perpendicular,
+the more remote is the greater.}
+
+\figc{032aa101}{Let $OC$ be perpendicular to $AB$, $OG$ and $OE$ two straight lines to
+$AB$, and $CE$ greater than $CG$.}
+
+\proveq{$OE$}{$> OG$.}
+
+\textbf{Proof.}  Take $CF$ equal to $CG$, and draw $OF$.
+
+\eq[\indent Then]{$OF$}{$= OG$,}{§~95}
+
+\pnote{(two straight lines drawn from a point in a $\perp$ to a line, cutting off on the
+line equal segments from the foot of the $\perp$, are equal).}
+
+\step{Produce $OC$ to $D$, making $CD = OC$.}{}
+
+\step{Draw $ED$ and $FD$.}{}
+
+\step[\indent Then]{$OE = ED$, and $OF = FD$.}{§~95}
+
+\eq[\indent But]{$OE + ED$}{$> OF + FD$,}{§~100}
+
+\eq{$\therefore 2OE > 2OF$, $OE$}{$> OF$, and $OE > OG$.}{\qed}
+
+\end{proof}
+
+\pp{\cor{Only two equal straight lines can be drawn
+from a point to a straight line; and of two unequal lines,
+the greater cuts off on the line the greater segment from the
+foot of the perpendicular.}}
+\scanpage{033.png}%
+
+
+\pagebreak
+\section{PARALLEL LINES.}
+
+
+\pp{\defn{Two \indexbf{parallel lines} are lines that lie in the same
+plane and cannot meet however far they are produced.}}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two straight lines in the same plane perpendicular
+to the same straight line are parallel.}
+
+\figc{033aa104}{Let $AB$ and $CD$ be perpendicular to $AC$.}
+
+\prove{$AB$ and $CD$ are parallel.}
+
+\textbf{Proof.} If $AB$ and $CD$ are not parallel, they will meet if
+sufficiently prolonged; and we shall have two perpendicular
+lines from their point of meeting to the same straight line;
+but this is impossible,~\hfill§~96
+
+\pnote{(only one perpendicular can be drawn to a given line from a given
+external point).}
+
+\step{$\therefore AB$ and $CD$ are parallel.}{\qed}
+
+\end{proof}
+
+\pp{\ax{Through a given point only one straight line
+can be drawn parallel to a given straight line.}\label{axiomparallel}}
+
+\begin{point}%
+\cor{Two straight lines in the same plane parallel to
+a third straight line are parallel to each other.}
+
+For if they could meet, we should have two straight lines
+from the point of meeting parallel to a straight line; but this
+is impossible.~\hfill§~105
+\end{point}
+\scanpage{034.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If a straight line is perpendicular to one of two
+  parallel lines, it is perpendicular to the other also.}
+
+\figc{034aa107}{Let $AB$ and $EF$ be two parallel lines, and let $HK$ be
+  perpendicular to $AB$, and cut $EF$ at $C$.}
+
+\proveq{$HK$}{is $\perp$ to $EF$.}
+
+\textbf{Proof.} Suppose $MN$ drawn through $C \perp$ to $HK$.
+
+\eq[\indent Then]{$MN$} {is $\parallel$ to $AB$.}{§~104}
+
+\eq[\indent But]{$EF$} {is $\parallel$ to $AB$.}{Hyp.}
+
+\step{$\therefore EF$ coincides with $MN$.}{§~105}
+
+\eq[\indent But]{$MN$} {is $\perp$ to $HK$.}{Const.}
+
+\eq{$\therefore EF$}{is $\perp$ to $HK$,}{}
+
+\eq[that is,]{$HK$}{is $\perp$ to $EF$.}{\qed}
+
+\end{proof}
+
+\pp{\defn{A straight line that cuts two or more straight
+lines is called a \indexbf{transversal} of those lines.}}
+
+\figc{034bb109}{}
+\begin{point}%
+If the transversal $EF$ cuts $AB$ and $CD$, the angles
+$a$, $d$, $g$, $f$ are called \emph{interior}\label{interior} angles; $b$, $c$, $h$,
+$e$ are called \emph{exterior}\label{exterior} angles.
+
+The angles $d$ and $f$, and $a$ and $g$, are called
+\emph{alternate-interior}\label{altint} angles; the angles $b$ and $h$, and $c$ and
+$e$, are called \emph{alternate-exterior}\label{altext} angles.
+
+The angles $b$ and $f$, $c$ and $g$, $e$ and $a$, $h$ and $d$, are
+called \emph{exterior-interior}\label{extint} angles.
+\end{point}
+\scanpage{035.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two parallel lines are cut by a transversal, the
+alternate-interior angles are equal.}
+
+\figc{035aa110}{Let $EF$ and $GH$ be two parallel lines cut by the transversal $BC$.}
+
+\proveq{$\angle EBC$}{$= \angle BCH$.}
+
+\textbf{Proof.} Through $O$, the middle point of $BC$, suppose $AD$
+drawn $\perp$ to $GH$.
+
+\step[\indent Then]{$AD$ is likewise $\perp$ to $EF$,}{§~107}
+
+\pnote{(a straight line $\perp$ to one of two $\parallel_s$ is $\perp$ to the other),}
+
+\step[that is,]{$CD$ and $BA$ are both $\perp$ to $AD$.}{}
+
+Apply the figure $COD$ to the figure $BOA$, so that $OD$ shall
+fall along $OA$.
+
+\step[\indent Then]{$OC$ will fall along $OB$,}{§~93}
+
+\pnote{(since $\angle COD = \angle BOA$, being vertical $\angle_s$);}
+
+\step[and]{$C$ will fall on $B$,}{}
+
+\pnote{(since $OC = OB$, by construction).}
+
+\step[\indent Then]{the $\perp CD$ will fall along the $\perp BA$,}{§~96}
+
+\pnote{(only one $\perp$ can be drawn to a given line from a given external point).}
+
+\step{$\therefore \angle OCD$ coincides with $\angle OBA$, and is equal to it,}{§~60}
+
+\pnote{(two angles are equal, if their vertices coincide and their sides coincide, each
+with each).}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{036.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} When two straight lines in the same
+plane are cut by a transversal, if the alternate-interior
+angles are equal, the two straight lines are parallel.}
+
+\figc{036aa111}{Let $EF$ cut the straight lines $AB$ and $CD$ in the points $H$ and $K$,
+and let the angles $AHK$ and $HKD$ be equal.}
+
+\proveq{$AB$ is}{$\parallel$ to $CD$.}
+
+\textbf{Proof}. Suppose $MN$ drawn through $H \parallel$ to $CD$.
+
+\eq[\indent Then]{$\angle MHK$}{$= \angle HKD$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq[\indent But]{$\angle AHK$}{$= \angle HKD$.}{Hyp.}
+
+\eq{$\therefore \angle MHK$}{$= \angle AHK$.}{Ax.~1}
+
+\step{$\therefore MN$ and $AB$ coincide.}{§~60}
+
+\eq[\indent But]{$MN$ is}{$\parallel$ to $CD$.}{Const.}
+
+\step{$\therefore AB$, which coincides with $MN$, is $\parallel$ to $CD$.}{\qed}
+
+\end{proof}
+
+\ex{Find the complement and the supplement of an angle that contains
+$37°\ 53'\ 49''$.}
+
+\ex{If the complement of an angle is one third of its supplement,
+how many degrees does the angle contain?}
+\scanpage{037.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two parallel lines are cut by a transversal, the
+exterior-interior angles are equal.}
+
+\figc{037aa112}{Let $AB$ and $CD$ be two parallel lines cut by the transversal $EF$,
+in the points $H$ and $K$.}
+
+\proveq{$\angle EHB$}{$= \angle HKD$.}
+
+\eq[\indent\textbf{Proof.}]{$\angle EHB$}{$= \angle AHK$,}{§~93}
+
+\pnote{(being vertical $\angle_s$).}
+
+\eq{$\angle AHK$}{$= \angle HKD$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq{$\therefore \angle EHB$}{$= \angle HKD$.}{Ax.~1}
+
+\eq[\indent In like manner]{$\angle EHA$}{$= \angle HKC$.}{\qed}
+
+\end{proof}
+
+\pp{\cor{The alternate-exterior angles $EHB$ and $CKF$,
+and also $AHE$ and $DKF$, are equal.}}
+
+\proposition{Theorem.}
+
+\begin{point}%
+\obs{\textsc{Conversely:} When two straight lines in a plane
+are cut by a transversal, if the exterior-interior angles
+are equal, these two straight lines are parallel.}
+
+(Proof similar to that in §~111.)
+\end{point}
+\scanpage{038.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two parallel lines are cut by a transversal,
+  the two interior angles on the same side of the transversal are
+  supplementary.}
+
+\figc{038aa115}{Let $AB$ and $CD$ be two parallel lines cut by the transversal
+  $EF$ in the points $H$ and $K$.}
+
+\prove{$\angle_s BHK$ and $HKD$ are supplementary.}
+
+\step[\indent\textbf{Proof.}]{$\angle EHB + \angle BHK = \text{a st.\ }\angle$,}{§~86}
+
+\pnote{(being sup.-adj.~$\angle_s$).}
+
+\step[\indent But]{$\angle EHB = \angle HKD$,}{§~112}
+
+\pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\step{\( \therefore \angle BHK + \angle HKD = \text{a st.\ }\angle \).}{}
+
+\step{$\therefore \angle_s BHK$ and $HKD$ are supplementary.}{§~77}
+
+\hfill\qed
+
+\end{proof}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} When two straight lines in a
+  plane are cut by a transversal, if two interior angles on the same
+  side of the transversal are supplementary, the two straight lines
+  are parallel.}
+
+(Proof similar to that in §~111.)
+
+\end{proof}
+\scanpage{039.png}%
+
+\section{TRIANGLES.}
+
+\begin{point}%
+A \indexbf{triangle} is a portion of a plane bounded by three
+straight lines; as, $ABC$ (Fig.~1).
+
+\figcc{039aa117}{039bb118}
+The bounding lines are called the
+\textbf{sides}\label{trisides} of the triangle, and their sum is
+called its \indexbf{perimeter}; the angles included
+by the sides are called the \textbf{angles} of the
+triangle\label{anglestri}, and the vertices of these angles,
+the \textbf{vertices} of the triangle\label{trivertices}.
+\end{point}
+
+\begin{point}%
+\textbf{Adjacent angles}\label{adjacent2} of a rectilinear
+figure are two angles that have one side
+of the figure common; as, angles $A$
+and $B$ (Fig.~2).
+\end{point}
+
+\begin{point}%
+An \textbf{exterior angle} of a triangle\label{exteriortri} is an angle included by
+one side and another side produced; as, $ACD$ (Fig.~2). The
+interior angle $ACB$ is adjacent to the exterior angle; the interior
+angles, $A$ and $B$, are called \textbf{opposite interior angles}.
+\end{point}
+
+\figc{039ce119}{}
+
+\begin{point}%
+A triangle is called a \indexbf{scalene triangle} when no two of
+its sides are equal; an \indexbf{isosceles triangle}, when two of its sides
+are equal; an \indexbf{equilateral triangle}, when its three sides are equal.
+
+\figc{039fi120}{}
+\end{point}
+\scanpage{040.png}%
+
+\begin{point}%
+A triangle is called a \indexbf{right triangle}, when one of its
+angles is a right angle; an \indexbf{obtuse triangle}, when one of its
+angles is an obtuse angle; an \textbf{acute triangle}, when all three
+of its angles are acute angles; an \indexbf{equiangular triangle}, when
+its three angles are equal.
+\end{point}
+
+\begin{point}%
+In a right triangle, the side opposite the right angle is
+called the \indexbf{hypotenuse}, and the other two sides the \indexbf{legs}.
+\end{point}
+
+\begin{point}%
+The side on which a triangle is supposed to stand is
+called the base\label{basetri} of the triangle. In the isosceles triangle, the
+equal sides are called the legs, and the other side, the base\label{baseiso}; in
+other triangles, any one of the sides may be taken as the base.
+\end{point}
+
+\begin{point}%
+The angle opposite the base of a triangle is called the
+\indexbf{vertical angle}, and its vertex, the \textbf{vertex} of the triangle\label{trivertex}.
+\end{point}
+
+\begin{point}%
+The \textbf{altitude} of a triangle\label{alttri} is the perpendicular from the
+vertex to the base, or to the base produced; as, $AD$ (Fig.~1).
+\end{point}
+
+\begin{point}%
+The three perpendiculars from the vertices of a triangle
+to the opposite sides (produced if necessary) are called the
+\textbf{altitudes} of the triangle; the three bisectors of the angles are
+called the \textbf{bisectors} of the triangle\label{tribisectors}; and the three lines from
+the vertices to the middle points of the opposite sides are
+called the \textbf{medians}\label{trimedians} of the triangle.
+\end{point}
+
+\begin{point}%
+If two triangles have the angles of the one equal, respectively,
+to the angles of the other, the equal angles are called
+\indexbf{homologous angles}, and the sides opposite the equal angles are
+called \indexbf{homologous sides}.
+\end{point}
+
+\begin{point}%
+Two triangles are equal in all respects if they can be
+made to coincide (§~60). The homologous sides of \indexemph{equal triangles}
+are equal, and the homologous angles are equal.
+\end{point}
+\scanpage{041.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The sum of the three angles of a triangle is equal
+to two right angles.}
+
+\figc{041aa129}{Let $A$, $B$, and $BCA$ be the angles of the triangle $ABC$.}
+
+\prove{$\angle A+\angle B+\angle BCA = 2$ rt.~$\angle_s$.}
+
+\textbf{Proof.} Suppose $CE$ drawn $\parallel$ to $AS$, and prolong $AC$ to $F$.
+
+\step[\indent Then]{$\angle ECF + \angle ECB + \angle BCA = 2$ rt.~$\angle_s$,}{§~89}
+
+\pnote{(the sum of all the $\angle_s$ about a point on the same side of a straight line
+passing through the point is equal to $2$~rt.~$\angle_s$).}
+
+\eq[\indent But]{$\angle A$}{$= \angle ECF$,}{§~112}
+
+\pnote{(being ext.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$),}
+
+\eq[and]{$\angle B$}{$= \angle BCE$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of the $\parallel$ lines $AB$ and $CE$).}
+
+Put for the $\angle_s ECF$ and $BCE$ their equals, the $\angle_s A$ and $B$.
+
+\step[\indent Then]{$\angle A +\angle B + \angle BCA = 2$ rt.~$\angle_s$.}{\qed}
+
+\end{proof}
+
+\pp{\cor[1]{The sum of two angles of a triangle is less
+than two right angles.}}
+
+\pp{\cor[2]{If the sum of two angles of a triangle is
+taken from two right angles, the remainder is equal to the
+third angle.}}
+
+\pp{\cor[3]{If two triangles have two angles of the one
+equal to two angles of the other, the third angles are equal.}}
+\scanpage{042.png}%
+
+\pp{\cor[4]{If two right triangles have an acute angle of
+the one equal to an acute angle of the other, the other acute
+angles are equal.}}
+
+\pp{\cor[5]{In a triangle there can be but one right angle,
+or one obtuse angle.}}
+
+\pp{\cor[6]{In a right triangle the two acute angles are
+together equal to one right angle, or~$90°$.}}
+
+\pp{\cor[7]{In an equiangular triangle, each angle is one
+third of two right angles, or~$60°$.}}
+
+\pp{\cor[8]{An exterior angle of a triangle is equal to
+the sum of the two opposite interior angles, and therefore
+greater than either of them.}}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The sum of two sides of a triangle is greater than
+the third side, and their difference is less than the third
+side.}
+
+\figc{042aa138}{In the triangle $ABC$, let $AC$ be the longest side.}
+
+\prove{$AB + BC > AC$, and $AC - BC < AB$}.
+
+\step[\indent\textbf{Proof.}]{$AB + BC > AC$,}{§~49}
+
+\pnote{(a straight line is the shortest line from one point to another).}
+
+\step{Take away $BC$ from both sides.}{}
+
+\step[\indent Then]{$AB > AC - BC$,}{Ax.~5}
+
+\step[or]{$AC - BC < AB$.}{\qed}
+
+\end{proof}
+\scanpage{043.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two triangles are equal if two angles and the
+included side of the one are equal, respectively, to two
+angles and the included side of the other.}
+
+\figc{043ab139}{In the triangles $ABC$, $DEF$, let the angle $A$ be equal to the angle
+$D$, $B$ to $E$, and the side $AB$ to $DE$.}
+
+\proveq{$\triangle ABC$}{$= \triangle DEF$.}
+
+\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall
+coincide with its equal, $DE$.
+
+\step[\indent Then]{$AC$ will fall along $DF$, and $BC$ along $EF$,}{}
+
+\pnote{(for $\angle A = \angle D$, and $\angle B = \angle E$, by hyp.).}
+
+\step{$\therefore C$ will fall on $F$,}{§~48}
+
+\pnote{(two straight lines can intersect in only one point).}
+
+\step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{§~60}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{Two triangles are equal if a side and any two
+angles of the one are equal to the homologous side and two
+angles of the other.}~\hfill§~132}
+
+\pp{\cor[2]{Two right triangles are equal if the hypotenuse
+and an acute angle of the one are equal, respectively, to
+the hypotenuse and an acute angle of the other.}~\hfill§~133}
+
+\pp{\cor[3]{Two right triangles are equal if a leg and
+an acute angle of the one are equal, respectively, to a leg
+and the homologous acute angle of the other.}~\hfill§~133}
+\scanpage{044.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two triangles are equal if two sides and the included
+angle of the one are equal, respectively, to two sides
+and the included angle of the other.}
+
+\figc{044ab143}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$,
+and the angle $A$ to the angle $D$.}
+
+\proveq{$\triangle ABC$}{$= \triangle DEF$.}
+
+\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle DEF$ so that $AB$ shall
+coincide with its equal, $DE$.
+
+\step{Then $AC$ will fall along $DF$,}{}
+
+\pnote{(for $\angle A = \angle D$, by hyp.);}
+
+\step{and $C$ will fall on $F$,}{}
+
+\pnote{(for $AC = DF$, by hyp.).}
+
+\step{$\therefore CB = FE$,}{}
+
+\pnote{(their extremities being the same points).}
+
+\step{$\therefore$ the two $\triangle_s$ coincide, and are equal.}{\qed}
+
+\end{proof}
+
+\pp{\cor{Two right triangles are equal if their legs are
+equal, each to each.}}
+
+\note{In §~139 we have given two angles and the included side, in
+§~143 two sides and the included angle; hence, by interchanging the
+words \emph{sides} and \emph{angles}, either theorem is changed to the other. This
+is called the \emph{Principle of Duality}\label{princduality}, or the \emph{Principle of Reciprocity}\label{princreciprocity}. The
+reciprocal of a theorem is not always true, just as the converse of a
+theorem is not always true.}
+\scanpage{045.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\textit{In an isosceles triangle the angles opposite the
+equal sides are equal.}
+
+\figc{045aa145}{Let $ABC$ be an isosceles triangle, having $AB$ and $AC$ equal.}
+
+\proveq{$\angle B$}{$= \angle C$.}
+
+\textbf{Proof.} Suppose $AD$ drawn so as to bisect the $\angle BAC$.
+
+\step{In the $\triangle_s ADB$ and $ADC$,}{}
+
+\eq{$AB$}{$=AC$,}{Hyp.}
+
+\eq{$AD$}{$=AD$,}{Iden.}
+
+\eq{and $\angle BAD$}{$= \angle CAD$.}{Const.}
+
+\eq{$\therefore \triangle ADB$}{$= \triangle ADC$,}{§~143}
+
+\pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$ of the one are equal,
+respectively, to two sides and the included $\angle$ of the other).}
+
+\eq{$\therefore \angle B$}{$= \angle C$,}{§~128}
+
+\pnote{(being homologous angles of equal triangles).}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{An equilateral triangle is equiangular, and each
+angle is two thirds of a right angle.}}
+
+\ex{If the equal sides of an isosceles triangle are produced, the
+angles on the other side of the base are equal.}
+\scanpage{046.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two angles of a triangle are equal, the sides
+opposite the equal angles are equal, and the triangle is
+isosceles.}
+
+\figc{046aa147}{In the triangle $ABC$, let the angle $B$ be equal to the angle~$C$.}
+
+\proveq{$AB$}{$= AC$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $AD$ drawn $\perp$ to $BC$.}{}
+
+In the rt.~$\triangle_s ADB$ and $ADC$,
+
+\eq{$AD$}{$= AD$,}{Iden.}
+
+\eq{and $\angle B$}{$= \angle C$.}{Hyp.}
+
+\eq{$\therefore$ rt.~$\triangle ADB$}{$=$ rt.~$\triangle ADC$,}{§~142}
+
+\pnote{(having a leg and an acute $\angle$ of the one equal, respectively, to a leg and
+the homologous acute $\angle$ of the other).}
+
+\eq{$\therefore AB$}{$= AC$,}{§~128}
+
+\pnote{(being homologous sides of equal $\triangle_s$).}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{An equiangular triangle is also equilateral.}}
+
+\pp{\cor[2]{The perpendicular from the vertex to the
+base of an isosceles triangle bisects the base, and bisects the
+vertical angle of the triangle.}}
+\scanpage{047.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two triangles are equal if the three sides of the
+one are equal, respectively, to the three sides of the other.}
+
+\figc{047ab150}{In the triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, $AC$ to
+$A'C'$, $BC$ to $B'C'$.}
+
+\proveq{$\triangle ABC$}{$= \triangle A'B'C'$.}
+
+\textbf{Proof.} Place $\triangle A'B'C'$ in the position $\triangle AB'C$ having its
+greatest side $\triangle A'C'$ in coincidence with its equal $\triangle AC$, and its
+vertex at $B'$, opposite $B$; and draw $BB'$.
+
+\eq[\indent Since]{$AB$}{$=AB'$}{Hyp.}
+
+\eq{$\angle ABB'$}{$= \angle AB'B$}{§~145}
+
+\pnote{(in an isosceles $\triangle$ the $\angle_s$ opposite the
+equal sides are equal).}
+
+\eq[\indent Since]{$CB$}{$= CB'$,}{Hyp.}
+
+\eq{$\angle CBB'$}{$= \angle CB'B$.}{§~145}
+
+\eq{$\therefore \angle ABB' + \angle CBB'$}{$= \angle AB'B + \angle CB'B$.}{Ax.~2}
+
+\eq[\indent Hence,]{$\angle ABC$}{$= \angle AB'C$.}{}
+
+\eq{$\therefore \triangle ABC$}{$= \triangle AB'C$,}{§~143}
+
+\pnote{(two $\triangle_s$ are equal if two sides and the included $\angle$
+ of the one are equal, respectively, to two sides and the included $\angle$ of the other).}
+
+\eq{$\therefore \triangle ABC$}{$= \triangle A'B'C'$.}{\qed}
+
+\end{proof}
+\scanpage{048.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two right triangles are equal if a leg and the
+hypotenuse of the one are equal, respectively, to a leg
+and the hypotenuse of the other.}
+
+\figc{048ac151}{In the right triangles $ABC$ and $A'B'C'$, let $AB$ be equal to $A'B'$, and
+$AC$ to $A'C'$.}
+
+\proveq{$\triangle ABC$}{$= \triangle A'B'C'$.}
+
+\textbf{Proof.} Apply the $\triangle ABC$ to the $\triangle A'B'C'$, so that $AB$ shall
+coincide with $A'B'$, $A$ falling on $A'$, $B$ on $B'$, and $C$ and $C'$ on
+opposite sides of $A'B'$.
+
+\step[\indent Then]{$BC$ will fall along $C'B'$ produced,}{}
+
+\pnote{(for $\angle ABC = \angle A'B'C'$, each being a rt.~$\angle$.).}
+
+\eq[\indent Since]{$AC$}{$= A'C'$,}{Hyp.}
+
+\step{the $\triangle A'CC'$ is an isosceles triangle.}{§~120}
+
+\eq{$\therefore \angle C$}{$= \angle C'$,}{§~145}
+
+\pnote{($\angle_s$ opposite the equal sides of an isosceles $\triangle$ are equal).}
+
+\step{$\therefore \triangle_s ABC$ and $A'B'C'$ are equal,}{§~141}
+
+\pnote{(two right $\triangle_s$ are equal if they have the hypotenuse and an acute $\angle$ of, the
+one equal to the hypotenuse and an acute $\angle$ of the other).}
+
+\hfill\qed
+
+\end{proof}
+
+\ex{How many degrees are there in each of the acute angles of an
+isosceles right triangle?}
+\scanpage{049.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two sides of a triangle are unequal, the angles
+opposite are unequal, and the greater angle is opposite
+the greater side.}
+
+\figc{049aa152}{In the triangle $ACB$, let $AB$ be greater than~$AC$.}
+
+\prove{$\angle ACB$ is greater than $\angle B$.}
+
+\step[\indent\textbf{Proof.}]{On $AB$ take $AE$ equal to $AC$.}{}
+
+\step{Draw $EC$.}{}
+
+\eq{$\angle AEC$}{$= \angle ACE$}{§~145}
+
+\pnote{(being $\angle_s$ opposite equal sides).}
+
+\step[\indent But]{$\angle AEC$ is greater than $\angle B$}{§~137}
+
+\pnote{(an exterior $\angle$ of a $\triangle$ is greater than either opposite interior $\angle$),}
+
+\step[and]{$\angle ACB$ is greater than $\angle ACE$.}{Ax.~8}
+
+\step{Substitute for $\angle ACE$ its equal $\angle AEC$,}{}
+
+\step[then]{$\angle ACB$ is greater than $\angle AEC$.}{}
+
+Since $\angle AEC$ is greater than $\angle B$, and $\angle ACB$ is greater
+than $\angle AEC$,
+
+\step{$\angle ACB$ is greater than $\angle B$.}{\qed}
+
+\end{proof}
+
+\ex{If any angle of an isosceles triangle is equal to two thirds of a
+right angle~($60°$), what is the value of each of the two remaining angles?}
+
+\ex{One angle of a triangle is~$34°$. Find the other angles, if one
+of them is twice the other.}
+\scanpage{050.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Reciprocally:} If two angles of a triangle are
+unequal, the sides opposite are unequal, and the greater
+side is opposite the greater angle.}
+
+\figc{050aa153}{In the triangle $ACB$, let the angle $C$ be greater than the angle~$B$.}
+
+\prove{$AB > AC$.}
+
+\step[\indent\textbf{Proof.}]{Now $AB=AC$, or $< AC$, or $>AC$.}{}
+\label{41}
+
+\step{But $AB$ is not equal to $AC$;}{}
+
+\step{for then the $\angle C$ would be equal to the $\angle B$,}{§~145}
+
+\pnote{(being $\angle_s$ opposite equal sides).}
+
+\step{And $AB$ is not less than $AC$;}{}
+
+\step{for then the $\angle C$ would be less than the $\angle B$.}{§~152}
+
+Both these conclusions are contrary to the hypothesis that
+the $\angle C$ is greater than the $\angle B$.
+
+\step{Hence, $AB$ cannot be equal to $AC$ or less than $AC$.}{}
+
+\step{$\therefore AB > AC$.}{\qed}
+
+\end{proof}
+
+\ex{If the vertical angle of an isosceles triangle is equal to~$30°$,
+find the exterior angle included by a side and the base produced.}
+
+\ex{If the vertical angle of an isosceles triangle is equal to~$36°$,
+find the angle included by the bisectors of the base angles.}
+\scanpage{051.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two triangles have two sides of the one equal,
+respectively, to two sides of the other, but the included
+angle of the first triangle greater than the included
+angle of the second, then the third side of the first is
+greater than the third side of the second.}
+
+\figc{051ac154}{In the triangles $ABC$ and $ABE$, let $AB$ be equal to $AB$, $BC$ to $BE$;
+but let the angle $ABC$ be greater than the angle $ABE$.}
+
+\proveq{$AC$}{$> AE$.}
+
+\textbf{Proof.} Place the $\triangle_s$ so that $AB$ of the one shall fall on
+$AB$ of the other, and $BE$ within the $\angle ABC$.
+
+Suppose $BF$ drawn to bisect the $\angle EBC$, and draw $EF$.
+
+The $\triangle_s EBF$ and $CBF$ are equal.~\hfill§~143
+
+\eq[\indent For]{$BF$}{$= BF$,}{Iden.}
+
+\eq{$BE$}{$=BC$,}{Hyp.}
+
+\eq[and]{$\angle EBF$}{$=\angle CBF$.}{Const.}
+
+\eq{$\therefore EF$}{$=FC$.}{§~128}
+
+\eq[\indent Now]{$AF+FE$}{$> AE$.}{§~138}
+
+\eq{$\therefore AF+FC$}{$> AE$.}{}
+
+\eq{$\therefore AC$}{$> AE$.}{\qed}
+
+\end{proof}
+\scanpage{052.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely}: If two sides of a triangle are equal,
+respectively, to two sides of another, but the third
+side of the first triangle is greater than the third side
+of the second, then the angle opposite the third side of
+the first triangle is greater than the angle opposite the
+third side of the second.}
+
+\figc{052ab155}{In the triangles $ABC$ and $DEF$, let $AB$ be equal to $DE$, $AC$ to $DF$;
+but let $BC$ be greater than EF.}
+
+\prove{the $\angle A$ is greater than the $\angle D$.}
+
+\textbf{Proof}. Now the $\angle A$ is equal to the $\angle D$, or less than the $\angle D$,
+or greater than the $\angle D$.
+
+\step{But the $\angle A$ is not equal to the $\angle D$;}{}
+
+\step{for then the $\triangle ABC$ would be equal to the $\triangle DEF$,}{§~143}
+
+\pnote{(having two sides and the included $\angle$ of the one equal, respectively, to two
+sides and the included $\angle$ of the other),}
+
+\step{and $BC$ would be equal to $EF$.}{}
+
+And the $\angle A$ is not less than the $\angle D$, for then $BC$ would
+be less than $EF$.~\hfill§~154
+
+Both these conclusions are contrary to the hypothesis that
+$BC$ is greater than $EF$.
+
+Since the $\angle A$ is not equal to the $\angle D$ or less than the $\angle D$,
+
+\step{the $\angle A$ is greater than the $\angle D$.}{\qed}
+
+\end{proof}
+\scanpage{053.png}%
+
+
+\section{LOCI OF POINTS.}
+
+\begin{point}%
+If it is required to find a point which shall fulfil a
+\emph{single} geometric condition, the point may have an \emph{unlimited
+number of positions}. If, however, all the points are in the
+same plane, the required point will be confined to a \emph{particular
+line}, or \emph{group of lines}.
+
+A point in a plane at a given distance from a fixed straight
+line of indefinite length in that plane, is evidently in one of
+two straight lines, so drawn as to be everywhere at the given
+distance from the fixed line, one on one side of the fixed line,
+and the other on the other side of it.
+
+A point in a plane equidistant from two parallel lines in
+that plane is evidently in a straight line drawn between the
+two given parallel lines and everywhere equidistant from them.
+\end{point}
+
+\begin{point}%
+All points in a plane that satisfy a single geometrical
+condition lie, in general, in a line or group of lines; and this
+line or group of lines is called the \textbf{locus} of the points that
+satisfy the given condition.
+\end{point}
+
+\begin{point}%
+To prove \emph{completely} that a certain line is the locus
+of points that fulfil a given condition, it is necessary to
+prove
+
+1. \textit{Any point in the line satisfies the given condition;
+and any point not in the line does not satisfy the given condition.}
+
+Or, to prove
+
+2. \textit{Any point that satisfies the given condition lies in the
+line; and any point in the line satisfies the given condition}.
+\end{point}
+
+\note{The word \emph{locus} (pronounced lo\'{ }kus) is a Latin word that signifies
+\emph{place}. The plural of locus is loci (pronounced lo\'{ }si).}
+
+\pp{\defn{A line which bisects a given line and is perpendicular
+to it is called the \textbf{perpendicular bisector} of the line.}}
+\scanpage{054.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The perpendicular bisector\label{perpbisector} of a given line is the
+locus of points equidistant from the extremities of the
+line.}
+
+\figc{054aa160}{Let $PR$ be the perpendicular bisector of the line $AB$, $O$ any point in
+$PR$, and $C$ any point not in $PR$.}
+
+\step{Draw $OA$ and $OB$, $CA$ and $CB$.}{}
+
+\prove[To prove ]{$OA$ and $OB$ equal, $CA$ and $CB$ unequal.}
+
+\eq[\indent\textbf{Proof.}]{\textbf{1. }$\triangle OPA$}{$= \triangle OPB$,}{§~144}
+
+\step{for $PA = PB$ by hypothesis, and $OP$ is common,}{}
+
+\pnote{(two right $\triangle_s$ are equal if their legs are equal, each to each).}
+
+\eq{$\therefore OA$}{$= OB$.}{§~128}
+
+\textbf{2.~}Since $C$ is not in the $\perp$, $CA$ or $CB$ will cut the $\perp$.
+
+\step{Let $CA$ cut the $\perp$ at $D$, and draw $DB$.}{}
+
+Then, by the first part of the proof $DA = DB$.
+
+\step[\indent But]{$CB < CD + DB$.}{§~138}
+
+\step{$\therefore CB < CD + DA$.}{}
+
+\step[\indent That is, ]{$CB < CA$.}{}
+
+$\therefore PR$ is the locus of points equidistant from $A$ and $B$.~\hfill§~158,1
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{Two points each equidistant from the extremities
+of a line determine the perpendicular bisector of the line.}}
+\scanpage{055.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The bisector of a given angle is the locus of
+  points equidistant from the sides of the angle.}
+
+\vspace{-1ex}
+\figc{055aa162}{Let $O$ be any point equidistant from the sides of the angle $PAQ$.}
+
+\prove{$O$ is in the bisector of the $\angle PAQ$.}
+
+\step [\indent\textbf{Proof.}] {Draw $AO$.} {}
+
+\step {Suppose $OF$ drawn $\perp$ to $AP$ and $OG \perp$
+  to $AQ$.} {}
+
+\step {In the rt.~$\triangle_s AFO$ and $AGO$,} {}
+
+\eq {$AO $}{$= AO$,} {Iden.}
+
+\eq {$OF $}{$= OG$,} {Hyp.}
+
+\eq {$\therefore \triangle AFO $} {$= \triangle AGO$.} {§~151}
+
+\eq {$\therefore \angle FAO $} {$= \angle GAO$.} {§~128}
+
+\step {$\therefore O$ is in the bisector of the $\angle PAQ$.}{}
+
+\textbf{Let $\mathbf{D}$ be any point in the bisector of the angle $\mathbf{PAQ}$.}
+
+\prove {$D$ is equidistant from $AP$ and $AQ$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $DB$ drawn $\perp$ to $AP$ and $DC \perp$ to~$AQ$.}{}
+
+\step {In the rt.~$\triangle_s ABD$ and $ACD$,} {}
+
+\eq {$AD $} {$= AD$,} {Iden.}
+
+\eq {$\angle DAB $} {$= \angle DAC$,} {Hyp.}
+
+\eq {$\therefore \triangle ABD $} {$= \triangle ACD$.} {§~141}
+
+\eq {$\therefore DB $} {$= DC$.} {§~128}
+
+\step{$\therefore D$ is equidistant from $AP$ and $AQ$.}{}
+
+\step{$\therefore$ the bisector of the $\angle PAQ$ is the locus of points that are
+equidistant from its sides.}{§~158, 2}
+\end{proof}
+\scanpage{056.png}%
+
+
+\section{QUADRILATERALS.}
+
+\begin{point}%
+A \indexbf{quadrilateral} is a portion of a plane bounded by
+four straight lines. The bounding lines are the \textbf{sides}, the
+angles formed by these sides are the \textbf{angles}, and the vertices
+of these angles are the \textbf{vertices}, of the quadrilateral.
+\end{point}
+
+\begin{point}%
+A \indexbf{trapezium} is a quadrilateral which has no two sides
+parallel.
+\end{point}
+
+\begin{point}%
+A \indexbf{trapezoid} is a quadrilateral which has two sides, and
+only two sides, parallel.
+\end{point}
+
+\begin{point}%
+A \indexbf{parallelogram} is a quadrilateral which has its opposite
+sides parallel.
+\end{point}
+
+\figc{056ac166}{}
+
+\begin{point}%
+A \indexbf{rectangle} is a parallelogram which has its angles
+right angles.
+\end{point}
+
+\begin{point}%
+A \indexbf{square} is a rectangle which has its sides equal.
+\end{point}
+
+\begin{point}%
+A \indexbf{rhomboid} is a parallelogram which has its angles
+oblique angles.
+\end{point}
+
+\begin{point}%
+A \indexbf{rhombus} is a rhomboid which has its sides equal.
+\end{point}
+
+\figc{056dg170}{}
+
+\begin{point}%
+The side upon which a parallelogram stands, and the
+opposite side, are called its lower and upper \emph{bases}\label{basepar}.
+\end{point}
+\scanpage{057.png}%
+
+\begin{point}%
+Two parallel sides of a trapezoid are called its \textbf{bases}\label{basetrap},
+the other two sides its \textbf{legs}\label{legstrap}, and the line joining the middle
+points of the legs is called the \textbf{median} of the trapezoid\label{mediantrap}.
+\end{point}
+
+\figc{057aa174}{}
+\begin{point}%
+A trapezoid is called an \indexbf{isosceles
+trapezoid} if its legs are equal.
+\end{point}
+
+\begin{point}%
+The \textbf{altitude} of a parallelogram\label{altpar}
+or trapezoid\label{alttrap} is the perpendicular distance
+between its bases, as $PQ$.
+\end{point}
+
+\begin{point}%
+A \textbf{diagonal}\label{diagonal1} of a quadrilateral is a straight line joining
+two opposite vertices, as $AC$.
+\end{point}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two angles whose sides are parallel, each to each,
+are either equal or supplementary.}
+
+\figc{057bb176}{Let $BA$ be parallel to $HD$, and $BC$ be parallel to $MN$.}
+
+\prove[To prove ]{$\angle_s a$, $a'$ and $c$ equal; $a$ and $c'$ supplementary.}
+
+\step[\indent\textbf{Proof.}]{Let $HD$ and $BC$ prolonged intersect at $x$.}{}
+
+\step[\indent Then]{$\angle a = \angle x$, and $\angle a' = \angle x$.}{§~112}
+
+\step{$\therefore \angle a = \angle a'$.}{Ax.~1}
+
+\step[\indent Also]{$\angle c = \angle a'$ (§~93). $\therefore \angle c = \angle a$.}{Ax.~1}
+
+\step[\indent Now]{$\angle a'$ and $\angle c'$ are supplementary.}{§~89}
+
+\step{Put $\angle a$ for its equal, $\angle a'$.}{}
+
+\step[\indent Then]{$\angle a$ and $\angle c'$ are supplementary.}{\qed}
+
+\end{proof}
+
+\pp{\cor{The opposite angles of a parallelogram are
+equal, and the adjacent angles are supplementary.}}
+\scanpage{058.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The opposite sides of a parallelogram are equal.}
+
+\figc{058aa178}{Let the figure $ABCE$ be a parallelogram.}
+
+\prove[To prove ]{$BC = AE$, and $AB = EC$.}
+
+\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}
+
+\step{$\triangle ABC = \triangle CEA$.}{§~139}
+
+\step{For $AC$ is common,}{}
+
+\step{$\angle BAC = \angle ACE$, and $\angle ACB = \angle CAE$,}{§~110}
+
+\pnote{(being alt-int.  $\angle_s$ of $\parallel$ lines).}
+
+\step{$\therefore BC = AE$, and $AB = CE$,}{§~128}
+
+\pnote{(being homologous sides of equal $\triangle_s$).}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{A diagonal divides a parallelogram into two
+equal triangles.}}
+
+\pp{\cor[2]{Parallel lines comprehended between parallel
+lines are equal.}}
+
+\figc{058bb181}{}
+\begin{point}%
+\cor[3]{Two parallel lines
+are everywhere equally distant.}
+
+For if $AB$ and $DC$ are parallel,
+$\perp_s$ dropped from \emph{any} points in $AB$ to $DC$, are equal, §~180.
+Hence, \emph{all} points in $AB$ are equidistant from~$DC$.
+\end{point}
+\scanpage{059.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If the opposite sides of a quadrilateral are
+  equal, the figure is a parallelogram.}
+
+\figc{059aa182}{Let the figure $ABCE$ be a quadrilateral, having $BC$ equal to
+  $AE$ and $AB$ to $EC$.}
+
+\prove{the figure $ABCE$ is a $\Par$.}
+
+\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}
+
+In the $\triangle_s ABC$ and $CEA$,
+
+\eq{$BC$}{$= AE$,}{Hyp.}
+
+\eq{$AB$}{$= CE$,}{Hyp.}
+
+\eq[and]{$AC$}{$= AC$,}{Iden.}
+
+\eq{$\therefore \triangle ABC$}{$= \triangle CEA$,}{§~150}
+
+\pnote{(having three sides of the one equal, respectively,
+to the three sides of the other).}
+
+\eq{$\therefore \angle ACB$}{$= \angle CAE$,}{§~128}
+
+\eq[and]{$\angle BAC$}{$= \angle ACE$,}{}
+
+\pnote{(being homologous $\angle_s$ of equal $\triangle_s$).}
+
+\eq{$\therefore BC$}{is $\parallel$ to $AE$,}{}
+
+\eq[and]{$AB$}{is $\parallel$ to $EC$,}{§~111}
+
+\pnote{(two lines in the same plane cut by a transversal are parallel,
+if the alt.-int.~$\angle_s$ are equal).}
+
+\step{$\therefore$ the figure $ACBE$ is a $\Par$,}{§~166}
+
+\pnote{(having its opposite sides parallel).}\hfill\llap{\qed}
+
+\end{proof}
+\scanpage{060.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two sides of a quadrilateral are equal and parallel, then the
+other two sides are equal and parallel, and the figure is a parallelogram.}
+
+\figc{060aa183}{Let the figure $ABCE$ be a quadrilateral, having the side $AE$ equal and
+parallel to $BC$.}
+
+\prove{$AB$ is equal and parallel to $EC$.}
+
+\step[\indent\textbf{Proof.}]{Draw $AC$.}{}
+
+The $\triangle_s ABC$ and $CEA$ are equal,~\hfill§~143
+
+\pnote{(having two sides and the included $\angle$ of each equal,
+respectively).}
+
+\step[\indent For]{$AC$ is common,}{}
+
+\eq{$BC$}{$=AE$}{Hyp.}
+
+\eq[and]{$\angle BCA$}{$= \angle CAE$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq{$\therefore AB$}{$=EC$,}{}
+
+\eq[and]{$\angle BAC$}{$= \angle ACE$,}{§~128}
+
+\pnote{(being homologous parts of equal $\triangle_s$).}
+
+\step{$\therefore AB$ is $\parallel$ to $EC$,}{§~111}
+
+\pnote{(two lines are $\parallel$, if the alt.-int. $\angle_s$ are equal).}
+
+\step{$\therefore$ the figure $ABCE$ is a $\Par$,}{§~166}
+
+\pnote{(the opposite sides being parallel).}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{061.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The diagonals of a parallelogram bisect each other.}
+
+\figc{061aa184}{Let the figure $ABCE$ be a parallelogram, and let the diagonals $AC$
+and $BE$ cut each other at $O$.}
+
+\prove{$AO = OC$, and $BO = OE$.}
+
+\textbf{Proof.} In the $\triangle_s AOE$ and $COB$,
+
+\eq{$AE$}{$=BC$,}{§~178}
+
+\pnote{(being opposite sides of a $\Par$).}
+
+\eq{$\angle OAE$}{$=\angle OCB$,}{§~110}
+
+\eq{and $\angle OEA$}{$= \angle OBC$,}{}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq{$\therefore\triangle AOE$}{$=\triangle COB$,}{§~139}
+
+\pnote{(having two $\angle_s$ and the included side of the one equal, respectively, to two
+$\angle_s$ and the included side of the other).}
+
+\step{$\therefore AO=OC$, and $BO=OE$,}{§~128}
+\pnote{(being homologous sides of equal $\triangle_s$).}
+
+\hfill\qed
+
+\end{proof}
+
+\ex{The median from the vertex to the base of an isosceles triangle
+is perpendicular to the base, and bisects the vertical angle.}
+
+\ex{If two straight lines are cut by a transversal so that the alternate-exterior
+angles are equal, the two straight lines are parallel.}
+
+\ex{If two parallel lines are cut by a transversal, the two exterior
+angles on the same side of the transversal are supplementary.}
+
+\ex{If two straight lines are cut by a transversal so as to make the
+exterior angles on the same side of the transversal supplementary, the two
+lines are parallel.}
+\scanpage{062.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two parallelograms are equal, if two sides and
+the included angle of the one are equal, respectively, to
+two sides and the included angle of the other.}
+
+\figc{062ab185}{In the parallelograms $ABCD$ and $A'B'C'D'$, let $AB$ be equal to $A'B'$,
+$AD$ to $A'D'$, and angle $A$ to $A'$.}
+
+\prove{the $\Par_s$ are equal}.
+
+\textbf{Proof.} Place the $\Par$ $ABCD$ on the $\Par$ $A'B'C'D'$, so that $AD$
+will fall on and coincide with its equal, $A'D'$.
+
+\step{Then $AB$ will fall on $A'B'$, and $B$ on $B'$;}{}
+
+\pnote{(for $\angle A = \angle A'$, and $AB = A'B'$, by hyp.)}
+
+Now, $BC$ and $B'C'$ are both $\parallel$ to $A'D'$ and drawn through $B'$.
+
+\step{$\therefore BC$ and $B'C'$ coincide,}{§~105}
+
+\pnote{(through a given point only one line can be drawn $\parallel$
+  to a given line).}
+
+Also $DC$ and $D'C'$ are $\parallel$ to $A'B'$ and drawn through $D'$.
+
+\step{$\therefore DC$ and $D'C'$ coincide.}{§~105}
+
+\step{$\therefore C$ falls on $C'$,}{§~48}
+
+\pnote{(two lines can intersect in only one point),}
+
+\step{$\therefore$ the two $\Par_s$ coincide, and are equal.}{\qed}
+
+\end{proof}
+
+\pp{\cor{Two rectangles having equal bases and altitudes
+are equal.}}
+\scanpage{063.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If three or more parallels intercept equal parts
+on one transversal, they intercept equal parts on every
+transversal.}
+
+\figc{063aa187}{Let the parallels $AH$, $BK$, $CM$, $DP$ intercept equal parts $HK$, $KM$,
+$MP$ on the transversal $HP$.}
+
+\prove{they intercept equal parts $AB$, $BC$, $CD$ on
+the transversal~$AD$.}
+
+\textbf{Proof.} Suppose $AH$, $BF$, and $CG$ drawn $\parallel$ to $HP$.
+
+\step{$\angle_s$ $AEB$, $BFC$, etc.\ $=\angle_s$ $HKE$, $KMF$, etc., respectively.}{§~112}
+
+\step{But $\angle_s$ $HKE$, $KMF$, etc.\ are equal.}{§~112}
+
+\step{$\therefore \angle_s$ $AEB$, $BFC$, etc.\ are equal.}{Ax.~1}
+
+\step{Also $\angle_s$ $BAE$, $CBF$, etc.\ are equal.}{§~112}
+
+\step{Now $AE = HK$, $BF = KM$, $CG = MP$,}{§~180}
+
+\pnote{(parallels comprehended between parallels are equal).}
+
+\step{$\therefore AE = BF = CG$.}{Ax.~1}
+
+\step{$\therefore \triangle ABE = \triangle BCF = \triangle CDG$,}{§~139}
+
+\pnote{(having two $\angle_s$ and the included side of each respectively equal).}
+
+\step{$\therefore AB = BC = CD$.}{§~128}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{064.png}%
+
+\figc{064aa188}{}
+\begin{point}%
+\cor[1]{If a line is parallel to the base of a triangle
+and bisects one side, it bisects the other
+side also.}
+
+Let $DE$ be $\parallel$ to $BC$ and bisect $AB$. Suppose
+a line is drawn through $A \parallel$ to $BC$.
+Then this line is $\parallel$ to $DE$, by §~106. The
+three parallels by hypothesis intercept
+equal parts on the transversal $AB$, and therefore, by §~187,
+they intercept equal parts on the transversal $AC$; that is, the
+line $DE$ bisects $AC$.
+\end{point}
+
+\begin{point}%
+\cor[2]{The line which joins the middle points of two
+sides of a triangle is parallel to the third side, and is equal
+to half the third side.}
+
+A line drawn through $D$, the middle point of $AB$, $\parallel$ to $BC$,
+passes through $E$, the middle point of $AC$, by §~188. Therefore
+the line joining $D$ and $E$ coincides with this parallel and
+is $\parallel$ to $BC$. Also, since $EF$ drawn $\parallel$ to $AB$ bisects $AC$, it
+bisects $BC$, by §~188; that is, $BF=FC = \frac{1}{2}BC$. But $BDEF$
+is a $\Par$ by §~166, and therefore $DE = BF = \frac{1}{2}BC$.
+\end{point}
+
+\figc{064bb190}{}
+\begin{point}%
+\cor[3]{The median of a trapezoid is parallel to the
+bases, and is equal to half the sum
+of the bases.}
+
+Draw the diagonal $DB$. In the
+$\triangle ADB$ join $E$, the middle point of
+$AD$, to $F$, the middle point of $DB$.
+Then, by §~189, $EF$ is $\parallel$ to $AB$ and $= \frac{1}{2}AB$. In the $\triangle DBC$
+join $F$ to $G$, the middle point of $BC$. Then $FG$ is $\parallel$ to $DC$
+and $=\frac{1}{2}DC$. $AB$ and $FG$, being $\parallel$ to $DC$, are $\parallel$ to each other.
+But only one line can be drawn through $F \parallel$ to $AB$ (§~105).
+Therefore $FG$ is the prolongation of $EF$. Hence, $EFG$ is
+parallel to $AB$ and $DC$, and equal to $\frac{1}{2} (AB + DC)$.
+\end{point}
+\scanpage{065.png}%
+
+
+\section{POLYGONS IN GENERAL.}
+
+\begin{point}%
+A \indexbf{polygon} is a portion of a plane bounded by straight
+lines.
+
+The bounding lines are the sides\label{polysides}, and their sum, the \textbf{perimeter}\label{perimeter2}
+of the polygon. The angles included by the adjacent
+sides are the \textbf{angles}\label{polyangles} of the polygon, and the vertices of these
+angles are the \textbf{vertices} of the polygon\label{polyvertices}. The number of sides
+of a polygon is evidently equal to the number of its angles.
+\end{point}
+
+\begin{point}%
+A \textbf{diagonal}\label{diagonal2} of a polygon is a line joining the vertices
+of two angles not adjacent; as, $AC$ (Fig.~1).
+
+\figc{065ac192}{}
+\end{point}
+
+\begin{point}%
+An \indexbf{equilateral polygon} is a polygon which has all its
+sides equal.
+\end{point}
+
+\begin{point}%
+An \indexbf{equiangular polygon} is a polygon which has all its
+angles equal.
+\end{point}
+
+\begin{point}%
+A \indexbf{convex polygon} is a polygon of which no side, when
+produced, will enter the polygon.
+\end{point}
+
+\begin{point}%
+A \indexbf{concave polygon} is a polygon of which two or more
+sides, if produced, will enter the polygon.
+\end{point}
+
+\begin{point}%
+Each angle of a convex polygon (Fig.~2) is called a
+\emph{salient}\label{salient} angle, and is less than a straight angle.
+\end{point}
+
+\begin{point}%
+The angle $EDF$ of the concave polygon (Fig.~3) is
+called a \emph{re-entrant} angle, and is greater than a straight angle.
+
+When the term polygon is used, a \emph{convex} polygon is meant.
+\end{point}
+\scanpage{066.png}%
+
+\begin{point}%
+Two polygons are \emph{equal} when they can be divided by
+diagonals into the same number of triangles, equal each to each,
+and similarly placed; for if the polygons are applied to each
+other, the corresponding triangles will coincide, and hence the
+polygons will coincide and be equal.
+\end{point}
+
+\begin{point}%
+Two polygons are \indexemph{mutually equiangular}, if the angles
+of the one are equal to the angles of the other, each to each,
+when taken in the same order. Figs.~1 and 2.
+\end{point}
+
+\begin{point}%
+The equal angles in mutually equiangular polygons are
+called \emph{homologous} angles\label{homangles}; and the sides which are included
+by homologous angles are called \emph{homologous} sides\label{homsides}.
+\end{point}
+
+\begin{point}%
+Two polygons are \indexemph{mutually equilateral}, if the sides of
+the one are equal to the sides of the other, each to each, when
+taken in the same order. Figs.~1 and 2.
+
+\figc{066ad203}{}
+\end{point}
+
+\begin{point}%
+Two polygons may be mutually equiangular without
+being mutually equilateral; as, Figs.~4 and 5.
+
+And, \emph{except in the case of triangles}, two polygons may be
+mutually equilateral without being mutually equiangular; as,
+Figs.~6 and 7.
+
+If two polygons are mutually equilateral and mutually equiangular
+\emph{they are equal}, for they can be made to coincide.
+\end{point}
+
+\begin{point}%
+A polygon of three sides is called a \emph{triangle}\label{triangle2}; one of
+four sides, a \emph{quadrilateral}\label{quadrilateral2}; one of five sides, a \indexemph{pentagon}; one
+of six sides, a \indexemph{hexagon}; one of seven sides, a \indexemph{heptagon}; one
+of eight sides, an \indexemph{octagon}; one of ten sides, a \indexemph{decagon}; one of
+twelve sides, a \indexemph{dodecagon}.
+\end{point}
+\scanpage{067.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The sum of the interior angles of a polygon is
+equal to two right angles, taken as many times less two
+as the figure has sides.}
+
+\figc{067aa205}{Let the figure $ABCDEF$ be a polygon, having $n$ sides.}
+
+\prove{$\angle A + \angle B + \angle C$, etc.\ $= (n-2)2$ rt.~$\angle_s$.}
+
+\textbf{Proof.} From $A$ draw the diagonals $AC$, $AD$, and $AE$.
+
+The sum of the $\angle_s$ of the $\triangle_s$ is equal to the sum of the $\angle_s$ of
+the polygon.
+
+\step{Now, there are $(n-2)$~$\triangle_s$,}{}
+
+\step{and the sum of the $\angle_s$ of each $\triangle = 2$ rt.~$\angle_s$.}{§~129}
+
+$\therefore$ the sum of the $\angle_s$ of the $\triangle_s$, that is, the sum of the $\angle_s$ of
+the polygon is equal to $(n-2) 2$ rt.~$\triangle_s$.~\hfill\qed
+
+\end{proof}
+
+\pp{\cor{The sum of the angles of a quadrilateral equals
+4 right angles; and if the angles are all equal, each is a right
+angle. In general, each angle of an equiangular polygon of
+$n$ sides is equal to $\displaystyle \frac{2(n-2)}{n}$ right angles.}}
+
+\ex{How many diagonals can be drawn in a polygon of $n$ sides?}
+\scanpage{068.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The exterior angles of a polygon, made by
+  producing each of its sides in succession, are together equal to
+  four right angles.}
+
+\figc{068aa207}{Let the figure $ABCDE$ be a polygon, having its sides produced
+  in succession.}
+
+\prove[To prove ]{the sum of the ext.~$\angle_s = 4$ rt.~$\angle_s$.}
+
+\textbf{Proof.} Denote the int.~$\angle_s$ of the polygon by $A$, $B$,
+$C$, $D$, $E$, and the corresponding ext.~$\angle_s$ by $a$, $b$, $c$,
+$d$, $e$.
+
+\eq{$\angle A + \angle a$}{$= 2$ rt.~$\angle_s$,}{§~89}
+
+\eq[and]{$\angle B + \angle b$}{$= 2$ rt.~$\angle_s$,}{}
+
+\pnote{(being sup.-adj.~$\angle_s$).}
+
+In like manner each pair of adj.~$\angle_s = 2$ rt.~$\angle_s$.
+
+$\therefore$ the sum of the interior and exterior $\angle_s$ of a
+polygon of $n$ sides is equal to $2n$ rt.~$\angle_s$.
+
+%[** TN: ad hoc visual formatting]
+But the sum of the interior~$\angle_s = (n-2) 2$ rt.~$\angle_s$\hfill§~205 \\
+$\phantom{\text{\indent But the sum of the interior}~\angle_s} = 2 n$ rt.~$\angle_s - 4$ rt.~$\angle_s$.
+
+\step{$\therefore$ the sum of the exterior $\angle_s = 4$ rt.~$\angle_s$.}{\qed}
+
+\end{proof}
+
+\ex{How many sides has a polygon if the sum of its
+interior $\angle_s$ is twice the sum of its exterior $\angle_s$?  ten
+times the sum of its exterior $\angle_s$?}
+\scanpage{069.png}%
+
+
+\section{SYMMETRY.}
+
+\label{symmetry}
+\begin{point}%
+Two points are said to be \textbf{symmetrical} with respect to
+a third point, called the \textbf{centre of symmetry}\label{centresym}, if this third point
+bisects the straight line which joins them.
+
+\figc{069ac208}{}
+
+Two points are said to be \emph{symmetrical} with respect to a
+straight line, called the \textbf{axis of symmetry}\label{axissym}, if this straight line
+bisects at right angles the straight line which joins them.
+
+Thus, $P$ and $P'$ are symmetrical with respect to $O$ as a centre, and $XX'$
+as an axis, if $O$ bisects the line $PP'$, and if $XX'$ bisects $PP'$ at right angles.
+\end{point}
+
+\begin{point}%
+A figure is symmetrical with respect to a point as a
+centre of symmetry, if the point bisects every straight line
+drawn through it and terminated by the boundary of the figure.
+\end{point}
+
+\begin{point}%
+A figure is symmetrical with respect to a line as an
+axis of symmetry if one of the parts of the figure coincides,
+point for point, with the other part when it is folded over on
+that line as an axis.
+\end{point}
+
+\figc{069dd211}{}
+\begin{point}%
+Two figures are said to be symmetrical
+with respect to an axis if every point
+of one has a corresponding symmetrical
+point in the other.
+
+Thus, if every point in the figure $A'B'C'$ has a
+symmetrical point in $ABC$, with respect to $XX'$
+as an axis, the figure $A'B'C'$ is symmetrical to
+$ABC$ with respect to $XX'$ as an axis.
+\end{point}
+\scanpage{070.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{A quadrilateral which has two adjacent sides
+  equal, and the other two sides equal, is symmetrical with respect to
+  the diagonal joining the vertices of the angles formed by the equal
+  sides, and the diagonals are perpendicular to each other.}
+
+\figc{070aa212}{Let $ABCD$ be a quadrilateral, having $AB$ equal to $AD$, and
+  $CB$ equal to $CD$, and having the diagonals $AC$ and $BD$.}
+
+\prove{the diagonal $AC$ is an axis of symmetry, and that
+  it is $\perp$ to the diagonal $BD$.}
+
+\textbf{Proof.} In the $\triangle_s ABC$ and $ADC$,
+
+\step{$AB = AD$, and $BC = DC$,}{Hyp.}
+
+\eq[and]{$AC$}{$= AC$.}{Iden.}
+
+\eq{$\therefore \triangle ABC$}{$= \triangle ADC$.}{§~150}
+
+\step{$\therefore \angle BAC = \angle DAC$, and
+        $\angle BCA = \angle DCA$.}{}
+
+Hence, if $ABC$ is turned on $AC$ as an axis until it falls on $ADC$,
+$AB$ will fall upon $AD$, $CB$ on $CD$, and $OB$ on $OD$.
+
+\step{$\therefore$ the $\triangle ABC$ will coincide with the $\triangle
+  ADC$.}{}
+
+\step{$\therefore AC$ is an axis of symmetry (§~210) and is $\perp$ to
+  $BD$.}{§~208}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{071.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If a figure is symmetrical with respect to two
+axes perpendicular to each other, it is symmetrical with
+respect to their intersection as a centre.}
+
+\figc{071aa213}{Let the figure $ABCDEFGH$ be symmetrical with respect to the two
+perpendicular axes $XX'$, $YY'$, which intersect at $O$.}
+
+\prove{$O$ is the centre of symmetry of the figure.}
+
+\textbf{Proof.}  Let $N$ be any point in the perimeter.
+
+\step{Suppose $NMI$ drawn $\perp$ to $YY'$, $IKL \perp$ to $XX'$.}{}
+
+\step[\indent Then]{$NI$ is $\parallel$ to $XX'$ and $IL$ is $\parallel$ to $YY'$.}{§~104}
+
+\step{Draw $LO$, $ON$, and $KM$.}{}
+
+\eq[\indent Now]{$KI$}{$= KL$,}{§~208}
+
+\pnote{(the figure being symmetrical with respect to $XX'$).}
+
+\eq[\indent But]{$KI$}{$=OM$.}{§~180}
+
+\step{$\therefore KL=OM$, and $KLOM$ is a $\Par$.}{§~183}
+
+\step{$\therefore LO$ is equal and parallel to $KM$.}{§~183}
+
+\step{In like manner $ON$ is equal and parallel to $KM$.}{}
+
+\step{$\therefore LON$ is a straight line.}{§~105}
+
+$\therefore O$ bisects $LN$, \emph{any} straight line and therefore \emph{every} straight
+line drawn through $O$ and terminated by the perimeter.
+
+\step{$\therefore O$ is the centre of symmetry of the figure.}{\qed}
+
+\end{proof}
+\scanpage{072.png}%
+
+\subsection{REVIEW QUESTIONS ON BOOK I.}
+
+\begin{myenum}
+
+\item What is the subject-matter of Geometry?
+
+\item What is a geometric magnitude?
+
+\item What is an axiom? a theorem? a converse theorem? an opposite
+theorem? a contradictory theorem?
+
+\item Define a straight line; a curved line; a broken line; a plane surface;
+a curved surface.
+
+\item How many points are necessary to determine a straight line?
+
+\item How many straight lines are necessary to determine a point?
+
+\item On what does the magnitude of an angle depend?
+
+\item Define a straight angle; a right angle; an oblique angle.
+
+\item Define adjacent angles; complementary angles; supplementary
+angles; conjugate angles.
+
+\item Define parallel lines and give the axiom of parallels.
+
+\item If two lines in the same plane are parallel and cut by a transversal,
+what pairs of angles are equal?  what pairs are supplementary?
+
+\item Define a right triangle; an isosceles triangle; a scalene triangle.
+
+\item To how many right angles is the sum of the angles of a triangle
+equal? the sum of the acute angles of a right triangle?
+
+\item To what angles is the exterior angle of a triangle equal?
+
+\item What is the test of equality of two geometric magnitudes?
+
+\item How does a reciprocal theorem differ from a converse theorem?
+
+\item State the three cases in which two triangles are equal.
+
+\item State the cases in which two right triangles are equal.
+
+\item What is meant by a locus of points?
+
+\item Where are the points located in a plane that are each equidistant
+from two given points? from two intersecting lines?
+
+\item Define a parallelogram; a trapezoid; an isosceles trapezoid.
+
+\item When is a figure symmetrical with respect to a centre?
+
+\item When is a figure symmetrical with respect to an axis?
+
+\item Must a triangle be equiangular if equilateral? must a triangle be
+equilateral if equiangular?
+
+\item When are two polygons said to be mutually equiangular?
+
+\item When are two polygons said to be mutually equilateral?
+
+\item Can two polygons of more than three sides be mutually equiangular
+without being mutually equilateral? mutually equilateral without being
+mutually equiangular?
+
+\item What line do two points each equidistant from the extremities of
+a given straight line determine?
+
+\end{myenum}
+\scanpage{073.png}%
+
+\subsection{METHODS OF PROVING THEOREMS.}
+
+\begin{point}%
+There are \emph{three} general methods of proving theorems,
+the \textbf{synthetic}, the \textbf{analytic}, and the \textbf{indirect} methods.
+
+The \emph{synthetic} method is the method employed in most of
+the theorems already given, and consists in putting together
+known truths in order to obtain a new truth.
+
+The \emph{analytic} method is the reverse of the synthetic method.
+It asserts that the conclusion is true if another proposition is
+true, and so on step by step, until a known truth is reached.
+Thus, proposition $A$ is true if proposition $B$ is true, and $B$ is
+true if $C$ is true; but $C$ \emph{is} true, hence $A$ and $B$ are true.
+
+If a known truth \emph{suggests} the required proof, it is best to
+use the synthetic form at once. If no proof occurs to the
+mind, it is necessary to use the analytic method to \emph{discover}
+the proof, and then the synthetic proof may be given.
+
+The \emph{indirect} method, or the method of \emph{reductio ad absurdum},
+is illustrated on page \pageref{41}. It consists in proving a theorem to
+be true by proving its contradictory to be false.
+\end{point}
+
+\begin{point}%
+Generally auxiliary lines are required, as a line \emph{connecting
+two points}; a line \emph{parallel to or perpendicular to a given
+line}; a line \emph{produced by its own length}; a line \emph{making with
+another line an angle equal to a given angle.}
+
+\textbf{Two lines are proved equal} by proving them \emph{homologous sides
+of equal triangles}; or \emph{legs of an isosceles triangle}; or \emph{opposite
+sides of a parallelogram.}
+
+\textbf{Two angles are proved equal} by proving them \emph{alternate-interior
+angles or exterior-interior angles of parallel lines}; or \emph{homologous
+angles of equal triangles}; or \emph{base angles of an isosceles triangle};
+or \emph{opposite angles of a parallelogram.}
+
+Two suggestions are of special importance to the beginner:
+\begin{myenum}
+\item \emph{Draw as accurate figures as possible.}
+\item \emph{Draw as general figures as possible.}
+\end{myenum}
+\end{point}
+\scanpage{074.png}%
+
+\section{EXERCISES.}
+
+\exheader{Prove by the analytic method:}
+
+\figc{074aaZ19}{}
+\begin{proofex}%
+\obs{A median of a triangle is less than half the sum of the two adjacent
+sides.}
+
+\prove[To prove ]{the median $AD < \frac{1}{2}(AB + AC)$.}
+
+\eq[\indent Now]{}{$AD < \frac{1}{2}(AB + AC)$,}{}
+
+\eq[if]{}{$2AD < AB + AC$.}{}
+
+This suggests producing $AD$ by its own length to $E$,
+and joining $BE$.
+
+\eq[\indent Then]{}{$AE=2AD$,}{}
+
+\step[and]{$2AD<AB+AC$ if $AE<AB+AC$.}{}
+
+\step[\indent But]{$AE<AB+BE$.}{§~138}
+
+\step{$\therefore AE<AB+AC$ if $AC = BE$.}{}
+
+\step[\indent And]{$AC=BE$ if $\triangle ACD = \triangle EBD$.}{§~128}
+
+\eq[\indent But]{$\triangle ACD$}{$=\triangle EBD$.}{§~143}
+
+\eq[\indent For]{$CD$}{$=DB$,}{Hyp.}
+
+\eq{$AD$}{$=DE$,}{Const.}
+
+\eq[and]{$\angle ADC$}{$=\angle BDE$.}{§~93}
+
+\step{$\therefore AE < AB + AC$.}{}
+
+\step{$\therefore AD < \frac{1}{2}(AB+AC)$.}{}
+
+
+\end{proofex}
+
+
+\figc{074bbZ20}{}
+\begin{proofex}%
+\obs{A straight line which bisects two sides of a triangle is parallel
+to the third side.}
+
+If $AD = DB$ and $AE = EC$, to prove $DE\parallel$ to $BC$.
+
+Draw $CG\parallel$ to $BA$, and produce $DE$ to meet it at $G$.
+
+\step{$DE$ is $\parallel$ to $BC$ if $BCGD$ is a $\Par$.}{§~166}
+
+\step{$BCGD$ is a $\Par$ if $CG=BD$.}{§~183}
+
+\step{$CG=BD$ if each is equal to $AD$.}{Ax.~1}
+
+\eq[\indent Now]{$BD$}{$=AD$.}{Hyp.}
+
+\eq[\indent And]{$CG$}{$=AD$ if $\triangle CGE = \triangle ADE$.}{§~128}
+
+\eq[\indent But]{$\triangle CGE$}{$=\triangle ADE$.}{§~139}
+
+\eq[\indent For]{$EC$}{$=AE$.}{Hyp.}
+
+\eq{$\angle GEC$}{$=\angle AED$.}{§~93}
+
+\eq{$\angle ECG$}{$=\angle DAE$.}{§~110}
+
+\step{$\therefore DE$ is $\parallel$ to $BC$.}{}
+
+\end{proofex}
+\scanpage{075.png}%
+
+%\pagebreak
+Prove by the synthetic method:
+
+\begin{proofex}%
+\obs{The middle point of the hypotenuse of a
+right triangle is equidistant from the three vertices.}
+
+From $D$, the middle point, draw $DE \perp$ to $CB$.
+
+$DE$ is $\parallel$ to $AC$ (why?), and $DE$ bisects $CB$ (why?).
+
+$\therefore D$ is equidistant from $B$, $A$, and $C$.    (Why?)
+
+\end{proofex}
+
+\figcc{075aaZ21}{075bbZ22}
+\begin{proofex}%
+\obs{If one acute angle of a right triangle is double the other, the
+hypotenuse is double the shorter leg.}
+
+The median $CD = BD = AD$ (Ex.~21).
+
+Then $\angle b = \angle a$; and $\angle c = \angle 2a$.    (Why?)
+
+Now $a + 2 a = 90°$.    (Why?)
+
+$\therefore \angle a = 30°$; $\angle 2a = 60°$; $\angle c = 60°$.
+
+$\therefore \triangle ACD$ is equilateral (why?), and $AD$, half of
+$AB = AC$.  $\therefore AB = 2AC$.
+
+\end{proofex}
+
+\begin{proofex}%
+\obs{If two triangles have two sides of the one equal, respectively, to
+two sides of the other, and the angles opposite two equal sides equal, the
+angles opposite the other two equal sides are equal or supplementary, and
+if equal the triangles are equal.}
+
+Let $AC = A'C'$, $BC = B'C'$, and $\angle B = \angle B'$.
+
+Place $\triangle A'B'C'$ on $\triangle ABC$ so that $B'C'$ shall coincide with $BC$, and
+$\angle A'$ and $\angle A$ shall be on the same side of $BC$.
+
+\figc{075ceZ23}{}
+
+Since $\angle B'= \angle B$, $B'A'$ will fall along $BA$, and $A'$ will fall at $A$ or at
+some other point in $BA$, as $D$. If $A'$ falls at $A$, the $\triangle_s A'B'C'$ and $ABC$
+coincide and are equal.
+
+If $A'$ falls at $D$, the $\triangle_s A'B'C'$ and $DBC$ coincide and are equal.
+
+Since $CD = C'A'= CA$, $\angle A = \angle CDA$.    (Why?)
+
+But $\angle_s CDA$ and $CDB$ are supplements.    (Why?)
+
+$\therefore \angle_s A$ and $CDB$ are supplements.    (Why?)
+
+Draw figures and show that the triangles are equal:
+
+1. If the given angles $B$ and $B'$ are both right or both obtuse angles.
+
+2. If the required angles $A$ and $A'$ are both acute, both right, or both
+obtuse.
+
+3. If $AC$ and $A'C'$ are not less than $BC$ and $B'C'$, respectively.
+\end{proofex}
+\scanpage{076.png}%
+
+\filbreak
+\figcc{076aaZ24}{076bbZ25}
+\begin{proofex}%
+\obs{The bisectors of the angles of a triangle meet
+  in a point which is equidistant from the sides of the triangle.}
+
+Let the bisectors $AD$ and $BE$ intersect at $O$.  Then $O$ being in
+$AD$ is equidistant from $AC$ and $AB$. (Why?)  And $O$ being in $BE$
+is equidistant from $BC$ and $AB$.  Hence, $O$ is equidistant from
+$AC$ and $BC$, and therefore in the bisector $CF$. (Why?)
+
+\end{proofex}
+
+\begin{proofex}%
+\obs{The perpendicular bisectors of the sides of a
+  triangle meet in a point which is equidistant from the vertices of
+  the triangle.}
+
+Let the $\perp$ bisectors $EE'$ and $DD'$ intersect at $O$.  Then $O$
+being in $EE'$ is equidistant from $A$ and $C$. (Why?)  And $O$ being
+in $DD'$ is equidistant from $A$ and $B$.  Hence, $O$ is equidistant
+from $B$ and $C$, and therefore is in the $\perp$ bisector
+$FF'$. (Why?)
+
+\end{proofex}
+
+%\pagebreak
+\begin{proofex}%
+\obs{The perpendiculars from the vertices of a
+  triangle to the opposite sides meet in a point.}
+
+Let the $\perp_s$ be $AH$, $BP$, and $CK$.  Through $A$, $B$, $C$
+suppose $B'C'$, $A'C'$, $A'B'$, drawn $\parallel$ to $BC$, $AC$, $AB$,
+respectively.  Then $AH$ is $\perp$ to $B'C'$.  (Why?)  Now $ABCB'$
+and $ACBC'$ are $\Par_s$ (why?) and $AB' = BC$, and $AC' = BC$. (Why?)
+That is, $A$ is the middle point of $B'C'$.  In the same way, $B$ and
+$C$ are the middle points of $A'C'$ and $A'B'$, respectively.
+Therefore, $AH$, $BP$, and $CK$ are the $\perp$ bisectors of the sides
+of the $\triangle A'B'C'$.  Hence, they meet in a point. (Why?)
+
+\end{proofex}
+
+\figcc{076ccZ26}{076ddZ27}
+\begin{proofex}%
+\obs{The medians of a triangle meet in a point which
+  is two thirds of the distance from each vertex to the middle of the
+  opposite side.}
+
+Let the two medians $AD$ and $CE$ meet in $O$.  Take $F$ the middle
+point of $OA$, and $G$ of $OC$.  Join $GF$, $FE$, $ED$, and $DG$.  In
+$\triangle AOC$, $GF$ is $\parallel$ to $AC$ and equal to
+$\frac{1}{2}AC$. (Why?)  $DE$ is $\parallel$ to $AC$ and equal to
+$\frac{1}{2}AC$. (Why?)  Hence, $DGFE$ is a $\Par$.  (Why?)  Hence,
+$AF = FO = OD$, and $CG = GO = OE$.  (Why?)  Hence, \emph{any median}
+cuts off \emph{on any other median} two thirds of the distance from
+the vertex to the middle of the opposite side.  Therefore, the median
+from $B$ will cut off $AO$, two thirds of $AD$; that is, will pass
+through $O$.
+
+\end{proofex}
+
+\note{If \emph{three} or more lines pass through the same
+point, they are called \emph{concurrent} lines\label{concurrent}.}
+\scanpage{077.png}%
+
+\filbreak
+\ex{If an angle is bisected, and if a line is drawn through the vertex
+perpendicular to the bisector, this line forms equal angles with the sides
+of the given angle.}
+
+\figc{077adZ28}{}
+
+\ex{The bisectors of two supplementary adjacent angles are perpendicular
+to each other.}
+
+\ex{If the bisectors of two adjacent angles are perpendicular to
+each other, the adjacent angles are supplementary.}
+
+\ex{The bisector of one of two vertical angles bisects the other.}
+
+\ex{The bisectors of two vertical angles form one line.}
+
+\ex{The bisectors of the two pairs of vertical angles formed by two
+intersecting lines are perpendicular to each other.}
+
+\filbreak
+\begin{proofex}%
+The bisector of the vertical angle of an isosceles
+triangle bisects the base, and is perpendicular to the base.
+
+\step{$\triangle ADC = \triangle BDC$ (§~143)}{}
+
+\end{proofex}
+
+\figcc{077eeZ34}{077ffZ35}
+
+\ex{The perpendicular bisector of the base of an
+isosceles triangle passes through the vertex and bisects
+the angle at the vertex (§~160).}
+
+\ex{If the perpendicular bisector of the base of a
+triangle passes through the vertex, the triangle is isosceles.}
+
+\ex{Any point in the bisector of the vertical angle of an isosceles
+triangle is equidistant from the extremities of the base (Ex.~34, §~160).}
+
+\ex{If the bisector of an angle of a triangle is perpendicular to the
+opposite side, the triangle is isosceles.}
+
+\ex{If two isosceles triangles are on the same base, a straight line
+passing through their vertices is perpendicular to the base, and bisects
+the base (§~161).}
+\scanpage{078.png}%
+
+\ex{Two isosceles triangles are equal when a side and an angle of
+the one are equal, respectively, to the homologous side and angle of the
+other.}
+
+\figc{078aaZ41}{}
+\ex{The bisector of an exterior angle of an isosceles
+\phantomsection\label{page:69}% [** TN: Ref. to Exercise 41, p. 69]
+triangle, formed by producing one of the legs through the
+vertex, is parallel to the base. Why does $\angle DAC = \angle B +
+\angle C$? Why is $\angle DAE = \angle ABC$? Why is $AE \parallel$ to $BC$?}
+
+\ex{If the bisector of an exterior angle of a triangle is parallel to
+one side, the triangle is isosceles.}
+
+\figc{078bbZ43}{}
+\begin{proofex}%
+If one of the legs of an isosceles triangle is produced
+through the vertex by its own length, the line joining
+the end of the leg produced to the nearer end of the base is
+perpendicular to the base.
+
+\step{$\angle CBA = \angle A$, and $\angle CBD = \angle D$. (Why?)}{}
+
+\step{$\therefore \angle ABD = \angle A + \angle D$.}{}
+
+\end{proofex}
+
+\ex{A line drawn from the vertex of the right angle of a right triangle
+to the middle point of the hypotenuse divides the triangle into two
+isosceles triangles.}
+
+\ex{If the equal sides of an isosceles triangle are produced through
+the vertex so that the external segments are equal, the extremities of
+these segments will be equally distant from the extremities of the base,
+respectively.}
+
+\figc{078ccZ46}{}
+\ex{If through any point in the bisector of an
+angle a line is drawn parallel to either of the sides of
+the angle, the triangle thus formed is isosceles.}
+
+\ex{Through any point $C$ in the line $AB$ an intersecting line is
+drawn, and from any two points in this line equidistant from $C$ perpendiculars
+are dropped on $AB$ or $AB$ produced.  Prove that these perpendiculars
+are equal.}
+
+\ex{If the median drawn from the vertex of a triangle to the base
+is equal to half the base, the vertical angle is a right angle.}
+
+\figc{078ddZ49}{}
+\ex{The lines joining the middle points of the sides of
+a triangle divide the triangle into four equal triangles.}
+\scanpage{079.png}%
+
+\begin{proofex}%
+The altitudes upon the legs of an isosceles triangle are equal.
+
+\step{Rt.~$\triangle BEC =$ rt.~$\triangle CDB$ (§~141).}{}
+
+\end{proofex}
+
+\begin{proofex}%
+If the altitudes upon two sides of a triangle are equal, the triangle
+is isosceles.
+
+\step{Rt.~$\triangle BEC =$ rt.~$\triangle CDB$ (§~151).}{}
+\end{proofex}
+
+\figc{079adZ51}{}
+
+\begin{proofex}%
+The medians drawn to the legs of an isosceles triangle are equal.
+
+\step{$\triangle BEC = \triangle CDB$ (§~143).}{}
+
+\end{proofex}
+
+\begin{proofex}%
+If the medians to two sides of a triangle are equal, the triangle
+is isosceles.
+
+\step{$BO = CO$, and $OE = OD$ (Ex.~27).}{}
+
+\step{$\angle BOE = \angle COD$. \quad $\therefore \triangle BOE = \triangle COD$ (§~143).}{}
+
+\end{proofex}
+
+\begin{proofex}%
+The bisectors of the base angles of an isosceles triangle are
+equal.
+
+\step{$\triangle BEC = \triangle CDB$ (§~139).}{}
+
+\end{proofex}
+
+\begin{proofex}%
+\textsc{Opposite Theorem.} If a triangle is not isosceles, the bisectors
+of the base angles are not equal.
+
+Let $\angle ABC$ be greater than $\angle ACB$; then $KC > KB$. (Why?)
+
+Now $CD > BE$, if $KD$ is greater than or equal to $KE$.
+
+But suppose $KD < KE$. Lay off $KH = KD$ and $KG = KB$, join $HG$,
+and draw $GF \parallel$ to $BE$.
+
+$\triangle KDB = \triangle KHG$. (Why?) $\therefore \angle KHG = \angle KDB$. (Why?)
+
+$\therefore \angle KEC$ is greater than $\angle KHG$. (Why?) $\therefore GF > HE$. (Why?)
+
+$\angle GFC$ is greater than $\angle FCG$ ($\frac{1}{2}ACB$). $\therefore CG > GF$, and $> HE$.
+
+$\therefore KC - KG > KE - KH$, or $KC + KD > KB + KE$, or $CD > BE$.
+
+\end{proofex}
+
+\ex{State the converse theorem of Ex.~54. Is the converse theorem
+true?}
+
+\figc{079eeZ57}{}
+\begin{proofex}%
+The perpendiculars dropped from the middle
+point of the base upon the legs of an isosceles triangle are
+equal.
+
+\step{$\triangle BED = \triangle CFD$ (§~141).}{}
+
+\end{proofex}
+
+\begin{proofex}%
+State and prove the converse.
+
+\step{$\triangle BED = \triangle CFD$ (§~151).}{}
+
+\end{proofex}
+\scanpage{080.png}%
+
+\filbreak
+\begin{proofex}%
+The difference of the distances from any
+point in the base produced of an isosceles triangle to
+the equal sides of the triangle is constant.
+
+Rt. $\triangle DGC=$ rt.~$\triangle DFC$. (Why?) $\therefore DF = DG$.
+
+$\therefore DE - DF = DE - DG = EG$, the $\perp$ distance
+between the two $\parallel_s$, $BA$ and $CH$.
+
+\end{proofex}
+
+\figcc{080aaZ59}{080bbZ60}
+
+\begin{proofex}%
+The sum of the perpendiculars dropped from any point in the
+base of an isosceles triangle to the legs is constant, and equal to the altitude
+upon one of the legs.
+
+Let $PE$ and $PD$ be the $\perp_s$ and $BF$ the altitude.
+
+Draw $PG \perp$ to $BF$.
+
+$EPGF$ is a parallelogram. (Why?) $\therefore GF = PE$.
+It remains to prove $GB = PD$.
+
+The rt.~$\triangle PGB =$ the rt.~$\triangle BDP$. (Why?)
+\end{proofex}
+
+%\pagebreak
+\begin{proofex}%
+The sum of the perpendiculars dropped from any point within
+an equilateral triangle to the three sides is constant,
+and equal to the altitude.
+
+$AD$ is the altitude, $PE$, $PG$, and $PF$ the three perpendiculars.
+Through $P$ draw $HK \parallel$ to $BC$, meeting
+$AD$ at $M$.
+
+\eq[\indent Then]{$MD$}{$= PE$. (Why?)}{}
+
+\eq{$PG + PF$}{$= AM$ (Ex.~60).}{}
+
+
+\end{proofex}
+
+\figcc{080ccZ61}{080ddZ62}
+
+\ex{$ABC$ and $ABD$ are two triangles on the same base
+$AB$, and on the same side of it, the vertex of each triangle
+being without the other. If $AC$ equals $AD$, show that $BC$
+cannot equal $BD$ (§~154).}
+
+\ex{The sum of the lines which join a point
+within a triangle to the three vertices is less than the
+perimeter, but greater than half the perimeter.}
+
+\figcc{080eeZ63}{080ffZ64}
+\ex{If from any point in the base of an isosceles triangle
+parallels to the legs are drawn, a parallelogram is
+formed whose perimeter is constant, being equal to the sum
+of the legs of the triangle.}
+\scanpage{081.png}%
+
+%\pagebreak
+\ex{The bisector of the vertical angle $A$ of a triangle
+$ABC$, and the bisectors of the exterior angles at
+the base formed by producing the sides $AB$ and $AC$,
+meet in a point which is equidistant from the base and
+the sides produced (§~162).}
+
+\figcc{081aaZ65}{081bbZ66}
+
+\begin{proofex}%
+If the bisectors of the base angles of a triangle
+are drawn, and through their point of intersection a line
+is drawn parallel to the base, the length of this parallel
+between the sides is equal to the sum of the segments of
+the sides between the parallel and the base.
+
+\step{$\angle EOB = \angle OBC = \angle OBE$. \quad $\therefore BE = EO$.}{}
+\end{proofex}
+
+\begin{proofex}%
+The bisector of the vertical angle of a triangle makes with the
+perpendicular from the vertex to the base an angle equal
+to half the difference of the base angles.
+
+Let $\angle B$ be greater than $\angle A$.
+
+\eq{$\angle DCE$}{$= 90° - \angle A - \angle ACD$.}{}
+
+\eq{$\angle ACD$}{$= 90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B$.}{}
+
+\step{$\therefore \angle DCE = 90° - \angle A -
+    (90° - \frac{1}{2}\angle A - \frac{1}{2}\angle B) =
+  \frac{1}{2}\angle B - \frac{1}{2}\angle A$.}{}
+
+\end{proofex}
+
+\figcc{081ccZ67}{081ddZ68}
+\ex{If the diagonals of a quadrilateral bisect each
+other, the figure is a parallelogram.
+
+Prove $\triangle AOB = \triangle COD$.}
+
+\figc{081eeZ69}{}
+\ex{The diagonals of a rectangle are equal.
+
+Prove $\triangle ABC = \triangle BAD$.}
+
+\ex{If the diagonals  of a parallelogram are
+equal, the figure is a rectangle.}
+
+\ex{The diagonals of a rhombus are perpendicular to each other,
+and bisect the angles of the rhombus.}
+
+\ex{The diagonals of a square are perpendicular to each other, and
+bisect the angles of the square.}
+
+\figc{081ffZ73}{}
+\begin{proofex}%
+Lines from two opposite vertices of a parallelogram to the
+middle points of the opposite sides trisect the diagonal.
+
+\step{$EBFD$ is a $\Par$ (why?), and $DF$ is $\parallel$ to $EB$.}{}
+
+\step{$AM = MN$, and $MN = CN$ (§~188).}{}
+\end{proofex}
+\scanpage{082.png}%
+
+\begin{proofex}%
+The lines joining the middle points of the sides of any quadrilateral,
+taken in order, enclose a parallelogram.
+
+Prove $HG$ and $EF \parallel$ to $AC$; and $FG$ and $EH \parallel$ to $BD$ (§~189).
+
+Then $HG$ and $EF$ are each equal to $\frac{1}{2}AC$.
+\end{proofex}
+
+\figc{082adZ74}{}
+
+\ex{The lines joining the middle points of the sides of a rhombus,
+taken in order, enclose a rectangle. (Proof similar to that of Ex.~74.)}
+
+\ex{The lines joining the middle points of the sides of a rectangle
+(not a square), taken in order, enclose a rhombus.}
+
+\ex{The lines joining the middle points of the sides of a square,
+taken in order, enclose a square.}
+
+\begin{proofex}%
+The lines joining the middle points of the sides of an isosceles
+trapezoid, taken in order, enclose a rhombus or a square.
+
+$SHR$ and $QFP$ drawn $\perp$ to $AB$ are parallel. $\therefore PQSR$ is a $\Par$, and by
+Const.~is a rectangle or a square.
+
+$\therefore EFGH$ is a rhombus or a square (Exs.~76, 77).
+
+\figc{082ehZ78}{}
+
+
+\end{proofex}
+
+\ex{The bisectors of the angles of a rhomboid enclose a rectangle.}
+
+\ex{The bisectors of the angles of a rectangle enclose a square.}
+
+\ex{If two parallel lines are cut by a transversal, the bisectors of
+the interior angles form a rectangle.}
+
+\filbreak
+\figc{082iiZ82}{}
+\begin{proofex}%
+The median of a trapezoid passes through the
+middle points of the two diagonals.
+
+The median $EF$ is $\parallel$ to $AB$ and bisects $AD$ (§~190).
+
+$\therefore$ it bisects $DB$.
+
+Likewise $EF$ bisects $BC$ and $BD$.
+
+\end{proofex}
+\scanpage{083.png}%
+
+\begin{proofex}%
+The lines joining the middle points of the diagonals of a trapezoid
+is equal to half the difference of the bases.
+
+\step{$\triangle BFG = \triangle DFC$. (Why?)
+  $\therefore EF = \frac{1}{2}AG$ (§~180).}{}
+
+\step{$CF=FG$, $DC=BG$.}{}
+
+\step{$\therefore AG=AB-DC$. $\therefore EF=\frac{1}{2}(AB-DC)$}{}
+
+\figc{083adZ83}{}
+
+\end{proofex}
+
+\begin{proofex}%
+In an isosceles trapezoid each base makes equal angles with the legs.
+
+Draw $CE \parallel$ to $DB$.  $CE=DB$. (Why?)  $\angle A = \angle CEA$,
+$\angle B = \angle CEA$, $\angle_s C$ and $D$ have equal supplements.
+
+\end{proofex}
+
+\ex{If the angles at the base of a trapezoid are equal, the other angles are equal, and the trapezoid is isosceles.}
+
+
+\begin{proofex}%
+In an isosceles trapezoid the opposite angles are supplementary:
+
+\step{$\angle C = \angle D$ (Ex.~84)}{}
+\end{proofex}
+
+
+\begin{proofex}%
+The diagonals on an isosceles trapezoidal are equal.
+
+Prove $\triangle ACD = \triangle BDC$.
+\end{proofex}
+
+%\pagebreak
+\begin{proofex}%
+If the diagonals of a trapezoid are equal, the trapezoid is isosceles.
+
+Draw $CE$ and $DF \perp$ to $AB$.
+
+\eq{$\triangle ADF$}{$= \triangle BCE$.}{(Why?)}
+
+\eq{$\therefore \angle ADF$}{$= \angle CBA$.}{}
+
+\eq{$\triangle ABC$}{$= \triangle BAD$.}{}
+
+\end{proofex}
+
+\figcc{083eeZ88}{083ffZ89}
+
+\begin{proofex}%
+If from the diagonal $DB$, of a square $ABCD$, $BE$
+is cut off equal to $BC$, and $EF$ is drawn perpendicular
+to $BD$ meeting $DC$ at $F$, then $DE$ is equal to $EF$ and
+also to $FC$.
+
+$\angle EDF = 45°$, and $\angle DFE = 45°$; and $DE=DF$.
+Rt.~$\triangle BEF =$ rt.~$\triangle BCF$ (§~151); and $EF=FC$.
+
+\end{proofex}
+
+\ex{Two angles whose sides are so perpendicular, each to each, are either equal or supplementary.}
+\scanpage{084.png}%
+
+
+\chapter{BOOK II\@. THE CIRCLE.}
+
+\section{DEFINITIONS.}
+
+\begin{point}%
+A \textbf{circle}\label{circle} is a portion of a plane bounded by a curved
+line, all points of which are equally distant from a point within
+called the \textbf{centre}\label{centrecirc}. The bounding line is called the \textbf{circumference}\label{circumference}
+of the circle.
+\end{point}
+
+\begin{point}%
+A \textbf{radius} is a straight line from the centre to the circumference;
+and a \textbf{diameter}\label{diameter} is a straight line through the
+centre, with its ends in the circumference.
+
+By the definition of a circle, \emph{all its radii are equal}. All its
+diameters are equal, since a diameter is equal to two radii.
+\end{point}
+
+\begin{point}%
+\textbf{Postulate.} A circumference can be described from any
+point as a centre, with any given radius.
+\end{point}
+
+\begin{point}%
+A \indexbf{secant} is a straight line of unlimited length which
+intersects the circumference in two points; as, $AD$ (Fig.~1).
+\end{point}
+
+\figc{084aa220}{}
+
+\begin{point}%
+A \indexbf{tangent} is a straight line of unlimited length which
+has one point, and only one, in common
+with the circumference; as, $BC$ (Fig.~1).
+In this case the circle is said to be tangent
+to the straight line. The common
+point is called the \indexbf{point of contact}, or
+\indexbf{point of tangency}.
+\end{point}
+
+\begin{point}%
+Two \emph{circles} are tangent to each
+other, if both are tangent to a straight line at the same point;
+and are said to be tangent \emph{internally} or \emph{externally}, according
+as one circle lies wholly \emph{within} or \emph{without} the other.
+\end{point}
+\scanpage{085.png}%
+
+\pp{An \textbf{arc}\label{arc} is any part of the circumference; as, $BC$ (Fig.~3).
+Half a circumference is called a \indexbf{semicircumference}. Two arcs
+are called \textbf{conjugate arcs}, if their sum is a circumference.}
+
+\pp{A \textbf{chord}\label{chord} is a straight line that has its extremities in
+the circumference; as, the straight line $BC$ (Fig.~3).}
+
+\pp{A chord subtends two conjugate arcs. If the arcs are
+unequal, the less is called the \indexbf{minor} arc, and the greater the
+\indexbf{major} arc. A minor arc is generally called simply an arc.}
+
+\figc{085ac224}{}
+
+\pp{A \indexbf{segment} of a circle is a portion of the circle bounded
+by an arc and its chord (Fig.~2).}
+
+\pp{A \indexbf{semicircle} is a segment equal to half the circle (Fig.~2).}
+
+\pp{A \indexbf{sector} of a circle is a portion of the circle bounded
+by two radii and the arc which they intercept. The angle
+included by the radii is called the \emph{angle of the sector} (Fig.~2).}
+
+\pp{A \indexbf{quadrant} is a sector equal to a quarter of the circle
+(Fig.~2).}
+
+\pp{An angle is called a \textbf{central angle}\label{central}, if its vertex is at the
+centre and its sides are radii of the circle; as, $\angle AOD$ (Fig.~2).}
+
+\begin{point}%
+An angle is called an \textbf{inscribed angle}\label{inscribedcirc}, if its vertex is in
+the circumference and its sides are chords; as, $\angle ABC$ (Fig.~3).
+
+An angle is \emph{inscribed in a segment}\label{inscribedseg}, if its vertex is in the
+arc of the segment and its sides pass through the extremities
+of the arc.
+\end{point}
+\scanpage{086.png}%
+
+\pp{A polygon is \emph{inscribed in a circle}\label{polyinscribed}, if its sides are chords;
+and a circle is \emph{circumscribed about a polygon}\label{circcircumscribed}, if all the vertices
+of the polygon are in the circumference (Fig.~3).}
+
+\pp{A circle is \emph{inscribed in a polygon}\label{circinscribed}, if the sides of the
+polygon are tangent to the circle; and a polygon is \emph{circumscribed
+about}\label{polycircumscribed} a circle if its sides are tangents (Fig.~4).}
+
+\begin{point}%
+\emph{Two circles are equal, if they have equal radii.}
+
+For they will coincide, if their centres are made to coincide.
+
+\textsc{Conversely:} \emph{Two equal circles have equal radii.}
+\end{point}
+
+\pp{\emph{Two circles are concentric}\label{concentric}, if they have the same centre.}
+
+\filbreak
+\section{ARCS, CHORDS, AND TANGENTS.}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{A straight line cannot meet the circumference of
+a circle in more than two points}.
+
+\figc{086aa235}{Let $HK$ be any line meeting the circumference $HKM$ in $H$ and $K$.}
+
+\prove{$HK$ cannot meet the circumference in any
+other point}.
+
+\textbf{Proof.} If possible, let $HK$ meet the circumference in $P$.
+
+\step{Then the radii $OH$, $OP$, and $OK$ are equal.}{§~217}
+
+\step{$\therefore P$ does not lie in the straight line $HK$.}{§~102}
+
+\step{$\therefore HK$ meets the circumference in only two points.}{\llap{\qed}}
+
+\end{proof}
+\scanpage{087.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In the same circle or in equal circles, equal
+central angles intercept equal arcs; and of two unequal
+central angles the greater intercepts the greater arc.}
+
+\figc{087ab236}{In the equal circles whose centres are $O$ and $O'$, let the angles
+$AOB$ and $A'O'B'$ be equal, and angle $AOC$ be greater than angle  $A'O'C'$.}
+
+\prove{1. $\arc AB = \arc A'B'$;}
+
+\prove[\phantom{To prove that~}]{2. $\arc AC > \arc A'B'$.}
+
+\textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that the
+$\angle A'O'B'$ shall coincide with its equal, the $\angle AOB$.
+
+\step{Then $A'$ falls on $A$, and $B'$ on $B$.}{§~233}
+
+\step{$\therefore$ $\arc A'B'$ coincides with $\arc AB$.}{§~216}
+
+\textbf{2.~} Since the $\angle AOC$ is greater than the $\angle A'O'B'$,
+it is greater than the $\angle AOB$, the equal of the $\angle A'O'B'$.
+
+\step{Therefore, $OC$ falls without the $\angle AOB$.}{}
+
+\step{$\therefore$ $\arc AC > \arc AB$.}{Ax.~8}
+
+\step{$\therefore$ $\arc AC > \arc A'B'$, the equal of $\arc AB$.}{\qed}
+
+\end{proof}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} In the same circle or in equal circles,
+equal arcs subtend equal central angles; and of two unequal
+arcs the greater subtends the greater central angle.}
+\scanpage{088.png}%
+
+\proveq{\textup{1.} $\angle AOB$}{$=\angle A'O'B'$;}
+
+\proveq[\indent]{\settowidth{\TmpLen}{\textit{To prove that}}\rule{\TmpLen}{0pt}\textup{2.} $\angle AOC$}{is greater than $\angle A'O'B'$.}
+
+\textbf{Proof. 1.} Place the $\odot A'B'P'$ on the $\odot ABP$ so that $O'A$
+shall fall on its equal $OA$, and the arc $A'B'$ on its equal $AB$.
+
+\step{Then $O'B'$ will coincide with $OB$.}{§~47}
+
+\step{$\therefore \angle A'O'B'=\angle AOB$.}{§~60}
+
+\textbf{2.~}Since $\arc AC>A'B'$, it is greater than $\arc AB$, the equal
+of $A'B'$, and $OB$ will fall within the $\angle AOC$.
+
+\eq{}{$\therefore \angle AOC$ is greater than $\angle AOB$.}{Ax.~8}
+
+\eq{}{$\therefore \angle AOC$ is greater than $\angle A'O'B'$.}{\qed}
+\end{proof}
+
+
+\pp{\cor[1]{In the same circle or in equal circles, two sectors
+that have equal angles are equal; two sectors that have
+unequal angles are unequal, and the greater sector has the
+greater angle.}}
+
+\pp{\cor[2]{In the same circle or in equal circles, equal
+sectors have equal angles; and of two unequal sectors the
+greater has the greater angle.}}
+
+\begin{point}%
+\textbf{Law of Converse Theorems.}\label{converse2} It was stated in §~32 that the converse
+of a theorem is not necessarily true. If, however, a theorem is in
+fact a group of three theorems, and if \emph{one of the hypotheses} of the group
+\emph{must} be true, and \emph{no two of the conclusions can be true at the same time},
+then the converse of the theorem is \emph{necessarily} true.
+
+Proposition II. is a group of three theorems. It asserts that the arc
+$AB$ is equal to the arc $A'B'$, if the angle $AOB$ is equal to the angle
+$A'O'B'$; that the arc $AB$ is greater than the arc $A'B'$, if the angle $AOB$
+is greater than the angle $A'O'B'$; that the arc $AB$ is less than the arc
+$A'B'$, if the angle $AOB$ is less than the angle $A'O'B'$.
+
+One of these hypotheses must be true; for the angle $AOB$ must be
+equal to, greater than, or less than, the angle $A'O'B'$.
+
+No two of the conclusions can be true at the same time, for the arc $AB$
+cannot be both equal to and greater than the arc $A'B'$; nor can it be both
+equal to and less than the arc $A'B'$; nor both greater than and less than
+the arc $A'B'$. In such a case, the converse theorem is \emph{necessarily} true,
+and no proof like that given in the text is required to establish it.
+\end{point}
+\scanpage{089.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In the same circle or in equal circles, equal arcs
+  are subtended by equal chords; and of two unequal arcs the greater
+  is subtended by the greater chord.}
+
+\figc{089ab241}{In the equal circles whose centres are $O$ and $O'$, let the arcs
+  $AB$ and $A'B'$ be equal, and the arc $AF$ greater than arc $A'B'$.}
+
+\proveq{\textup{1. }chord $AB$}{$=$ chord $A'B'$;}
+
+\proveq[]{\textup{2. }chord $AF$}{$>$ chord $A'B'$.}
+
+\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{}
+
+\step[\indent 1.]{The $\triangle_s AOB$ and $A'O'B'$ are equal.}{§~143}
+
+\step{For $OA = O'A'$, and $OB = O'B'$,}{§~233}
+
+\pnote{(radii of equal circles),}
+
+\step{and $\angle AOB = \angle A'O'B'$,}{§~237}
+
+\pnote{(in equal $\odot_s$ equal arcs subtend equal central $\angle_s$).}
+
+\step{$\therefore$ chord $AB = $ chord $A'B'$.}{§~128}
+
+\step{}{}
+
+\step[\indent 2.]{In the $\triangle_s AOF$ and $A'O'B'$,}{}
+
+\step{$OA = O'A'$, and $OF = O'B'$.}{§~233}
+
+\step{But the $\angle AOF$ is greater than the $\angle A'O'B'$,}{§~237}
+
+\pnote{(in equal $\odot_s$, the greater of two unequal arcs subtends the
+  greater $\angle$).}
+
+\step{$\therefore$ chord $AF >$ chord $A'B'$.}{§~154}
+
+\hfill\qed
+
+\end{proof}
+
+
+\pp{\cor{In the same circle or in equal
+  circles, the greater of two unequal major arcs is subtended by the
+  less chord.}}
+\scanpage{090.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} In the same circle or in equal
+circles, equal chords subtend equal arcs; and of two
+unequal chords the greater subtends the greater arc.}
+
+\figc{090ab243}{In the equal circles whose centres are $O$ and $O'$, let the chords $AB$
+and $A'B'$ be equal, and the chord $AF$ greater than $A'B'$.}
+
+\prove{\quad\upshape{1.} $\arc AB = \arc A'B'$;}
+
+\prove[\phantom{To prove that~}]{\quad\upshape{2.} $\arc AF > \arc A'B'$.}
+
+\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OF$, $O'A'$, $O'B'$.}{}
+
+\step[\indent 1.]{The $\triangle_s OAB$ and $O'A'B'$ are equal.}{§~150}
+
+\step{For $OA = O'A'$, and $OB = O'B'$,}{§~233}
+
+\step{and chord $AB =$ chord $A'B'$.}{Hyp.}
+
+\step{$\therefore \angle AOB = \angle A'O'B'$.}{§~128}
+
+\step{$\therefore \arc AB = \arc A'B'$,}{§~236}
+
+\pnote{(in equal $\odot_s$ equal central $\angle_s$ intercept equal arcs).}
+
+
+\step[\indent 2.]{In the $\triangle_s OAF and O'A'B'$,}{}
+
+\step{$OA = O'A'$ and $OF = O'B'$.}{§~233}
+
+\step{But chord $AF >$ chord $A'B'$.}{Hyp.}
+
+\step{$\therefore$ the $\angle AOF$ is greater than the $\angle A'O'B'$.}{§~155}
+
+\step{$\therefore \arc AF > \arc A'B'$,}{§~236}
+
+\pnote{(in equal $\odot_s$ the greater central $\angle$ intercepts the greater arc).}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{In the same circle or in equal circles, the greater
+of two unequal chords subtends the less major arc.}}
+\scanpage{091.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{A diameter perpendicular to a chord bisects the
+  chord and the arcs subtended by it.}
+
+\figc{091aa245}{Let $ES$ be a diameter perpendicular to the chord $AB$ at $M$.}
+
+\prove{$AM = BM$, $AS = BS$, and $AE = BE$.}
+
+\textbf{Proof.} Draw $OA$ and $OB$ from $O$, the centre of the circle.
+
+\step{The rt.~$\triangle_s OAM$ and $OBM$ are equal.}{§~151}
+
+\eq[\indent For]{$OM$}{$= OM$,}{Iden.}
+
+\eq[and]{$OA$}{$= OB$.}{§~217}
+
+\step{$\therefore AM = BM$, and $\angle AOS = \angle BOS$.}{§~128}
+
+\eq[\indent Likewise]{$\angle AOE$}{$= \angle BOE$.}{§~85}
+
+\step{$\therefore AS = BS$, and $AE = BE$.}{§~236}
+
+\hfill\qed
+
+\end{proof}
+
+
+\pp{\cor[1]{A diameter bisects the
+  circumference and the circle.}}
+
+\pp{\cor[2]{A diameter which bisects a chord
+  is perpendicular to it.}}
+
+\pp{\cor[3]{The perpendicular bisector of a
+  chord passes through the centre of the circle, and bisects the arcs
+  of the chord.}}
+\scanpage{092.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In the same circle or in equal circles, equal
+  chords are equally distant from the centre. \textsc{Conversely:}
+  Chords equally distant from the centre are equal.}
+
+\figc{092aa249}{Let $AB$ and $CF$ be equal chords of the circle $ABFC$.}
+
+\prove{$AB$ and $CF$ are equidistant from the centre $O$.}
+
+\textbf{Proof.} Draw $OP \perp$ to $AB$, $OH \perp$ to $CF$, and join
+$OA$ and $OC$.
+
+\step{$OP$ bisects $AB$, and $OH$ bisects $CF$.}{§~245}
+
+\step{The rt.~$\triangle_s OPA$ and $OHC$ are equal.}{§~151}
+
+\eq{$AP$}{$= CH$,}{Ax.~7}
+
+\eq[and]{$OA$}{$= OC$,}{§~217}
+
+\eq[\indent Hence,]{$OP$}{$= OH$.}{§~128}
+
+\step{$\therefore AB$ and $CF$ are equidistant from $O$.}{}
+
+\step{}{}
+
+\eq[\indent \textsc{Conversely:}]{\textbf{Let }$OP$}{$=OH$.}{}
+
+\proveq[\indent To prove]{$AB$}{$= CF$.}
+
+\textbf{Proof.} The rt.~$\triangle_s OPA$ and $OHC$ are equal.~\hfill§~151
+
+\eq[\indent For]{$OA$}{$= OC$,}{§~217}
+
+\eq[and]{$OP$}{$= OH$,}{Hyp.}
+
+\eq[\indent Hence,]{$AP$}{$= CH$.}{§~128}
+
+\eq{$\therefore AB$}{$= CF$.}{Ax.~6}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{093.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In the same circle or in equal circles, if two chords
+are unequal, they are unequally distant from the centre;
+and the greater chord is at the less distance.}
+
+\figc{093aa250}{In the circle whose centre is $O$, let the chords $AB$ and $CD$ be
+unequal, and $AB$ the greater; and let $OE$ be perpendicular to $AB$ and
+$OF$ perpendicular to $CD$.}
+
+\proveq{$OE$}{$< OF$.}
+
+\textbf{Proof.} Suppose $AG$ drawn equal to $CD$, and $OH \perp$ to $AG$.
+
+\step{Draw $EH$.}{}
+
+\step{$OE$ bisects $AB$, and $OH$ bisects $AG$.}{§~245}
+
+\eq[By hypothesis,]{$AB$}{$> CD$.}{}
+
+\step{$\therefore AB > AG$, the equal of $CD$.}{}
+
+\eq{$\therefore AE$}{$> AH$.}{Ax.~7}
+
+\step{$\therefore \angle AHE$ is greater than $\angle AEH$.}{§~152}
+
+$\therefore \angle OHE$, the complement of $\angle AHE$, is less than $\angle OEH$,
+the complement of $\angle AEH$.\hfill~Ax.~5
+
+\eq{$\therefore OE$}{$< OH$.}{§~153}
+
+\eq[\indent But]{$OH$}{$=OF$.}{§~249}
+
+\eq{$\therefore OE$}{$< OF$.}{\qed}
+
+\end{proof}
+
+\ex{The perpendicular bisectors of the sides of an inscribed polygon
+are concurrent (pass through the same point).}
+\scanpage{094.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} In the same circle or in
+  equal circles, if two chords are unequally distant from the centre,
+  they are unequal; and the chord at the less distance is the greater.}
+
+\figc{094aa251}{In the circle whose centre is $O$, let $AB$ and $CD$ be
+  unequally distant from $O$; and let $OE$, the perpendicular to $AB$,
+  be less than $OF$, the perpendicular to $CD$.}
+
+\proveq{$AB$}{$> CD$.}
+
+\textbf{Proof.} Suppose $AG$ drawn equal to $CD$,
+  and $OH \perp$ to $AG$.
+
+\eq[\indent Then]{$OH$}{$= OF$}{§~249}
+
+\eq[\indent Hence,]{$OE$}{$< OH$.}{}
+
+\step{Draw $EH$.}{}
+
+\step{$\angle OHE$ is less than $\angle OEH$.}{§~152}
+
+$\therefore \angle AHE$, the complement of $\angle OHE$, is greater
+than $\angle AEH$, the complement of $\angle OEH$.\hfill~Ax.~5
+
+\eq{$\therefore AE$}{$> AH$.}{§~153}
+
+\step[\indent But]{$AE = \frac{1}{2}AB$, and $AH = \frac{1}{2}AG$.}{§~245}
+
+\eq{$\therefore AB$}{$> AG$.}{Ax.~6}
+
+\eq[\indent But]{$CD$}{$= AG$.}{Const.}
+
+\eq{$\therefore AB$}{$> CD$.}{\qed}
+
+
+\end{proof}
+
+\pp{\cor{A diameter of a circle is greater
+  than any other chord.}}
+\scanpage{095.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{A straight line perpendicular to a radius at its
+  extremity is a tangent to the circle.}
+
+\figc{095aa253}{Let $MB$ be perpendicular to the radius $OA$ at $A$.}
+
+\prove{$MB$ is a tangent to the circle.}
+
+\textbf{Proof.} From $O$ draw any other line to $MB$, as $OH$.
+
+\eq[\indent Then]{$OH$}{$> OA$.}{§~97}
+
+\step{$\therefore$ the point $H$ is without the circle.}{§~216}
+
+Hence, \emph{every point}, except $A$, of the line $MB$ is without the
+circle, and therefore $MB$ is a tangent to the circle at $A$.~\hfill§~220
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor[1]{A tangent to a circle is
+  perpendicular to the radius drawn to the point of contact.}
+
+For $OA$ is the shortest line from $O$ to $MB$, and is therefore
+$\perp$ to $MB$ (§~98); that is, $MB$ is $\perp$ to $OA$.
+\end{point}
+
+\begin{point}%
+\cor[2]{A perpendicular to a tangent at
+  the point of contact passes through the centre of the circle.}
+
+For a radius is $\perp$ to a tangent at the point of contact, and
+therefore a $\perp$ erected at the point of contact coincides with
+this radius and passes through the centre.
+\end{point}
+
+\pp{\cor[3]{A perpendicular from the centre
+  of a circle to a tangent passes through the point of contact.}}
+\scanpage{096.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Parallels intercept equal arcs on a circumference.}
+
+\figc{096ac257}{\textnormal{\textsc{Case 1.~}} Let $AB$ \textnormal{(Fig.~1)} be a tangent at
+  $F$ parallel to $CD$, a secant.}
+
+\proveq{$\arc CF$}{$= \arc DF$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $FF'$ drawn $\perp$ to $AB$.}{}
+
+\step{Then $FF'$ is a diameter of the circle.}{§~255}
+
+\step{And $FF'$ is also $\perp$ to $CD$.}{§~107}
+
+\step{$\therefore CF = DF$, and $CF' = DF'$.}{§~245}
+
+\step{}{}
+
+\textsc{Case 2.~}\textbf{Let $AB$ and $CD$} (Fig.~2) \textbf{be
+  parallel secants.}
+
+\proveq{$\arc AC$}{$= \arc BD$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $EF \parallel$ to $CD$ and tangent to the
+circle at $M$.}{}
+
+\eq[\indent Then]{$\arc AM$}{$= \arc BM$,}{Case~1}
+
+\eq[and]{$\arc CM$}{$= \arc DM$.}{}
+
+%proofrule
+\eq{$\therefore \arc AC$}{$= \arc BD$.}{Ax.~3}
+
+\step{}{}
+
+\textsc{Case 3.~}\textbf{Let $AB$ and $CD$} (Fig.~3) \textbf{be
+  parallel tangents at $E$ and $F$.}
+
+\proveq{$\arc EGF$}{$= \arc EHF$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $GH$ drawn $\parallel$ to $AB$.}{}
+
+\eq[\indent Then]{$\arc EG$}{$= \arc EH$,}{Case~1}
+
+\eq[and]{$\arc GF$}{$= \arc HF$.}{}
+
+%proofrule
+\eq{$\therefore \arc EGF$}{$= \arc EHF$.}{Ax.~2}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{097.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Through three points not in a straight line one
+circumference, and only one, can be drawn.}
+
+\figc{097aa258}{Let $A$, $B$, $C$ be three points not in a straight line.}
+
+\prove{one circumference, and only one, can be drawn
+through $A$, $B$, and~$C$.}
+
+\step[\indent\textbf{Proof.}]{Draw $AB$ and $BC$.}{}
+
+At the middle points of $AB$ and $BC$ suppose $\perp_s$ erected.
+
+These $\perp_s$  will intersect at some point $O$, since $AB$ and $BC$
+are not in the same straight line.
+
+The point $O$ is in the perpendicular bisector of $AB$, and is
+therefore equidistant from $A$ and $B$; the point $O$ is also in
+the perpendicular bisector of $BC$, and is therefore equidistant
+from $B$ and $C$.~\hfill§~160
+
+Therefore, $O$ is equidistant from $A$, $B$, and $C$; and a circumference
+described from $O$ as a centre, with a radius $OA$,
+will pass through the three given points.
+
+The centre of a circumference passing through the three
+points must be in both perpendiculars, and hence at their
+intersection. As two straight lines can intersect in only one
+point, $O$ is the centre of the only circumference that can pass
+through the three given points.~\hfill\qed
+
+\end{proof}
+
+\pp{\cor{Two circumferences can intersect in only two
+points. \textup{For, if two circumferences have three points common,
+they coincide and form one circumference.}}}
+\scanpage{098.png}%
+
+\pp{\defn{A \textbf{tangent from an external point
+  to a circle}\label{tangent2} is the part of the tangent between the external point
+  and the point of contact.}}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The tangents to a circle drawn from an external
+  point are equal, and make equal angles with the line joining the
+  point to the centre.}
+
+\figc{098aa261}{Let $AB$ and $AC$ be tangents from $A$ to the circle whose
+  centre is $O$, and let $AO$ be the line joining $A$ to the centre
+  $O$.}
+
+\prove{$AB = AC$, and $\angle BAO = \angle CAO$.}
+
+\step[\indent\textbf{Proof.}]{Draw $OB$ and $OC$.}{}
+
+\step{$AB$ is $\perp$ to $OB$, and $AC \perp$ to $OC$,}{§~254}
+
+\pnote{(a tangent to a circle is $\perp$
+    to the radius drawn to the point of contact).}
+
+\step{The rt.~$\triangle_s OAB$ and $OAC$ are equal.}{§~151}
+
+For $OA$ is common, and the radii $OB$ and $OC$ are equal.~\hfill§~217
+
+\step{$\therefore AB=AC$, and $\angle BAO = \angle CAO$.}{§~128}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\defn{The line joining the centres of two
+circles is called the \indexbf{line of centres}.}}
+
+\pp{\defn{A tangent to two circles is called a
+\indexbf{common external tangent} if it does not cut the line of
+centres, and a \indexbf{common internal tangent} if it cuts the line of centres.}}
+\scanpage{099.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two circles intersect each other, the line of
+  centres is perpendicular to their common chord at its middle point.}
+
+\figc{099aa264}{Let $C$ and $C'$ be the centres of the two circles, $AB$ the
+  common chord, and $CC'$ the line of centres.}
+
+\prove{$CC'$ is $\perp$ to $AB$ at its middle point.}
+
+\step[\indent\textbf{Proof.}]{Draw $CA$, $CB$, $C'A$, and $C'B$.}{}
+
+\step{$CA = CB$, and $C'A = C'B$.}{§~217}
+
+\step{$\therefore C$ and $C'$ are two points, each
+  equidistant from $A$ and $B$.}{}
+
+\step{$\therefore CC'$ is the perpendicular bisector of $AB$.}{§~161}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{proofex}%
+Describe the relative position of two circles if the
+line of centres:
+\begin{myenum}
+\item is greater than the sum of the radii;
+\item is equal to the sum of the radii;
+\item is less than the sum but greater than the difference of the
+  radii;
+\item is equal to the difference of the radii;
+\item is less than the difference of the radii.
+\end{myenum}
+
+Illustrate each case by a figure.
+
+\end{proofex}
+
+\ex{The straight line drawn from the middle point of a
+chord to the middle point of its subtended arc is perpendicular to the
+chord.}
+
+\ex{The line which passes through the middle points of
+two parallel chords passes through the centre of the circle.}
+\scanpage{100.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two circles are tangent to each other, the line
+of centres passes through the point of contact.}
+
+\figc{100aa265}{Let the two circles, whose centres are $C$ and $C'$, be tangent to the
+straight line $AB$ at $Q$, and $CC'$ the line of centres.}
+
+\prove{$O$ is in the straight line $CC'$.}
+
+\textbf{Proof.}  A $\perp$ to $AB$, drawn through the point $O$, passes through
+the centres $C$ and $C'$,~\hfill§~255
+
+\pnote{(a $\perp$ to a tangent at the point of contact passes through the centre
+of the circle).}
+
+$\therefore$ the line $CC'$, having two points in common with this $\perp$
+must coincide with it.~\hfill§~47
+
+\step{$\therefore O$ is in the straight line $CC'$.}{\qed}
+
+\end{proof}
+
+\begin{proofex}%
+Describe the relative position of two circles if they may have:
+\begin{myenum}
+\item two common external and two common internal tangents;
+\item two common external tangents and one common internal tangent;
+\item two common external tangents and no common internal tangent;
+\item one common external and no common internal tangent;
+\item no common tangent.
+\end{myenum}
+
+Illustrate each case by a figure.
+
+\end{proofex}
+
+\ex{The line drawn from the centre of a circle to the point of intersection
+of the two tangents is the perpendicular bisector of the chord joining
+the points of contact.}
+\scanpage{101.png}%
+
+
+\section{MEASUREMENT.}
+
+\begin{point}%
+To \textbf{measure} a quantity of any kind is to find \emph{the number
+of times} it contains a known quantity of the \emph{same kind}, called
+the \textbf{unit of measure}.
+
+The \emph{number} which shows the number of times a quantity
+contains the unit of measure is called the \indexbf{numerical measure} of
+that quantity.
+\end{point}
+
+\begin{point}%
+No quantity is great or small except by comparison
+with another quantity of the \emph{same kind}. This comparison is
+made by finding the numerical measures of the two quantities
+in terms of a common unit, and then dividing one of the
+measures by the other.
+
+The quotient is called their \indexbf{ratio}. In other words the ratio
+of two quantities of the same kind is the \emph{ratio} of their \emph{numerical
+measures} expressed in terms of a common unit.
+
+The ratio of $a$ to $b$ is written $a : b$, or $\dfrac{a}{b}$.
+\end{point}
+
+\begin{point}%
+Two quantities that can be expressed in \emph{integers} in
+terms of a common unit are said to be \textbf{commensurable}\label{commensurable}, and the
+exact value of their ratio can be found. The common unit is
+called their \emph{common measure}, and each quantity is called a
+\emph{multiple} of this common measure.
+
+Thus, a common measure of $2\frac{1}{2}$~feet and $3\frac{2}{3}$~feet is $\frac{1}{6}$~of a foot, which is
+contained $15$~times in $2\frac{1}{2}$~feet, and $22$~times in $3\frac{2}{3}$~feet. Hence, $2\frac{1}{2}$~feet
+and $3\frac{2}{3}$~feet are multiples of $\frac{1}{6}$~of a foot, since $2\frac{1}{2}$~feet may be obtained by
+taking $\frac{1}{6}$~of a foot $15$~times, and $3\frac{2}{3}$~feet by taking $\frac{1}{6}$~of a foot $22$~times. The
+ratio of $2\frac{1}{2}$~feet to $3\frac{2}{3}$~feet is expressed by the fraction~$\frac{15}{22}$.
+\end{point}
+
+\begin{point}%
+Two quantities of the same kind that cannot \emph{both} be
+expressed in \emph{integers} in terms of a common unit, are said to be
+\textbf{incommensurable}, and the \emph{exact value} of their ratio cannot be
+found. But by taking the unit sufficiently small, an \emph{approximate
+value} can be found that shall differ from the true value
+of the ratio by less than any assigned value, however small.
+\scanpage{102.png}%
+
+Thus, suppose the ratio, $\dfrac{a}{b} = \sqrt{2}$.
+
+Now $\sqrt{2} = 1.41421356\cdots$, a value greater than $1.414213$,
+but less than $1.414214$.
+
+If, then, a \emph{millionth part} of $b$ is taken as the unit of measure,
+the value of $\dfrac{a}{b}$ lies between $1.414213$ and $1.414214$, and therefore
+differs from either of these values by less than $0.000001$.
+
+By carrying the decimal further, an approximate value may
+be found that will differ from the true value of the ratio by
+less than \emph{a billionth, a trillionth, or any other assigned value}.
+
+In general, if $\dfrac{a}{b} > \dfrac{m}{n}$ but $< \dfrac{m+1}{n}$, then the error in taking
+either of these values for $\dfrac{a}{b}$ is less than $\dfrac{1}{n}$, the difference
+between these two fractions.     But by increasing $n$ indefinitely,
+$\dfrac{1}{n}$ can be decreased indefinitely, and a value of the ratio
+can be found within any required degree of accuracy.
+\end{point}
+
+\pp{The ratio of two incommensurable quantities is called
+an \indexbf{incommensurable ratio}; and is a \emph{fixed value} which its successive
+approximate values constantly approach.}
+
+
+\section[THEORY OF LIMITS.]{THE THEORY OF LIMITS.}
+
+\begin{point}%
+When a quantity is regarded as having a \emph{fixed} value
+throughout the same discussion, it is called a \indexbf{constant}; but
+when it is regarded, under the conditions imposed upon it, as
+having \emph{different successive} values, it is called a \indexbf{variable}.
+
+If a variable, by having different successive values, can be
+made to differ from a given constant by less than any assigned
+value, however small, but cannot be made absolutely equal to
+the constant, that constant is called the \indexbf{limit} of the variable,
+and the variable is said to \textbf{approach the constant as its limit}.
+\end{point}
+\scanpage{103.png}%
+
+\figc{103aa272}{}
+\begin{point}%
+Suppose a point to move from $A$ toward $B$, under
+the conditions that the
+first second it shall
+move one half the distance from $A$ to $B$, that is, to $M$; the
+next second, one half the remaining distance, that is, to $M'$;
+and so on indefinitely.
+
+Then it is evident that the moving point \emph{may approach as
+near to $B$ as we choose, but will never arrive at $B$}. For, however
+near it may be to $B$ at any instant, the next second it
+will pass over half the distance still remaining; it must,
+therefore, approach nearer to $B$, since \emph{half} the distance still
+remaining is \emph{some} distance, but will not reach $B$, since \emph{half}
+the distance still remaining is not the \emph{whole} distance.
+
+Hence, the distance from $A$ to the moving point is an
+increasing variable, which indefinitely approaches the constant
+$AB$ as its \emph{limit}; and the distance from the moving point
+to $B$ is a decreasing variable, which indefinitely approaches
+the \emph{constant zero} as its \emph{limit}.
+\end{point}
+
+\figc{103bb273}{}
+\begin{point}%
+Again, suppose a square $ABCD$ inscribed in a circle,
+and $E$, $F$, $H$, $K$ the middle points of the arcs subtended by the
+sides of the square. If we draw the
+lines $AE$, $EB$, $BF$, etc., we shall have an
+inscribed polygon of double the number
+of sides of the square.
+
+The length of the perimeter of this
+polygon, represented by the dotted lines,
+is greater than that of the square, since
+two sides replace each side of the square
+and form with it a triangle, and two
+sides of a triangle are together greater than the third side;
+but less than the length of the circumference, for it is made
+up of straight lines, each one of which is less than the part of
+the circumference between its extremities.
+\end{point}
+\scanpage{104.png}%
+
+By continually doubling the number of sides of each resulting
+inscribed figure, the length of the perimeter will increase
+with the increase of the number of sides, but will not become
+equal to the length of the circumference.
+
+The difference between the perimeter of the inscribed polygon
+and the circumference of the circle can be made less than
+any assigned value, but cannot be made equal to zero.
+
+The length of the circumference is, therefore, the \emph{limit} of the
+length of the perimeter as the \emph{number of sides} of the inscribed
+figure is \emph{indefinitely increased.}~\hfill§~271
+
+\begin{point}%
+Consider the decimal $0.333 \cdots$ which may be written
+
+\centerline{\( \frac{3}{10} + \frac{3}{100} + \frac{3}{1000} + \cdots \)}
+
+The value of each fraction after the first is one tenth of the
+preceding fraction, and by continuing the series we shall reach
+a fraction less than \emph{any} assigned value, that is, the values of
+the successive fractions \emph{approach zero as a limit.}
+
+The \emph{sum} of these fractions is less than $\frac{1}{3}$; but the more terms
+we take, the nearer does the sum \emph{approach $\frac{1}{3}$ as a limit.}
+\end{point}
+
+\begin{point}%
+\textbf{Test for a limit.} In order to prove that a variable
+approaches a constant as a limit, it is necessary to prove that
+the difference between the variable and the constant:
+
+\begin{myenum}
+\item \emph{Can be made less than any assigned value, however small.}
+\item \emph{Cannot be made absolutely equal to zero.}
+\end{myenum}
+\end{point}
+
+\begin{point}%
+\thm{If the limit of a variable $x$ is zero, then the
+limit of $kx$, the product of the variable by any finite constant
+$k$, is zero.}
+
+1. Let $q$ be any assigned quantity, however small.
+
+Then $\dfrac{q}{k}$ is not~$0$.  Hence $x$, which may differ as little as we please from~$0$,
+may be taken less than $\dfrac{q}{k}$, and then $kx$ will be less than $q$.
+
+2. Since $x$ cannot be~$0$, $kx$ cannot be~$0$.
+
+\step{Therefore, the limit of $kx=0$}{§~275}
+\end{point}
+\scanpage{105.png}%
+
+\begin{point}%
+\cor{If the limit of a variable~$x$ is zero, then the
+limit of the quotient of the variable by any finite constant~$k$,
+is also zero.}
+
+For $\dfrac{x}{k} = \dfrac{1}{k} × x$, which by §~276 can be made less than any assigned
+value, however small, but cannot be made equal to zero.
+\end{point}
+
+\begin{point}%
+\thm{The limit of the sum of a finite number of
+variables \mbox{$x$, $y$, $z$, $\cdots$} is equal to the sum of their respective
+limits \mbox{$a$, $b$, $c$, $\cdots$.}}
+
+\sloppy
+Let \mbox{$d$, $d'$, $d''$, $\cdots$} denote the differences between
+\mbox{$x$, $y$, $z$, $\cdots$} and \mbox{$a$, $b$, $c$, $\cdots$,}
+respectively. Then \mbox{$d+d'+d''+\cdots$} can be made less than any
+assigned quantity $q$.
+
+\fussy
+For, if \mbox{$d$, $d'$, $d''$, $\cdots$} are $n$ in number and $d$ is the largest,
+
+\step{$d+d'+d''+\cdots < nd$.}{(1)}
+
+Since $d$ may be diminished at pleasure, we may make $d$ so small that
+
+\step{$ d < \dfrac{q}{n}$; and therefore $nd < q$.}{}
+
+But by (1), $d+d'+d''+\cdots < nd$, and therefore $< q$.
+
+Therefore, the difference between ($x + y + z + \cdots$) and $(a + b + c + \cdots)$
+can be made less than any assigned quantity, but not zero.
+
+Therefore, the limit of $(x + y + z + \cdots) = a + b + c + \cdots$.~\hfill§~275
+\end{point}
+
+\begin{point}%
+\thm{If the limit of a variable~$x$ is not zero, and
+if $k$~is any finite constant, the limit of the product~$kx$ is
+equal to the limit of~$x$ multiplied by~$k$.}
+
+1. If $a$ denotes the limit of $x$, then $x$ cannot be equal to $a$.~\hfill§~271
+
+\step{Therefore, $kx$ cannot be equal to $ka$.}{}
+
+2. The limit of $(a - x) = 0$. Hence, the limit of $ka - kx=0$.~\hfill§~276
+
+\step{Therefore, the limit of $kx = ka$.}{§~275}
+\end{point}
+
+\begin{point}%
+\cor{The limit of the quotient of a variable $x$ by
+any finite constant $k$ is the limit of $x$ divided by $k$.}
+
+For $\dfrac{x}{k} = \dfrac{1}{k} × x$, and $\dfrac{\text{the
+ limit of }x}{k} = \dfrac{1}{k}$ $×$ the limit of $x$.
+\end{point}
+\scanpage{106.png}%
+
+\begin{point}%
+\thm{The limit of the product of two or more
+variables is the product of their respective limits, provided
+no one of these limits is zero.}
+
+If $x$ and $y$ are variables, $a$ and $b$ their respective limits, we may put
+$x = a - d$, $y = b - d'$; then $d$ and $d'$ are variables which can be made
+less than any assigned quantity, but not zero.~\hfill§~275
+
+Now, %[**TN: ad hoc visual formatting]
+\vspace*{-1.5\baselineskip}
+\begin{align*}
+xy &= (a - d) (b - d') \\
+   &= ab - ad' - bd + dd' \\
+\therefore ab - xy &= ad' + bd - dd'.
+\end{align*}
+
+Since every term on the right contains $d$ or $d'$, the whole right member
+can be made less than any assigned quantity, but not zero.~\hfill§~278
+
+Hence, $ab - xy$ can be made less than any assigned quantity, but not
+zero.
+
+\step{Therefore, the limit of $xy = ab$.}{§~275}
+
+\step{Similarly, for three or more variables.}{}
+\end{point}
+
+\begin{point}%
+\cor[1]{The limit of the nth power of a variable is
+the nth power of its limit.}
+
+For the limit of the product of the variables $x$, $y$, $z$, $\cdots$ to $n$ factors is
+the product of their respective limits, the constants $a$, $b$, $c$, $\cdots$ to $n$ factors~(§~281).
+If the $n$ factors $xyz\cdots$ are each equal to $x$, and the $n$ factors
+$abc\cdots$ are each equal to $a$, we have $xyz\cdots = x^n$, and $abc\cdots = a^n$.
+
+\step{Therefore, the limit of $x^n = a^n$.}{}
+\end{point}
+
+\begin{point}%
+\cor[2]{The limit of the nth root of a variable is the
+nth root of its limit.}
+
+For if the limit of $x = a$, we may put this in the following form,
+
+\step{the limit of \( \sqrt[n]{x^n} = \sqrt[n]{a^n} \);}{}
+
+\noindent that is, the limit of \( \sqrt[n]{xxx\cdots\text{ to $n$~factors }} \) is
+  \( \sqrt[n]{aaa\cdots\text{ to $n$~factors}} \).
+
+Now, $xxx\cdots$ is a variable since each factor is a variable,
+and $aaa\cdots$ is a constant since each factor is a constant.
+
+If we denote $xxx\cdots$ to $n$~factors by the variable~$y$, and $aaa\cdots$ to $n$
+factors by the constant~$b$, we have
+
+% the vphantoms make the two sqrt boxes the same height, to line up better
+\step{the limit of \( \displaystyle \sqrt[n]{\vphantom{b}y} = \sqrt[n]{\vphantom{y}b} \).}{}
+\end{point}
+\scanpage{107.png}%
+
+\begin{point}%
+\thm{If two variables are constantly equal, and
+each approaches a limit, the limits are equal.}
+
+Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits, $d$ and $d'$
+the respective differences between the variables and their limits. Then,
+if the variables are \emph{increasing} toward their limits
+
+\step{$a = x + d$, and $b = y + d'$.}{}
+
+Since the equation $x = y$ is always true, we have by subtraction
+
+\step{$a - b = d - d'$.}{}
+
+Since $a$ and $b$ are constants, $a - b$ is a constant; therefore, $d - d'$,
+which is equal to $a - b$, is a constant.
+
+But the only constant which is less than any assigned value is~$0$.
+Therefore, $d - d' = 0$. Therefore, $a - b = 0$, and $a = b$.
+
+If the variables $x$ and $y$ are \emph{decreasing} toward their limits $a$ and $b$,
+respectively, then
+
+\step{$a = x - d$ and $b = y - d'$.}{}
+
+Therefore, by subtraction
+
+\step{$a - b = d' - d$.}{}
+
+Therefore, by the same proof as for increasing variables
+
+\step{$a = b$.}{}
+\end{point}
+
+
+\begin{point}%
+\thm{If two variables have a constant ratio, and
+each approaches a limit that is not zero, the limits have the
+same ratio.}
+
+Let $x$ and $y$ be two variables, $a$ and $b$ their respective limits.
+
+\eq[\indent Let]{$\dfrac{x}{y}$}{$= r$, a constant; then $x = ry$.}{}
+
+Since $x$ and $ry$ are two variables that are always equal,
+
+\eq{the limit of $x$}{$=$ the limit of $ry$.}{§~284}
+
+\eq[\indent Now,]{the limit of $ry$}{$= r$ $×$ limit of $y$.}{§~279}
+
+But the limit of $x$ is $a$, and the limit of $y$ is $b$.
+
+\eq[\indent Therefore,]{$a$}{$= rb$; that is, $\dfrac{a}{b} = r$.}{}
+\end{point}
+\scanpage{108.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the ratio of two straight lines.}
+
+\figc{108aa286}{Let $AB$ and $CD$ be two straight lines.}
+
+\prove[To find ]{the ratio of $AB$ and $CD$.}
+
+\step{Apply $CD$ to $AB$ as many times as possible.}{}
+
+\step{Suppose twice, with a remainder $EB$.}{}
+
+\step{Then apply $EB$ to $CD$ as many times as possible.}{}
+
+\step{Suppose three times, with a remainder $FD$.}{}
+
+\step{Then apply $FD$ to $EB$ as many times as possible.}{}
+
+\step{Suppose once, with a remainder $HB$.}{}
+
+\step{Then apply $HB$ to $FD$ as many times as possible.}{}
+
+\step{Suppose once, with a remainder $KD$.}{}
+
+\step{Then apply $KD$ to $HB$ as many times as possible.}{}
+
+\step{Suppose $KD$ is contained just twice in $HB$.}{}
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq[\indent Then]{$HB$}{$= 2 KD$;}{}
+
+\eq{$FD$}{$= HB + KD = 3 KD$;}{}
+
+\eq{$EB$}{$= FD + HB = 5 KD$;}{}
+
+\eq{$CD$}{$= 3 EB + FD = 18 KD$;}{}
+
+\eq{$AB$}{$= 2 CD + EB = 41 KD$;}{}
+
+\eq{$\therefore \dfrac{AB}{CD}$}{$=
+          \dfrac{41 KD}{18 KD} = \dfrac{41}{18}$.}{\qef}
+
+\setlength{\eqalign}{.5\dentwidth}
+
+\end{proof}
+
+\note{By the same process the ratio of two arcs of the same circle or of equal circles can be found.
+\par
+If the lines or arcs are incommensurable, an approximate value of the ratio can be found by the same method.}
+\scanpage{109.png}%
+
+\clearpage
+\section{MEASURE OF ANGLES.}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In the same circle or in equal circles, two
+  central angles have the same ratio as their intercepted arcs.}
+
+\figc{109ac287}{In the equal circles whose centres are $C$ and $C'$, let $ACB$
+  and $A'C'B'$ be the angles, $AB$ and $A'B'$ the intercepted arcs.}
+
+\proveq{$\dfrac{\angle A'C'B'}{\angle ACB}$}
+     {$=\dfrac{\arc A'B'}{\arc AB}$.}
+
+\textsc{Case~1.} \emph{When the arcs are commensurable} (Figs.~1 and
+2).
+
+\textbf{Proof.} Let the arc~$m$ be a common measure of $A'B'$ and
+$AB$.
+
+\step{Suppose $m$ to be contained $4$~times in $A'B'$,}{}
+
+\step{and $7$~times in $AB$.}{}
+
+\eq[\indent Then]{$\dfrac{\arc A'B'}{\arc AB}$}{$=\dfrac{4}{7}$.}{}
+
+At the several points of division on $AB$ and $A'B'$ draw radii.
+
+These radii will divide $\angle ACB$ into $7$~parts, and $\angle A'C'B'$
+into $4$~parts, equal each to each,~\hfill§~237
+
+\pnote{(in the same $\odot$, or equal
+    $\odot_s$, equal arcs subtend equal central $\angle_s$).}
+
+\eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$}
+  {$=\dfrac{4}{7}$.}{}
+
+\eq{$\therefore \dfrac{\angle A'C'B'}{\angle ACB}$}
+  {$=\dfrac{\arc A'B'}{\arc AB}$.}{Ax.~1}
+\scanpage{110.png}%
+
+\step{}{}
+
+\filbreak
+\textsc{Case 2.}  \emph{When the arcs are incommensurable} (Figs.~2 and 3).
+
+\textbf{Proof.} Divide $AB$ into any number of equal parts, and apply
+one of these parts to $A'B'$ as many times as $A'B'$ will contain it.
+
+Since $AB$ and $A'B'$ are incommensurable, a certain number
+of these parts will extend from $A'$ to some point, as $D$, leaving
+a remainder $DB'$ less than one of these parts. Draw $C'D$.
+
+\step{By construction $AB$ and $A'D$ are commensurable.}{}
+
+\eq{$\therefore \dfrac{\angle A'C'D}{\angle ACB}$}
+  {$=\dfrac{\arc A'D}{\arc AB}$.}{Case~1}
+
+By increasing the \emph{number} of equal parts into which $AB$ is
+divided we can diminish at pleasure the \emph{length} of each part,
+and therefore make $DB'$ less than any assigned value, however
+small, since $DB'$ is always less than one of the equal parts into
+which $AB$ is divided.
+
+We cannot make $DB'$ equal to zero, since, by hypothesis, $AB$
+and $A'B'$ are incommensurable.~\hfill§~269
+
+Hence, $DB'$ approaches zero as a limit, if the number of
+parts of $AB$ is indefinitely increased.~\hfill§~275
+
+And the corresponding angle $DC'B'$ approaches zero as a
+limit.
+
+Therefore, the arc $A'D$ approaches the arc $A'B'$ as a limit,~\hfill§~271\linebreak
+and the $\angle A'C'D$ approaches the $\angle A'C'B'$ as a limit.
+
+\step[\indent Therefore,]{$\dfrac{\arc A'D}{\arc AB}$ approaches
+  $\dfrac{\arc A'B'}{\arc AB}$ as a limit,}{§~280}
+
+\step[and]{$\dfrac{\angle A'C'D}{\angle ACB}$ approaches
+  $\dfrac{\angle A'C'B'}{\angle ACB}$ as a limit.}{§~280}
+
+\step[\indent But]{$\dfrac{\angle A'C'D}{\angle ACB}$ is constantly equal to
+  $\dfrac{\arc A'D}{\arc AB}$,}{}
+
+\step{as $A'D$ varies in value and approaches $A'B'$ as a limit.}{}
+
+\step{$\therefore \dfrac{\angle A'C'B'}{\angle ACB} =
+  \dfrac{\arc A'B'}{\arc AB}$.}{§~284}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{111.png}%
+
+\begin{point}%
+A circumference is divided into $360$~equal parts, called
+  \emph{degrees}; and therefore a unit angle at the centre intercepts
+  a unit arc on the circumference.  Hence, the \emph{numerical measure
+  of a central angle} expressed in terms of the unit angle is equal to
+  the \emph{numerical measure of its intercepted arc} expressed in
+  terms of the unit arc.  This must be understood to be the meaning
+  when it is said that
+
+\emph{A central angle is measured by its intercepted arc.}
+\end{point}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An inscribed angle is measured by half the arc
+  intercepted between its sides.}
+
+\figc{111ac289}{1. Let the centre $C$ \textnormal{(Fig.~1)} be in one of the sides of
+  the angle.}
+
+\prove{the $\angle B$ is measured by $\frac{1}{2}$ the
+  arc $PA$.}
+
+\step[\indent\textbf{Proof.}]{Draw $CA$.}{}
+
+\eq{$CA$}{$= CB$.}{§~217}
+
+\eq{$\therefore \angle B$}{$= \angle A$.}{§~145}
+
+\eq[\indent But]{$\angle PCA$}{$= \angle B + \angle A$.}{§~137}
+
+\eq{$\therefore \angle PCA$}{$= 2 \angle B$.}{}
+
+\step[\indent But]{$\angle PCA$ is measured by $\arc PA$,}{§~288}
+
+\pnote{(a central $\angle$ is measured
+    by its intercepted arc).}
+
+\step{$\therefore \angle B$ is measured by $\frac{1}{2} \arc PA$.}{}
+\scanpage{112.png}%
+
+\step{}{}
+
+\textbf{2. Let the centre $C$ \textnormal{(Fig.~2)} fall within the angle $PBA$.}
+
+\prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.}
+
+\step[\indent\textbf{Proof.}]{Draw the diameter $BCE$.}{}
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq[\indent Then]{$\angle EBA$}{is measured by $\frac{1}{2} \arc AE$,}{}
+
+\eq[and]{$\angle EBP$}{is measured by $\frac{1}{2} \arc EP$.}{Case~1}
+
+\eq{$\therefore \angle EBA + \angle EBP$}{is measured by $\frac{1}{2}(\arc AE +\arc EP)$,}{}
+
+\eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{}
+
+
+\step{}{}
+
+\textbf{3. Let the centre $C$} (Fig.~3) \textbf{fall without the angle $PBA$.}
+
+\prove{the $\angle PBA$ is measured by $\frac{1}{2}$ the arc $PA$.}
+
+\step[\indent\textbf{Proof.}]{Draw the diameter $BCF$.}{}
+
+\eq[\indent Then]{$\angle FBA$}{is measured by $\frac{1}{2} \arc FA$,}{}
+
+\eq[and]{$\angle FBP$}{is measured by $\frac{1}{2} \arc FP$.}{Case~1}
+
+\eq{$\therefore \angle FBA - \angle FBP$}{is measured by $\frac{1}{2}(\arc FA - \arc FP)$,}{}
+
+\eq[or]{$\angle PBA$}{is measured by $\frac{1}{2} \arc PA$.}{\qed}
+
+\setlength{\eqalign}{.5\dentwidth}
+
+
+\end{proof}
+
+\figc{112ac290}{}
+
+\pp{\cor[1]{An angle inscribed in a semicircle is a right
+angle. \textup{For it is measured by half a semicircumference (Fig.~4).}}}
+
+\pp{\cor[2]{An angle inscribed in a segment greater than
+a semicircle is an acute angle. \textup{For it is measured by an arc
+less than half a semicircumference; as, $\angle CAD$ (Fig.~5).}}}
+
+\pp{\cor[3]{An angle inscribed in a segment less than a
+semicircle is an obtuse angle. \textup{For it is measured by an arc
+greater than half a semicircumference; as, $\angle CBD$ (Fig.~5).}}}
+
+\pp{\cor[4]{Angles inscribed in the same segment or in
+equal segments are equal \textup{(Fig.~6).}}}
+\scanpage{113.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An angle formed by two chords intersecting
+within the circumference is measured by half the sum
+of the intercepted arcs.}
+
+\figc{113aa294}{Let the angle $COD$ be formed by the chords $AC$ and $BD$.}
+
+\prove{the $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $AE$ drawn $\parallel$ to $BD$.}{}
+
+\eq{}{Then $\arc AB = \arc DE$,}{§~257}
+
+\pnote{(parallels intercept equal arcs on a circumference).}
+
+\eq{}{Also $\angle COD = \angle CAE$,}{§~112}
+
+\pnote{(ext.-int.~$\angle_s$ of $\parallel_s$).}
+
+\step{But $\angle CAE$ is measured by $\frac{1}{2}(CD + DE)$,}{§~289}
+
+\pnote{(an inscribed $\angle$ is measured by half its intercepted arc).}
+
+Put $\angle COD$ for its equal, the $\angle CAE$, and $\arc AB$ for its
+equal, the arc $DE$; then $\angle COD$ is measured by $\frac{1}{2}(CD + AB)$.
+
+\hfill\qed
+
+\end{proof}
+
+\ex{The opposite angles of an inscribed quadrilateral are supplementary.}
+
+\ex{If through a point within a circle two perpendicular chords
+are drawn, the sum of either pair of the opposite arcs which they intercept
+is equal to a semicircumference.}
+
+\ex{The line joining the centre of the square described upon the
+hypotenuse of a right triangle to the vertex of the right angle bisects the
+right angle.}
+\scanpage{114.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An angle included by a tangent and a chord drawn
+  from the point of contact is measured by half the intercepted arc.}
+
+\figc{114aa295}{Let $MAH$ be the angle included by the tangent $MO$ to the
+  circle at $A$ and the chord $AH$.}
+
+\prove{the $\angle MAH$ is measured by $\frac{1}{2}$ the
+  arc $AEH$.}
+
+\step[\indent\textbf{Proof.}]{Suppose $HF$ drawn $\parallel$ to $MO$.}{}
+
+\step{Then $\arc AF = \arc AEH$,}{§~257}
+
+\pnote{(parallels intercept equal arcs
+    on a circumference).}
+
+\step{Also $\angle MAH = \angle AHF$,}{§~110}
+
+\pnote{(alt.-int.~$\angle_s$ of $\parallel_s$).}
+
+\step{$\angle AHF$ is measured by $\frac{1}{2} AF$,}{§~289}
+
+\pnote{(an inscribed $\angle$ is measured by half its intercepted arc).}
+
+Put $\angle MAH$ for its equal, the $\angle AHF$, and $\arc AEH$ for
+its equal, the arc $AF$; then $\angle MAH$ is measured by
+$\frac{1}{2} \arc AEH$.
+
+Likewise, the $\angle OAH$, the supplement of the $\angle MAH$, is
+measured by half the arc $AFH$, the conjugate of the arc $AEH$.
+
+\hfill\qed
+
+\end{proof}
+
+\ex{Two circles are tangent externally at $A$, and a
+common external tangent touches them at $B$ and $C$, respectively.
+Show that angle $BAC$ is a right angle.}
+\scanpage{115.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An angle formed by two secants, two tangents, or a
+  tangent and a secant, drawn to a circle from an external point, is
+  measured by half the difference of the intercepted arcs.}
+
+\figc{115ac296}{}
+
+The proof of this theorem is left as an exercise for the student.
+
+\end{proof}
+
+\figcc{115dd297}{115ee297}
+\begin{point}%
+\textbf{Positive and Negative Quantities.} In
+measurements it is convenient to mark the distinction between two
+quantities that are measured in \emph{opposite directions}, by calling
+one of them \indexbf{positive} and the other \indexbf{negative}.
+
+Thus, if $OA$ is considered positive, then $OC$ may be considered
+negative, and if $OR$ is considered positive, then $OD$ may be
+considered negative.
+
+When this distinction is applied to angles, an angle is considered to
+be \emph{positive}, if the rotating line that describes it moves in
+the opposite direction to the hands of a clock (counter clockwise),
+and to be \emph{negative}, if the rotating line moves in the same
+direction as the hands of a clock (clockwise).
+
+Arcs corresponding to positive angles are considered \emph{positive},
+and arcs corresponding to negative angles are considered
+\emph{negative}.
+
+Thus, the angle $ACB$ described by a line rotating about $C$ from $CA$
+to $CB$ is positive, and the arc $AB$ is positive; the angle $ACB'$
+described by the line rotating about $C$ from $CA$ to $CB'$ is
+negative, and the arc $AB'$ is negative.
+\end{point}
+\scanpage{116.png}%
+
+\pp{\textbf{The Principle of Continuity.}\label{princcont} By marking the distinction between
+quantities measured in opposite directions, a theorem may often be so
+stated as to include two or more particular theorems.}
+
+The following theorem furnishes a good illustration:
+
+\begin{point}%
+\textit{The angle included between two lines of unlimited length that cut
+or touch a circumference is measured by half the sum of the intercepted arcs.}
+
+Here the word \emph{sum} means the algebraic sum and includes both the arithmetical
+sum and the arithmetical difference of two quantities.
+
+\figc{116ag299}{}
+
+1. If the lines intersect at the centre, the two intercepted arcs are equal,
+and half the sum will be one of the arcs (§~288).
+
+2. If the lines intersect between the centre and the circumference, the
+angle is measured by half the sum of the arcs (§~294).
+
+3. If the lines intersect on the circumference, one of the arcs becomes
+zero and we have an inscribed angle (§~289), or an angle formed by a
+tangent and a chord (§~295). In each case the angle is measured by half
+the sum of the intercepted arcs.
+
+4. If the lines intersect without the circumference, then the arc $ab$
+is negative and the algebraic sum is the arithmetical difference of the
+included arcs.
+
+When the reasoning employed to prove a theorem is continued in the
+manner just illustrated to include two or more theorems, we are said to
+reason by the \emph{Principle of Continuity.}
+\end{point}
+
+\subsection{REVIEW QUESTIONS ON BOOK II.}
+
+\begin{myenum}
+\item  What do we call the locus of points in a plane that are equidistant
+from a fixed point in the plane?
+
+\item What does the chord of a segment become when the segment is a
+semicircle?
+
+\item What kind of an angle do the radii of a sector include when the
+sector is a semicircle?
+
+\item  What is the difference between a chord and a secant?
+
+\item What part of a tangent is meant by a tangent to a circle from an
+external point?
+\scanpage{117.png}%
+
+\item Two chords are equal in equal circles under either of two conditions.
+What are the two conditions?
+
+\item Points that lie in a straight line are called \emph{collinear}; points that
+lie in a circumference are called \emph{concyclic}. How many collinear points
+can be concyclic?
+
+\item What is meant by the statement that a central angle is measured
+by the arc intercepted between its sides?
+
+\item What is an inscribed angle?   What is its measure?
+
+\item What kind of an angle is the angle inscribed in a semicircle? in a
+segment less than a semicircle? in a segment greater than a semicircle?
+
+\item What is the measure of an angle included by two intersecting
+chords? by two secants intersecting without the circle?
+
+\item What is the measure of an angle included by a tangent and a
+chord drawn to the point of contact?
+
+\item When are two quantities of the same kind incommensurable?
+
+\item When are two quantities of the same kind commensurable?
+
+\item Define a variable and the limit of a variable.
+
+\item Does the series $\frac{1}{2}$ in., $\frac{3}{4}$ in., $\frac{7}{8}$ in., $\frac{15}{16}$ in., etc., constitute a variable?
+Is the variable increasing or decreasing?
+
+\item What is the limit of this variable?
+
+\item What is the test of a limit?
+\end{myenum}
+
+%\section{EXERCISES.}
+
+\subsection{THEOREMS.}
+
+\ex{An angle formed by a tangent and a chord is equal to the
+angle inscribed in the opposite segment.}
+
+\ex{Two chords drawn perpendicular to a third chord at its
+extremities are equal.}
+
+\ex{The sum of two opposite sides of a circumscribed quadrilateral
+is equal to the sum of the other two sides.}
+
+\figcc{117aa104}{117bb105}
+\begin{proofex}%
+If the sum of two opposite angles of a quadrilateral
+is equal to two right angles, a circle may be circumscribed
+about the quadrilateral.
+
+Let $\angle A + \angle C = 2 \text{ rt.\ } \angle_s$. Pass a circumference through
+$D$, $A$, and $B$, and prove that this circumference passes through $C$.
+
+\end{proofex}
+
+\begin{proofex}%
+The shortest line that can be drawn from a point within a
+circle to the circumference is the shorter segment of the
+diameter through that point.
+
+Let $A$ be the given point. Prove $AB$ shorter than any
+other line $AD$ from $A$ to the circumference.
+
+\end{proofex}
+\scanpage{118.png}%
+
+\ex{The longest line that can be drawn from a point within a
+circle to the circumference is the longer segment of the diameter through
+that point.}
+
+\figc{118aa107}{}
+\ex{The shortest line that can be drawn
+from a point without a circle to the circumference
+will pass through the centre of the circle if produced.}
+
+\ex{The longest line that can be drawn from a point without a
+circle to the concave arc of the circumference passes through the centre of
+the circle.}
+
+\figc{118be108}{}
+
+\ex{The shortest chord that can be drawn through a point within
+a circle is perpendicular to the diameter at that point.}
+
+\begin{proofex}%
+If two intersecting chords make equal angles with the diameter
+drawn through the point of intersection, the two chords are equal.
+
+% special case, two centered columns
+\hspace{\stretch{1}}Rt.~$\triangle COM =$ rt.~$\triangle CON$.
+\hspace{\stretch{1}}$\therefore OM = ON$.
+\hspace{\stretch{1}}
+
+
+\end{proofex}
+
+\ex{The angles subtended at the centre of a circle by any two
+opposite sides of a circumscribed quadrilateral are supplementary.}
+
+\begin{proofex}%
+The radius of a circle inscribed in an equilateral triangle is
+equal to one third the altitude of the triangle.
+
+$\triangle OEF$ is equiangular and equilateral; $\angle FEA = \angle FAE$.
+
+% special case, two centered columns
+\hspace{\stretch{1}}$\therefore AF = EF$.
+\hspace{\stretch{1}}$\therefore AF = FO = OD$.
+\hspace{\stretch{1}}
+
+\end{proofex}
+
+\ex{The radius of a circle circumscribed about an equilateral
+triangle is equal to two thirds the altitude of the triangle (Ex.~27).}
+
+\ex{A parallelogram inscribed in a circle is a rectangle.}
+
+\ex{A trapezoid inscribed in a circle is an isosceles trapezoid.}
+
+\ex{All chords of a circle which touch an interior concentric
+circle are equal, and are bisected at the point of contact.}
+
+\ex{If the inscribed and circumscribed circles of a triangle are
+concentric, the triangle is equilateral (Ex.~116).}
+\scanpage{119.png}%
+
+\ex{If two circles are tangent to each other the tangents to them
+from any point of the common internal tangent are equal.}
+
+\ex{If two circles touch each other and a line is drawn through
+the point of contact terminated by the circumferences, the tangents at its
+ends are parallel.}
+
+\figcc{119aa120}{119bb121}
+\begin{proofex}%
+If two circles touch each other and two
+lines are drawn through the point of contact terminated
+by the circumferences, the chords joining the
+ends of these lines are parallel.
+
+$\angle A = \angle MPC$ and $\angle B = \angle NPD$. \quad $\therefore \angle A = \angle B$.
+
+\end{proofex}
+
+\ex{If two circles intersect and a line is drawn
+through each point of intersection terminated by the circumferences,
+the chords joining the ends of these lines
+are parallel.}
+
+\figcc{119cc122}{119dd123}
+\begin{proofex}%
+Through one of the points of intersection of two circles a
+diameter of each circle is drawn.    Prove that the line
+joining the ends of the diameters passes through the other
+point of intersection.
+
+\step{$\angle ABC = \angle ABD = 90°$}{§~290}
+
+\end{proofex}
+
+\ex{If two common external tangents or two common internal
+tangents are drawn to two circles, the segments
+intercepted between the points of contact are equal.}
+
+\ex{The diameter of the circle inscribed in
+a right triangle is equal to the difference between
+the sum of the legs and the hypotenuse.}
+
+\figcc{119ee125}{119ff126}
+\begin{proofex}%
+If one leg of a right triangle is the diameter
+of a circle, the tangent at the point where the circumference
+cuts the hypotenuse bisects the other leg.
+
+\begin{center}
+$\angle BOE = \angle DOE$.\qquad $\angle BOE = \angle OAD$.
+
+$\therefore OE$ and $AC$ are $\parallel$.\qquad $\therefore BE = EC$ (§~188).
+\end{center}
+
+
+\end{proofex}
+
+\ex{If, from any point in the circumference of
+a circle, a chord and a tangent are drawn, the perpendiculars
+dropped on them from the middle point of the
+subtended arc are equal. $\angle BAM = \angle CAM$.}
+
+\ex{The median of a trapezoid circumscribed about a circle is equal
+to one fourth the perimeter of the trapezoid (Ex.~103).}
+\scanpage{120.png}%
+
+\ex{Two fixed circles touch each other externally and a circle of
+variable radius touches both externally. Show that the difference of the
+distances from the centre of the variable circle to the centres of the fixed
+circles is constant.}
+
+\ex{If two fixed circles intersect, and circles are drawn to touch
+both, show that either the sum or the difference of the distances of their
+centres from the centres of the fixed circles is constant, according as they
+touch (i)~one internally and one externally, (ii)~both internally or both
+externally.}
+
+\figcc{120aa130}{120bb131}
+\begin{proofex}%
+If two straight lines are drawn through
+any point in a diagonal of a square parallel to the sides
+of the square, the points where these lines meet the
+sides lie on the circumference of a circle whose centre
+is the point of intersection of the diagonals.
+
+$\triangle POE = \triangle POF$ (§~143). \quad
+  $\therefore OE = OF$. \quad $\triangle POE' = \triangle POF'$.
+
+\end{proofex}
+
+\begin{proofex}%
+If $ABC$ is an inscribed equilateral triangle and
+$P$ is any point in the arc $BC$, then $PA = PB + PC$.
+
+Take $PM = PB$. $\triangle ABM = \triangle CBP$ (§~143) and $AM
+= PC$.
+
+\end{proofex}
+
+\ex{The tangents drawn through the vertices of an inscribed
+rectangle, which is not a square, enclose a rhombus.}
+
+\figc{120cd132}{}
+
+\begin{proofex}%
+The bisectors of the angles included by the opposite sides
+(produced) of an inscribed quadrilateral intersect at right angles.
+
+\eq{Arc $AF - \arc BM$}{$= \arc DF - \arc CM$}{}
+
+\eq{and $\arc AH - \arc DN$}{$= \arc BH - \arc CN$.}{}
+
+\eq{$\therefore \arc FH + \arc MN$}{$= \arc HM + \arc FN$.}{}
+
+\eq{$\therefore \angle FIH$}{$= \angle HIM$.}{}
+
+\textbf{Discussion.}  This problem is impossible, if any two sides of the quadrilateral
+are parallel.
+
+\end{proofex}
+\scanpage{121.png}%
+
+
+\section{PROBLEMS OF CONSTRUCTION.}
+
+\note{Hitherto we have supposed the figures constructed. We now
+proceed to explain the methods of constructing simple problems, and afterwards
+to apply these methods to the solution of more difficult problem.}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To let fall a perpendicular upon a given line from
+a given external point.}
+
+\figc{121aa300}{Let $AB$ be the given straight line, and $C$ the given external point.}
+
+\prove[]{To let fall a $\perp$ to the line $AB$ from the point $C$}.
+
+From $C$ as a centre, with a radius sufficiently great, describe
+an arc cutting $AB$ in two points, $H$ and $K$.
+
+From $H$ and $K$ as centres, with equal radii greater than $\frac{1}{2}HK$,
+
+\step{describe two arcs intersecting at $O$.}{}
+
+\step{Draw $CO$,}{}
+
+\step{and produce it to meet $AB$ at $M$.}{}
+
+\step{$CM$ is the $\perp$ required.}{}
+
+\textbf{Proof.} Since $C$ and $O$ are two points each equidistant from
+$H$ and $K$, they determine a $\perp$ to $HK$ at its middle point.~\hfill§~161
+
+\hfill\qef
+
+\end{proof}
+
+\note{\emph{Given} lines of the figures are represented by full lines, \emph{resulting}
+lines by long-dashed, and \emph{auxiliary} lines by short-dashed lines.}
+\scanpage{122.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{At a given point in a straight line, to erect a
+perpendicular to that line.}
+
+\figc{122ab301}{1. Let $O$ be the given point in $AC$. \textnormal{Fig.~1.}}
+
+\step{Take $OH$ and $OB$ equal.}{}
+
+From $H$ and $B$ as centres, with equal radii greater than $OB$,
+describe two arcs intersecting at $R$. Join $OR$.
+
+Then the line $OR$ is the $\perp$ required.
+
+\textbf{Proof.} $O$ and $R$, two points each equidistant from $H$ and $B$,
+determine the perpendicular bisector of $HB$.~\hfill§~161
+
+\lett{2. Let $B$ be the given point. \textnormal{Fig.~2.}}
+
+Take any point $C$ without $AB$; and from $C$ as a centre,
+with the distance $CB$ as a radius, describe an arc intersecting
+$AB$ at $E$.
+
+Draw $EC$, and prolong it to meet the arc again at $D$.
+
+Join $BD$, and $BD$ is the $\perp$ required.
+
+\step[\indent\textbf{Proof.}]{The $\angle B$ is a right angle.}{§~290}
+
+\step{$\therefore BD$ is $\perp$ to $AB$.}{\qef}
+
+\textbf{Discussion.} The point $C$ must be so taken that it will not
+be in the required perpendicular.
+
+\end{proof}
+\scanpage{123.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To bisect a given straight line.}
+
+\figc{123aa302}{To bisect the given straight line $AB$.}
+
+From $A$ and $B$ as centres, with equal radii greater than
+$\frac{1}{2} AB$, describe arcs intersecting at $C$ and $E$.
+
+\step{Join $CE$.}{}
+
+\step{Then $CE$ bisects $AB$.}{§~161}
+
+\hfill\qef
+
+\end{proof}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To bisect a given arc.}
+
+\figc{123bb303}{To bisect the given arc $AB$.}
+
+\step{Draw the chord $AB$.}{}
+
+From $A$ and $B$ as centres, with equal radii greater than
+$\frac{1}{2} AB$, describe arcs intersecting at $D$ and $E$.
+\scanpage{124.png}%
+
+\step{Draw $DE$.}{}
+
+\step{Then $DE$ is the $\perp$ bisector of the chord $AB$.}{§~161}
+
+\step{$\therefore DE$ bisects the arc $ACB$.}{§~248}
+
+\hfill\qef
+
+\end{proof}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To bisect a given angle.}
+
+\figc{124aa304}{Let $AEB$ be the given angle.}
+
+From $E$ as a centre, with any radius, as $EA$, describe an
+arc cutting the sides of the $\angle E$ at $A$ and $B$.
+
+From $A$ and $B$ as centres, with equal radii greater than half
+the distance from $A$ to $B$, describe two arcs intersecting at $D$.
+
+\step{Draw $DE$.}{}
+
+\step{Then $DE$ bisects the arc $AB$ at $C$.}{§~303}
+
+\step{$\therefore DE$ bisects the angle $E$.}{§~237}
+
+\hfill\qef
+
+\end{proof}
+
+\ex{To construct an angle of $45°$; of $135°$.}
+
+\ex{To construct an equilateral triangle, having given one side.}
+
+\ex{To construct an angle of $60°$; of $150°$.}
+
+\ex{To trisect a right angle.}
+\scanpage{125.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{At a given point in a given straight line, to
+construct an angle equal to a given angle.}
+
+\figc{125ab305}{At $C$ in the line $CM$, construct an angle equal to the given angle $A$.}
+
+From $A$ as a centre, with any radius, $AE$, describe an arc
+cutting the sides of the $\angle A$ at $E$ and $F$.
+
+\step{From $C$ as a centre, with a radius equal to $AE$,}{}
+
+\step{describe an arc $HG$ cutting $CM$ at $H$.}{}
+
+\step{From $H$ as a centre, with a radius equal to the chord $EF$,}{}
+
+\step{describe an arc intersecting the arc $HG$ at $O$.}{}
+
+\step{Draw $CO$, and $\angle$ $HCO$ is the required angle.}{Why?}
+
+\hfill\qef
+
+\end{proof}
+
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To draw a straight line parallel to a given
+straight line through a given external point.}
+
+\figc{125cc306}{Let $AB$ be the given line, and $C$ the given point.}
+\scanpage{126.png}%
+
+\step{Draw $ECD$, making any convenient $\angle EDB$.}{}
+
+\step{At the point $C$ construct $\angle ECF$ equal to $\angle EDB$.}{§~305}
+
+\step{Then the line $HCF$ is $\parallel$ to $AB$.}{Why?}
+
+\hfill\qef
+
+
+\end{proof}
+
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To divide a given straight line into a given
+number of equal parts.}
+
+\figc{126aa307}{Let $AB$ be the given straight line.}
+
+From $A$ draw the line $AO$, making any convenient angle
+with $AB$.
+
+Take any convenient length, and apply it to $AO$ as many
+times as the line $AB$ is to be divided into parts.
+
+From $C$, the last point thus found on $AO$, draw $CB$.
+
+Through the points of division on $AO$ draw parallels to
+the line $CB$.~\hfill§~306
+
+\step{These lines will divide $AB$ into equal parts.}{§~187}
+
+\hfill\qef
+
+\end{proof}
+
+\ex{To construct an equilateral triangle, having given the perimeter.}
+
+\ex{To divide a line into four equal parts by two different
+methods.}
+
+\begin{proofex}%
+Through a given point to draw a line which shall make equal
+angles with the two sides of a given angle.
+
+Through the given point draw a $\perp$ to the bisector of the given $\angle$.
+
+\end{proofex}
+
+\ex{To draw a line through a given point, so that it shall form
+with the sides of a given angle an isosceles triangle (Ex.~140).}
+\scanpage{127.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the third angle of a triangle when two
+of the angles are given.}
+
+\figc{127ab308}{Let $A$ and $B$ be the two given angles.}
+
+\step{At any point $H$ in any line $EF$,}{}
+
+\step{construct $\angle a$ equal to $\angle A$, and $\angle b$ equal to $\angle B$.}{§~305}
+
+\step[\indent Then]{$\angle c$ is the $\angle$ required.}{Why?}
+
+\hfill\qef
+
+\end{proof}
+
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a triangle when two sides and the
+included angle are given.}
+
+\figc{127cd309}{Let $b$ and $c$ be the two sides of the triangle and $E$ the included angle.}
+
+\step{Take $AB$ equal to the side $c$.}{}
+
+At $A$, construct $\angle BAD$ equal to the given $\angle E$.~\hfill§~305
+\scanpage{128.png}%
+
+\step{On $AD$ take $AC$ equal to $b$, and draw $CB$.}{}
+
+\step{Then $\triangle ACB$ is the $\triangle$ required.}{\qef}
+
+
+\end{proof}
+
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a triangle when a side and two
+angles of the triangle are given.}
+
+\figc{128ab310}{Let $c$ be the given side, $A$ and $B$ the given angles.}
+
+\step{Take $EC$ equal to the side $c$.}{}
+
+\step{At $E$ construct the $\angle CEH$ equal to $\angle A$.}{§~305}
+
+\step{At $C$ construct the $\angle ECK$ equal to $\angle B$.}{}
+
+\step{Produce $EH$ and $CK$ until they intersect at $O$.}{}
+
+\step{Then $\triangle COE$ is the $\triangle$ required.}{\qef}
+
+\textsc{Remark.} If one of the given angles is opposite to the given side, find
+the third angle by §~308, and proceed as above.
+
+\textbf{Discussion.} The problem is impossible when the two given
+angles are together equal to or greater than two right angles.
+
+\end{proof}
+
+\ex{To construct an equilateral triangle, having given the altitude.}
+
+\exheader{To construct an isosceles triangle, having given:}
+
+\ex{The base and the altitude.}
+
+\ex{The altitude and one of the legs.}
+
+\ex{The angle at the vertex and the altitude.}
+\scanpage{129.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a triangle when two sides and the
+angle opposite one of them are given.}
+
+\figc{129ab311}{Let $a$ and $b$ be the given sides, and $A$ the angle opposite $a$.}
+
+\textsc{Case 1.} \emph{If $a$~is less than~$b$.}
+
+\step{Construct $\angle DAE$ equal to the given $\angle A$}{§~305}
+
+\step{On $AD$ take $AB$ equal to $b$.}{}
+
+\step{From $B$ as a centre, with a radius equal to $a$,}{}
+
+\step{describe an arc intersecting the line $AE$ at $C$ and $C'$.}{}
+
+\step{Draw $BC$ and $BC'$.}{}
+
+Then both the $\triangle_s ABC$ and $ABC'$ fulfil the conditions, and
+hence we have two constructions.
+
+This is called the \emph{ambiguous} case.
+
+\figcc{129cc311}{129dd311}
+\textbf{Discussion.} If the side $a$ is equal to
+the $\perp BH$, the arc described from $B$ will touch $AE$, and there will be but
+one construction, the right $\triangle ABH$.
+
+If the given side $a$ is less than the
+$\perp$ from $B$, the arc described from $B$
+ will not intersect or touch $AE$, and
+hence the problem is impossible.
+\scanpage{130.png}%
+
+If the $\angle A$ is right or obtuse, the problem is impossible; for
+the side opposite a right or obtuse angle is the greatest side.~\hfill§~153
+\filbreak
+\textsc{Case 2.} \emph{If $a$ is equal to $b$.}
+
+\figc{130aa311}{}
+If the $\angle A$ is acute, and $a = b$, the arc described from $B$ as
+a centre, and with a radius equal to $a$, will
+cut the line $AE$ at the points $A$ and $C$.
+There is therefore but one solution: the
+isosceles $\triangle ABC$.
+
+\textbf{Discussion.} If the $\angle A$ is right or obtuse, the problem is
+impossible; for equal sides of a $\triangle$ have equal $\angle_s$ opposite
+them, and a $\triangle$ cannot have two right $\angle_s$ or two obtuse $\angle_s$.
+
+\figccc{130bb311}{130cc311}{130dd311}
+\textsc{Case 3.} \emph{If $a$ is greater than $b$.}
+
+If the given $\angle A$ is acute, the arc described from $B$ will cut
+the line $ED$ on opposite sides of $A$, at $C$ and $C'$. The $\triangle ABC$
+answers the required conditions, but the
+$\triangle$ $ABC'$ does not, for it does not contain
+the acute $\angle A$. There is then only one
+solution; namely, the $\triangle ABC$.
+
+If the $\angle A$ is right, the arc described
+from $B$ cuts the line $ED$ on opposite
+sides of $A$, and we have two \emph{equal} right
+$\triangle_s$ which fulfil the required conditions.
+
+If the $\angle A$ is obtuse, the arc described
+from $B$ cuts the line $ED$ on opposite
+sides of $A$, at the points $C$ and $C'$. The
+$\triangle ABC$ answers the required conditions,
+but the  $\triangle ABC'$ does not, for it does not contain the obtuse
+$\angle A$. There is then only one solution; namely, the $\triangle ABC$.~\hfill\qef
+
+\end{proof}
+\scanpage{131.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a triangle when the three sides of
+the triangle are given.}
+
+\figc{131aa312}{Let the three sides be $c$, $a$, and $b$.}
+
+Take $AB$ equal to $c$. From $A$ as a centre, with a radius
+equal to $b$, describe an arc. From $B$ as a centre, with a radius
+equal to $a$, describe an arc, intersecting the other arc at $C$.
+
+\step{Draw $CA$ and $CB$.}{}
+
+\step{$\triangle CAB$ is the $\triangle$ required.}{\qef}
+
+\textbf{Discussion.} The problem is impossible when one side is equal
+to or greater than the sum of the other two sides.
+
+\end{proof}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a parallelogram when two sides
+and the included angle are given}.
+
+\figc{131bb313}{Let $m$ and $o$ be the two sides, and $C$ the included angle.}
+
+\step{Take $AB$ equal to $o$.}{}
+
+\step{At $A$ construct $\angle BAD$ equal to $\angle C$.}{§~305}
+\scanpage{132.png}%
+
+Take $AH$ equal to $m$. From $H$ as a centre, with a radius
+equal to $o$, describe an arc, and from $B$ as a centre, with a
+radius equal to $m$, describe an arc, intersecting the other arc
+at $E$; and draw $EH$ and $EB$.
+
+\step{The quadrilateral $ABEH$ is the $\Par$ required.}{§~182}
+
+\hfill\qef
+
+\end{proof}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To circumscribe a circle about a given triangle.}
+
+\figc{132aa314}{Let $ABC$ be the given triangle.}
+
+\step{Bisect $AB$ and $BC$.}{§~302}
+
+\step{At $E$ and $D$, the points of bisection, erect $\perp_s$.}{§~301}
+
+Since $BC$ is not the prolongation of $AB$, these $\perp_s$ will intersect
+at some point $O$.
+
+From $O$, with a radius equal to $OB$, describe a circle.
+
+\step{The $\odot ABC$ is the $\odot$ required.}{}
+
+\step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$,}{}
+
+\step{and also is equidistant from $B$ and $C$.}{§~160}
+
+\step{$\therefore$ the point $O$ is equidistant from $A$, $B$, and $C$,}{}
+
+\noindent and a $\odot$ described from $O$ as a centre, with a radius equal to
+$OB$, will pass through the vertices $A$, $B$, and $C$.~\hfill\qef
+
+\end{proof}
+
+The same construction serves to describe a circumference
+which shall pass through three points not in the same straight
+line; also to find the centre of a given circle or of a given arc.
+
+\note{The point $O$ is called the \emph{circum-centre}\label{circum-centre} of the triangle.}
+\scanpage{133.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe a circle in a given triangle.}
+
+\figc{133aa315}{Let $ABC$ be the given triangle.}
+
+\step{Bisect the $\angle_s A$ and $C$.}{§~304}
+
+\step{From $E$, the intersection of the bisectors,}{}
+
+\step{draw $EH \perp$ to the side $AC$.}{§~300}
+
+\step{From $E$ as centre, with radius $EH$, describe the $\odot KHM$.}{}
+
+\step{The $\odot KHM$ is the $\odot$ required.}{}
+
+\textbf{Proof.} Since $E$ is in the bisector of the $\angle A$, it is equidistant
+from the sides $AB$ and $AC$; and since $E$ is in the bisector of
+the $\angle C$, it is equidistant from the sides $AC$ and $BC$.~\hfill§~162
+
+$\therefore$ a $\odot$ described from $E$ as centre, with a radius equal to $EH$,
+will touch the sides of the $\triangle$ and be inscribed in it.~\hfill\qef
+
+\end{proof}
+
+\note{The point $E$ is called the \indexemph{in-centre} of the triangle.}
+
+\figc{133bb316}{}
+\begin{point}%
+The intersections of the
+bisectors of the exterior angles of
+a triangle are the centres of three
+circles, each of which will touch
+one side of the triangle, and the
+two other sides produced. These
+three circles are called \emph{escribed}\label{escribed}
+circles; and their centres are called
+the \indexemph{ex-centres} of the triangle.
+\end{point}
+\scanpage{134.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{Through a given point, to draw a tangent to a
+given circle.}
+
+\textsc{Case 1.} \emph{When the given point is on the circumference.}
+
+\figc{134aa317}{Let $C$ be the given point on the circumference whose centre is~$O$.}
+
+\step{From the centre $O$ draw the radius $OC$.}{}
+
+\step{Through $C$ draw $AM \perp$ to $OC$.}{§~301}
+
+\step{Then $AM$ is the tangent required.}{§~253}
+
+\textsc{Case 2.} \emph{When the given point is without the circle.}
+
+\lett{Let $O$ be the centre of the given circle, $E$ the given point.}
+
+\step{Draw $OE$.}{}
+
+On $OE$ as a diameter, describe a circumference intersecting
+the given circumference at the points $M$ and $H$.
+
+\step{Draw $OM$ and $EM$.}{}
+
+\step{Then $EM$ is the tangent required.}{}
+
+\step[\indent\textbf{Proof.}]{$\angle OME$ is a right angle.}{§~290}
+
+\step{$\therefore EM$ is tangent to the circle at $M$.}{§~253}
+
+In like manner, we may prove $EH$ tangent to the given $\odot$.
+
+\hfill\qef
+
+\end{proof}
+
+\ex{To draw a tangent to a given circle, so that it shall be parallel
+to a given straight line.}
+\scanpage{135.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{Upon a given straight line, to describe a segment
+of a circle in which a given angle may be
+inscribed.}
+
+\figc{135aa318}{Let $AB$ be the given line, and $M$ the given angle.}
+
+\step{Construct the $\angle ABE$ equal to the $\angle M$.}{§~305}
+
+\step{Bisect the line $AB$ by the $\perp OF$.}{§~302}
+
+\step{From the point $B$ draw $BO \perp$ to $EB$.}{§~301}
+
+From $O$, the point of intersection of $FO$ and $BO$, as a centre
+with a radius equal to $OB$, describe a circumference.
+
+\step{The segment $AKB$ is the segment required.}{}
+
+\step[\indent\textbf{Proof.}]{The point $O$ is equidistant from $A$ and $B$.}{§~160}
+
+\step{$\therefore$ the circumference will pass through $A$.}{}
+
+\step{But $BE$ is $\perp$ to $OB$.}{Const.}
+
+\step{$\therefore BE$ is tangent to the $\odot$,}{§~253}
+
+\pnote{(a straight line $\perp$ to a radius at its extremity is tangent to the $\odot$).}
+
+\step{$\therefore \angle ABE$ is measured by $\frac{1}{2} \arc AB$,}{§~295}
+
+\pnote{(being an $\angle$ formed by a tangent and a chord).}
+
+But any $\angle$ as $\angle K$ inscribed in the segment $AKB$ is measured
+by $\frac{1}{2} \arc AB$.\hfill§~289
+
+$\therefore$ the $\angle M$ may be inscribed in the segment $AKB$.
+
+\hfill\qef
+
+\end{proof}
+\scanpage{136.png}%
+
+
+\subsection{SOLUTION OF PROBLEMS.}
+
+\pp{If a problem is so simple that the solution is obvious
+from a known theorem, we have only to make the construction
+according to the theorem, and then give a synthetic proof,
+if a proof is necessary, that the construction is correct, as in
+the examples of the fundamental problems already given.}
+
+\begin{point}%
+But problems are usually of a more difficult type.
+The application of known theorems to their solution is not
+immediate, and often far from obvious. To discover the mode
+of application is the first and most difficult part of the solution.
+The best way to attack such problems is by a method
+resembling the analytic proof of a theorem, called the \textbf{analysis}
+of the problem.
+
+1. \textbf{Suppose the construction made}, and let the figure represent
+all parts concerned, both given and required.
+
+2. Study the relations among the parts with the aid of
+known theorems, and try to find some relation that will suggest
+the construction.
+
+3. If this attempt fails, introduce new relations by drawing
+auxiliary lines, and study the new relations. If this attempt
+fails, make a new trial, and so on till a clue to the right construction
+is found.
+\end{point}
+
+\pp{A problem is \emph{determinate} if it has a \emph{definite} number
+of solutions, \emph{indeterminate} if it has an \emph{indefinite} number of
+solutions, and \emph{impossible} if it has \emph{no} solution. A problem is
+sometimes determinate for certain relative positions or magnitudes
+of the given parts, and indeterminate for other positions
+or magnitudes of the given parts.}
+
+\pp{The \textbf{discussion} of a problem consists in examining the
+problem with reference to all possible conditions, and in determining
+the conditions necessary for its solution.}
+\scanpage{137.png}%
+
+\begin{proofex}%
+\obs{\textsc{Problem.} To construct a circle that
+  shall pass through a given point and cut chords of a given length
+  from two parallels.}
+
+\figc{137aa147}{}
+\textbf{Analysis.} Suppose the problem solved.  Let $A$ be the given
+point, $BC$ and $DE$ the given parallels, $MN$ the given length, and
+$O$ the centre of the required circle.
+
+Since the circle cuts equal chords from two parallels its centre must
+be equidistant from them.  Therefore, one locus for $O$ is $FG
+\parallel$ to $BC$ and equidistant from $BC$ and $DE$.
+
+Draw the $\perp$ bisector of $MN$, cutting $FG$ in $P$.  $PM$ is the
+radius of the circle required.  With $A$ as centre and radius $PM$
+describe an arc cutting $FG$ at $O$.  Then $O$ is the centre of the
+required circle.
+
+\textbf{Discussion.} The problem is impossible if the distance from
+$A$ to $FG$ is greater than $PM$.
+
+\end{proofex}
+
+\figc{137bb148}{}
+\begin{proofex}%
+\obs{\textsc{Problem.} To construct a triangle,
+  having given the perimeter, one angle, and the altitude from the
+  vertex of the given angle.}
+
+\textbf{Analysis.} Suppose the problem solved, and let $ABC$ be the
+$\triangle$ required, $ACB$ the given $\angle$, and $CD$ the given
+altitude.
+
+Produce $AB$ both ways, and take $AE=AC$, and $BF=BC$, then $EF=$ the
+given perimeter.  Join $CE$ and $CF$, forming the isosceles
+$\triangle_s CAE$ and $CBF$.
+
+In the $\triangle ECF$, $\angle E+\angle F+\angle ECF=180°$ (why?),
+but $\angle ECF=\angle ECA+\angle FCB+\angle ACB$.
+
+Since $\angle E=\angle ECA$ and $\angle F=\angle FCB$, we have $\angle
+ECF=\angle E+\angle F+\angle ACB$.\quad $\therefore 2\angle E+2\angle
+F+\angle ACB = 180°$.
+
+\( \therefore \angle E+\angle F+\frac{1}{2}\angle ACB = 90° \), and\
+\( \angle E+\angle F = 90° - \frac{1}{2}\angle ACB \).
+
+By substitution, \(\angle ECF = 90° + \frac{1}{2}\angle ACB \).
+
+$\therefore \angle ECF$ is known.
+
+\end{proofex}
+
+\textbf{Construction.} To find the point $C$, construct on $EF$ a
+segment that will contain the $\angle ECF$ (§~318), and draw a
+parallel to $EF$ at the distance $CD$, the given altitude.
+
+To find the points $A$ and $B$, draw the $\perp$ bisectors of the
+lines $CE$ and $CF$, and the points $A$ and $B$ will be vertices of
+the required $\triangle$.  Why?
+\scanpage{138.png}%
+
+
+\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}
+
+\ex{Find the locus of a point at a given distance from a given
+circumference.}
+
+\exheader{Find the locus of the centre of a circle:}
+
+\ex{Which has a given radius $r$ and passes through a given point $P$.}
+
+\ex{Which has a given radius $r$ and touches a given line $AB$.}
+
+\ex{Which passes through two given points $P$ and $Q$.}
+
+\ex{Which touches a given straight line $AB$ at a given point $P$.}
+
+\ex{Which touches each of two given parallels.}
+
+\ex{Which touches each of two given intersecting lines.}
+
+\begin{proofex}%
+To find in a given line a point $X$ which is equidistant from
+two given points.
+
+The required point is the intersection of the given line with the perpendicular
+bisector of the line joining the two given points (§~160).
+
+\end{proofex}
+
+\ex{To find a point $X$ equidistant from three given points.}
+
+\figc{138aa158}{}
+\ex{To find a point $X$ equidistant from two given
+points and at a given distance from a third given point.}
+
+\ex{To construct a circle which has a given radius
+and passes through two given points.}
+
+\ex{To find a point $X$ at given distances from two given points.}
+
+\ex{To construct a circle which has its centre in a given line and
+passes through two given points.}
+
+\ex{To find a point $X$ equidistant from two given points and also
+equidistant from two given intersecting lines (§§~160 and 162).}
+
+\ex{To find a point $X$ equidistant from two given points and also
+equidistant from two given parallel lines.}
+
+\ex{To find a point $X$ equidistant from two given intersecting
+lines and also equidistant from two given parallels.}
+
+\figc{138bb165}{}
+\ex{To find a point $X$ equidistant from two given
+intersecting lines and at a given distance from a given point.}
+
+\ex{To find a point $X$ which lies in one side of a
+given triangle and is equidistant from the other two sides.}
+\scanpage{139.png}%
+% 169 = c
+% 174 = d
+% 175 = e
+% 177 = f
+% 176 = g
+
+\filbreak
+\figcc{139aa167}{139bb168}
+\ex{A straight railway passes two miles from a
+town.  A place is four miles from the town and one
+mile from the railway. To find by construction the
+places that answer this description.}
+
+\ex{In a triangle $ABC$, to draw $DE$ parallel to
+the base $BC$, cutting the sides of the triangle in $D$ and
+$E$, so that $DE$ shall equal $DB + EC$ (§~162).}
+
+\figc{139cc169}{}
+\begin{proofex}%
+To draw through two sides of a triangle a line
+parallel to the third side so that the part intercepted between
+the sides shall have a given length.
+\null
+\step{Take $BD = d$.}{}
+
+\end{proofex}
+
+\ex{Prove that the locus of the vertex of a right triangle, having
+a given hypotenuse as base, is the circumference described upon the given
+hypotenuse as diameter (§~290). }
+
+\ex{Prove that the locus of the vertex of a triangle, having a given
+base and a given angle at the vertex, is the arc which forms with the base
+a segment capable of containing the given angle (§~318).}
+
+\ex{Find the locus of the middle point of a chord of a given length
+that can be drawn in a given circle.}
+
+\ex{Find the locus of the middle point of a chord drawn from a
+given point in a given circumference.}
+
+\figccc{139dd174}{139ee175}{139gg176}
+\ex{Find the locus of the middle point of a straight line drawn
+from a given exterior point to a given circumference.}
+
+\ex{A straight line moves so that it remains parallel to a given
+line, and touches at one end a given circumference. Find the locus of
+the other end.}
+
+\ex{A straight rod moves so that its ends constantly
+touch two fixed rods which are perpendicular to
+each other.   Find the locus of its middle point. }
+\scanpage{140.png}%
+
+\ex{In a given circle let $AOB$ be a diameter, $OC$ any radius, $CD$
+the perpendicular from $C$ to $AB$.   Upon $OC$ take $OM$ equal to $CD$. Find
+the locus of the point $M$ as $OC$ turns about $O$.}
+
+\figcc{139ff177}{140aa178}
+\ex{To construct an equilateral triangle, having
+given the radius of the circumscribed circle.}
+
+\exheader{To construct on isosceles triangle, having given:}
+
+\ex{The angle at the vertex and the base (§~160 and §~318).}
+
+\ex{The base and the radius of the circumscribed circle.}
+
+\ex{The base and the radius of the inscribed circle.}
+
+\figc{140bb182}{}
+\begin{proofex}%
+The perimeter and the altitude.
+
+Let $ABC$ be the $\triangle$ required, $EF$ the given perimeter.
+The altitude $CD$ passes through the middle
+of $EF$, and the  $\triangle_s AEC$, $BFC$ are isosceles.
+
+\end{proofex}
+
+\exheader{To construct a right triangle, having given:}
+
+\ex{The hypotenuse and one leg.}
+
+\ex{One leg and the altitude upon the hypotenuse.}
+
+\ex{The median and the altitude drawn from the vertex of the
+right angle.}
+
+\ex{The hypotenuse and the altitude upon the hypotenuse.}
+
+\ex{The radius of the inscribed circle and one leg.}
+
+\ex{The radius of the inscribed circle and an acute angle.}
+
+\ex{An acute angle and the sum of the legs.}
+
+\ex{An acute angle and the difference of the legs.}
+
+\figc{140cc191}{}
+\ex{To construct an equilateral triangle, having
+given the radius of the inscribed circle.}
+
+\exheader{To construct a triangle, having given:}
+
+\ex{The base, the altitude, and an angle at the base.}
+
+\ex{The base, the altitude, and the $\angle$ at the vertex.}
+
+\ex{The base, the corresponding median, and the $\angle$ at the vertex.}
+
+\ex{The perimeter and the angles.}
+
+\ex{One side, an adjacent $\angle$, and the sum of the other sides.}
+\scanpage{141.png}%
+
+\exheader{To construct a triangle, having given:}
+
+\ex{One side, an adjacent $\angle$, and the difference of the other sides.}
+
+\ex{The sum of two sides and the angles.}
+
+\ex{One side, an adjacent $\angle$, and the radius of the circumscribed
+circle.}
+
+\ex{The angles and the radius of the circumscribed circle.}
+
+\ex{The angles and the radius of the inscribed circle. }
+
+\ex{An angle, and the bisector and the altitude drawn from the
+vertex of the given angle.}
+
+\ex{Two sides and the median corresponding to the other side.}
+
+\ex{The three medians.}
+
+\exheader{To construct a square, having given:}
+
+\ex{The diagonal.}
+
+\begin{proofex}%
+The sum of the diagonal and one side.
+
+Let $ABCD$ be the square required, $CA$ the diagonal.
+Produce $CA$, making $AE = AB$. $\triangle_s ABC$ and $ABE$ are
+isosceles and $\angle BAC = \angle BCA = 45°$.
+
+\end{proofex}
+
+\figcc{141aa206}{141bb207}
+\ex{Given two perpendiculars, $AB$ and $CD$,
+intersecting in $O$, and a straight line intersecting
+these perpendiculars in $E$ and $F$; to construct a
+square, one of whose angles shall coincide with one
+of the right angles at $O$, and the vertex of the opposite
+angle of the square shall lie in $EF$. (Two solutions.)}
+
+\exheader{To construct a rectangle, having given:}
+
+\ex{One side and the angle between the diagonals.}
+
+\ex{The perimeter and the diagonal.}
+
+\ex{The perimeter and the angle between the diagonals.}
+
+\ex{The difference of two adjacent sides and the angle between
+the diagonals.}
+
+\exheader{To construct a rhombus, having given:}
+
+\ex{The two diagonals.}
+
+\ex{One side and the radius of the inscribed circle.}
+\scanpage{142.png}%
+
+\ex{One angle and the radius of the inscribed circle.}
+
+\ex{One angle and one of the diagonals.}
+
+\exheader{To construct a rhomboid, having given:}
+
+\ex{One side and the two diagonals.}
+
+\ex{The diagonals and the angle between them.}
+
+\ex{One side, one angle, and one diagonal.}
+
+\ex{The base, the altitude, and one angle.}
+
+\exheader{To construct an isosceles trapezoid, having given:}
+
+\ex{The bases and one angle.}
+
+\ex{The bases and the altitude.}
+
+\ex{The bases and the diagonal.}
+
+\figc{142aa223}{}
+\begin{proofex}%
+The bases and the radius of the circumscribed circle.
+
+Let $ABCD$ be the isosceles trapezoid required, $O$ the
+centre of the circumscribed $\odot$. A diameter $\perp$ to $AB$ is $\perp$ to
+$CD$, and bisects both $AB$ and $CD$. Draw $CG$ $\parallel$ to $FE$.
+Then $EG = FC = \frac{1}{2}DC$.
+
+\end{proofex}
+
+\exheader{To construct a trapezoid, having given:}
+
+\ex{The four sides.}
+
+\ex{The two bases and the two diagonals.}
+
+\ex{The bases, one diagonal, and the $\angle$ between the diagonals.}
+
+\exheader{To construct a circle which has the radius $r$ and which also:}
+
+\ex{Touches each of two intersecting lines $AB$ and $CD$.}
+
+\ex{Touches a given line $AB$ and a given circle~$K$.}
+
+\ex{Passes through a given point $P$ and touches a given line~$AB$.}
+
+\ex{Passes through a given point~$P$ and touches a given circle~$K$.}
+
+\exheader{To construct a circle which shall:}
+
+\ex{Touch two given parallels and pass through a given point~$P$.}
+
+\ex{Touch three given lines two of which are parallel.}
+
+\ex{Touch a given line $AB$ at $P$ and pass through a given point~$Q$.}
+
+\ex{Touch a given circle at $P$ and pass through a given point~$Q$.}
+
+\ex{Touch two given lines and touch one of them at a given point~$P$.}
+\scanpage{143.png}%
+
+\ex{Touch a given line and touch a given circle at a point $P$.}
+
+\ex{Touch a given line $AB$ at $P$ and also touch a given circle.}
+
+\ex{To inscribe a circle in a given sector.}
+
+\ex{To construct within a given circle three equal circles, so that
+each shall touch the other two and also the given circle.}
+
+\ex{To describe circles about the vertices of a given triangle as
+centres, so that each shall touch the two others.}
+
+\figc{143aa241}{}
+\begin{proofex}%
+To bisect the angle formed by two lines, without
+producing the lines to their point of intersection.
+
+Draw any line $EF \parallel$ to $BA$. Take $EG = EH$, and produce
+$GH$ to meet $BA$ at $I$. Draw the $\perp$ bisector of $GI$.
+
+\figccc{143bb242}{143cc243}{143dd244}
+
+
+\end{proofex}
+
+\ex{To draw through a given point $P$ between the sides of an
+angle $BAC$ a line terminated by the sides of the angle and bisected at $P$.}
+
+\begin{proofex}%
+Given two points $P$, $Q$, and a line $AB$; to draw lines from $P$
+and $Q$ which shall meet on $AB$ and make equal angles with $AB$.
+
+Make use of the point which forms with $P$ a pair of points symmetrical
+with respect to $AB$.
+
+\end{proofex}
+
+\ex{To find the shortest path from $P$ to $Q$ which shall touch a line~$AB$.}
+
+\figc{143ee245}{}
+\begin{proofex}%
+To draw a common tangent to two given circles.
+
+Let $r$ and $r'$ denote the radii of the circles, $O$ and $O'$ their centres.
+With centre $O$ and radius
+$r-r'$ describe a $\odot$.
+From $O'$ draw the tangents
+$O'M$, $O'N$. Produce
+$OM$ and $ON$ to
+meet the circumference
+at $A$ and $C$. Draw the
+radii $O'B$ and $O'D \parallel$,
+respectively, to $OA$ and $OC$. Draw $AB$ and $CD$.
+
+To draw the internal tangents use an auxiliary $\odot$ of radius $r + r'$.
+
+\end{proofex}
+\scanpage{144.png}%
+
+
+\chapter{BOOK III\@.  PROPORTION\@.  SIMILAR POLYGONS.}
+
+\section[THEORY OF PROPORTION.]{THE THEORY OF PROPORTION.}
+
+\pp{A \indexbf{proportion} is an expression of equality between two
+equal ratios; and is written in one of the following forms:
+\[ a:b = c:d; \qquad a:b::c:d; \qquad \frac{a}{b}=\frac{c}{d}. \]
+This proportion is read, ``$a$ is to $b$ as $c$ is to $d$''; or ``the ratio
+of $a$ to $b$ is equal to the ratio of $c$ to $d$.''}
+
+\begin{point}%
+The \indexbf{terms} of a proportion are the four quantities compared;
+the \emph{first} and \indexemph{third} terms are called the \indexbf{antecedents}, the
+\emph{second} and \indexemph{fourth} terms, the \indexbf{consequents}; the \emph{first} and \emph{fourth}
+terms, the \indexbf{extremes}, the \emph{second} and \emph{third} terms, the \indexbf{means}.
+
+Thus, in the proportion $a : b = c : d$; $a$ and $c$ are the antecedents, $b$ and
+$d$ the consequents, $a$ and $d$ the extremes, $b$ and $c$ the means.
+\end{point}
+
+\begin{point}%
+The fourth proportional to three given quantities is the
+fourth term of the proportion which has for its first three
+terms the three given quantities \emph{taken in order.}
+
+Thus, $d$ is the fourth proportional to $a$, $b$, and $c$ in the proportion
+\[ a:b = c:d. \]
+\end{point}
+
+\begin{point}%
+The quantities $a$, $b$, $c$, $d$, $e$, are said to be in \textbf{continued
+proportion}\label{continuedprop}, if $a:b = b:c = c:d = d:e$.
+
+If three quantities are in continued proportion, the second
+is called the \indexbf{mean proportional} between the other two, and the
+third is called the \textbf{third proportional} to the other two.
+
+Thus, in the proportion $a:b = b:c$; $b$ is the mean proportional between
+$a$ and $c$; and $c$ is the third proportional to $a$ and $b$.
+\end{point}
+\scanpage{145.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In every proportion the product of the extremes is
+equal to the product of the means.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{§~323}
+
+\eq[\indent Whence]{$ad$}{$= bc$.}{\qed}
+\end{proof}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The mean proportional between two quantities is
+  equal to the square root of their product.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= b:c}$.}{}
+
+\eq[\indent Then]{$b^2$}{$= ac$.}{§~327}
+
+Whence, extracting the square root,
+
+\eq{$b$}{$= \sqrt{ac}$.}{\qed}
+\end{proof}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If the product of two quantities is equal to the
+  product of two others, either two may be made the extremes of the
+  proportion in which the other two are made the means.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{ad}$}{$\mathbf{= bc}$.}{}
+
+\proveq{$a:b$}{$= c:d$}
+
+Divide both members of the given equation by $bd$.
+
+\eq[\indent Then]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}
+
+\eq[\indent Or]{$a:b$}{$= c:d$.}{\qed}
+
+\end{proof}
+\scanpage{146.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If four quantities are in proportion, they are in
+proportion by \textnormal{\textbf{alternation}}\label{alternation}; that is, the first term is to
+the third as the second is to the fourth.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\proveq{$a:c$}{$=b:d$.}
+
+\eq[\indent Now]{$ad$}{$=bc$.}{§~327}
+
+Divide each member of the equation by $cd$.
+
+\eq[\indent Then]{$\dfrac{a}{c}$}{$= \dfrac{b}{d}$.}{}
+
+\eq[\indent Or]{$a:c$}{$= b:d$.}{\qed}
+\end{proof}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If four quantities are in proportion, they are in
+proportion by \textnormal{\indexbf{inversion}}; that is, the second term is to
+the first as the fourth is to the third.}
+
+\eq[\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\proveq{$b:a$}{$= d:c$.}
+
+\eq[\indent Now]{$bc$}{$= ad$.}{§~327}
+
+Divide each member of the equation by $ac$.
+
+\eq[\indent Then]{$\dfrac{b}{a}$}{$=\dfrac{d}{c}$.}{}
+
+\eq[\indent Or]{$b:a$}{$= d:c$.}{\qed}
+\end{proof}
+\scanpage{147.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}
+\obs{If four quantities are in proportion, they are in
+proportion by \textnormal{\textbf{composition}}\label{composition} that is, the sum of the first
+two terms is to the second term as the sum of the last
+two terms is to the fourth term.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\proveq{$a+b:b$}{$= c+d:d$.}
+
+\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}
+
+\eq[\indent Then]{$\dfrac{a}{b}+1$}{$= \dfrac{c}{d}+1$;}{}
+
+\eq[that is,]{$\dfrac{a+b}{b}$}{$=\dfrac{c+d}{d}$.}{}
+
+\eq[\indent Or]{$a+b:b$}{$= c+d:d$.}{}
+
+\eq[\indent In like manner]{$a+b:a$}{$= c+d:c$.}{\qed}
+\end{proof}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If four quantities are in proportion, they are in
+proportion by \textnormal{\textbf{division}}\label{division}; that is, the difference of the first
+two terms is to the second term as the difference of the
+last two terms is to the fourth term.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\proveq{$a-b:b$}{$= c-d:d$.}
+
+\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}
+
+\eq[\indent Then]{$\dfrac{a}{b}-1$}{$= \dfrac{c}{d}-1$;}{}
+
+\eq[that is,]{$\dfrac{a-b}{b}$}{$= \dfrac{c-d}{d}$.}{}
+
+\eq[\indent Or]{$a-b:b$}{$= c-d:d$.}{}
+
+\eq[\indent In like manner]{$a-b:a$}{$= c-d:c$.}{\qed}
+
+\end{proof}
+\scanpage{148.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If four quantities are in proportion, they are in
+proportion by \textnormal{\textbf{composition and division}}; that is, the sum
+of the first two terms is to their difference as the sum of
+the last two terms is to their difference.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$}{}
+
+\eq[\indent Then]{$\dfrac{a+b}{a}$}{$=\dfrac{c+d}{c}$.}{§~332}
+
+\eq[\indent And]{$\dfrac{a-b}{a}$}{$=\dfrac{c-d}{c}$.}{§~333}
+
+\eq[\indent Divide,]{$\dfrac{a+b}{a-b}$}{$=\dfrac{c+d}{c-d}$.}{}
+
+\eq[\indent Or]{$a+b:a-b$}{$ = c+d:c-d$.}{\qed}
+
+
+\end{proof}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In a series of equal ratios, the sum of the antecedents
+is to the sum of the consequents as any antecedent
+is to its consequent.}
+
+\step[\indent\textbf{Let}]{$\mathbf{a:b = c:d = e:f = g:h}$.}{}
+
+\prove{$a+c+e+g : b+d+f+h = a:b$.}
+
+\step[\indent Let]{$r = \dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{g}{h}$.}{}
+
+\step[\indent Then]{$a=br$, $c=dr$, $e=fr$, $g=hr$.}{}
+
+\step[\indent And]{$a+c+e+g = ( b+d+f+h )r$.}{}
+
+Divide by $( b+d+f+h )$.
+
+\step[\indent Then]{$\dfrac{a+c+e+g}{b+d+f+h} = r = \dfrac{a}{b}$.}{}
+
+\step[\indent Or]{$a+c+e+g : b+d+f+h = a:b$.}{\qed}
+
+\end{proof}
+\scanpage{149.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The products of the corresponding terms of two
+or more proportions are in proportion.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b = c:d, \; e:f}$}
+ {$\mathbf{= g:h, \; k:l = m:n}$.}{}
+
+\proveq{$aek:bfl$}{$= cgm:dhn$.}
+
+\eq[\indent Now]{$\dfrac{a}{b} = \dfrac{c}{d}$, $\dfrac{e}{f}$}
+  {$= \dfrac{g}{h}$, $\dfrac{k}{l} = \dfrac{m}{n}$.}{}
+
+The products of the first members and of the second members
+of these equations give
+
+\eq{$\dfrac{aek}{bfl}$}{$= \dfrac{cgm}{dhn}$.}{}
+
+\eq[\indent Or]{$aek:bfl$}{$= cgm:dhn$.}{\qed}
+
+\end{proof}
+
+\pp{\cor{If three quantities are in continued proportion,
+the first is to the third as the square of the first is to the
+square of the second.}}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Like powers of the terms of a proportion are in
+proportion.}
+
+\eq[\indent\textbf{Let}]{$\mathbf{a:b}$}{$\mathbf{= c:d}$.}{}
+
+\proveq{$a^n:b^n$}{$= c^n:d^n$.}
+
+\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{c}{d}$.}{}
+
+\eq[\indent Raise to the $n$th power,]
+  {$\dfrac{a^n}{b^n}$}{$= \dfrac{c^n}{d^n}$.}{}
+
+\eq[\indent Or]{$a^n:b^n$}{$= c^n:d^n$.}{\qed}
+
+\end{proof}
+
+
+\pp{\defn{\textbf{Equimultiples}\label{equimultiples} of two quantities are the products
+obtained by multiplying each of them by the same number.
+Thus, $ma$ and $mb$ are equimultiples of $a$ and $b$.}}
+\scanpage{150.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Equimultiples of two quantities are in the same
+ratio as the quantities themselves.}
+
+\textbf{Let $a$~and~$b$ be any two quantities.}
+
+\proveq{$ma : mb$}{$= a:b$.}
+
+\eq[\indent Now]{$\dfrac{a}{b}$}{$= \dfrac{a}{b}$.}{}
+
+Multiply both terms of the first fraction by $m$.
+
+\eq[\indent Then]{$\dfrac{ma}{mb}$}{$= \dfrac{a}{b}$.}{}
+
+\eq[\indent Or]{$ma:mb$}{$= a:b$.}{\qed}
+
+\end{proof}
+
+
+
+\begin{point}%
+\textsc{Scholium.} In the treatment of proportion, it is assumed
+that the \emph{quantities} involved are expressed by their
+\emph{numerical measures}. It is evident that the ratio of two quantities
+of the same kind may be represented by a fraction, if the
+two quantities are expressed in \emph{integers} in terms of a \emph{common
+unit}. If there is no unit in terms of which both quantities can
+be expressed in \emph{integers}, it is still possible by taking the unit
+of measure sufficiently small to find a fraction that will represent
+the ratio \emph{to any required degree of accuracy.}~\hfill§~269
+
+If we speak of the product of two quantities, it must be
+understood that we mean simply \emph{the product of the numbers
+which represent them when they are expressed in terms of a
+common unit.}
+
+In order that four quantities, $a$, $b$, $c$, $d$, may form a proportion,
+$a$ and $b$ must be quantities of the same kind; and $c$ and $d$
+must be quantities of the same kind; though $c$ and $d$ need not
+be of the same kind as $a$ and $b$. In alternation, however, \emph{the
+four quantities must be of the same kind.}
+\end{point}
+\scanpage{151.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If a line is drawn through two sides of a triangle
+parallel to the third side, it divides those sides proportionally.}
+
+\figc{151aa342}{In the triangle $ABC$, let $EF$ be drawn parallel to $BC$.}
+
+\proveq{$EB : AE$}{$= FC: AF$.}
+
+\textsc{Case 1.} \textit{When $AE$ and $EB$ \textup{(Fig.~1)} are commensurable.}
+
+\textbf{Proof.} Find a common measure of $AE$ and $EB$, as $MB$.
+
+Let $MB$ be contained $m$ times in $EB$, and $n$ times in $AE$.
+
+\eq[\indent Then]{$EB:AE$}{$= m:n$.}{}
+
+At the points of division on $BE$ and $AE$ draw lines $\parallel$ to $BC$.
+These lines will divide $AC$ into $m + n$ equal parts, of which $FC$
+will contain $m$, and $AF$ will contain $n$.~\hfill§~187
+
+\eq{$\therefore FC:AF$}{$= m:n$.}{}
+
+\eq{$\therefore EB:AE$}{$=FC:AF$.}{Ax.~1}
+
+
+\textsc{Case 2.}  \textit{When $AE$ and $EB$ \textup{(Fig.~2)} are incommensurable.}
+
+\textbf{Proof.} Divide $AE$ into any number of equal parts, and apply
+one of these parts to $EB$ as many times as $EB$ will contain it.
+
+Since $AE$ and $EB$ are incommensurable, a certain number
+of these parts will extend from $E$ to some point $K$, leaving a
+remainder $KB$ less than one of these parts. Draw $KH \parallel BC$.
+
+\eq[\indent Then]{$EK:AE$}{$= FH:AF$}{Case~1}
+\scanpage{152.png}%
+
+By increasing the \emph{number} of equal parts into which $AE$ is
+divided, we can make the \emph{length} of each part less than any
+assigned value, however small, but not zero.
+
+Hence, $KB$, which is less than one of these equal parts, has
+zero for a limit.~\hfill§~275
+
+And the corresponding segment $HC$ has zero for a limit.
+
+Therefore, $EK$ approaches $EB$ as a limit,~\hfill§~271
+
+and $FH$ approaches $FC$ as a limit.
+
+\step{$\therefore$ the variable $\dfrac{EK}{AE}$ approaches
+  $\dfrac{EB}{AE}$ as a limit,}{§~280}
+
+\step{and the variable $\dfrac{FH}{AF}$ approaches $\dfrac{FC}{AF}$ as a limit.}{}
+
+\step[\indent But]{$\dfrac{EK}{AE}$ is constantly equal to $\dfrac{FH}{AF}$}{Case~1}
+
+\step{$\therefore \dfrac{EB}{AE} = \dfrac{FC}{AF}$.}{§~284}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor[1]{One side of a triangle is to either part cut
+off by a straight line parallel to the base as the other side is
+to the corresponding part.}
+
+\eq[\indent For]{$AE:EB$}{$=AF:FC$.}{}
+
+\eq[\indent By composition,]{$AE+EB:AE$}{$= AF+FC:AF$.}{§~332}
+
+\eq[\indent Or]{$AB:AE$}{$= AC:AF$.}{}
+\end{point}
+
+\begin{point}%
+\cor[2]{If two lines are cut by any number of parallels
+the corresponding intercepts are proportional.}
+
+\figc{152aa344}{}
+Draw $AN \parallel$ to $CD$. Then
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq{$AL=CG$, $LM$}{$=GK$, $MN=KD$.}{§~180}
+
+\eq[\indent Now]{$AH:AM$}{$=AF:AL=FH:LM$}{}
+
+\eq{}{$=HB:MN$.}{§~343}
+
+\eq[\indent Or]{$AF:CG$}{$=FH:GK=HB:KD$.}{}
+
+\setlength{\eqalign}{.5\dentwidth}
+\end{point}
+\scanpage{153.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If a straight line divides two sides of a triangle proportionally, it is parallel to the third side.}
+
+\figc{153aa345}{In the triangle $ABC$, let $EF$ be drawn so that}
+
+\eq{$\dfrac{AB}{AE}$}{$= \dfrac{AC}{AF}$.}{}
+
+\proveq{$EF$ is}{$\parallel$ to $BC$.}
+
+\step[\indent\textbf{Proof.}]{From $E$ draw $EH \parallel$ to $BC$.}{}
+
+\eq[\indent Then]{$AB:AE$}{$= AC:AH$,}{§~343}
+
+\pnote{(one side of a triangle is to either part cut off by a line parallel to the base
+as the other side to the corresponding part).}
+
+\eq[\indent But]{$AB:AE$}{$= AC:AF$.}{Hyp.}
+
+\eq{$\therefore AC:AF$}{$= AC:AH$.}{Ax.~1}
+
+\eq{$\therefore AF$}{$= AH$.}{}
+
+\step{$\therefore EF$ and $EH$ coincide.}{§~47}
+
+\eq[\indent But]{$EH$ is}{$\parallel$ to $BC$.}{Const.}
+
+\step{$\therefore EF$, which coincides with $EH$, is $\parallel$ to $BC$.}{\qed}
+
+\end{proof}
+
+
+\ex{Find the fourth proportional to $91$,~$65$, and~$133$.}
+
+\ex{Find the mean proportional between $39$~and~$351$.}
+
+\ex{Find the third proportional to $54$~and~$3$.}
+\scanpage{154.png}%
+
+\begin{point}%
+If a given line $AB$ is divided at $M$, a point between
+the extremities $A$ and $B$, it is said to be divided \textbf{internally}
+into the segments $MA$ and $MB$; and if it is divided at $M'$,
+a point in the prolongation of $AB$, it is said to be divided
+\textbf{externally} into the segments $M'A$ and $M'B$.
+
+\figc{154aa346}{}
+
+In either case the segments are the \emph{distances} from the point
+of division to the extremities of the line. If the line is divided
+internally, the \emph{sum} of the segments is equal to the line; and
+if the line is divided externally, the \emph{difference} of the segments
+is equal to the line.
+
+Suppose it is required to divide the given line $AB$ \textbf{internally
+and externally in the same ratio}; as, for example, the ratio of
+the two numbers $3$~and~$5$.
+
+\figc{154bb346}{}
+
+We divide $AB$ into $5 + 3$, or~$8$, equal parts, and take $3$~parts
+from $A$; we then have the point $M$, such that
+
+\eq{$MA:MB$}{$= 3:5$.}{(1)}
+
+Secondly, we divide $AB$ into $5-3$, or~$2$, equal parts, and lay
+off on the prolongation of $AB$, to the left of $A$, three of these
+equal parts; we then have the point $M'$, such that
+
+\eq{$M'A:M'B$}{$= 3:5$.}{(2)}
+
+Comparing (1) and (2),
+
+\eq{$MA:MB$}{$= M'A:M'B$.}{}
+\end{point}
+
+\pp{\defn{If a given straight line is divided internally and
+externally into segments having the same ratio, the line is
+said to be \indexbf{divided harmonically}.}}
+\scanpage{155.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The bisector of an angle of a triangle divides the
+opposite side into segments which are proportional to
+the adjacent sides.}
+
+\figc{155aa348}{Let $CM$ bisect the angle $C$ of the triangle $CAB$.}
+
+\proveq{$MA:MB$}{$= CA:CB$.}
+
+\textbf{Proof.} Draw $AE \parallel$ to $MC$, meeting $BC$ produced at $E$.
+
+\eq[\indent Then]{$MA:MB$}{$= CE:CB$,}{§~342}
+
+\pnote{(if a line is drawn through two sides of a $\triangle$ parallel to the third side, it
+divides those sides proportionally).}
+
+\eq[\indent Also,]{$\angle ACM$}{$= \angle CAE$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines);}
+
+\eq[and]{$\angle BCM$}{$= \angle CEA$,}{§~112}
+
+\pnote{(being ext.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq[\indent But]{$\angle ACM$}{$= \angle BCM$.}{Hyp.}
+
+\eq{$\therefore \angle CAE$}{$= \angle CEA$.}{Ax.~1}
+
+\eq{$\therefore CE$}{$=CA$.}{§~147}
+
+Put $CA$ for its equal, $CE$, in the first proportion.
+
+\eq[\indent Then]{$MA:MB$}{$=CA:CB$.}{\qed}
+
+
+\end{proof}
+
+\ex{In a triangle $ABC$, $AB=12$, $AC=14$, $BC=13$. Find the
+segments of $BC$ made by the bisector of the angle $A$.}
+
+\ex{In a triangle $CAB$, $CA=6$, $CB=12$, $AB=15$. Find the
+segments of $AB$ made by the bisector of the angle $C$.}
+\scanpage{156.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The bisector of an exterior angle of a triangle
+divides the opposite side externally into segments which
+are proportional to the adjacent sides.}
+
+\figc{156aa349}{Let $CM'$ bisect the exterior angle $ACE$ of the triangle $CAB$, and
+meet $BA$ produced at $M'$.}
+
+\proveq{$M'A:M'B$}{$=CA:CB$.}
+
+\step[\indent\textbf{Proof.}]{Draw $AF \parallel$ to $M'C$, meeting $BC$ at $F$.}{}
+
+\eq[\indent Then]{$M'A:M'B$}{$CF:CB$.}{§~343}
+
+\eq[\indent Now]{$\angle M'CE$}{$=\angle AFC$,}{§~112}
+
+\eq[and]{$\angle M'CA$}{$=\angle CAF$,}{§~110}
+
+\pnote{(being alt.-int.~$\angle_s$ of $\parallel$ lines).}
+
+\eq[\indent But]{$\angle M'CE$}{$=\angle M'CA$.}{Hyp.}
+
+\eq{$\therefore \angle AFC$}{$= \angle CAF$.}{Ax.~1}
+
+\eq{$\therefore CA$}{$=CF$.}{§~147}
+
+Put $CA$ for its equal, $CF$, in the first proportion.
+
+\eq[\indent Then]{$M'A:M'B$}{$=CA:CB$.}{\qed}
+
+\end{proof}
+
+\textbf{Question.} To what case does this theorem not apply? (See
+\hyperref[page:69]{Ex.~41, page~\pageref{page:69}}.)
+
+\pp{\cor{The bisectors of an interior angle and an exterior
+angle at one vertex of a triangle meeting the opposite side
+divide that side harmonically.}~\hfill§~347}
+\scanpage{157.png}%
+
+
+\section{SIMILAR POLYGONS.}
+
+\begin{point}%
+\defn{\indexbf{Similar polygons} are polygons that have their
+homologous angles equal, and their homologous sides proportional.}
+
+\figc{157aa351}{}
+
+Thus, the polygons $ABCDE$ and $A'B'C'D'E'$ are similar, if
+the $\angle_s A$, $B$, $C$, etc., are equal, respectively, to the $\angle_s A'$, $B'$, $C'$,
+etc., and if
+
+\step{$AB:A'B' = BC:B'C' = CD:C'D'$, etc.}{}
+\end{point}
+
+\pp{\defn{\indexbf{Homologous lines} in similar polygons are lines
+similarly situated.}}
+
+\pp{\defn{The ratio of any two homologous lines in similar
+polygons, is called the \indexbf{ratio of similitude} of the polygons.}}
+
+The primary idea of similarity is \textbf{likeness of form}. The two
+conditions \emph{necessary} to similarity are:
+
+\begin{myenum}
+\item \emph{For every angle in one of the figures there must be an
+equal angle in the other.}
+
+\item \emph{The homologous sides must be proportional.}
+\end{myenum}
+
+Thus, $Q$ and $Q'$ are not similar; the homologous sides are
+proportional, but the homologous angles are not equal. Also
+$R$ and $R'$ are not similar; the homologous angles are equal,
+but the sides are not proportional.
+
+\figc{157bb353}{}
+
+In the case of \emph{triangles}, either condition involves the other
+(see §~354 and §~358).
+\scanpage{158.png}%
+
+\proposition{Theorem.}
+
+\label{similar triangles}
+\begin{proof}%
+\obs{Two mutually equiangular triangles are similar.}
+
+\figc{158aa354}{In the triangles $ABC$ and $A'B'C'$, let the angles $A$, $B$, $C$ be equal
+to the angles $A'$, $B'$, $C'$, respectively.}
+
+\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}
+
+Since the $\triangle_s$ are mutually equiangular, we have only to
+prove that
+
+\step{$AB:A'B' = AC:A'C' = BC:B'C'$.}{§~351}
+
+\textbf{Proof.}  Place the $\triangle A'B'C'$ on the $\triangle ABC$ so that $\angle A'$ shall
+coincide with its equal, the $\angle A$; and $B'C'$ take the position $EH$.
+
+\eq[\indent Then]{$\angle AEH$}{$= \angle B$}{Hyp.}
+
+\eq{$\therefore EH$ is}{$\parallel$ to $BC$.}{§~114}
+
+\eq{$\therefore AB:AE$}{$= AC:AH$.}{§~343}
+
+\eq[\indent That is,]{$AB:A'B'$}{$= AC:A'C'$.}{}
+
+Similarly, by placing $\triangle A'B'C'$ on $\triangle ABC$, so that $\angle B'$
+ shall coincide with its equal, the $\angle B$, we may prove that
+
+\eq{$AB:A'B'$}{$= BC:B'C'$}{\qed}
+
+\end{proof}
+
+\pp{\cor[1]{Two triangles are similar if two angles of the
+one are equal, respectively, to two angles of the other.}}
+
+\pp{\cor[2]{Two right triangles are similar if an acute
+angle of the one is equal to an acute angle of the other.}}
+\scanpage{159.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two triangles have an angle of the one equal
+to an angle of the other, and the including sides proportional,
+they are similar}.
+
+\figc{159aa357}{In the triangles $ABC$ and $A'B'C'$, let $\angle A$ = $\angle A'$, and let}
+
+\eq{$\mathbf{AB : A'B'}$}{$\mathbf{= AC : A'C'}$.}{}
+
+\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}
+
+In this case we prove the $\triangle_s$ similar by proving them mutually
+equiangular.
+
+
+\textbf{Proof.}  Place the $\triangle A'B'C'$ on the $\triangle ABC$, so that the $\angle A'$
+shall coincide with its equal, the $\angle A$.
+
+Then the $\triangle A'B'C'$ will take the position of $\triangle AEH$.
+
+
+\eq[\indent Now]{$\dfrac{AB}{A'B'}$}{$=\dfrac{AC}{A'C'}$.}{Hyp.}
+
+\eq[\indent That is,]{$\dfrac{AB}{AE}$}{$=\dfrac{AC}{AH}$.}{}
+
+\step{$\therefore EH$ is $\parallel$ to $BC$,}{§~345}
+
+\pnote{(if a line divides two sides of a $\triangle$ proportionally, it is $\parallel$ to the third side).}
+
+\step{$\therefore\angle AEH = \angle B$, and $\angle AHE = \angle C$.}{§~112}
+
+\step{$\therefore\triangle AEH$ is similar to $\triangle ABC$.}{§~354}
+
+\step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed}
+
+\end{proof}
+\scanpage{160.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two triangles have their sides respectively proportional,
+they are similar}.
+
+\figc{160aa358}{In the triangles $ABC$ and $A'B'C'$, let}
+
+\step{$\mathbf{AB : A'B' = AC : A'C' = BC : B'C'}$.}{}
+
+\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar.}
+
+
+\textbf{Proof.}   Upon $AB$ take $AE$ equal to $A'B'$, and upon $AC$ take
+$AH$ equal to $A'C'$; and draw $EH$.
+
+\eq[\indent Now]{$AB:A'B'$}{$= AC: A'C'$.}{Hyp.}
+
+\eq[\indent Or, since]{$AE$}{$ = A'B'$ and $AH = A'C'$,}{}
+
+\eq{$AB:AE$}{$= AC:AH$.}{}
+
+\step{$\therefore\triangle_s ABC$ and $AEH$ are similar.}{§~357}
+
+\eq{$\therefore AB:AE$}{$= BC:EH$;}{§~351}
+
+\eq[that is,]{$AB:A'B'$}{$= BC:EH$.}{}
+
+\eq[\indent But]{$AB:A'B'$}{$= BC: B'C'$.}{Hyp.}
+
+\eq{$\therefore BC:EH$}{$= BC:B'C'$.}{Ax.~1}
+
+\eq{$\therefore EH$}{$= B'C'$.}{}
+
+\step{Hence, the $\triangle_s AEH$ and $A'B'C'$ are equal.}{§~150}
+
+\step[\indent But]{$\triangle AEH$ is similar to $\triangle ABC$.}{}
+
+\step{$\therefore\triangle A'B'C'$ is similar to $\triangle ABC$.}{\qed}
+
+\end{proof}
+\scanpage{161.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two triangles which have their sides respectively
+parallel, or respectively perpendicular, are similar.}
+
+\figc{161aa359}{Let $ABC$ and $A'B'C'$ have their sides respectively parallel; and
+$DEF$ and $D'E'F'$ have their sides respectively perpendicular.}
+
+\prove{the $\triangle_s ABC$ and $A'B'C'$ are similar; and
+that the $\triangle_s DEF$ and $D'E'F'$ are similar.}
+
+\textbf{Proof.} 1.~Prolong $BC$ and $AC$ to $B'A'$, forming $\angle_s x$ and $y$.
+
+\step{Then $\angle B' = \angle x$ (§~112), and $\angle B = \angle x$.}{§~110}
+
+\eq[\indent Therefore,]{$\angle B'$}{$= \angle B$}{Ax.~1}
+
+\eq[\indent In like manner,]{$\angle A'$}{$=  \angle A$.}{}
+
+\step{Therefore, $\triangle A'B'C'$ is similar to $\triangle ABC$.}{§~355}
+
+2.~Prolong $DE$ and $FD$ to meet $D'E'$ at $H$ and $D'F'$ at $K$.
+
+The quadrilateral $EHE'O$ has $\angle_s EHE'$ and $E'OE$ right
+angles, by hypothesis.
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq[\indent Therefore,]{}{$\angle E'$ and $\angle OEH$ are supplementary.}{§~206}
+
+\eq[\indent But]{}{$\angle DEF$ and $\angle OEH$ are supplementary.}{§~86}
+
+\setlength{\eqalign}{.5\dentwidth}
+
+\eq{Therefore, $\angle DEF$}{$= \angle E'$.}{§~85}
+
+\eq[\indent In like manner,]{$\angle EDF$}{$= \angle D'$.}{}
+
+\step{Therefore, $\triangle DEF$ is similar to $\triangle D'E'F'$.}{§~355}
+
+\hfill\qed
+
+
+\end{proof}
+
+
+\pp{\cor{The parallel sides and the perpendicular sides
+are homologous sides of the triangles.}}
+\scanpage{162.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The homologous altitudes of two similar triangles
+have the same ratio as any two homologous sides.}
+
+\figc{162aa361}{In the two similar triangles $ABC$ and $A'B'C'$, let $CO$ and $C'O'$ be
+homologous altitudes.}
+
+\prove{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}
+
+\textbf{Proof.}  In the rt.~$\triangle_s COA$ and $C'O'A'$,
+
+\step{$\angle A = \angle A'$,}{§~351}
+
+\pnote{(being homologous $\triangle_s$ of the similar $\triangle_s ABC$ and $A'B'C'$).}
+
+\step{$\therefore\triangle_s COA$ and $C'O'A'$ are similar,}{§~356}
+
+\pnote{(two rt.~$\triangle_s$ having an acute $\angle$ of the one equal to an acute $\angle$ of the other
+are similar).}
+
+\step{$\therefore\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}$.}{§~351}
+
+In the similar $\triangle_s ABC$ and $A'B'C'$,
+
+\step{$\dfrac {AC}{A'C'}=\dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{§~351}
+
+\step[Therefore,]{$\dfrac {CO}{C'O'}=\dfrac {AC}{A'C'}=
+  \dfrac {AB}{A'B'}=\dfrac {BC}{B'C'}$.}{\qed}
+
+
+\end{proof}
+
+\ex{The base and altitude of a triangle are $7$~feet $6$~inches and
+$5$~feet $6$~inches, respectively. If the homologous base of a similar triangle
+is $5$~feet $6$~inches, find its homologous altitude.}
+\scanpage{163.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two parallels are cut by three or more transversals
+that pass through the same point, the corresponding
+segments are proportional}.
+
+\figc{163aa362}{Let the two parallels $AE$ and $A'E'$ be cut by the transversals $OA$,
+$OB$, $OC$, $OD$, $OE$ in $A$, $A'$, $B$, $B'$, etc.}
+
+\prove{$\dfrac{AB}{A'B'}= \dfrac{BC}{B'C'}= \dfrac{CD}{C'D'}= \dfrac{DE}{D'E'}$.}
+
+\textbf{Proof.}   Since $A'E'$ is $\parallel$ to $AE$, the pairs of  $\triangle_s OAB$ and $OA'B'$,
+$OBC$ and $OB'C'$, etc., are similar.~\hfill§~354
+
+\eq{$\therefore\dfrac{AB}{A'B'} = \dfrac{OB}{OB'}$}
+  {and $\dfrac{BC}{B'C'} = \dfrac{OB}{OB'}$.}{§~351}
+
+\pnote{(homologous sides of similar $\triangle_s$ are proportional).}
+
+\eq{$\therefore \dfrac{AB}{A'B'}$}{$= \dfrac{BC}{B'C'}$.}{Ax.~1}
+
+In a similar way it may be shown that
+
+\eq{$\dfrac{BC}{B'C'} = \dfrac{CD}{C'D'}$}
+  {and $\dfrac{CD}{C'D'} = \dfrac{DE}{D'E'}$.}{\qed}
+
+
+\end{proof}
+
+\note{A condensed form of writing the above is
+\par
+\step{\( \dfrac{AB} {A'B'}=\left(\dfrac{OB} {OB'}\right)=\dfrac{BC} {B'C'}=\left(\dfrac{OC} {OC'}\right)=\dfrac{CD} {C'D'}=\left(\dfrac{OD} {OD'}\right)=\dfrac{DE} {D'E'} \). }{}
+\par
+A parenthesis about a ratio signifies that this ratio is used to prove the
+equality of the ratios immediately preceding and following it.}
+\scanpage{164.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} If three or more non-parallel straight
+lines intercept proportional segments upon two parallels,
+they pass through a common point.}
+
+\figc{164aa363}{Let $AB$, $CD$, $EF$ cut the parallels $AE$ and $BF$ so that}
+
+\eq{$\mathbf{AC : BD}$}{$\mathbf{= CE : DF}$.}{}
+
+\prove{$AB$, $CD$, $EF$ prolonged meet in a point.}
+
+\textbf{Proof.}   Prolong $AB$ and $CD$ until they meet in $O$.
+
+\step{Draw $OE$.}{}
+
+\step{Designate by $F'$ the point where $OE$ cuts $BF$.}{}
+
+\eq[\indent Then]{$AC:BD$}{$=CE:DF'$.}{§~362}
+
+\eq[\indent But]{$AC:BD$}{$=CE:DF$.}{Hyp.}
+
+\eq{$\therefore CE:DF'$}{$= CE:DF$.}{Ax.~1}
+
+\eq{$\therefore DF'$}{$= DF$.}{}
+
+\step{$\therefore F'$ coincides with $F$.}{}
+
+\step{$\therefore EF$ coincides with $EF'$.}{§~47}
+
+\step{$\therefore EF$ prolonged passes through $O$.}{}
+
+\step{$\therefore AB$, $CD$, and $EF$ prolonged meet in the point $O$.}{\qed}
+
+\end{proof}
+\scanpage{165.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The perimeters of two similar polygons have the
+same ratio as any two homologous sides.}
+
+\figc{165aa364}{Let the two similar polygons be $ABCDE$ and $A'B'C'D'E'$, and let
+$P$ and $P'$ represent their perimeters.}
+
+\proveq{$P:P'$}{$= AB: A'B'$.}
+
+\step[\indent\textbf{Proof.}]{$AB : A'B' = BC : B'C' = CD : C'D'$, etc.}{§~351}
+
+\step{$\therefore AB + BC + \text{etc.}\ : A'B' + B'C' + \text{etc.}\ = AB : A'B'$,}{§~335}
+
+\pnote{(in a series of equal ratios the sum of the antecedents is to the sum of the
+consequents as any antecedent is to its consequent).}
+
+\eq[\indent That is,]{$P : P'$}{$= AB : A'B'$.}{\qed}
+
+\end{proof}
+
+\ex{If the line joining the middle points of the bases of a trapezoid
+is produced, and the two legs are also produced, the three lines will
+meet in the same point.}
+
+\ex{$AB$ and $AC$ are chords drawn from any point $A$ in the circumference
+of a circle, and $AD$ is a diameter. The tangent to the circle
+at $D$ intersects $AB$ and $AC$ at $E$ and $F$, respectively. Show that the
+triangles $ABC$ and $AEF$ are similar.}
+
+\ex{$AD$ and $BE$ are two altitudes of the triangle $CAB$. Show
+that the triangles $CED$ and $CAB$ are similar.}
+
+\ex{If two circles are tangent to each other, the chords formed
+by a straight line drawn through the point of contact have the same ratio
+as the diameters of the circles.}
+\scanpage{166.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two polygons are similar, they are composed
+of the same number of triangles, similar each to each,
+and similarly placed.}
+
+\figc{166aa365}{Let the polygons $ABCDE$ and $A'B'C'D'E'$ be similar.}
+
+From two homologous vertices, as $E$ and $E'$, draw diagonals
+$EB$, $EC$, and $E'B'$, $E'C'$.
+
+\prove{the $\triangle_s EAB$, $EBC$, $ECD$ are similar,
+respectively, to the $\triangle_s E'A'B'$, $E'B'C'$, $E'C'D'$.}
+
+\textbf{Proof.} The $\triangle_s EAB$ and $E'A'B'$ are similar.~\hfill§~357
+
+\eq[\indent For]{$\angle A$}{$= \angle A'$,}{§~351}
+
+\eq[and]{$AE:A'E'$}{$=AB:A'B'$.}{§~351}
+
+\eq[\indent Now]{$\angle ABC$}{$= \angle A'B'C'$,}{§~351}
+
+\eq[and]{$\angle ABE$}{$=\angle A'B'E'$.}{§~351}
+
+%proofrule
+
+\eq[\indent By subtracting,]{$\angle EBC$}{$=\angle E'B'C'$.}{Ax.~3}
+
+\eq[\indent Now]{$EB:E'B'$}{$=AB:A'B'$}{§~351}
+
+\eq[and]{$BC:B'C'$}{$=AB:A'B'$}{§~351}
+
+\eq{$\therefore EB:E'B'$}{$=BC:B'C'$.}{Ax.~1}
+
+\step{$\therefore \triangle_s EBC$ and $E'B'C'$ are similar.}{§~357}
+
+In like manner $\triangle_s ECD$ and $E'C'D'$ are similar.~\hfill\qed
+
+\end{proof}
+\scanpage{167.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{\textsc{Conversely:} If two polygons are composed of
+the same number of triangles, similar each to each, and
+similarly placed, the polygons are similar.}
+
+\figc{167aa366}{In the two polygons $ABCDE$ and $A'B'C'D'E'$, let the triangles
+$AEB$, $BEC$, $CED$ be similar, respectively, to the triangles $A'E'B'$,
+$B'E'C'$, $C'E'D'$; and similarly placed.}
+
+\prove{$ABCDE$ is similar to $A'B'C'D'E'$.}
+
+\eq[\indent\textbf{Proof.}]{$\angle A$}{$= \angle A'$}{§~351}
+
+\eq[\indent Also,]{$\angle ABE$}{$= \angle A'B'E'$,}{}
+
+\eq[and]{$\angle EBC$}{$= \angle E'B'C'$.}{§~351}
+
+%proofrule
+
+\eq[\indent By adding,]{$\angle ABC$}{$= \angle A'B'C'$.}{Ax.~2}
+
+In like manner, $\angle BCD = \angle B'C'D'$, $\angle CDE=\angle C'D'E'$, etc.
+
+\step{Hence, the polygons are mutually equiangular.}{}
+
+Also, \( \dfrac{AB}{A'B'} = \left(\dfrac{EB}{E'B'}\right) =
+  \dfrac{BC}{B'C'} = \left(\dfrac{EC}{E'C'}\right) =
+  \dfrac{CD}{C'D'} \), etc.~\hfill§~351
+
+
+Hence, the polygons have their homologous sides proportional.
+
+\step{Therefore, the polygons are similar.}{§~351}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{168.png}%
+
+
+\pagebreak
+\section[EXERCISES.]{THEOREMS.}
+
+\ex{If two circles are tangent to each other externally, the corresponding
+segments of two lines drawn through the point of contact and
+terminated by the circumferences are proportional.}
+
+\ex{In a parallelogram $ABCD$, a line $DE$ is drawn, meeting the }
+diagonal $AC$ in $F$, the side $BC$ in $G$, and the side $AB$ produced in $E$.
+Prove that $\overline{DF}^2 = FG × FE$.
+
+\ex{Two altitudes of a triangle are inversely proportional to the
+corresponding bases.}
+
+\ex{Two circles touch at $P$. Through $P$ three lines are drawn,
+meeting one circle in $A$, $B$, $C$, and the other in $A'$, $B'$, $C'$, respectively.
+Prove that the triangles $ABC$, $A'B'C'$ are similar.}
+
+\begin{proofex}%
+Two chords $AB$, $CD$ intersect at $M$, and $A$ is the middle point
+of the arc $CD$.   Prove that the product $AB × AM$ is constant if the chord
+$AB$ is made to turn about the fixed point $A$.
+
+Draw the diameter $AE$, and draw $BE$.
+
+\end{proofex}
+
+\begin{proofex}%
+If two circles touch each other, their common external tangent
+is the mean proportional between their diameters.
+
+Let $AB$ be the common tangent. Draw the diameters $AC$, $BD$. Join
+the point of contact $P$ to $A$, $B$, $C$, and $D$. Show that $APD$ and $BPC$ are
+straight lines $\perp$ to each other, and that $\triangle_s CAB$, $ABD$ are similar.
+
+\end{proofex}
+
+
+\begin{proofex}%
+If two circles are tangent internally, all chords of the greater
+circle drawn from the point of contact are divided proportionally by the
+circumference of the smaller circle.
+
+Draw any two of the chords, and join the points where they meet the
+circumferences. The $\triangle_s$ thus formed are similar (Ex.~120).
+
+\end{proofex}
+
+
+\figc{168aa263}{}
+\begin{proofex}%
+In an inscribed quadrilateral, the product of
+the diagonals is equal to the sum of the products of the
+opposite sides.
+
+Draw $DE$, making $\angle CDE = \angle ADB$. The $\triangle_s ABD$ and
+$ECD$ are similar; and the $\triangle_s BCD$ and $AED$ are similar.
+
+\end{proofex}
+
+\ex{Two isosceles triangles with equal vertical angles are similar.}
+
+\begin{proofex}%
+The bisector of the vertical angle $A$ of the triangle $ABC$ intersects
+the base at $D$ and the circumference of the circumscribed circle at $E$.
+
+Show that $ AB × AC = AD × AE$.
+
+\end{proofex}
+\scanpage{169.png}%
+
+\clearpage
+\section{NUMERICAL PROPERTIES OF FIGURES.}
+
+\proposition{Theorem.}
+\label{160}
+
+\begin{proof}%
+\obs{If in a right triangle a perpendicular is drawn
+from the vertex of the right angle to the hypotenuse:}
+
+\begin{myenum}
+\item \emph{The triangles thus formed are similar to the given
+triangle, and to each other.}
+
+\item \emph{The perpendicular is the mean proportional between
+the segments of the hypotenuse.}
+
+\item \emph{Each leg of the right triangle is the mean proportional
+between the hypotenuse and its adjacent segment.}
+\end{myenum}
+
+\figc{169aa367}{In the right triangle $ABC$, let $CF$ be drawn from the vertex of
+the right angle $C$, perpendicular to $AB$.}
+
+\prove[\textup{\textbf{1.~}}To prove that ]{$\triangle$'s $BCA$, $CFA$, $BFC$ are similar.}
+
+\textbf{Proof.} The rt.~$\triangle_s CFA$ and $BCA$ are similar,~\hfill§~356
+
+\step{since the $\angle a'$ is common.}{}
+
+The rt.~$\triangle_s BFC$ and $BCA$ are similar,~\hfill§~356
+
+\step{since the $\angle b$ is common.}{}
+
+Since the $\triangle_s CFA$ and $BFC$ are each similar to $\triangle BCA$,
+they are similar to each other.~\hfill§~354
+
+\proveq[\indent\textup{\textbf{2.~}}To prove that]{$AF:CF$}{$=CF:FB$.}
+
+\textbf{Proof.} In the similar $\triangle_s CFA$ and $BFC$,
+
+\eq{$AF:CF$}{$=CF:FB$.}{§~351}
+\scanpage{170.png}%
+
+\filbreak
+\label{161}
+
+\proveq[\indent\textup{\textbf{3.~}}To prove that]{$AB:AC$}{$= AC:AF$,}
+
+\eq[\emph{and}]{$AB:BC$}{$= BC:BF$.}{}
+
+\textbf{Proof.} In the similar $\triangle_s BCA$ and $CFA$,
+
+\eq{$AB:AC$}{$= AC:AF$}{§~351}
+
+In the similar $\triangle_s BCA$ and $BFC$,
+
+\eq{$AB:BC$}{$=BC:BF$.}{§~351}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor[1]{The squares of the two legs of a right triangle
+are proportional to the adjacent segments of the hypotenuse.}
+
+From the proportions in §~367,3,
+
+\step{$\overline{AC}^2=AB × AF$, and $\overline{BC}^2=AB × BF$.}{§~327}
+
+\step[\indent Hence,]{\( \dfrac{\overline{AC}^2}{\overline{BC}^2} =
+  \dfrac{AB × AF}{AB × BF} =
+  \dfrac{AF}{BF} \).}{}
+\end{point}
+
+\begin{point}%
+\cor[2]{The squares of the hypotenuse and either leg
+are proportional to the hypotenuse and the adjacent segment.}
+
+\step[\indent For]{\( \dfrac{\overline{AB}^2}{\overline{AC}^2} =
+  \dfrac{AB × AB}{AB × AF} =
+  \dfrac{AB}{AF} \).}{}
+\end{point}
+
+\figc{170aa370}{}
+\begin{point}%
+\cor[3]{The perpendicular from any point in the
+circumference to the diameter of a circle
+is the mean proportional between the segments
+of the diameter.
+
+The chord drawn from any point in
+the circumference to either extremity of
+the diameter is the mean proportional between the diameter
+and the adjacent segment.}
+
+\step[\indent For]{the $\angle ACB$ is a rt.~$\angle$.}{§~290}
+\end{point}
+\scanpage{171.png}%
+
+\proposition{Theorem.}
+\label{162}
+
+\begin{proof}%
+\obs{The sum of the squares of the two legs of a right
+triangle is equal to the square of the hypotenuse.}
+
+\figc{171aa371}{Let $ABC$ be a right triangle with its right angle at $C$.}
+
+\proveq{$\overline{AC}^2 + \overline{CB}^2$}{$= \overline{AB}^2$.}
+
+\eq[\indent\textbf{Proof.}]{Draw $CF$}{$\perp$ to $AB$.}{}
+
+\eq[\indent Then]{$\overline{AC}^2$}{$= AB × AF$,}{}
+
+\eq[and]{$\overline{CB}^2$}{$= AB × BF$.}{§~367}
+
+%proofrule
+
+\eq[\indent By adding,]{$\overline{AC}^2 + \overline{CB}^2$}
+  {$= AB(AF + BF) = \overline{AB}^2$}{\qed}
+
+\end{proof}
+
+\pp{\cor[1]{The square of either leg of a right triangle is
+equal to the difference of the square of the hypotenuse and
+the square of the other leg.}}
+
+\figcc{171bb373}{171cc374}
+\begin{point}%
+\cor[2]{The  diagonal and a side of a
+square are incommensurable.}
+
+\step[\indent For]{$\overline{AC}^2 = \overline{AB}^2 + \overline{BC}^2 = 2 \overline{AB}^2$.}{}
+
+\step{$\therefore AC = AB \sqrt{2}$.}{}
+\end{point}
+
+\pp{\defn{The \textbf{projection} of any line
+upon a second line is the segment of
+the second line included between the
+perpendiculars drawn to it from the
+extremities of the first line. Thus,
+$PR$ is the projection of $CD$ upon $AB$.}}
+\scanpage{172.png}%
+
+\proposition{Theorem.}
+\label{163}
+
+\begin{proof}%
+\obs{In any triangle, the square of the side opposite an
+acute angle is equal to the sum of the squares of the
+other two sides diminished by twice the product of one of
+those sides by the projection of the other upon that side.}
+
+\figc{172aa375}{Let $C$ be an acute angle of the triangle $ABC$, and $DC$ the projection
+of $AC$ upon $BC$.}
+
+\prove{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.}
+
+\textbf{Proof.} If $D$ falls upon the base (Fig.~1),
+
+\eq{$DB$}{$= BC - DC$.}{}
+
+If $D$ falls upon the base produced (Fig.~2),
+
+\eq{$DB$}{$= DC - BC$.}{}
+
+In either case,
+
+\step{$\overline{DB}^2 = \overline{BC}^2 + \overline{DC}^2 - 2 BC × DC$.}{}
+
+Add $\overline{AD}^2$ to both sides of this equality, and we have
+
+\step{\( \overline{AD}^2 + \overline{DB}^2 =
+  \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 - 2 BC × DC \).}{}
+
+\eq[\indent But]{$\overline{AD}^2 + \overline{DB}^2$}{$= \overline{AB}^2$}{}
+
+\eq[and]{$\overline{AD}^2 + \overline{DC}^2$}{$= \overline{AC}^2$}{§~371}
+
+Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality.
+
+\step[\indent Then]{$\overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 - 2 BC × DC$.}{\qed}
+
+\end{proof}
+\scanpage{173.png}%
+
+\proposition{Theorem.}
+\label{164}
+
+\begin{proof}%
+\obs{In any obtuse triangle, the square of the side
+opposite the obtuse angle is equal to the sum of the
+squares of the other two sides increased by twice the
+product of one of those sides by the projection\label{projection} of
+the other upon that side.}
+
+\figc{173aa376}{Let $C$ be the obtuse angle of the triangle $ABC$, and $CD$ be the projection
+of $AC$ upon $BC$ produced.}
+
+\prove{\( \overline{AB}^2 = \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).}
+
+\step[\indent\textbf{Proof.}]{$DB = BC + DC$.}{}
+
+\step[\indent Squaring,]{\( \overline{DB}^2 =
+  \overline{BC}^2 + \overline{DC}^2 + 2 BC × DC \).}{}
+
+Add $\overline{AD}^2$ to both sides, and we have
+
+\step{\( \overline{AD}^2 + \overline{DB}^2 =
+  \overline{BC}^2 + \overline{AD}^2 + \overline{DC}^2 + 2 BC × DC \).}{}
+
+\step[\indent But]{\( \overline{AD}^2 + \overline{DB}^2 = \overline{AB}^2 \text{, and }
+  \overline{AD}^2 + \overline{DC}^2 = \overline{AC}^2 \).}{§~371}
+
+Put $\overline{AB}^2$ and $\overline{AC}^2$ for their equals in the above equality.
+
+\step[\indent Then]{\( \overline{AB}^2 =
+  \overline{BC}^2 + \overline{AC}^2 + 2 BC × DC \).}{\qed}
+
+\end{proof}
+
+\note[1]{By the Principle of Continuity the last three theorems may
+be included in one theorem. Let the student explain.}
+
+\note[2]{The last three theorems enable us to compute the lengths of
+the altitudes of a triangle if the lengths of the three sides are known.}
+\scanpage{174.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+1.~\obs{The sum of the squares of two sides of a triangle
+is equal to twice the square of half the third side
+increased by twice the square of the median upon that
+side.}
+
+2.~\obs{The difference of the squares of two sides of a triangle
+is equal to twice the product of the third side by
+the projection of the median upon that side.}
+
+\figc{174aa377}{In the triangle $ABC$, let $AM$ be the median and $MD$ the projection
+of $AM$ upon the side $BC$.   Also, let $AB$ be greater than $AC$.}
+
+\proveq{\textup{1.} $\overline{AB}^2+\overline{AC}^2$}{$=
+  2\overline{BM}^2+2\overline{AM}^2$.}
+
+\proveq[]{\textup{2.} $\overline{AB}^2-\overline{AC}^2$}{$=
+  2BC × MD$.}
+
+\textbf{Proof.}   Since $AB > AC$, the $\angle AMB$ will be obtuse, and the
+$\angle AMC$ will be acute.~\hfill§~155
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq[\indent Then]{$\overline{AB}^2$}{$=\overline{BM}^2+\overline{AM}^2+2BM × MD$,}{§~376}
+
+\eq[and]{$\overline{AC}^2$}{$=\overline{MC}^2+\overline{AM}^2-2MC × MD$.}{§~375}
+
+\setlength{\eqalign}{.5\dentwidth}
+
+Add these two equalities, and observe that $BM=MC$.
+
+
+\eq[\indent Then]{$\overline{AB}^2+\overline{AC}^2$}{$=2 \overline{BM}^2+2 \overline{AM}^2$.}{}
+
+Subtract the second equality from the first.
+
+\eq[\indent Then]{$\overline{AB}^2-\overline{AC}^2$}{$=2 BC × MD$.}{\qed}
+
+\end{proof}
+
+\note{This theorem enables us to compute the lengths of the medians
+of a triangle if the lengths of the three sides are known.}
+\scanpage{175.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If two chords intersect in a circle, the product
+of the segments of one is equal to the product of the
+segments of the other.}
+
+\figc{175aa378}{Let any two chords $MN$ and $PQ$ intersect at $O$.}
+
+\proveq{$OM × ON$}{$= OQ × OP$.}
+
+\step[\indent\textbf{Proof.}]{Draw $MP$ and $NQ$.}{}
+
+\eq{$\angle a$}{$= \angle a'$,}{§~289}
+
+\pnote{(each being measured by $\frac{1}{2} \arc PN$).}
+
+\eq[\indent And]{}{$\angle c = \angle c'$,}{§~289}
+
+\pnote{(each being measured by $\frac{1}{2} \arc MQ$).}
+
+\step{$\therefore$ the $\triangle_s NOQ$ and $POM$ are similar.}{§~355}
+
+\eq{$\therefore OQ:OM$}{$=ON:OP$.}{§~351}
+
+\eq{$\therefore OM × ON$}{$= OQ × OP$.}{§~327}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}\textsc{Scholium.} This proportion may be written
+
+\step{$\dfrac{OM}{OQ} = \dfrac{OP}{ON}$, or
+  $\dfrac{OM}{OQ} = \dfrac{1}{\dfrac{ON}{OP}}$;}{}
+
+that is, the ratio of two corresponding segments is equal to the
+\emph{reciprocal} of the ratio of the other two segments.
+
+Hence, these segments are said to be \emph{reciprocally proportional}.
+\end{point}
+\scanpage{176.png}%
+
+\pp{\defn{\textbf{A secant from a point to a circle}\label{secant2} is understood to
+mean the segment of the secant lying between the point and
+the \emph{second point} of intersection of the secant and circumference.}}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If from a point without a circle a secant and a
+tangent are drawn, the tangent is the mean proportional
+between the whole secant and its external segment.}
+
+\figc{176aa381}{Let $AD$ be a tangent and $AC$ a secant drawn from the point $A$ to
+the circle $BCD$.}
+
+\prove{$AC : AD = AD : AB$.}
+
+\step[\indent\textbf{Proof.}]{Draw $DC$ and $DB$.}{}
+
+\step{The $\triangle_s ADC$ and $ABD$ are similar.}{§~355}
+
+\step{For $\angle b$ is common; and $\angle a' = \angle a$,}{§§~289,~295}
+
+\pnote{(each being measured by $\frac{1}{2} \arc BD$).}
+
+\step{$\therefore AC : AD = AD : AB$.}{§~351}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor{If from a fixed point without a circle a secant
+is drawn, the product of the secant and its external segment
+is constant in whatever direction the secant is drawn.}
+
+\step[\indent For]{$AC × AB = \overline{AD}^2$.}{§~327}
+\end{point}
+\scanpage{177.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The square of the bisector of an angle of a triangle
+is equal to the product of the sides of this angle
+diminished by the product of the segments made by
+the bisector upon the third side of the triangle.}
+
+\figc{177aa383}{Let $NO$ bisect the angle $MNP$ of the triangle $MNP$.}
+
+\prove{$\overline{NO}^2 = NM × NP - OM × OP$.}
+
+\step[\indent\textbf{Proof.}]{Circumscribe the $\odot MNP$ about the $\triangle MNP$.}{§~314}
+
+Produce $NO$ to meet the circumference in $Q$, and draw $PQ$.
+
+\step{The $\triangle_s NQP$ and $NMO$ are similar.}{§~355}
+
+\eq[\indent For]{$\angle b$}{$= \angle b'$}{Hyp.}
+
+\eq[and]{$\angle a$}{$= \angle a'$}{§~289}
+
+\eq[\indent Whence]{$NQ:NM$}{$= NP:NO$.}{§~351}
+
+\eq{$\therefore NM × NP$}{$= NQ × NO$}{}
+
+\eq{}{$= (NO+OQ)NO$}{}
+
+\eq{}{$= \overline{NO}^2 + NO × OQ$.}{}
+
+\eq[\indent But]{$NO × OQ$}{$= MO × OP$.}{§~378}
+
+\eq{$\therefore MN × NP$}{$= \overline{NO}^2 + MO × OP$.}{}
+
+\step[\indent Whence]{$\overline{NO}^2 = NM × NP = MO × OP$.}{Ax.~3}
+
+\hfill\qed
+
+\end{proof}
+
+\note{This theorem enables us to compute the lengths of the bisectors
+of the angles of a triangle if the lengths of the sides are known.}
+\scanpage{178.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{In any triangle the product of two sides is equal
+to the product of the diameter of the circumscribed circle
+by the altitude upon the third side.}
+
+\figc{178aa384}{Let NMQ be a triangle, NO the altitude, and QNMP the circle circumscribed
+about the triangle NMQ.}
+
+\step{Draw the diameter $NP$, and draw $PQ$.}{}
+
+\prove{$NM × NQ = NP × NO$.}
+
+\textbf{Proof.} In the $\triangle_s NOM$ and $NQP$,
+
+\step{$\angle NOM$ is a rt.~$\angle$,}{Hyp.}
+
+\step{$\angle NQP$ is a rt.~$\angle$,}{§~290}
+
+\eq{and $\angle a$}{$= \angle a'$,}{§~289}
+
+\pnote{(each being measured by $\frac{1}{2} \arc NQ$).}
+
+\step{$\therefore \triangle_s NOM$ and $NQP$ are similar.}{§~356}
+
+\eq[\indent Whence]{$NM:NP$}{$= NO:NQ$.}{§~351}
+
+\eq{$\therefore NM × NQ$}{$= NP × NO$.}{§~327}
+
+\hfill\qed
+
+\end{proof}
+
+\note{This theorem enables us to compute the length of the radius of
+a circle circumscribed about a triangle, if the lengths of the three sides of
+the triangle are known. }
+
+\ex{If $OE$, $OF$, $OG$ are the perpendiculars from any point $O$
+within the triangle $ABC$ upon the sides $AB$, $BC$, $CA$, respectively, show
+that \( \overline{AE}^2 + \overline{BF}^2 +  \overline{CG}^2 =  \overline{EB}^2 +  \overline{FC}^2 +  \overline{GA}^2 \).}
+\scanpage{179.png}%
+
+
+\section[EXERCISES.]{THEOREMS.}
+
+\begin{proofex}%
+The sum of the squares of the segments of two perpendicular
+chords is equal to the square of the diameter of the circle.
+
+If $AB$, $CD$ are the chords, draw the diameter $BE$, draw $AC$, $ED$, $BD$.
+Prove that $AC = ED$, and apply §~371.
+
+\end{proofex}
+
+\ex{The tangents to two intersecting circles drawn from any point
+in their common chord produced, are equal. (§~381.)}
+
+\ex{The common chord of two intersecting circles, if produced,
+will bisect their common tangents. (§~381.)}
+
+\figc{179aa270}{}
+\begin{proofex}%
+If three circles intersect one another, the common chords all
+pass through the same point.
+
+Let two of the chords $AB$ and $CD$ meet at $O$.   Join
+the point of intersection $E$ to $O$, and suppose that $EO$
+produced meets the same two circles at two different
+points $P$ and $Q$.    Then prove that $OP = OQ$ (§~378);
+hence, that the points $P$ and $Q$ coincide.
+
+\end{proofex}
+
+\ex{If two circles are tangent to each other, the common internal
+tangent bisects the two common external tangents.}
+
+\begin{proofex}%
+If the perpendiculars from the vertices of the triangle $ABC$
+upon the opposite sides intersect at $D$, show that
+
+\step{\( \overline{AB}^2-\overline{AC}^2 = \overline{BD}^2-\overline{CD}^2 \).}{}
+
+\end{proofex}
+
+\ex{In an isosceles triangle, the square of a leg is equal to the
+square of any line drawn from the vertex to the base, increased by the
+product of the segments of the base.}
+
+\ex{The squares of two chords drawn from the same point in a
+circumference have the same ratio as the projections of the chords on the
+diameter drawn from the same point.}
+
+\ex{The difference of the squares of two sides of a triangle is
+equal to the difference of the squares of the segments of the third side,
+made by the perpendicular on the third side from the opposite vertex.}
+
+\begin{proofex}%
+$E$ is the middle point of $BC$, one of the parallel sides of the
+trapezoid $ABCD$; $AE$ and $DE$ produced meet $DC$ and $AB$ produced at
+$F$ and $G$, respectively. Show that $FG$ is parallel to $DA$.
+
+$\triangle_s AGD$ and $BGE$ are similar; and $\triangle_s AFD$ and $EFC$ are similar.
+
+\end{proofex}
+\scanpage{180.png}%
+
+\ex{If two tangents are drawn to a circle at the extremities of a
+diameter, the portion of a third tangent intercepted between them is
+divided at its point of contact into segments whose product is equal to the
+square of the radius.}
+
+\ex{If two exterior angles of a triangle are bisected, the line
+drawn from the point of intersection of the bisectors to the opposite angle
+of the triangle bisects that angle.}
+
+\ex{The sum of the squares of the diagonals of a quadrilateral is
+equal to twice the sum of the squares of the lines that join the middle
+points of the opposite sides.}
+
+\figcc{180aa280}{180bb281}
+\begin{proofex}%
+The sum of the squares of the four sides of any quadrilateral
+is equal to the sum of the squares of the diagonals, increased
+by four times the square of the line joining the
+middle points of the diagonals.
+
+Apply §~377 to the $\triangle_s$ $ABC$ and $ADC$, add the results,
+and eliminate $\overline{BE}^2 + \overline{DE}^2$ by applying §~377 to the $\triangle BDE$.
+
+\end{proofex}
+
+\begin{proofex}%
+The square of the bisector of an exterior angle of a triangle is
+equal to the product of the external segments determined
+by the bisector upon one of the sides, diminished by the
+product of the other two sides.
+
+Let $CD$ bisect the exterior $\angle BCH$ of the $\triangle ABC$.
+$\triangle_s$ $ACD$ and $FCB$ are similar (§~355).    Apply §~382.
+
+\end{proofex}
+
+\ex{If a point $O$ is joined to the vertices of a triangle $ABC$;
+through any point $A'$ in $OA$ a line parallel to $AB$ is drawn, meeting $OB$
+at $B'$; through $B'$ a line parallel to $BC$, meeting $OC$ at $C'$; and $C'$ is
+joined to $A'$; the triangle $A'B'C'$ is similar to the triangle $ABC$.}
+
+\ex{If the line of centres of two circles meets the circumferences
+at the consecutive points $A$, $B$, $C$, $D$, and meets the common external tangent
+at $P$, then $PA × PD = PB × PC$.}
+
+\begin{proofex}%
+The line of centres of two circles meets the common external
+tangent at~$P$, and a secant is drawn from~$P$, cutting the circles at the
+consecutive points $E$, $F$, $G$,~$H$. Prove that $PE × PH = PF × PG$.
+
+Draw radii to the points of contact, and to $E$, $F$, $G$, $H$. Let fall $\perp_s$ on
+$PH$ from the centres of the $\odot_s$. The various pairs of $\triangle_s$ are similar.
+
+\end{proofex}
+
+\ex{If a line drawn from a vertex of a triangle divides the opposite
+side into segments proportional to the adjacent sides, the line bisects
+the angle at the vertex.}
+\scanpage{181.png}%
+
+\clearpage
+\section{PROBLEMS OF CONSTRUCTION.}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To divide a given straight line into parts
+  proportional to any number of given lines.}
+
+\figc{181aa385}{Let $AB$, $m$, $n$, and $p$ be given straight lines.}
+
+\prove[To divide ]{$AB$ into parts proportional to $m$, $n$, and $p$.}
+
+\step{Draw $AX$, making any convenient $\angle$ with $AB$.}{}
+
+\step{On $AX$ take $AC$ equal to $m$, $CE$ to $n$, $EF$ to $p$.}{}
+
+\step{Draw $BF$.}{}
+
+\step{From $E$ and $C$ draw $EK$ and $CH \parallel$ to $FB$.}{}
+
+\step{Through $A$ draw a line $\parallel$ to $BF$.}{}
+
+\step{$K$ and $H$ are the division points required.}{}
+
+\step[\indent\textbf{Proof.}]{$\dfrac{AH}{AC} = \dfrac{HK}{CE} = \dfrac{KB}{EF}$,}{§~344}
+
+\pnote{(if two lines are cut by any number of parallels, the corresponding
+intercepts are proportional).}
+
+Substitute $m$, $n$, and $p$ for their equals $AC$, $CE$, and $EF$.
+
+\step[\indent Then]{$\dfrac{AH}{m} = \dfrac{HK}{n} = \dfrac{KB}{p}$.}{\qef}
+\end{proof}
+
+\ex{Divide a line $12$~inches long into three parts
+proportional to the numbers $3$,~$5$,~$7$.}
+\scanpage{182.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the fourth proportional to three given
+  straight lines.}
+
+\figc{182aa386}{Let the three given lines be $m$, $n$, and $p$.}
+
+\prove[To find ]{the fourth proportional to $m$, $n$, and $p$.}
+
+
+\step {Draw $Ax$ and $Ay$ containing any convenient angle.} {}
+
+\step {On $Ax$ take $AB$ equal to $m$, $BC$ to $n$.} {}
+
+\step {On $Ay$ take $AD$ equal to $p$.} {}
+
+\step {Draw $BD$.} {}
+
+\step {From $C$ draw $CF \parallel$ to $BD$, meeting $Ay$ at $F$.} {}
+
+\step {$DF$ is the fourth proportional required.} {}
+
+\eq [\indent\textbf{Proof.}] {$AB:BC$ } {$ = AD:DF$,} {§~342}
+
+\pnote{(a line drawn through two sides of a $\triangle \parallel$ to the third side divides those sides
+    proportionally).}
+
+\step{Substitute $m$, $n$, and $p$ for their equals $AB$, $BC$, and $AD$.} {}
+
+\eq [\indent Then] {$m:n$} {$ = p:DF$} {\qef}
+
+
+
+\end{proof}
+
+\ex {The square of the altitude of an equilateral
+triangle is equal to three fourths of the square of one side of the
+triangle.}
+\scanpage{183.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the third proportional to two given
+straight lines.}
+
+\figc{183aa387}{Let $m$ and $n$ be the two given straight lines.}
+
+\prove[To find ]{the third proportional to $m$ and $n$.}
+
+\step{Construct any convenient angle $A$,}{}
+
+\step{and take $AB$ equal to $m$, $AC$ equal to $n$.}{}
+
+\step{Produce $AB$ to $D$, making $BD$ equal to $AC$.}{}
+
+\step{Draw $BC$.}{}
+
+\step{Through $D$ draw $DE \parallel$ to $BC$, meeting $AC$ produced at $E$.}{}
+
+\step{$CE$ is the third proportional required.}{}
+
+\eq[\indent\textbf{Proof.}]{$AB:BD$}{$= AC:CE$,}{§~342}
+
+\pnote{(a line drawn through two sides of a $\triangle$ parallel to the third side divides
+those sides proportionally).}
+
+Substitute, in the above proportion, $AC$ for its equal $BD$.
+
+\eq[\indent Then]{$AB:AC$}{$= AC:CE$,}{}
+
+\eq[that is,]{$m:n$}{$=n:CE$.}{\qef}
+
+\end{proof}
+
+\begin{proofex}%
+Construct $x$, if (1) $x=\dfrac{ab}{c}$, (2) $x = \dfrac{a^2}{c}$.
+
+Special cases: (1)~$a = 2$, $b = 8$, $c = 4$; (2)~$a = 3$, $b = 7$, $c = 11$; (3)~$a = 2$,
+$c = 3$; (4)~$a = 3$, $c = 5$; (5)~$a = 2c$.
+
+\end{proofex}
+\scanpage{184.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the mean proportional between two given
+straight lines.}
+
+\figc{184aa388}{Let the two given lines be $m$ and $n$.}
+
+\prove[To find ]{the mean proportional between $m$ and $n$.}
+
+\step{On the straight line $AE$}{}
+
+\step{take $AC$ equal to $m$, and $CB$ equal to $n$.}{}
+
+\step{On $AB$ as a diameter describe a semicircumference.}{}
+
+\step{At $C$ erect the $\perp$ $CH$ meeting the circumference at $H$.}{}
+
+\step{$CH$ is the mean proportional between $m$ and $n$.}{}
+
+\eq[\indent\textbf{Proof.}]{$AC:CH$}{$=CH:CB$}{§~370}
+
+\pnote{(the $\perp$ let fall from a point in a circumference to the diameter of a circle is
+the mean proportional between the segments of the diameter).}
+
+Substitute for $AC$ and $CB$ their equals $m$ and $n$.
+
+\eq[\indent Then]{$m:CH$}{$=CH:n$.}{\qef}
+
+\end{proof}
+
+
+\pp{\defn{A straight line is divided \textbf{in extreme and mean
+ratio}\label{extrememean}, when one of the segments is the mean proportional
+between the whole line and the other segment.}}
+
+\begin{proofex}%
+Construct $x$, if $ x=\sqrt{ab} $.
+
+Special cases: (1)~$a = 2$, $b = 3$; (2)~$a = 1$, $b = 6$; (3)~$a = 3$, $b = 7$.
+
+\end{proofex}
+\scanpage{185.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To divide a given line in extreme and mean ratio.}
+
+\figc{185aa390}{Let $AB$ be the given line.}
+
+\prove[To divide ]{$AB$ in extreme and mean ratio.}
+
+\step{At $B$ erect a $\perp BE$ equal to half of $AB$.}{}
+
+\step{From $E$ as a centre, with a radius equal to $EB$, describe a $\odot$.}{}
+
+\step{Draw $AE$, meeting the circumference in $F$ and $G$.}{}
+
+\step{On $AB$ take $AC$ equal to $AF$.}{}
+
+\step{On $BA$ produced take $AC'$ equal to $AG$.}{}
+
+Then $AB$ is divided internally at $C$ and externally at $C'$ in
+extreme and mean ratio.
+
+\step{$AG:AB = AB:AF$.}{§~381}
+
+\begin{center}
+\begin{tabular}{r@{}l@{}l | r@{}l@{}l}
+$\overline{AB}^2$& $= AF × AG$ &&
+$\overline{AB}^2$& $= AG × AF$ \\
+
+& $= AC(AF+AG)$ &&
+& $= C'A(AG-AF)$ \\
+
+& $= AC(AC+AB)$ &&
+& $= C'A(C'A-AB)$ \\
+
+& $= \overline{AC}^2 + AB × AC$. &&
+& $= \overline{C'A}^2 - AB × C'A$. \\
+
+& $\therefore \overline{AB}^2-AB × AC$ & $=\overline{AC}^2$. &
+& $\therefore \overline{AB}^2+AB × C'A$ & $=\overline{C'A}^2$. \\
+
+& $\therefore AB(AB-AC)$ & $=\overline{AC}^2$. &
+& $\therefore AB(AB+C'A)$ & $=\overline{C'A}^2$. \\
+
+& $\therefore AB × CB$ & $=\overline{AC}^2$. &
+& $\therefore AB × C'B$ & $=\overline{C'A}^2$.
+\end{tabular}
+\end{center}
+
+\hfill\qef
+
+\end{proof}
+\scanpage{186.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{Upon a given line homologous to a given side of
+a given polygon, to construct a polygon similar to the
+given polygon.}
+
+\figc{186aa391}{Let $A'E'$ be the given line homologous to $AE$ of the given polygon
+$ABCDE$.}
+
+\prove[To construct ]{on $A'E'$ a polygon similar to the given polygon.}
+
+\step{From $E$ draw the diagonals $EB$ and $EC$.}{}
+
+\step{From $E'$ draw $E'B'$, $E'C'$, and $E'D'$,}{}
+
+\step{making $ \triangle$'s $A'E'B'$, $B'E'C'$, and $C'E'D'$ equal, respectively, to}{}
+
+\step{$ \triangle_s AEB$, $BEC$, and $CED$.}{}
+
+\step{From $A'$ draw $A'B'$, making $\angle E'A'B'$ equal to $\angle EAB$,}{}
+
+\step{and meeting $E'B'$ at $B'$.}{}
+
+\step{From $B'$ draw $B'C'$, making $\angle E'B'C'$ equal to $\angle EBC$,}{}
+
+\step{and meeting $E'C'$ at $C'$.}{}
+
+\step{From $C'$ draw $C'D'$, making $\angle E'C'D'$ equal to $\angle ECD$,}{}
+
+\step{and meeting $E'D'$ at $D'$.}{}
+
+\step{Then $A'B'C'D'E'$ is the required polygon.}{}
+
+\step[\indent\textbf{Proof.}]{The $\triangle_s ABE$, $A'B'E'$, etc., are similar.}{§~354}
+
+\step{Therefore, the two polygons are similar.}{§~366}
+
+\hfill\qef
+
+\end{proof}
+\scanpage{187.png}%
+
+
+\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}
+
+\ex{To divide one side of a given triangle into segments proportional
+to the adjacent sides (§~348).}
+
+\figcccc{187aa291}{187bb292}{187cc293}{187dd294}
+
+\begin{proofex}%
+To find in one side of a given triangle a point whose distances
+from the other sides shall be to each other in the given ratio $m : n$.
+
+Take $AG = m \perp$ to $AC$, $GH=n \perp$ to $BC$.   Draw $CD \parallel$ to $OG$.
+
+\end{proofex}
+
+\ex{Given an obtuse triangle; to draw a line from the vertex of
+the obtuse angle to the opposite side which shall be the mean proportional
+between the segments of that side.}
+
+\begin{proofex}%
+Through a given point $P$ within a given circle to draw a chord
+$AB$ so that the ratio $AP: BP$ shall equal the given ratio $m : n$.
+
+Draw $OPC$ so that $OP:PC = n:m$.    Draw $CA$ equal to the fourth
+proportional to $n$, $m$, and the radius of the circle.
+
+\end{proofex}
+
+\begin{proofex}%
+To draw through a given point $P$ in the arc subtended by a
+chord $AB$ a chord which shall be bisected by $AB$.
+
+On radius $OP$ take $CD$ equal to $CP$.   Draw $DE \parallel$ to $BA$.
+\end{proofex}
+
+\figcccc{187ee295}{187ff296}{187gg297}{187hh298}
+\begin{proofex}%
+To draw through a given external point $P$ a secant $PAB$ to a
+given circle so that the ratio $PA:AB$ shall equal the given ratio $m : n$.
+\[ PD:DC = m:n. \quad PD:PA = PA:PC. \]
+
+\end{proofex}
+
+\begin{proofex}%
+To draw through a given external point $P$ a secant $PAB$ to a
+given circle so that $\overline{AB}^2 = PA × PB$.
+\[ PC:CD = CD:PD. \quad PA = CD. \]
+
+\end{proofex}
+\scanpage{188.png}%
+
+\ex{To find a point $P$ in the arc subtended by a given chord $AB$
+so that the ratio $PA:PB$ shall equal the given ratio $m : n$.}
+
+\ex{To draw through one of the points of intersection of two
+circles a secant so that the two chords that are formed shall be in the
+given ratio $m:n$.}
+
+\ex{Having given the greater segment of a line divided in extreme
+and mean ratio, to construct the line.}
+
+\ex{To construct a circle which shall pass through two given points
+and touch a given straight line.}
+
+\ex{To construct a circle which shall pass through a given point
+and touch two given straight lines.}
+
+\ex{To inscribe a square in a semicircle.}
+
+\figc{188aa303}{}
+\begin{proofex}%
+To inscribe a square in a given triangle.
+
+Let $DEFG$ be the required inscribed square. Draw $CM \parallel$ to $AB$, meeting
+$AF$ produced in $M$. Draw $CH$ and $MN \perp$ to $AB$, and
+produce $AB$ to meet $MN$ at $N$. The $\triangle_s ACM$, $AGF$ are
+similar; also, the $\triangle_s AMN$, $AFE$ are similar. By these
+triangles show that the figure $CMNH$ is a square. By
+constructing this square, the point $F$ can be found.
+
+\end{proofex}
+
+\ex{To inscribe in a given triangle a rectangle similar to a given
+rectangle.}
+
+\ex{To inscribe in a circle a triangle similar to a given triangle.}
+
+\ex{To inscribe in a given semicircle a rectangle similar to a given
+rectangle.}
+
+\ex{To circumscribe about a circle a triangle similar to a given
+triangle.}
+
+\ex{To construct the expression, $x = \dfrac{2abc}{de}$;
+  that is, $\dfrac{2ab}{d} × \dfrac{c}{e}$.}
+
+\ex{To construct two straight lines, having given their sum and
+their ratio.}
+
+\ex{To construct two straight lines, having given their difference
+and their ratio.}
+
+\ex{Given two circles, with centres $O$ and $O'$, and a point $A$ in
+their plane, to draw through the point $A$ a straight line, meeting the circumferences
+at $B$ and $C$, so that $AB:AC=m:n$.}
+\scanpage{189.png}%
+
+
+\subsection{PROBLEMS OF COMPUTATION.}
+
+\begin{proofex}%
+To compute the altitudes of a triangle in terms of its sides.
+
+\figc{189aa312}{}
+
+At least one of the angles $A$ or $B$ is acute. Suppose $B$ is acute.
+
+\eq[\indent In the $\triangle$ $CDB$,]{$h^2$}{$=a^2- \overline{BD}^2$,}{§~372}
+
+\eq[\indent In the $\triangle$ $ABC$,]{$b^2$}{$=a^2+c^2 - 2c ×\ BD$.}{§~376}
+
+\eq[\indent Whence]{$BD$}{$=\dfrac{a^2+c^2-b^2}{2c}$.}{}
+
+\setlength{\eqalign}{0.25\dentwidth}
+\eq[\indent Hence,]{$h^2$}{\(=a^2-\dfrac{(a^2+c^2-b^2)^2} {4c^2}=
+  \dfrac{4a^2c^2-(a^2+c^2-b^2)^2} {4c^2} \)}{}
+
+\eq{}{\( =\dfrac{(2ac+a^2+c^2-b^2)(2ac-a^2-c^2+b^2)} {4c^2} \)}{}
+
+\eq{}{\( =\dfrac{\{(a+c)^2-b^2\}\{b^2-(a-c)^2\}} {4c^2} \)}{}
+
+\eq{}{\( =\dfrac{(a+b+c)(a+c-b)(b+a-c)(b-a+c)} {4c^2}. \)}{}
+
+\setlength{\eqalign}{0.5\dentwidth}
+
+\eq[\indent Let]{$a+b+c$}{$=2s$.}{}
+
+\eq[\indent Then]{$a+c-b$}{$=2(s-b)$,}{}
+
+\eq{$b+a-c$}{$= 2(s-c)$,}{}
+
+\eq{$b-a+c$}{$=2(s-a)$.}{}
+
+\setlength{\eqalign}{0.25\dentwidth}
+\eq[\indent Hence,]{$h^2$}{\( =\dfrac{2s × 2(s-a) × 2(s-b) × 2(2-c)} {4c^2} \).}{}
+
+By simplifying, and extracting the square root,
+
+\label{formtrialtitude}%
+\eq{$h$}{\( =\dfrac{2}{c} \sqrt{s(s-a)(s-b)(s-c)} \).}{}
+
+\setlength{\eqalign}{0.5\dentwidth}
+\end{proofex}
+
+
+\figc{189bb313}{}
+\begin{proofex}%
+To compute the medians of a triangle in terms of its sides.
+
+\setlength{\eqalign}{0.33\dentwidth}
+\eq[\indent By §~377,]{}{\( a^2+b^2 = 2m^2+2\left(\dfrac{c}{2}\right)^2 \).}{}
+
+\eq[\indent Whence]{$4m^2$}{$=2(a^2+b^2)-c^2$.}{}
+
+\label{formtrimedian}%
+\eq{$\therefore m$}{$=\dfrac{1}{2} \sqrt{2(a^2+b^2)-c^2}$.}{}
+
+\end{proofex}
+\setlength{\eqalign}{0.5\dentwidth}
+\scanpage{190.png}%
+
+\figc{190aa314}{}
+\begin{proofex}%
+To compute the bisectors of a triangle in terms of the sides.
+
+\setlength{\eqalign}{.33\dentwidth}
+\eq[\indent By §~383,]{$t^2$}{$=ab-AD × BD$.}{}
+
+\eq[\indent By §348,]{$\dfrac{AD}{B}$}{$=\dfrac{BD}{a}=\dfrac{AD+BD}{a+b}=\dfrac{c}{a+b}$.}{}
+
+\eq{$\therefore AD$}{$=\dfrac{bc}{a+b}$, and $BD=\dfrac{ac}{a+b}$.}{}
+
+\eq[\indent Whence]{$t^2$}{$= ab - \dfrac{abc^2}{(a+b)^2}$}{}
+
+\eq{}{$=ab\left[1-\dfrac{c^2}{(a+b)^2}\right]$}{}
+
+\eq{}{$=\dfrac{ab\{(a+b)^2-c^2\}}{(a+b)^2}$}{}
+
+\eq{}{$=\dfrac{ab(a+b+c)(a+b-c)}{(a+b)^2}$}{}
+
+\eq{}{$=\dfrac{ab × 2s × 2(s-c)}{(a+b^2)}$.}{}
+
+\label{formtribisector}%
+\eq[\indent Whence]{$t$}{$= \dfrac{2}{a+b} \sqrt{abs(s-c)}$.}{}
+
+\setlength{\eqalign}{.5\dentwidth}
+
+
+\end{proofex}
+
+\figc{190bb315}{}
+\begin{proofex}%
+To compute the radius of the circle circumscribed about a triangle in terms of the sides of the triangle.
+
+\eq[\indent By §384,]{$AC × AB$}{$= AE × AD$,}{}
+
+\eq[or,]{$bc$}{$= 2 B × AD$.}{}
+
+\setlength{\eqalign}{0.33\dentwidth}
+\eq[\indent But]{$AD$}{$=\dfrac{2}{a}\sqrt{s(s-a)(s-b)(s-c)}$.}{Ex.~312}
+
+\label{formradcircum}%
+\eq{$\therefore R$}{$=\dfrac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}}$.}{}
+
+\setlength{\eqalign}{0.5\dentwidth}
+
+\end{proofex}
+
+\ex{If the sides of a triangle are $3$,~$4$, and~$5$, is the angle
+opposite~$5$ right, acute, or obtuse?}
+
+\ex{If the sides of a triangle are $7$,~$9$, and~$12$, is the angle
+opposite~$12$ right, acute, or obtuse?}
+
+\ex{If the sides of a triangle are $7$,~$9$, and~$11$, is the angle
+opposite~$11$ right, acute, or obtuse?}
+
+\ex{The legs of a right triangle are $8$~inches and $12$~inches; find
+the lengths of the projections of these legs upon the hypotenuse, and the
+distance of the vertex of the right angle from the hypotenuse.}
+
+\ex{If the sides of a triangle are $6$~inches, $9$~inches, and
+$12$~inches, find the lengths (1)~of the altitudes; (2)~of the medians;
+(3)~of the bisectors; (4)~of the radius of the circumscribed circle.
+}
+\scanpage{191.png}%
+
+\ex{A line is drawn parallel to a side $AB$ of a triangle $ABC$,
+cutting $AC$ in $D$, $BC$ in $E$. If $AD:DC = 2:3$, and $AB = 20$ inches,
+find $DE$.}
+
+\ex{The sides of a triangle are $9$,~$12$,~$15$. Find the segments of
+the sides made by bisecting the angles.}
+
+\ex{A tree casts a shadow $90$~feet long, when a post $6$~feet high
+casts a shadow~$4$ feet long. How high is the tree?}
+
+\ex{The lower and upper bases of a trapezoid are $a$, $b$, respectively;
+and the altitude is~$h$.  Find the altitudes of the two triangles
+formed by producing the legs until they meet.}
+
+\ex{The sides of a triangle are $6$,~$7$,~$8$, respectively. In a similar
+triangle the side homologous to~$8$ is~$40$.  Find the other two sides.}
+
+\ex{The perimeters of two similar polygons are $200$~feet and $300$~feet.
+If a side of the first is $24$~feet, find the homologous side of the
+second.}
+
+\ex{How long a ladder is required to reach a window $24$~feet high,
+if the lower end of the ladder is $10$~feet from the side of the house?}
+
+\ex{If the side of an equilateral triangle is~$a$, find the altitude.}
+
+\ex{If the altitude of an equilateral triangle is~$h$, find the side.}
+
+\ex{Find the length of the longest chord and of the shortest chord
+that can be drawn through a point $6$~inches from the centre of a circle
+whose radius is $10$~inches.}
+
+\ex{The distance from the centre of a circle to a chord $10$~feet long
+is $12$~feet. Find the distance from the centre to a chord $24$~feet long.}
+
+\ex{The radius of a circle is $5$~inches. Through a point $3$~inches
+from the centre a diameter is drawn, and also a chord perpendicular to
+the diameter. Find the length of this chord, and the distance from one
+end of the chord to the ends of the diameter.}
+
+\ex{The radius of a circle is $6$~inches. Find the lengths of the
+tangents drawn from a point $10$~inches from the centre, and also the
+length of the chord joining the points of contact.}
+
+\ex{The sides of a triangle are $407$~feet, $368$~feet, and $351$~feet.
+Find the three bisectors and the three altitudes.
+}
+\scanpage{192.png}%
+
+\ex{If a chord $8$~inches long is $8$~inches distant from the centre of
+the circle, find the radius, and the chords drawn from the end of the chord
+to the ends of the diameter which bisects the chord.}
+
+\ex{From the end of a tangent $20$~inches long a secant is drawn
+through the centre of the circle. If the external segment of this secant is
+$8$~inches, find the radius of the circle.}
+
+\ex{The radius of a circle is $13$~inches. Through a point $5$~inches
+from the centre any chord is drawn. What is the product of the two
+segments of the chord? What is the length of the shortest chord that
+can be drawn through the point?}
+
+\ex{The radius of a circle is $9$~inches and the length of a tangent
+$12$~inches. Find the length of a line drawn from the extremity of the
+tangent to the centre of the circle.}
+
+\ex{Two circles have radii of $8$~inches and $3$~inches, respectively,
+and the distance between their centres is $15$~inches. Find the lengths of
+their common tangents.}
+
+\ex{Find the segments of a line $10$~inches long divided in extreme
+and mean ratio.}
+
+\ex{The sides of a triangle are $4$,~$5$,~$5$. Is the largest angle acute,
+right, or obtuse?}
+
+\ex{Find the third proportional to two lines whose lengths are
+$28$~feet and $42$~feet.}
+
+\ex{If the sides of a triangle are $a$,~$b$,~$c$, respectively, find the
+lengths of the three altitudes.}
+
+\ex{The diameter of a circle is $30$~feet and is divided into five
+equal parts. Find the lengths of the chords drawn through the points of
+division perpendicular to the diameter.}
+
+\ex{The radius of a circle is $2$~inches. From a point $4$~inches
+from the centre a secant is drawn so that the internal segment is $1$~inch.
+Find the length of the secant.}
+
+\ex{The sides of a triangular pasture are $1551$ yards, $2068$ yards,
+$2585$ yards. Find the median to the longest side.}
+
+\ex{The diagonal of a rectangle is $d$, and the perimeter is $p$.
+Find the sides.}
+
+\ex{The radius of a circle is $r$. Find the length of a chord whose
+distance from the centre is $\frac{1}{2} r$.}
+
+\scanpage{193.png}%
+
+
+\chapter{BOOK IV\@. AREAS OF POLYGONS.}
+\markboth{\Headings{BOOK IV\@. PLANE GEOMETRY.}}
+{\Headings{AREAS OF POLYGONS.}}%
+
+\pp{\defn{The \textbf{unit of surface} is a square whose side is a
+\emph{unit of length}.}}
+
+\pp{\defn{The \textbf{area of a surface}\label{area} is the \emph{number of units of
+surface} it contains.}}
+
+\pp{\defn{Plane figures that \emph{have equal areas but cannot
+be made to coincide} are called \textbf{equivalent}\label{equivalent2}.}}
+
+\note{In propositions relating to \emph{areas}, the words ``rectangle,'' ``triangle,''
+etc., are often used for ``area of rectangle,'' ``area of triangle,'' etc.}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two rectangles having equal altitudes are to
+each other as their bases.}
+
+\figc{193aa395}{Let the rectangles $AC$ and $AF$ have the same altitude $AD$.}
+
+\prove{$\rect AC: \rect AF = \base AB: \base AE$.}
+
+\textsc{Case 1.~} \emph{When $AB$ and $AE$ are commensurable.}
+
+\textbf{Proof.} Suppose $AB$ and $AE$ have a common measure, as
+$AO$, which is contained $m$ times in $AB$ and $n$ times in $AE$.
+
+\eq[\indent Then]{$AB:AE$}{$ = m:n$.}{}
+\scanpage{194.png}%
+
+\filbreak
+Apply $AO$ as a unit of measure to $AB$ and $AE$, and at the
+several points of division erect $\perp_s$.
+
+\step[\indent The]{$\rect AC$ is divided into $m$~rectangles,}{}
+
+\step[and the]{$\rect AF$ is divided into $n$~rectangles.}{§~107}
+
+\step{These rectangles are all equal.}{§~186}
+
+\eq[\indent Hence,]{}{$\rect AC: \rect AF = m:n$.}{}
+
+\eq[\indent Therefore,]{}{$\rect AC: \rect AF = AB:AE$.}{Ax.~1}
+
+\textsc{Case 2.} \emph{When $AB$ and $AE$ are incommensurable.}
+
+\figc{194aa395}{}
+
+\textbf{Proof.} Divide $AB$ into any number of equal parts, and apply
+one of them to $AE$ as many times as $AE$ will contain it.
+
+Since $AB$ and $AE$ are incommensurable, a certain number
+of these parts will extend from $A$ to some point $K$, leaving a
+remainder $KE$ less than one of the equal parts of $AB$.
+
+\step{Draw $KH \parallel$ to $EF$.}{}
+
+Then $AB$ and $AK$ are commensurable by construction.
+
+\step[\indent Therefore,]{\( \dfrac{\rect AH}{\rect AC} = \dfrac{AK}{AB}. \)}{Case~1}
+
+If the number of equal parts into which $AB$ is divided is
+indefinitely increased, the varying values of these ratios will
+continue equal, and approach for their respective limits the
+ratios
+
+\step{\( \dfrac{\rect AF}{\rect AC} \) and \( \dfrac{AE}{AB} \). (See §~287.)}{}
+
+\step{\( \therefore \dfrac{\rect AF}{\rect AC} = \dfrac{AE}{AB}. \)}{§~284}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{Two rectangles having equal bases are to each
+other as their altitudes.}}
+\scanpage{195.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two rectangles are to each other as the products
+of their bases by their altitudes.}
+
+\vspace{1ex}
+\figc{195aa397}{Let $R$ and $R'$ be two rectangles, having for their bases $b$ and $b'$,
+and for their altitudes $a$ and $a'$, respectively.}
+
+\proveq{$\dfrac{R}{R'}$}{$= \dfrac{a × b}{a' × b'}$.}
+
+\textbf{Proof.} Construct the rectangle $S$, with its base equal to that
+of $R$, and its altitude equal to that of $R'$.
+
+\eq[\indent Then]{$\dfrac{R}{S}$}{$=\dfrac{a}{a'}$,}{§~396}
+
+\eq[and]{$\dfrac{S}{R'}$}{$=\dfrac{b}{b'}$.}{§~395}
+
+The products of the corresponding members of these equations
+give
+
+\eq{$\dfrac{R}{R'}$}{$=\dfrac{a × b}{a' × b'}$.}{\qed}
+
+\end{proof}
+
+\ex{Find the ratio of a rectangular lawn $72$~yards by $49$~yards to
+a grass turf $18$~inches by $14$~inches.}
+
+\ex{Find the ratio of a rectangular courtyard $18\frac{1}{2}$~yards by $15\frac{1}{2}$~yards
+to a flagstone $31$~inches by $18$~inches.}
+
+\ex{A square and a rectangle have the same perimeter, $100$~yards.
+The length of the rectangle is $4$~times its breadth. Compare their areas.}
+
+\ex{On a certain map the linear scale is $1$~inch to $5$~miles. How
+many acres are represented on this map by a square the perimeter of
+which is $1$~inch?}
+\scanpage{196.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a rectangle is equal to the product
+of its base by its altitude.}
+
+\figc{196aa398}{Let $R$ be a rectangle, $b$ its base, and $a$ its altitude.}
+
+\proveq{the area of $R$}{$= a × b$.}
+
+\textbf{Proof.} Let $U$ be the unit of surface.
+
+\eq{\( \dfrac{R}{U} = \dfrac{a × b}{1 × 1} \)}{$= a × b$,}{}
+
+\pnote{(two rectangles are to each other as the products of their bases and altitudes).}
+
+\step[\indent But]{$\dfrac{R}{U} =$ the \emph{number} of units of surface in $R$.}{§~393}
+
+\label{formarearect}%
+\eq{$\therefore$ the area of $R$}{$= a × b$.}{\qed}
+
+\end{proof}
+
+\figc{196bb399}{}
+\begin{point}\textsc{Scholium.} When the base and altitude each contain
+the linear unit an integral number of times, this proposition
+is rendered evident by dividing the figure into squares, each
+equal to the unit of surface. Thus, if the base contains seven
+linear units, and the altitude four, the figure may be divided
+into twenty-eight squares, each equal to the unit of surface.
+\end{point}
+\scanpage{197.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a parallelogram is equal to the
+product of its base by its altitude.}
+
+\figc{197aa400}{Let $AEFD$ be a parallelogram, $b$ its base, and $a$ its altitude.}
+
+\prove{the area of the \textnormal{$\Par AEFD = a × b$.}}
+
+\textbf{Proof.} From $A$ draw $AB$ $\parallel$ to $DC$ to meet $FE$ produced.
+
+Then the figure $ABCD$ is a rectangle, with the same base
+and the same altitude as the $\Par AEFD$.
+
+\step{The rt.~$\triangle_s ABE$ and $DCF$ are equal.}{§~151}
+
+\step{For $AB = CD$, and $AE = DF$.}{§~178}
+
+From $ABFD$ take the $\triangle DCF$; the $\rect ABCD$ is left.
+
+From $ABFD$ take the $\triangle ABE$; the $\Par AEFD$ is left.
+
+\step{$\therefore \rect ABCD \Bumpeq \Par AEFD$}{Ax.~3}
+
+\step{But the area of the $\rect ABCD = a × b$.}{§~398}
+
+\label{formareapar}%
+\step{$\therefore$ the area of the $\Par AEFD = a × b$.}{Ax.~1}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{Parallelograms having equal bases and equal
+altitudes are equivalent.}}
+
+\pp{\cor[2]{Parallelograms having equal bases are to each
+other as their altitudes; parallelograms having equal altitudes
+are to each other as their bases; any two parallelograms
+are to each other as the products of their bases by
+their altitudes.}}
+\scanpage{198.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a triangle is equal to half the product of its base by its altitude.}
+
+\figc{198aa403}{Let $a$ be the altitude and $b$ the base of the triangle $ABC$.}
+
+\prove{the area of the $\triangle{}ABC=\frac{1}{2}a × b$.}
+
+\textbf{Proof.} Construct on $AB$ and $BC$ the parallelogram $ABCH$.
+
+\step[\indent Then]{$\triangle ABC=\frac{1}{2}\Par ABCH$.}{§~179}
+
+\step{The area of the $\Par ABCH=a × b$.}{§~400}
+
+\label{formareatri}%
+\step{Therefore, the area of $\triangle ABC=\frac{1}{2}a × b$.}{Ax.~7}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{Triangles having equal bases and equal altitudes are equivalent.}}
+
+\pp{\cor[2]{Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the products of their bases by their altitudes.}}
+
+\pp{\cor[3]{The product of the legs of a right triangle is equal to the product of the hypotenuse by the altitude from the vertex of the right angle.}}
+
+
+\ex{The lines which join the middle point of either diagonal of a quadrilateral to the opposite vertices divide the quadrilateral into two equivalent parts.}
+\scanpage{199.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a trapezoid is equal to half the sum of its bases multiplied by the altitude.}
+
+\figc{199aa407}{Let $b$ and $b'$ be the bases and $a$ the altitude of the trapezoid $ABCH$.}
+
+\prove{the area of the $ABCH=\frac{1}{2}a(b+b')$.}
+
+\step[\indent\textbf{Proof.}]{Draw the diagonal $AC$.}{}
+
+\eq[\indent Then]{the area of the $\triangle ABC$}{$=\frac{1}{2}a × b$,}{}
+
+\eq[and]{the area of the $\triangle AHC$}{$=\frac{1}{2}a × b'$.}{§~403}
+
+\label{formareatrap}%
+\eq{$\therefore$ the area of $ABCH$}{$=\frac{1}{2}a(b+b')$.}{Ax.~2}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{The area of a trapezoid is equal to the product of the median by the altitude.}~\hfill§~190}
+
+\figc{199bb409}{}
+\begin{point}%
+\textsc{Scholium.} The area of an irregular polygon may
+be found by dividing the polygon into triangles, and by finding the
+area of each of these triangles separately.  Or, we may draw the
+longest diagonal, and let fall perpendiculars upon this diagonal from
+the other vertices of the polygon.
+
+The sum of the areas of the right triangles, rectangles, and
+trapezoids thus formed is the area of the polygon.
+\end{point}
+\scanpage{200.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The areas of two triangles which have an angle of
+  the one equal to an angle of the other are to each other as the
+  products of the sides including the equal angles.}
+
+\figc{200aa410}{Let the triangles $ABC$ and $ADE$ have the common angle $A$.}
+
+\prove {$\dfrac{\triangle ABC} {\triangle ADE} = \dfrac{AB × AC} {AD × AE}$.}
+
+\step [\indent Proof.] {Draw $BE$.}{}
+
+\step [\indent Now] {$\dfrac{\triangle ABC}{\triangle ABE}= \dfrac{AC}{AE}$,} {}
+
+\step [and] {$\dfrac{\triangle ABE}{\triangle ADE}= \dfrac{AB}{AD}$.} {§~405}
+
+
+The products of the first members and of the second members of these
+equalities give
+
+\step {$\dfrac{\triangle ABC}{\triangle ADE}=\dfrac{AB × AC}{AD × AE}$.} {\qed}
+\end {proof}
+
+\ex{The areas of two triangles which have an angle of
+the one supplementary to an angle of the other are to each other as
+the products of the sides including the supplementary angles.}
+\scanpage{201.png}%
+
+
+\clearpage
+\section{COMPARISON OF POLYGONS.}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The areas of two similar triangles are to each
+as the squares of any two homologous sides.}
+
+\figc{201ab411}{Let the two similar triangles be $ACB$ and $A'C'B'$.}
+
+\proveq{$\dfrac{\triangle ACB}{\triangle A'C'B'}$}
+  {$= \dfrac{\overline{AB}^2}{\overline{A'B'}^2}$.}
+
+\textbf{Proof.} Draw the altitudes $CO$ and $C'O'$.
+
+\step[\indent Then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} =
+  \dfrac{AB × CO}{A'B' × C'O'} =
+  \dfrac{AB}{A'B'} × \dfrac{CO}{C'O'} \),}{§~405}
+
+\pnote{(two $\triangle_s$ are to each other as the products of their bases by their altitudes).}
+
+\eq[\indent But]{$\dfrac{AB}{A'B'}$}{$= \dfrac{CO}{C'O'}$.}{§~361}
+
+\pnote{(the homologous altitudes of two similar $\triangle_s$ have the same ratio as any two
+homologous sides).}
+
+Substitute, in the above equality, for $\dfrac{CO}{C'O'}$ its equal $\dfrac{AB}{A'B'}$;
+
+\step[then]{\( \dfrac{\triangle ACB}{\triangle A'C'B'} =
+  \dfrac{AB }{A'B'} × \dfrac{AB }{A'B'} =
+  \dfrac{\overline{AB}^2 }{\overline{A'B'}^2} \).}{\qed}
+
+\end{proof}
+
+\ex{Prove this proposition by §~410.}
+\scanpage{202.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The areas of two similar polygons are to each
+  other as the squares of any two homologous sides.}
+
+\figc{202ab412}{Let $S$ and $S'$ denote the areas of the two similar polygons
+  $ABC$ etc.\ and $A'B'C'$ etc.}
+
+\proveq{$S:S'$}{$=\overline{AB}^2:\overline{A'B'^2}$.}
+
+\textbf{Proof.} By drawing all the diagonals from any homologous
+vertices $E$ and $E'$, the two similar polygons are divided into
+similar triangles.~\hfill§~365
+
+\step{\( \displaystyle \therefore \frac{\overline{AB}^2}{\overline{A'B'^2}}=
+ \frac{\triangle ABE}{\triangle A'B'E'}=
+ \left(\frac{\overline{BE}^2}{\overline{B'E'^2}}\right)=
+ \frac{\triangle BCE}{\triangle B'C'E'}=\text{etc.} \)}{§~411}
+
+\step[\indent That is,]{\( \displaystyle \frac{\triangle ABE}{\triangle A'B'E'}=
+ \frac{\triangle BCE}{\triangle B'C'E'}=
+ \frac{\triangle CDE}{\triangle C'D'E'} \).}{}
+
+\( \displaystyle \therefore
+  \frac{\triangle ABE + \triangle BCE + \triangle CDE}
+     {\triangle A'B'E' + \triangle B'C'E' + \triangle C'D'E'}=
+  \frac{\triangle ABE}{\triangle A'B'E'}=
+  \frac{\overline{AB}^2}{\overline{A'B'^2}} \).\hsp§~335
+
+\step{\( \displaystyle \therefore
+  S:S'=\overline{AB}^2:\overline{A'B'^2} \)}{\qed}
+
+\end{proof}
+
+\pp{\cor[1]{The areas of two similar polygons
+  are to each other as the squares of any two homologous lines.}}
+
+\pp{\cor[2]{The homologous sides of two
+  similar polygons have the same ratio as the square roots of their areas.}}
+\scanpage{203.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The square on the hypotenuse of a right triangle
+  is equivalent to the sum of the squares on the two legs.}
+
+\figc{203aa415}{Let $BE$, $CH$, $AF$, be squares on the three sides of the
+  right triangle $ABC$.}
+
+\prove{$BE \Bumpeq CH + AF$.}
+
+\textbf{Proof.} Through $A$ draw $AL \parallel$ to $CE$, and draw $AD$
+and $CF$.
+
+Since $\angle_s{} BAC$, $BAG$, and $CAH$ are rt.\ $\angle_s$, $CAG$ and
+$BAH$ are straight lines.~\hfill§~90
+
+\eq[\indent The]{$\triangle ABD$}{$=\triangle FBC$.}{§~143}
+
+\eq[\indent For]{$BD$}{$=BC$,}{}
+
+\eq{$BA$}{$=BF$,}{§~168}
+
+\eq[and]{$\angle ABD$}{$=\angle FBC$,}{Ax.~2}
+
+\pnote{(each being the sum of a rt.\ $\angle$ and the $\angle ABC$).}
+
+\step{Now the rectangle $BL$ is double the $\triangle ABD$,}{}
+
+\pnote{(having the same base $BD$, and the same altitude, the distance
+  between the $\parallel_s AL$ and $BD$),}
+
+\step{and the square $AF$ is double the $\triangle FBC$,}{}
+
+\pnote{(having the same base $FB$, and the same altitude $AB$).}
+
+$\therefore$ the rectangle $BL$ is equivalent to the square
+  $AF$.\hfill~Ax.~6
+
+In like manner, by drawing $AE$ and $BK$, it may be proved that the
+rectangle $CL$ is equivalent to the square $CH$.
+
+Hence, the square $BE$, the sum of the rectangles $BL$ and $CL$, is
+equivalent to the sum of the squares $CH$ and $AF$.~\hfill\qed
+
+\end{proof}
+
+\pp{\cor{The square on either leg of a right
+  triangle is equivalent to the difference of the square on the
+  hypotenuse and the square on the other leg.}}
+\scanpage{204.png}%
+
+
+\section[EXERCISES.]{THEOREMS.}
+
+\figccc{204aa356}{204bb357}{204cc358}
+\begin{proofex}%
+The square constructed upon the sum of two straight
+lines is equivalent to the sum of the squares constructed upon these
+two lines, increased by twice the rectangle of these lines:
+
+Let $AB$ and $BC$ be the two straight lines, and $AC$ their sum.
+Construct the squares $ACGK$ and $ABED$ upon $AC$ and $AB$,
+respectively.  Prolong $BE$ and $DE$ until they meet $KG$ and $CG$,
+respectively.  Then we have the square $EFGH$, with sides each equal
+to $BC$.  Hence, the square $ACGK$ is the sum of the squares $ABED$
+and $EFGH$, and the rectangles $DEHK$ and $BCFE$, the dimensions of
+which are equal to $AB$ and $BC$.
+
+\end{proofex}
+
+\begin{proofex}%
+The square constructed upon the difference of two
+straight lines is equivalent to the sum of the squares constructed
+upon these two lines, diminished by twice the rectangle of these
+lines.
+
+Let $AB$ and $AC$ be the two straight lines, and $BC$ their
+difference.  Construct the square $ABFG$ upon $AB$, the square $ACKH$
+upon $AC$, and the square $BEDC$ upon $BC$ (as shown in the figure).
+Prolong $ED$ to meet $AG$ in $L$.
+
+The dimensions of the rectangles $LEFG$ and $HKDL$ are $AB$ and $AC$,
+and the square $BCDE$ is evidently the difference between the whole
+figure and the sum of these rectangles; that is, the square
+constructed upon $BC$ is equivalent to the sum of the squares
+constructed upon $AB$ and $AC$, diminished by twice the rectangle of
+$AB$ and $AC$.
+
+\end{proofex}
+
+\begin{proofex}%
+The difference between the squares constructed upon
+two straight lines is equivalent to the rectangle of the sum and
+difference of these lines.
+
+Let $ABDE$ and $BCFG$ be the squares constructed upon the two straight
+lines $AB$ and $BC$.  The difference between these squares is the
+polygon $ACGFDE$, which is composed of the rectangles $ACHE$ and
+$GFDH$.  Prolong $AE$ and $CH$ to $I$ and $K$, respectively, making
+$EI$ and $HK$ each equal to $BC$, and draw $IK$.  The rectangles
+$GFDH$ and $EHKI$ are equal.  The difference between the squares
+$ABDE$ and $BCGF$ is then equivalent to the rectangle $ACKI$, which
+has for dimensions $AI$, equal to $AB + BC$, and $EH$, equal to $AB - BC$.
+
+\end{proofex}
+\scanpage{205.png}%
+
+\ex{The area of a rhombus is equal to half the product of its
+diagonals.}
+
+\ex{Two isosceles triangles are equivalent if their legs are equal
+each to each, and the altitude of one is equal to half the base of the other.}
+
+\ex{The area of a circumscribed polygon is equal to half the
+product of its perimeter by the radius of the inscribed circle.}
+
+\ex{Two parallelograms are equal if two adjacent sides of the one
+are equal, respectively, to two adjacent sides of the other, and the included
+angles are supplementary.}
+
+\ex{If $ABC$ is a right triangle, $C$ the vertex of the right angle,
+$BD$ a line cutting $AC$ in $D$, then \( \overline{BD}^2 + \overline{AC}^2 =
+\overline{AB}^2 + \overline{DC}^2 \).}
+
+\ex{Upon the sides of a right triangle as homologous sides three
+similar polygons are constructed. Prove that the polygon upon the
+hypotenuse is equivalent to the sum of the polygons upon the legs.}
+
+\ex{If the middle points of two adjacent sides of a parallelogram
+are joined, a triangle is formed which is equivalent to one eighth of the
+parallelogram.}
+
+\ex{If any point within a parallelogram is joined to the four vertices,
+the sum of either pair of triangles having parallel bases is equivalent
+to half the parallelogram.}
+
+\ex{Every straight line drawn through the intersection of the
+diagonals of a parallelogram divides the parallelogram into two equal
+parts.}
+
+\ex{The line which joins the middle points of the bases of a trapezoid
+divides the trapezoid into two equivalent parts.}
+
+\ex{Every straight line drawn through the middle point of the
+median of a trapezoid cutting both bases divides the trapezoid into two
+equivalent parts.}
+
+\ex{If two straight lines are drawn from the middle point of either
+leg of a trapezoid to the opposite vertices, the triangle thus formed is
+equivalent to half the trapezoid.}
+
+\ex{The area of a trapezoid is equal to the product of one of the
+legs by the distance from this leg to the middle point of the other leg.}
+
+\ex{The figure whose vertices are the middle points of the sides
+of any quadrilateral is equivalent to half the quadrilateral.}
+\scanpage{206.png}%
+
+
+\clearpage
+\section{PROBLEMS OF CONSTRUCTION.}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a square equivalent to the sum of two given squares.}
+
+\figc{206aa417}{Let $R$ and $R'$ be two given squares.}
+
+\prove[To construct ]{a square equivalent to $R'+R$.}
+
+\step {Construct the rt.\ $\angle A$.} {}
+
+\step {Take $AC$ equal to a side of $R'$,} {}
+
+\step {and $AB$ equal to a side of $R$; and draw $BC$.} {}
+
+\step {Construct the square $S$, having each of its sides equal to $BC$.} {}
+
+\step [\indent Then] {$S$ is the square required.} {}
+
+\step [\indent Proof.] {$\overline{BC}^2 \Bumpeq \overline{AC}^2 + \overline{AB}^2$,} {§~415}
+
+\pnote {(the square on the hypotenuse of a rt.\ $\triangle$ is equivalent to the sum of the squares on the two legs).}
+
+\step {$\therefore S \Bumpeq R'+R$.} {}
+
+\hfill\qef
+
+\end{proof}
+
+\ex{If the perimeter of a rectangle is $72$~feet, and the
+length is equal to twice the width, find the area.}
+
+\ex{How many tiles $9$~inches long and $4$~inches wide will
+be required to pave a path $8$~feet wide surrounding a rectangular court
+$120$~feet long and $36$~feet wide?}
+
+\ex{The bases of a trapezoid are $16$~feet and $10$~feet;
+each leg is equal to $5$~feet.  Find the area of the trapezoid.}
+\scanpage{207.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a square equivalent to the difference
+  of two given squares.}
+
+\figc{207aa418}{Let $R$ be the smaller square and $R'$ the larger.}
+
+\prove[To construct ]{a square equivalent to $R'-R$.}
+
+\step{Construct the rt.\ $\angle A$.}{}
+
+\step{Take $AB$ equal to a side of $R$.}{}
+
+\step{From $B$ as a centre, with a radius equal to a side of $R'$,}{}
+
+\step{describe an arc cutting the line $AX$ at $C$.}{}
+
+\step{Construct the square $S$, having each of its sides equal to $AC$.}{}
+
+\step[\indent Then]{$S$ is the square required.}{}
+
+\step[\indent\textbf{Proof.}]{\( \overline{AC}^2 \Bumpeq
+  \overline{BC}^2-\overline{AB}^2 \),}{§~416}
+
+\pnote{(the square on either leg of a rt.\ $\triangle$ is equivalent to
+  the difference of the square on the hypotenuse and the square on the
+  other leg).}
+
+\step{$\therefore S \Bumpeq R'-R$.}{}
+
+\hfill\qef
+
+\end{proof}
+
+
+\ex{Construct a square equivalent to the sum of two squares
+whose sides are $3$~inches and $4$~inches.}
+
+\ex{Construct a square equivalent to the difference of two squares
+whose sides are $2\frac{1}{2}$~inches and $2$~inches.}
+
+\ex{Find the side of a square equivalent to the sum of
+two squares whose sides are $24$~feet and $32$~feet.}
+
+\ex{Find the side of a square equivalent to the
+difference of two squares whose sides are $24$~feet and $40$~feet.}
+
+\ex{A rhombus contains $100$~square feet, and the length
+of one diagonal is $10$~feet.  Find the length of the other diagonal.
+}
+\scanpage{208.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a polygon similar to two given
+  similar polygons and equivalent to their sum.}
+
+\figc{208aa419}{Let $R$ and $R'$ be two similar polygons, and $AB$ and $A'B'$
+  two homologous sides.}
+
+\prove[To construct ]{a similar polygon equivalent to $R+R'$.}
+
+\step{Construct the rt.\ $\angle P$.}{}
+
+\step{Take $PH$ equal to $A'B'$, and $PO$ equal to $AB$.}{}
+
+\step{Draw $OH$, and take $A''B''$ equal to $OH$.}{}
+
+\step{Upon $A''B''$, homologous to $AB$, construct $R''$ similar to $R$.}{}
+
+\step{Then $R''$ is the polygon required.}{}
+
+\eq[\indent\textbf{Proof.}]{$\overline{PO}^2 + \overline{PH}^2$}{$= \overline{OH}^2$.}{§~415}
+
+Put for $PO$, $PH$, and $OH$ their equals $AB$, $A'B'$, and $A''B''$.
+
+\eq[\indent Then]{$\overline{AB}^2 + \overline{A'B'^2}$}{$= \overline{A''B''^2}$.}{}
+
+\step[\indent Now]{$\dfrac{R}{R''} = \dfrac{\overline{AB}^2}{\overline{A''B''^2}}$,
+  and $\dfrac{R'}{R''} = \dfrac{\overline{A'B'^2}}{\overline{A''B''^2}}$.}
+  {§~412}
+
+\step[\indent By addition,]{$\dfrac{R+R'}{R''} =
+   \dfrac{\overline{AB}^2 + \overline{A'B'^2}}{\overline{A''B''^2}} = 1$.}
+   {Ax.~2}
+
+\step{$\therefore R'' \Bumpeq R+R'$.}{\qef}
+
+\end{proof}
+\scanpage{209.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a triangle equivalent to a given polygon.}
+
+\figc{209aa420}{Let $ABCDHE$ be the given polygon.}
+
+\prove[To construct ]{a triangle equivalent to the given polygon.}
+
+Let $D$, $H$, and $E$ be any three consecutive vertices of the
+polygon.  Draw the diagonal $DE$.
+
+\step{From $H$ draw $HF \parallel$ to $DE$.}{}
+
+\step{Produce $AE$ to meet $HF$ at $F$, and draw $DF$.}{}
+
+Again, draw $CF$, and draw $DK \parallel$ to $CF$ to meet $AF$
+produced at $K$, and draw $CK$.
+
+In like manner continue to reduce the number of sides of the polygon
+until we obtain the $\triangle CIK$.
+
+\step{Then $\triangle CIK$ is the triangle required.}{}
+
+\textbf{Proof.}  The polygon $ABCDF$ has one side less than the
+polygon \newline$ACBDHE$, but the two polygons are equivalent.
+
+\step{For the part $ACBDE$ is common,}{}
+
+\step{and the $\triangle DEF \Bumpeq \triangle DEH$,}{§~404}
+
+\pnote{(for the base $DE$ is common, and their vertices $F$ and $H$ are
+  in the line $FH \parallel$ to the base).}
+
+In like manner it may be proved that
+
+\step{$ABCK \Bumpeq ABCDF$, and $CIK \Bumpeq ABCK$.}{\qef}
+
+\end{proof}
+\scanpage{210.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a square equivalent to a given parallelogram.}
+
+\figc{210aa421}{Let $ABCD$ be the parallelogram, $b$ its base, and $a$ its altitude.}
+
+\prove[To construct ]{a square equivalent to the $\Par ABCD$.}
+
+\step{Upon a line $MX$ take $MN$ equal to $a$, $NO$ equal to $b$.}{}
+
+\step{Upon $MO$ as a diameter, describe a semicircle.}{}
+
+\step{At $N$ erect $NP \perp$ to $MO$, meeting the circumference at $P$.}{}
+
+Then the square $R$, constructed upon a line equal to $NP$, is
+equivalent to the $\Par ACBD$.
+
+\eq[\indent\textbf{Proof.}]{$MN:NP$}{$= NP:NO$,}{§~370}
+
+\pnote{(a $\perp$ let fall from any point of a
+  circumference to the diameter is the mean proportional between the
+  segments of the diameter).}
+
+\step{$\therefore \overline{NP}^2 = MN × NO = a × b$.}{§~327}
+
+\step[\indent Therefore,]{$R \Bumpeq \Par ABCD$.}{\qef}
+
+\end{proof}
+
+\pp{\cor[1]{A square may be constructed
+  equivalent to a given triangle, by taking for its side the mean
+  proportional between the base and half the altitude of the triangle.}}
+
+\pp{\cor[2]{A square may be constructed
+  equivalent to a given polygon, by first reducing the polygon to an
+  equivalent triangle, and then constructing a square equivalent to
+  the triangle.}}
+\scanpage{211.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a parallelogram equivalent to a given
+  square, and having the sum of its base and altitude equal to a given line.}
+
+\figc{211aa424}{Let $R$ be the given square, and let the sum of the base and
+  altitude of the required parallelogram be equal to the given line $MN$.}
+
+\prove[To construct ]{a $\Par$ equivalent to $R$, with the sum of its
+  base and altitude equal to $MN$.}
+
+\step{Upon $MN$ as a diameter, describe a semicircle.}{}
+
+At $M$ erect $MP$, a $\perp$ to $MN$, equal to a side of the given
+square $R$.
+
+\step{Draw $PQ \parallel$ to $MN$, cutting the circumference at $S$.}{}
+
+\step{Draw $SC \perp$ to $MN$.}{}
+
+Any $\Par$ having $CM$ for its altitude and $CN$ for its base is
+equivalent to~$R$.
+
+\eq[\indent\textbf{Proof.}]{$SC$}{$=PM$.}{§§~104, 180}
+
+\eq{$\therefore \overline{SC}^2$}{$= \overline{PM}^2 = R$.}{}
+
+\eq{$MC:SC$}{$= SC:CN$,}{§~370}
+
+\pnote{(a $\perp$ let fall from any point of a circumference to the
+  diameter is the mean proportional between the segments of the
+  diameter).}
+
+\step[\indent Then]{$\overline{SC}^2 \Bumpeq MC × CN$.}{§~327}
+
+\hfill\qef
+
+\end{proof}
+
+\note{This problem may be stated as follows:}
+
+\emph{To construct two straight lines the sum and product of which are
+  known.}
+\scanpage{212.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a parallelogram equivalent to a given
+  square, and having the difference of its base and altitude equal to
+  a given line.}
+
+\figc{212aa425}{Let $R$ be the given square, and let the difference of the
+  base and altitude of the required parallelogram be equal to the
+  given line $MN$.}
+
+\textit{To construct a $\Par$ equivalent to $R$, with the difference of its
+  base and altitude equal to $MN$.}
+
+Upon the given line $MN$ as a diameter, describe a circle.
+
+From $M$ draw $MS$, tangent to the $\odot$, and equal to a side of
+the given square $R$.
+
+Through the centre of the $\odot$ draw $SB$ intersecting the
+circumference at $C$ and $B$.
+
+Then any $\Par$, as $R'$, having $SB$ for its base and $SC$ for its
+altitude, is equivalent to $R$.
+
+\step[\indent\textbf{Proof.}]{$SB:SM=SM:SC$,}{§~381}
+
+\pnote{(if from a point without a $\odot$ a secant and a tangent are
+  drawn, the tangent is the mean proportional between the whole secant
+  and the external segment).}
+
+\step[\indent Then]{$\overline{SM}^2 \Bumpeq SB × SC$,}{§~327}
+
+\noindent and the difference between $SB$ and $SC$ is the diameter of the
+$\odot$, that is,~$MN$.
+
+\hfill\qef
+
+\end{proof}
+
+\note{This problem may be stated: \textit{To construct two
+  straight lines the difference and product of which are known.}}
+\scanpage{213.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a polygon similar to a given polygon
+  $P$ and equivalent to a given polygon $Q$.}
+
+\figc{213aa426}{Let $P$ and $Q$ be the two given polygons, and $AB$ a side of $P$.}
+
+\prove[To construct ]{a polygon similar to $P$ and equivalent to $Q$.}
+
+\step{Find squares equivalent to $P$ and $Q$,}{§~423}
+
+\step{and let $m$ and $n$ respectively denote their sides.}{}
+
+Find $A'B'$, the fourth proportional to $m$, $n$, and $AB$.~\hfill§~386
+
+Upon $A'B'$, homologous to $AB$, construct $P'$ similar to $P$.
+
+\eq[\indent Then]{$P'$}{$\Bumpeq Q$.}{}
+
+\eq[\indent\textbf{Proof.}]{$m:n$}{$= AB:A'B'$.}{Const.}
+
+\eq{$\therefore m^2:n^2$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~338}
+
+\eq[\indent But]{$P \Bumpeq m^2$,}{and $Q \Bumpeq n^2$. }{Const.}
+
+\eq{$\therefore P:Q = m^2$}{$:n^2 = \overline{AB}^2:\overline{A'B'}^2$.}{}
+
+\eq[\indent But]{$P:P'$}{$= \overline{AB}^2:\overline{A'B'}^2$.}{§~412}
+
+\eq{$\therefore P:Q$}{$= P:P'$.}{Ax.~1}
+
+\eq{$\therefore P'$}{$\Bumpeq Q$.}{\qef}
+
+\end{proof}
+
+\ex{To construct a square equivalent to the sum of any
+number of given squares.}
+
+\ex{To construct a polygon similar to two given similar
+polygons and equivalent to their difference.
+}
+\scanpage{214.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a square which shall have a given
+  ratio to a given square.}
+
+\figc{214aa427}{Let R be the given square, and $\dfrac{n}{m}$ the given ratio.}
+
+\textit{To construct a square which shall be to $R$ as $n$ is to $m$.}
+
+Take $AB$ equal to a side of $R$, and draw $Ay$, making any convenient
+angle with $AB$.
+
+On $Ay$ take $AE$ equal to $m$, $EF$ equal to $n$, and draw $EB$.
+
+\step{Draw $FC \parallel$ to $EB$ meeting $AB$ produced at $C$.}{}
+
+\step{On $AC$ as a diameter, describe a semicircle.}{}
+
+\step{At $B$ erect the $\perp BD$, meeting the semicircumference at $D$.}{}
+
+\step{Then $BD$ is a side of the square required.}{}
+
+\step[\indent\textbf{Proof.}]{Denote $AB$ by $a$, $BC$ by $b$, and $BD$ by $x$.}{}
+
+
+\eq[\indent Now]{$a:x$}{$= x:b$.}{§~370}
+
+\eq[\indent Therefore,]{$a^2:x^2$}{$= a:b$.}{§~337}
+
+\eq[\indent But]{$a:b$}{$= m:n$.}{§~342}
+
+\eq[\indent Therefore,]{$a^2:x^2$}{$= m:n$.}{Ax.~1}
+
+\eq[\indent By inversion,]{$x^2:a^2$}{$= n:m$.}{§~331}
+
+Hence, the square on $BD$ will have the same ratio to $R$ as $n$ has
+to $m$.
+
+\hfill\qef
+
+\end{proof}
+\scanpage{215.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To construct a polygon similar to a given polygon
+  and having a given ratio to it.}
+
+\figc{215aa428}{Let $R$ be the given polygon, and $\dfrac{n}{m}$ the given
+  ratio.}
+
+\prove[To construct ]{a polygon similar to $R$, which shall be to $R$ as
+  $n$ is to $m$.}
+
+Construct a line $A'B'$, such that the square on $A'B'$ shall be to
+the square on $AB$ as $n$ is to $m$.~\hfill§~427
+
+Upon $A'B'$, as a side homologous to $AB$, construct the polygon $S$
+similar to $R$.~\hfill§~391
+
+\step{Then $S$ is the polygon required.}{}
+
+\eq[\indent\textbf{Proof.}]{$S:R$}{$= \overline{A'B'}^2 : \overline{AB}^2$.}{§~412}
+
+\eq[\indent But]{$\overline{A'B'}^2 : \overline{AB}^2$}{$= n:m$.}{Const.}
+
+\eq[\indent Therefore,]{$S:R$}{$= n:m$.}{Ax.~1}
+
+\hfill\qef
+
+\end{proof}
+
+
+\ex{To construct a triangle equivalent to a given
+triangle, and having one side equal to a given length $l$.}
+
+\ex{To transform a triangle into an equivalent right
+triangle.}
+
+\ex{To transform a given triangle into an equivalent
+right triangle, having one leg equal to a given length.}
+
+\ex{To transform a given triangle into an equivalent
+right triangle, having the hypotenuse equal to a given length.}
+\scanpage{216.png}%
+
+
+\section[EXERCISES.]{PROBLEMS OF CONSTRUCTION.}
+
+\begin{proofex}%
+To transform a triangle $ABC$ into an equivalent triangle,
+having a side equal to a given length $l$, and an angle equal to angle $BAC$.
+
+Upon $AB$ (produced if necessary), take $AD$ equal to $l$, draw $BE \parallel$ to
+$CD$, meeting $AC$ (produced if necessary) at $E$.
+
+$\triangle BED \Bumpeq \triangle BEC$.
+
+\end{proofex}
+
+\ex{To transform a given triangle into an equivalent isosceles triangle,
+having the base equal to a given length.}
+
+
+\exheader{To construct a triangle equivalent to:}
+
+\ex{The sum of two given triangles.}
+
+\ex{The difference of two given triangles.}
+
+\ex{To transform a given triangle into an equivalent equilateral
+triangle.}
+
+
+\exheader{To transform a parallelogram into an equivalent:}
+
+\ex{Parallelogram having one side equal to a given length.}
+
+\ex{Parallelogram having one angle equal to a given angle.}
+
+\ex{Rectangle having a given altitude.}
+
+
+\exheader{To transform a square into an equivalent:}
+
+\ex{Equilateral triangle.}
+
+\ex{Right triangle having one leg equal to a given length.}
+
+\ex{Rectangle having one side equal to a given length.}
+
+
+\exheader{To construct a square equivalent to:}
+
+\ex{Five eighths of a given square.}
+
+\ex{Three fifths of a given pentagon.}
+
+\ex{To divide a given triangle into two equivalent parts by a line
+through a given point $P$ in one of the sides.}
+
+\ex{To find a point within a triangle, such that the lines joining
+this point to the vertices shall divide the triangle into three equivalent
+parts.}
+
+\ex{To divide a given triangle into two equivalent parts by a line parallel to one of the sides.}
+
+\ex{To divide a given triangle into two equivalent parts by a line
+perpendicular to one of the sides.}
+\scanpage{217.png}%
+
+\subsection{PROBLEMS OF COMPUTATION.}
+
+\figccc{217aa404}{217bb405}{217cc406}
+
+\begin{proofex}%
+To find the area of an equilateral triangle in terms of its side.
+
+Denote the side by $a$, the altitude by $h$, and the area by $S$.
+
+\setlength{\eqalign}{0.33\dentwidth}
+\eq[\indent Then]{$h^2$}{$a^2 - \dfrac{a^2}{4} = \dfrac{3a^2}{4} =
+   \dfrac{a^2}{4} × 3$.}{§~372}
+
+\eq{$\therefore h$}{$= \dfrac{a}{2}\sqrt{3}$.}{}
+
+
+\eq[\indent But]{$S$}{$= \dfrac{a × h}{2}$.}{§~403}
+
+\label{formareaequitri}%
+\eq{$\therefore S$}
+  {$=\dfrac{a}{2} × \dfrac{a\sqrt{3}}{2} = \dfrac{a^2\sqrt{3}}{4}$.}{}
+
+\setlength{\eqalign}{0.5\dentwidth}
+
+\end{proofex}
+
+
+\begin{proofex}%
+To find the area of a triangle in terms of its sides.
+
+
+\label{formareatri2}%
+\setlength{\eqalign}{0.33\dentwidth}
+\eq[\indent By Ex.~312,]{$h$}{$= \dfrac{2}{b} \sqrt{s(s - a)(s - b)(s - c)}$.}{}
+
+\eq[\indent Hence,]{$S$}{$= \dfrac{b}{2} × \dfrac{2}{b}
+  \sqrt{s(s - a)(s - b)(s - c)}$}{§~403}
+
+\eq{}{$= \sqrt{s(s - a)(s - b)(s - c)}$.}{}
+
+\setlength{\eqalign}{0.5\dentwidth}
+
+\end{proofex}
+
+\begin{proofex}%
+To find the area of a triangle in terms of the
+radius of the circumscribed circle.
+
+If $R$ denotes the radius of the circumscribed circle, and $h$ the
+altitude of the triangle, we have, by §~384,
+
+\eq{$b × c$}{$= 2 R × h$.}{}
+
+Multiply by $a$, and we have,
+
+\eq{$a × b × c$}{$= 2 R × a × h$.}{}
+
+\eq[\indent But]{$a × h$}{$= 2 S$.}{§~403}
+
+\eq{$\therefore a × b × c$}{$= 4 R × S$.}{}
+
+\label{formareatri3}%
+\eq{$\therefore S$}{$= \dfrac{abc}{4R}$.}{}
+
+Show that the radius of the circumscribed circle is equal to
+$\dfrac{abc}{4S}$.
+
+\end{proofex}
+\scanpage{218.png}%
+
+\ex{Find the area of a right triangle, if the length of the hypotenuse
+is $17$~feet and the length of one leg is $8$~feet.}
+
+\ex{Find the ratio of the altitudes of two equivalent triangles, if
+the base of one is three times that of the other.}
+
+\ex{The bases of a trapezoid are $8$~feet and $10$~feet, and the altitude
+is $6$~feet.  Find the base of the equivalent rectangle that has an
+equal altitude.}
+
+\ex{Find the area of a rhombus, if the sum of its diagonals is
+$12$~feet, and their ratio is~$3:5$.}
+
+\ex{Find the area of an isosceles right triangle, if the hypotenuse
+is $20$~feet.}
+
+\ex{In a right triangle the hypotenuse is $13$~feet, one leg is $5$~feet.
+Find the area.}
+
+\ex{Find the area of an isosceles triangle, if base $=b$, and leg $=c$.}
+
+\ex{Find the area of an equilateral triangle, if one side $=8$~feet.}
+
+\ex{Find the area of an equilateral triangle, if the altitude $=h$.}
+
+\ex{A house is $40$~feet long, $30$~feet wide, $25$~feet high to the roof,
+and $35$~feet high to the ridge-pole. Find the number of square feet in its
+entire exterior surface.}
+
+\ex{The sides of a right triangle are as $3:4:5$. The altitude upon
+the hypotenuse is $12$~feet. Find the area.}
+
+\ex{Find the area of a right triangle, if one leg $=a$, and the altitude
+upon the hypotenuse $=h$.}
+
+\ex{Find the area of a triangle, if the lengths of the sides are
+$104$~feet, $111$~feet, and $175$~feet.}
+
+\ex{The area of a trapezoid is $700$~square feet. The bases are
+$30$~feet and $40$~feet, respectively. Find the altitude.}
+
+\ex{$ABCD$ is a trapezium; $AB = 87$ feet, $BC = 119$ feet, $CD = 41$ feet, $DA = 169$ feet, $AC = 200$ feet. Find the area.}
+
+\ex{What is the area of a quadrilateral circumscribed about a
+circle whose radius is $25$~feet, if the perimeter of the quadrilateral is $400$
+feet?  What is the area of a hexagon that has a perimeter of $400$ feet and
+is circumscribed about the same circle of $25$~feet radius (Ex.~361)?}
+
+\ex{The base of a triangle is $15$~feet, and its altitude is $8$~feet.
+Find the perimeter of an equivalent rhombus, if the altitude is $6$~feet.}
+\scanpage{219.png}%
+
+\ex{Upon the diagonal of a rectangle $24$~feet by $10$~feet a triangle
+equivalent to the rectangle is constructed.  What is its altitude?}
+
+\ex{Find the side of a square equivalent to a trapezoid whose bases
+are $56$~feet and $44$~feet, and each leg is $10$~feet.}
+
+\ex{Through a point $P$ in the side $AB$ of a triangle $ABC$, a line
+is drawn parallel to $BC$ so as to divide the triangle into two equivalent
+parts. Find the value of $AP$ in terms of $AB$.}
+
+\ex{What part of a parallelogram is the triangle cut off by a line
+from one vertex to the middle point of one of the opposite sides?}
+
+\ex{In two similar polygons, two homologous sides are $15$~feet
+and $25$~feet. The area of the first polygon is $450$~square feet.  Find the
+area of the second polygon.}
+
+\ex{The base of a triangle is $32$~feet, its altitude $20$~feet. What is
+the area of the triangle cut off by a line parallel to the base at a distance
+of $15$~feet from the base?}
+
+\ex{The sides of two equilateral triangles are $3$~feet and $4$~feet.
+Find the side of an equilateral triangle equivalent to their sum.}
+
+\ex{If the side of one equilateral triangle is equal to the altitude
+of another, what is the ratio of their areas?}
+
+\ex{The sides of a triangle are $10$~feet, $17$~feet, and $21$~feet.  Find
+the areas of the parts into which the triangle is divided by the bisector of
+the angle formed by the first two sides.}
+
+\ex{In a trapezoid, one base is $10$~feet, the altitude is $4$~feet, the
+area is $32$~square feet.  Find the length of a line drawn between the legs
+parallel to the bases and distant $1$~foot from the lower base.}
+
+\ex{The diagonals of a rhombus are $90$~yards and $120$~yards,
+respectively. Find the area, the length of one side, and the perpendicular
+distance between two parallel sides.}
+
+\ex{Find the number of square feet of carpet that are required to
+cover a triangular floor whose sides are, respectively, $26$~feet, $35$~feet, and
+$51$~feet.}
+
+\ex{If the altitude $h$ of a triangle is increased by a length $m$, how
+much must be taken from the base $a$ that the area may remain the same?}
+
+\ex{Find the area of a right triangle, having given the segments
+$p$, $q$, into which the hypotenuse is divided by a perpendicular drawn to
+the hypotenuse from the vertex of the right angle.}
+\scanpage{220.png}%
+
+
+\chapter{BOOK V\@. REGULAR POLYGONS AND CIRCLES.}
+\markboth{\Headings{BOOK V\@. PLANE GEOMETRY.}}
+{\Headings{REGULAR POLYGONS AND CIRCLES.}}%
+
+\pp{\defn{A \indexbf{regular polygon} is a polygon
+which is both equilateral and equiangular.  The equilateral triangle
+and the square are examples.}}
+
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An equilateral polygon inscribed in a circle is a
+  regular polygon.}
+
+\figc{220aa430}{Let $ABC$ etc.\ be an equilateral polygon inscribed in a
+  circle.}
+
+\prove{the polygon $ABC$ etc.\ is a regular polygon.}
+
+\step[\indent\textbf{Proof.}]{The arcs $AB$, $BC$, $CD$, etc., are equal.}{§~243}
+
+\step{Hence, arcs $ABC$, $BCD$, etc., are equal.}{Ax.~2}
+
+\step{Therefore, arcs $CFA$, $DFB$, etc., are equal.}{Ax.~3}
+
+\step{Therefore, $\angle_s A$, $B$, $C$, etc., are equal.}{§~289}
+
+Therefore, the polygon $ABC$ etc.\ is a regular polygon, being
+equilateral and equiangular.~\hfill§~429
+
+\hfill\qed
+
+\end{proof}
+\scanpage{221.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{A circle may be circumscribed about, and a circle
+  may be inscribed in, any regular polygon.}
+
+\figc{221aa431}{Let $ABCDE$ be a regular polygon.}
+
+\prove[\textup{1.} To prove that ]{a circle may be circumscribed about $ABCDE$.}
+
+\textbf{Proof.} Let $O$ be the centre of the circle which may be passed
+through $A$, $B$, and $C$.~\hfill§~258
+
+\step{Draw $OA$, $OB$, $OC$, and $OD$.}{}
+
+\eq[\indent Then]{$\angle ABC$}{$= \angle BCD$,}{§~429}
+
+\eq[and]{$\angle OBC$}{$= \angle OCB$.}{§~145}
+
+%proofrule
+\eq[\indent By subtraction,]{$\angle OBA$}{$= \angle OCD$.}{Ax.~3}
+
+\step{The $\triangle_s OBA$ and $OCD$ are equal.}{§~143}
+
+\eq[\indent For]{$\angle OBA$}{$= \angle OCD$,}{}
+
+\eq{$OB$}{$= OC$,}{§~217}
+
+\eq[and]{$AB$}{$= CD$.}{§~429}
+
+\eq{$\therefore OA$}{$= OD$.}{§~128}
+
+$\therefore$ the circle passing through $A$, $B$, $C$, passes through
+$D$.
+
+In like manner it may be proved that the circle passing through $B$,
+$C$, and $D$ also passes through $E$; and so on.
+\scanpage{222.png}%
+
+Therefore, the circle described from $O$ as a centre, with a
+radius $OA$, will be circumscribed about the polygon.~\hfill§~231
+
+\prove[\textup{2.} To prove that ]{a circle may be inscribed in $ABCDE$.}
+
+\textbf{Proof.}  Since the sides of the regular polygon are equal
+chords of the circumscribed circle, they are equally distant
+from the centre.~\hfill§~249
+
+Therefore, the circle described from $O$ as a centre, with the
+distance from $O$ to a side of the polygon as a radius, will be
+inscribed in the polygon (§~232).~\hfill\qed
+
+\end{proof}
+
+\pp{\defn{The radius of the circumscribed circle, $OA$, is
+called the \textbf{radius} of the polygon\label{polyradius}.}}
+
+\pp{\defn{The radius of the inscribed circle, $OF$, is called
+the \textbf{apothem}\label{apothem} of the polygon.}}
+
+\pp{\defn{The common centre, $O$, of the circumscribed and
+inscribed circles is called the \textbf{centre} of the polygon\label{centrepoly}.}}
+
+\pp{\defn{The angle between radii drawn to the extremities
+of any side is called the \textbf{angle at the centre} of the polygon.}}
+
+By joining the centre to the vertices of a regular polygon,
+the polygon can be decomposed into as many equal isosceles
+triangles as it has sides.
+
+\pp{\cor[1]{The angle at the centre of a regular polygon\label{anglecentreregpoly}
+is equal to four right angles divided by the number of sides
+of the polygon.  Hence, the angles at the centre of any regular
+polygon are all equal.}}
+
+\pp{\cor[2]{The radius drawn to any vertex of a regular
+polygon bisects the angle at the vertex.}}
+
+\pp{\cor[3]{The angle at the centre of a regular polygon
+and an interior angle of the polygon are supplementary.}}
+
+\step[\indent For]{$\angle_s FOB$ and $FBO$ are complementary.}{§~135}
+
+$\therefore$ their doubles $AOB$ and $FBC$ are supplementary.\hfill~Ax.~6
+\scanpage{223.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If the circumference of a circle is divided into
+any number of equal arcs, the chords joining the successive
+points of division form a regular inscribed polygon;
+and the tangents drawn at the points of division
+form a regular circumscribed polygon.}
+
+\figc{223aa439}{Suppose the circumference divided into equal arcs $AB$, $BC$, etc.
+Let $AB$, $BC$, etc., be the chords, $FBG$, $GCH$, etc., the tangents.}
+
+1.~\prove{$ABCDE$ is a regular polygon.}
+
+\step[\indent\textbf{Proof.}]{The sides $AB$, $BC$, $CD$, etc., are equal.}{§~241}
+
+\step{Therefore, the polygon is regular.}{§~430}
+
+2.~\prove{To prove that $FGHIK$ is a regular polygon.}
+
+\textbf{Proof.} The $\triangle_s AFB$, $BGC$, $CHD$, etc., are all equal isosceles
+triangles.\hfill\allowbreak\null\nobreak\hfill\nobreak§§~295,139
+
+\step{$\therefore \angle_s F$, $G$, $H$, etc., are equal, and $FB$, $BG$, $GC$, etc., are equal.}{}
+
+\step{$\therefore FG=GH=HI$, etc.}{Ax.~6}
+
+\step{$\therefore FGHIK$ is a regular polygon.}{§~429}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor[1]{Tangents to a circle at the vertices of a regular
+inscribed polygon form a regular circumscribed polygon of
+the same number of sides as the inscribed polygon.}}
+\scanpage{224.png}%
+
+\figccc{224aa441}{224bb442}{224cc443}
+\begin{point}%
+\cor[2]{Tangents to a circle at the middle points of
+the arcs subtended by the sides of a regular inscribed polygon
+form a circumscribed regular polygon,
+whose sides are parallel to the sides of
+the inscribed polygon and whose vertices
+lie on the radii (prolonged) of the inscribed
+polygon.}
+
+For two corresponding sides, $AB$ and
+$A'B'$, are perpendicular to $OM$ (§§~248,
+254), and are parallel (§~104); and the tangents $MB'$ and $NB'$,
+intersecting at a point equidistant from $OM$ and $ON$ (§~261),
+intersect upon the bisector of the $\angle MON$ (§~162); that is,
+upon the radius $OB$.
+\end{point}
+
+\pp{\cor[3]{If the vertices of a regular inscribed polygon
+are joined to the middle points of the arcs subtended
+by the sides of the polygon, the joining
+lines form a regular inscribed polygon of
+double the number of sides.}}
+
+\pp{\cor[4]{Tangents at the middle points
+the arcs between adjacent points of contact
+of the sides of a regular circumscribed polygon
+form a regular circumscribed polygon of
+double the number of sides.}}
+
+\begin{point}%
+\cor[5]{The perimeter of an inscribed polygon is
+less than the perimeter of an inscribed polygon of double
+the number of sides; and the perimeter of a circumscribed
+polygon is greater than the perimeter of a circumscribed
+polygon of double the number of sides.}
+
+For two sides of a triangle are together greater than the
+third side.~\hfill§~138
+\end{point}
+\scanpage{225.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two regular polygons of the same number of sides
+are similar.}
+
+\figc{225aa445}{Let $Q$ and $Q'$ be two regular polygons, each having $n$
+sides.}
+
+\prove{$Q$ and $Q'$ are similar.}
+
+\textbf{Proof.}  The sum of the interior $\angle_s$ of each polygon is
+equal to
+
+\step{$(n-2)2$ rt.~$\angle_s$,}{§~205}
+
+\pnote{(the sum of the interior $\angle_s$ of a polygon is equal to 2
+  rt.\ $\angle_s$ taken as many times less two as the polygon has
+  sides).}
+
+\step{Each angle of either polygon $=
+  \dfrac{(n-2) 2 \text{ rt.\ } \angle_s}{n}$,}{§~206}
+
+\pnote{(for the $\angle_s$ of a regular polygon are all equal, and
+  hence each $\angle$ is equal to the sum of the $\angle_s$ divided by
+  their number).}
+
+Hence, the two polygons $Q$ and $Q'$ are mutually equiangular.
+
+\step{Since $AB = BC$, etc., and $A'B' = B'C'$, etc.,}{§~429}
+
+\step{\( AB:A'B' = BC:B'C' \), etc.}{}
+
+Hence, the two polygons have their homologous sides proportional.
+
+\step{Therefore the two polygons are similar.}{§~351}
+
+\hfill\qed
+
+\end{proof}
+
+\pp{\cor{The areas of two regular polygons
+  of the same number of sides are to each other as the squares of any
+  two homologous sides.}~\hfill§~412}
+\scanpage{226.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The perimeters of two regular polygons of the same
+  number of sides are to each other as the radii of their
+  circumscribed circles, and also as the radii of their inscribed circles.}
+
+\figc{226aa447}{Let $P$ and $P'$ denote the perimeters, $O$ and $O'$ the
+  centres, of the two regular polygons.}
+
+From $O$, $O'$ draw $OA$, $O'A'$, $OB$, $O'B'$, and the $\perp_s OM$,
+$O'M'$.
+
+\prove{$P:P' = OA:O'A' = OM:O'M'$.}
+
+\step[\indent\textbf{Proof.}]{Since the polygons are similar,}{§~445}
+
+\eq{$P:P'$}{$= AB:A'B'$.}{§~364}
+
+The $\triangle_s OAB$ and $O'A'B'$ are isosceles.~\hfill§~431
+
+\eq[\indent Now]{$\angle O$}{$= \angle O'$,}{§~436}
+
+\eq[and]{$OA:OB$}{$= O'A':O'B'$.}{}
+
+\step{$\therefore$ the $\triangle_s OAB$ and $O'A'B'$ are similar.}{§~357}
+
+\eq{$\therefore AB:A'B'$}{$= OA:O'A'$.}{§~351}
+
+\eq[\indent Also,]{$AB:A'B'$}{$= OM:O'M'$.}{§~361}
+
+\step{$\therefore P:P' = OA:O'A' = OM:O'M'$.}{Ax.~1}
+
+\hfill\qed
+
+\end{proof}
+
+
+\pp{\cor{The areas of two regular polygons
+  of the same number of sides are to each other as the squares of the
+  radii of the circumscribed circles, and of the inscribed
+  circles.}~\hfill§~413}
+\scanpage{227.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{If the number of sides of a regular inscribed
+polygon is indefinitely increased, the apothem of the
+polygon approaches the radius of the circle as its limit.}
+
+\figc{227aa449}{Let $AB$ be a side and $OP$ the apothem of a regular polygon of $n$
+sides inscribed in the circle whose radius is $OA$.}
+
+\prove{$OP$ approaches $OA$ as a limit, when $n$ increases
+indefinitely.}
+
+\eq[\indent\textbf{Proof.}]{$OP$}{$<OA$,}{§~97}
+
+\eq[and]{$OA-OP$}{$< AP$.}{§~138}
+
+\eq{$\therefore OA-OP$}{$<AB$, which is twice $AP$.}{§~245 }
+
+Now, if $n$ is taken sufficiently great, $AB$, and consequently
+$OA-OP$, can be made less than any assigned value, however
+small, but cannot be made zero.
+
+Since $OA-OP$ can be made less than any assigned value
+by increasing $n$, but cannot be made zero, $OA$ is the limit of
+$OP$ by the test for a limit.~\hfill§~275
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor{If the number of sides of a regular inscribed
+polygon is indefinitely increased, the square of the apothem
+approaches the square of the radius of the circle as a limit.}
+
+\step[\indent For]{$\overline{OA}^2 - \overline{OP}^2 = \overline{AP}^2$.}{§~372}
+
+But by taking $n$ sufficiently great, $AB$ and consequently $AP$,
+the half of $AB$, can be made less than any assigned value.
+\end{point}
+\scanpage{228.png}%
+
+Therefore, $\overline{AP}^2$, the product of $AP$ by $AP$, can be made
+less than any assigned value; for the product of two finite factors
+approaches zero as a limit, if \emph{either} factor approaches zero as
+a limit (§~276); and for a still stronger reason, the product
+approaches zero as a limit, if \emph{each} of the factors approaches
+zero as a limit.
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{An arc of a circle is less than any line which
+envelops it and has the same extremities.}
+
+\figc{228aa451}{Let $ACB$ be an arc of a circle, and $AB$ its chord.}
+
+\prove{the arc $ACB$ is less than any other line which
+  envelops this arc and terminates at $A$ and $B$.}
+
+% the following are steps in a proof, but do not agree with the formatting.
+\textbf{Proof.} Of all the lines that can be drawn, each to include
+the area $ACB$ between itself and the chord $AB$, there must be at
+least one shortest line; for all the lines are not equal.
+
+Now the enveloping line $ADB$ cannot be the shortest; for drawing
+$ECF$ tangent to the arc $ACB$ at $C$, the line $AECFB < AEDFB$, since
+$ECF < EDF$.~\hfill§~49
+
+In like manner it can be shown that no other enveloping line can be
+the shortest.  Therefore, $ACB$ is the shortest.
+
+\step {} {\qed}
+
+
+\end{proof}
+
+\pp{\cor[1]{The circumference of a circle
+  is less than the perimeter of any polygon circumscribed about it.}}
+
+\pp{\cor[2]{Any convex curve\label{convexcurve} is less than
+  the perimeter of a polygon circumscribed about it.}}
+\scanpage{229.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The circumference of a circle is the limit which
+  the perimeters of regular inscribed polygons and of similar
+  circumscribed polygons approach, if the number of sides of the
+  polygons is indefinitely increased; and the area of a circle is the
+  limit which the areas of these polygons approach.}
+
+\figc{229aa454}{Let $P$ and $P'$ denote the lengths of the perimeters, $AB$
+  and $A'B'$ two homologous sides, $R$ and $R'$ the radii, of the
+  polygons, and $C$ the circumference of the circle.}
+
+\prove{$C$ is the limit of $P$ and of $P'$, if
+  the number of sides of the polygons is indefinitely increased.}
+
+\step[\indent\textbf{Proof.}]{Since the polygons are similar by hypothesis,}{}
+
+\eq {$P':P $} {$= R':R$.} {§~447}
+
+\eq [\indent Therefore,] {$P'-P:P $} {$= R'-R:R$.} {§~333}
+
+\eq [\indent Whence,] {$R(P'-P) $} {$ = P(R'-R)$.} {§~327}
+
+\eq [\indent Therefore,] {$P'-P $} {$= \frac{P}{R}(R' - R)$.} {}
+
+\step {Now $P$ is always less than $C$.} {§~273}
+
+\eq {$\therefore P'-P $} {$< \frac{C}{R} (R'-R)$.}{}
+\scanpage{230.png}%
+
+But $R'-R$, which is less than $A'C$ (§~138), can be made less than
+any assigned quantity by increasing the number of sides of the
+polygons; and therefore $\dfrac{C}{R}(R'-R)$ can be made less than any
+assigned quantity.~\hfill§~276
+
+Hence, $P'-P$ can be made less than any assigned quantity.
+
+Since $P'$ is always greater than $C$ (§~452), and $P$ is always less
+than $C$ (§~273), the difference between $C$ and either $P'$ or $P$
+is less than the difference $P'-P$, and consequently can be made less
+than any assigned quantity, but cannot be made zero.
+
+Therefore, $C$ is the common limit of $P'$ and $P$.~\hfill§~275
+
+\lett{Let $K$ denote the area of the circle, $S$ the area of the
+  inscribed polygon, and $S'$ the area of the circumscribed polygon.}
+
+2. \prove{$K$ is the limit of $S$ and $S'$.}
+
+\eq[\indent\textbf{Proof.}]{$S':S$}{$= R'^2:R^2$.}{§~448}
+
+\eq[\indent By division,]{$S'-S:S$}{$= R'^2-R^2:R^2$.}{§~333}
+
+\eq[\indent Whence]{$S'-S$}{$= \dfrac{S}{R^2}(R'^2-R^2)$.}{}
+
+\step{Now $K$ is always greater than $S$.}{Ax.~8}
+
+\eq[\indent Therefore,]{$S'-S$}{$< \dfrac{K}{R^2}(R'^2-R^2)$.}{}
+
+But $R'^2 - R^2$, which is equal to $(R'+R)(R'-R)$, can be made less
+than any assigned quantity; and therefore $\dfrac{K}{R^2}(R'^2-R^2)$ can
+be made less than any assigned quantity.~\hfill§~276
+
+Hence, $S'-S$ can be made less than any assigned quantity.
+
+Since $S' > K$ always, and $S < K$ always (Ax.~8), the difference
+between $K$ and either $S'$ or $S$ is less than the difference $S'-S$,
+and consequently can be made less than any assigned quantity, but
+cannot be made zero.
+
+Therefore, $K$ is the common limit of $S'$ and $S$.~\hfill§~275
+
+\hfill\qed
+
+\end{proof}
+\scanpage{231.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Two circumferences have the same ratio as their
+radii.}
+
+\figc{231aa455}{Let $C$ and $C'$ be the circumferences, $R$ and $R'$ the radii, of the two
+circles $Q$ and $Q'$.}
+
+\proveq{$C:C'$}{$= R:R'$.}
+
+\textbf{Proof.}   Inscribe in the $\odot_s$ two similar regular polygons, and
+denote their perimeters by $P$ and $P'$.
+
+\eq[\indent Then]{$P:P'$}{$= R:R'$.}{§~447}
+
+Conceive the number of sides of these regular polygons to
+be indefinitely increased, the polygons continuing similar.
+
+Then $P$ and $P'$ will have $C$ and $C'$ as limits.~\hfill§~454
+
+But $P:P'$ will always be equal to $R:R'$.~\hfill§~447
+
+\eq{$\therefore C:C'$}{$= R:R'$.}{§~285}
+
+\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor{The ratio of the circumference of a circle to its
+diameter is constant.}
+
+\eq[\indent For]{$C:C'$}{$= R: R'$.}{§~455}
+
+\eq{$\therefore C:C'$}{$= 2R:2R'$.}{§~340}
+
+\eq[\indent By alternation,]{$C: 2R$}{$= C' : 2R'$.}{§~330}
+\end{point}
+
+\pp{\defn{The constant ratio of the circumference of a
+circle to its diameter is represented by the Greek letter $\pi$\label{pi}.}}
+
+\label{formcircum}%
+\pp{\cor{$\pi = \dfrac{C}{2R}$. $\therefore C=2\pi R$.}}
+\scanpage{232.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a regular polygon is equal to half
+the product of its apothem by its perimeter.}
+
+\figc{232aa459}{Let $P$ represent the perimeter, $R$ the apothem, and $S$ the area of
+the regular polygon $ABC$ etc.}
+
+\prove{$S = \frac{1}{2}R × P$.}
+
+\step[\indent\textbf{Proof.}]{Draw the radii $OA$, $OB$, $OC$, etc.}{}
+
+\step{The polygon is divided into as many $\triangle_s$ as it has sides.}{}
+
+\step{The apothem is the common altitude of these $\triangle_s$,}{}
+
+\step{and the area of each $\triangle=\frac{1}{2}R$ multiplied by the base.}{§~403}
+
+Hence, the area of all the $\triangle_s$ is equal to $\frac{1}{2}R$ multiplied by
+the sum of all the bases.
+
+But the sum of the areas of all the $\triangle_s$ is equal to the area of
+the polygon.\hfill~Ax.~9
+
+And the sum of all the bases of the $\triangle_s$ is equal to the perimeter
+of the polygon.\hfill~Ax.~9
+
+\label{formareapoly}%
+\step{$\therefore S = \frac{1}{2}R × P$.}{\qed}
+
+\end{proof}
+
+\pp{\defn{In different circles \indexbf{similar arcs}, \indexbf{similar sectors},
+and \indexbf{similar segments} are such as correspond to \emph{equal angles at
+the centre}.}}
+\scanpage{233.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The area of a circle is equal to half the product
+of its radius by its circumference.}
+
+\figc{233aa461}{Let $R$ represent the radius, $C$ the circumference, and $S$ the area,
+of the circle whose centre is $O$.}
+
+\proveq{$S$}{$= \frac{1}{2}R × C$.}
+
+\textbf{Proof.} Circumscribe any regular polygon about the circle,
+and denote its perimeter by $P$, and its area by $S'$.
+
+\eq[\indent Then]{$S'$}{$= \frac{1}{2} R × P$.}{§~459}
+
+Conceive the number of sides of the polygon to be indefinitely
+increased.
+
+\step{Then $P$ approaches $C$ as its limit,}{§~454}
+
+\step{$\frac{1}{2}R × P$ approaches $\frac{1}{2}R× C$ as its limit,}{§~279}
+
+\step{and $S'$ approaches $S$ as its limit.}{§~454}
+
+\step[\indent But]{$S' = \frac{1}{2} R × P$, always.}{§~459}
+
+\step{$\therefore S = \frac{1}{2}R× C$.}{§~284}
+
+\hfill\qed
+
+\end{proof}
+
+\label{formareasector}%
+\begin{point}%
+\cor[1]{The area of a sector is equal to half the
+product of its radius by its arc.}
+
+For the sector and its arc are like parts of the circle and
+its circumference, respectively.
+\end{point}
+
+\begin{point}%
+\cor[2]{The area of a circle is equal to $\pi$ times the
+square of its radius.}
+
+\label{formareacircle}%
+For the area of the \( \odot = \frac{1}{2} R × C =
+\frac{1}{2} R × 2\pi R = \pi R^2 \).
+
+\end{point}
+\scanpage{234.png}%
+
+\begin{point}%
+\cor[3]{The areas of two circles are to
+  each other as the squares of their radii.}
+
+For, if $S$ and $S'$ denote the areas, and $R$ and $R'$ the radii,
+
+\[ S:S' = \pi R^2:\pi R'^2 = R^2:R'^2. \]
+\end{point}
+
+\pp{\cor[4]{Similar arcs are to each other
+  as their radii; similar sectors are to each other as the squares of
+  their radii.}}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{The areas of two similar segments are to each
+  other as the squares of their radii.}
+
+\figc{234aa466}{Let $AC$ and $A'C'$ be the radii of the two similar sectors
+  $ACB$ and $A'C'B'$, and let $ABP$ and $A'B'P'$ be the corresponding
+  segments.}
+
+\proveq{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.}
+
+\step[\indent\textbf{Proof.}]{Sector $ACB :$ Sector $A'C'B' =
+ \overline{AC}^2:\overline{A'C'}^2$.}{§~465}
+
+\step{The $\triangle_s ACB$ and $A'C'B'$ are similar.}{§~357}
+
+\eq{$\therefore \triangle ACB:\triangle A'C'B'$}
+  {$=\overline{AC}^2:\overline{A'C'}^2$.}{§~411}
+
+\eq{$\therefore$ sector $ACB :$ sector $A'C'B'$}
+  {$=\triangle ACB : \triangle A'C'B'$.}{Ax.~1}
+
+\eq{$\therefore$ sector $ACB : \triangle ACB$}
+  {$=$ sector $A'C'B' : \triangle A'C'B'$.}{§~330}
+
+\step{\( \therefore
+  \dfrac{\text{sector } ACB-\triangle ACB}
+       {\text{sector } A'C'B'-\triangle A'C'B'} =
+  \dfrac{\triangle ACB}{\triangle A'C'B'} =
+  \dfrac{\overline{AC}^2}{\overline{A'C'}^2} \).}{§~333}
+
+\eq[\indent That is,]{$ABP:A'B'P'$}{$= \overline{AC}^2:\overline{A'C'}^2$.}{\qed}
+
+
+\end{proof}
+\scanpage{235.png}%
+
+
+\clearpage
+\section{PROBLEMS OF CONSTRUCTION.}
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe a square in a given circle.}
+
+\figc{235aa467}{Let $O$ be the centre of the given circle.}
+
+\prove[]{To inscribe a square in the given circle.}
+
+\step{Draw two diameters $AC$ and $BD \perp$ to each other.}{}
+
+\step{Draw $AB$, $BC$, $CD$, and $DA$.}{}
+
+\step{Then $ABCD$ is the square required.}{}
+
+\step[\indent\textbf{Proof.}]{The $\angle_s ABC$, $BCD$, etc., are rt.\ $\angle_s$,}{§~290}
+
+\pnote{(each being inscribed in a semicircle),}
+
+\step{and the sides $AB$, $BC$, etc., are equal,}{§~241}
+
+\pnote{(in the same $\odot$ equal arcs are subtended by equal chords).}
+
+\step{Hence the quadrilateral $ABCD$ is a square.}{§~168}
+
+\hfill\qef
+
+\end{proof}
+
+\pp{\cor{By bisecting the arcs $AB$, $BC$,
+  etc., a regular polygon of eight sides may be inscribed in the
+  circle; and, by continuing the process, regular polygons of sixteen,
+  thirty-two, sixty-four, etc., sides may be inscribed.}}
+
+
+\ex{The area of a circumscribed square is equal to
+twice the area of the inscribed square.}
+
+\ex{The area of a circular ring is equal to that of a
+circle whose diameter is a chord of the outer circle tangent to the
+inner circle.}
+\scanpage{236.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe a regular hexagon in a given circle.}
+
+\figc{236aa469}{Let $O$ be the centre of the given circle.}
+
+\prove[To inscribe ]{a regular hexagon in the given circle.}
+
+\step{From $O$ draw any radius, as $OC$.}{}
+
+\step{From $C$ as a centre, with a radius equal to $OC$,}{}
+
+\step{describe an arc intersecting the circumference at $F$.}{}
+
+\step{Draw $OF$ and $CF$.}{}
+
+\step{Then $CF$ is a side of the regular hexagon required.}{}
+
+\step[\indent\textbf{Proof.}]{The $\triangle OFC$ is equiangular,}{§~146}
+
+\pnote{(since it is equilateral by construction).}
+
+Hence, the $\angle FOC$ is $\frac{1}{3}$ of $2$~rt.~$\angle_s$, or
+$\frac{1}{6}$ of $4$~rt.~$\angle_s$.~\hfill§~136
+
+\step{$\therefore$ the arc $FC$ is $\frac{1}{6}$ of the circumference,}{}
+
+and the chord $FC$ is a side of a regular inscribed hexagon.
+
+Hence, to inscribe a regular hexagon apply the radius six times as a
+chord.~\hfill\qef
+
+\end{proof}
+
+\pp{\cor[1]{By joining the alternate
+  vertices $A$, $C$, $D$, an equilateral triangle is inscribed in the
+  circle.}}
+
+\pp{\cor[2]{By bisecting the arcs $AB$,
+  $BC$, etc., a regular polygon of twelve sides may be inscribed in
+  the circle; and, by continuing the process, regular polygons of
+  twenty-four, forty-eight, etc., sides may be inscribed.}}
+\scanpage{237.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe a regular decagon in a given circle.}
+
+\figc{237aa472}{Let $O$ be the centre of the given circle.}
+
+\prove[To inscribe ]{a regular decagon in the given circle.}
+
+\step{Draw any radius $OC$,}{}
+
+\step{and divide it in extreme and mean ratio, so that $OC$ shall}{}
+
+\step{be to $OS$ as $OS$ is to $SC$.}{§~389}
+
+\step{From $C$ as a centre, with a radius equal to $OS$,}{}
+
+\step{describe an arc intersecting the circumference at $B$.}{}
+
+\step{Draw $BC$.}{}
+
+\step{Then $BC$ is a side of the regular decagon required.}{}
+
+\step[\indent\textbf{Proof.}]{Draw $BS$ and $BO$.}{}
+
+\eq[\indent Now]{$OC:OS$}{$= OS:SC$,}{Const.}
+
+\eq[and]{$BC$}{$= OS$.}{Const.}
+
+\eq{$\therefore OC:BC$}{$= BC:SC$.}{}
+
+\eq[\indent Moreover,]{$\angle OCB$}{$= \angle SCB$.}{Iden.}
+
+\step{Hence, the $\triangle_s OCB$ and $BCS$ are similar.}{§~357}
+
+\step{But the $\triangle OCB$ is isosceles.}{§~217}
+
+\step{$\therefore \triangle BCS$, which is similar to the $\triangle OCB$,
+  is isosceles,}{}
+
+\step{and $CB = BS = SO$.}{§~120}
+\scanpage{238.png}%
+
+\step{$\therefore$ the $\triangle SOB$ is isosceles, and the $\angle O =
+  \angle SBO$.}{§~145}
+
+\step{But the ext.~$\angle CSB = \angle O + \angle SBO = 2 \angle O$.}{§~137}
+
+\eq[\indent Hence,]{$\angle SCB$}{$= 2\angle O$,}{}
+
+\eq[and]{$\angle OBC$}{$= 2\angle O$.}{}
+
+\step{$\therefore$ the sum of the $\angle_s$ of the $\triangle OCB = 5
+\angle O = 2 \text{rt.\ } \angle_s$,}{}
+
+\step[and]{$\angle O = \frac{1}{5}$ of $2$~rt.~$\angle_s$, or $\frac{1}{10}$ of $4$~rt.~$\angle_s$.}{}
+
+\step{Therefore, the arc $BC$ is $\frac{1}{10}$ of the circumference,}{}
+
+and the chord $BC$ is a side of a regular inscribed decagon.
+
+Therefore, to inscribe a regular decagon, divide the radius internally
+in extreme and mean ratio, and apply the greater segment ten times as
+a chord.~\hfill\qef
+
+\end{proof}
+
+\pp{\cor[1]{By joining the alternate vertices
+  of a regular inscribed decagon, a regular pentagon is inscribed.}}
+
+\pp{\cor[2]{By bisecting the arcs $BC$, $CF$,
+  etc., a regular polygon of twenty sides may be inscribed in the
+  circle; and, by continuing the process, regular polygons of forty,
+  eighty, etc., sides may be inscribed.}}
+
+
+If $R$ denotes the radius of a regular inscribed polygon, $r$ the
+apothem, $a$ one side, $A$ an interior angle, and $C$ the angle at the
+centre, show that
+
+\ex{In a regular inscribed triangle $a = R\sqrt{3}$,
+$r =\frac{1}{2}R$, $A = 60°$, $C = 120°$.}
+
+\ex{In an inscribed square $a = R\sqrt{2}$,
+$r = \frac{1}{2}R\sqrt{2}$, $A=90°$, $C=90°$.}
+
+\ex{In a regular inscribed hexagon $a = R$,
+$r = \frac{1}{2}R\sqrt{3}$, $A=120°$, $C=60°$.}
+
+\ex{In a regular inscribed decagon
+\[ a = \frac{R(\sqrt{5}-1)}{2},\
+   r = \frac{1}{4} R \sqrt{10+2\sqrt{5}},\
+   A = 144°,\
+   C = 36°. \]}
+\scanpage{239.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe in a given circle a regular
+  pentedecagon, or polygon of fifteen sides.}
+
+\figc{239aa475}{Let $Q$ be the given circle.}
+
+\prove[To inscribe ]{in $Q$ a regular pentedecagon.}
+
+\step{Draw $EH$ equal to the radius of the circle,}{}
+
+\step{and $EF$ equal to a side of the regular inscribed decagon.}{§~472}
+
+\step{Draw $FH$.}{}
+
+Then $FH$ is a side of the regular pentedecagon required.
+
+\step[\indent\textbf{Proof.}]{The arc $EH$ is $\frac{1}{6}$ of the circumference,}{§~469}
+
+\step{and the arc $EF$ is $\frac{1}{10}$ of the circumference.}{Const.}
+
+Hence, the arc $FH$ is $\frac{1}{6} - \frac{1}{10}$, or
+$\frac{1}{15}$, of the circumference.
+
+And the chord $FH$ is a side of a regular inscribed pentedecagon.
+
+By applying $FH$ fifteen times as a chord, we have the polygon
+required.~\hfill\qef
+
+\end{proof}
+
+\pp{\cor{By bisecting the arcs $FH$, $HA$,
+  etc., a regular polygon of thirty sides may be inscribed; and, by
+  continuing the process, regular polygons of sixty, one hundred
+  twenty, etc., sides may be inscribed.}}
+\scanpage{240.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To inscribe in a given circle a regular
+  polygon similar to a given regular polygon.}
+
+\figc{240aa477}{Let $ABC$ etc.\ be the given regular polygon, and $O'$ the
+  centre of the given circle.}
+
+\prove[To inscribe ]{in the circle a regular polygon similar to $ABC$
+  etc.}
+
+\step{From $O$, the centre of the given polygon,}{}
+
+\step{draw $OD$ and $OC$.}{}
+
+\step{From $O'$, the centre of the given circle,}{}
+
+\step{draw $O'C'$ and $O'D'$,}{}
+
+\step{making the $\angle O'$ equal to the $\angle O$.}{}
+
+\step{Draw $C'D'$.}{}
+
+\step{Then $C'D'$ is a side of the regular polygon required.}{}
+
+\textbf{Proof.} Each polygon has as many sides as the $\angle O$, or
+$\angle O'$, is contained times in $4$~rt.~$\angle_s$.
+
+Therefore, the polygon $C'D'E'$ etc.\ is similar to the polygon $CDE$
+etc.,~\hfill§~445
+
+\pnote{(two regular polygons of the same number of sides are similar).}
+
+\hfill\qef
+
+\end{proof}
+
+
+\ex{The area of an inscribed regular octagon is equal to
+that of the rectangle whose sides are equal to the sides of the
+inscribed and the circumscribed squares.
+}
+\scanpage{241.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{Given the side and the radius of a regular
+  inscribed polygon, to find the side of the regular inscribed polygon
+  of double the number of sides.}
+
+\figc{241aa478}{Let $AB$ be a side of the regular inscribed polygon.}
+
+\prove[To find ]{$AD$, a side of the regular inscribed polygon of double
+  the number of sides.}
+
+\step{Denote the radius by $R$, and $AB$ by $a$.}{}
+
+\step{From $D$ draw $DH$ through the centre $O$, and draw $OA$, $AH$.}{}
+
+\step{$DH$ is $\perp$ to $AB$ at its middle point $C$.}{§~161}
+
+\eq[\indent In the rt.~$\triangle OCA$,]{$\overline{OC}^2$}
+  {$= R^2 - \frac{1}{4}a^2$.}{§~372}
+
+\eq[\indent Therefore,]{$OC$}{$= \sqrt{R^2 - \frac{1}{4}a^2}$,}{}
+
+\eq[and]{$DC$}{$= R - \sqrt{R^2 - \frac{1}{4}a^2}$.}{}
+
+\step{The $\angle DAH$ is a rt.~$\angle$.}{§~290}
+
+In the rt.~$\triangle DAH$, \( \overline{AD}^2 = DH × DC \).~\hfill§~367
+
+But $DH = 2R$, and \( DC = R - \sqrt{R^2 - \frac{1}{4}a^2} \).
+
+\eq{$\therefore AD$}{$= \sqrt{2R(R - \sqrt{R^2 - \frac{1}{4}a^2})}$}{}
+
+\eq{}{$= \sqrt{R(2R - \sqrt{4R^2 - a^2})}$.}{\qef}
+
+\end{proof}
+
+\pp{\cor{If $R=1$, $AD =
+  \sqrt{2-\sqrt{4-a^2}}$.}}
+\scanpage{242.png}%
+
+\proposition{Problem.}
+
+\begin{proof}%
+\obs{To find the numerical value of the ratio of the
+  circumference of a circle to its diameter.}
+
+\figc{242aa480}{Let $C$ be the circumference, when the radius is unity.}
+
+\prove[To find ]{the numerical value of $\pi$.}
+
+By §~458, $2\pi R = C$. \qquad $\therefore \pi = \frac{1}{2}C$ when
+$R = 1$.
+
+Let $S_6$ be the length of a side of a regular polygon of $6$~sides,
+$S_{12}$ of $12$~sides, and so on.
+
+If $R=1$, by §~469, $S_6=1$ and by §~479 we have
+\[
+\begin{array}{lcc}
+\multicolumn{1}{c}{\text{\footnotesize Form of Computation.}} &
+\text{\footnotesize Length of Side.} &
+\text{\footnotesize Length of Perimeter.} \\
+%
+S_{12} = \sqrt{2 - \sqrt{4 - 1^2}} &
+0.51763809 &
+6.21165708 \\
+%
+S_{24} = \sqrt{2 - \sqrt{4 - (0.51763809)^2}} &
+0.26105238 &
+6.26525722 \\
+%
+S_{48} = \sqrt{2 - \sqrt{4 - (0.26105238)^2}} &
+0.13080626 &
+6.27870041 \\
+%
+S_{96} = \sqrt{2 - \sqrt{4 - (0.13080626)^2}} &
+0.06543817 &
+6.28206396 \\
+%
+S_{192} = \sqrt{2 - \sqrt{4 - (0.06543817)^2}} &
+0.03272346 &
+6.28290510 \\
+%
+S_{384} = \sqrt{2 - \sqrt{4 - (0.03272346)^2}} &
+0.01636228 &
+6.28311544 \\
+%
+S_{768} = \sqrt{2 - \sqrt{4 - (0.01636228)^2}} &
+0.00818121 &
+6.28316941 \\
+\end{array}
+\]
+$\therefore C = 6.28317$ approximately; that is, $\pi = 3.14159$
+nearly.
+\hfill\qef
+
+\end{proof}
+
+\begin{point}%
+\textsc{Scholium.~}$\pi$ is incommensurable.  We
+generally take
+\[ \pi = 3.1416, \text{ and } \frac{1}{\pi} = 0.31831. \]
+\end{point}
+\scanpage{243.png}%
+
+
+\section{MAXIMA AND MINIMA.}
+
+\begin{point}%
+\defn{Among geometrical magnitudes which satisfy
+given conditions, the \emph{greatest} is called the \indexbf{maximum}; and
+the \emph{smallest} is called the \indexbf{minimum}.}
+
+Thus, the diameter of a circle is the maximum among all chords; and
+the perpendicular is the minimum among all lines drawn to a given line
+from a given external point.
+\end{point}
+
+\pp{\defn{\indexbf{Isoperimetric} polygons are polygons which have
+equal \newline perimeters.}}
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of all triangles having two given sides, that in
+which these sides include a right angle is the maximum.}
+
+\figc{243aa484}{Let the triangles $ABC$ and $EBC$ have the sides $AB$ and $BC$ equal
+to $EB$ and $BC$, respectively; and let the angle $ABC$ be a right angle.}
+
+\proveq{$\triangle ABC$}{$> \triangle EBC$.}
+
+\step[\indent\textbf{Proof.}]{From $E$ draw the altitude $ED$.}{}
+
+The $\triangle_s ABC$ and $EBC$, having the same base, $BC$, are to
+each other as their altitudes $AB$ and $ED$.~\hfill§~405
+
+\eq[\indent Now]{$EB$}{$> ED$.}{§~97}
+
+\eq[\indent But]{$EB$}{$= AB$.}{Hyp.}
+
+\eq{$\therefore AB$}{$> ED$.}{}
+
+\eq{$\therefore \triangle ABC$}{$> \triangle EBC$.}{§~405}
+
+\hfill\qed
+
+\end{proof}
+\scanpage{244.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of all isoperimetric triangles having the same
+  base the isosceles triangle is the maximum.}
+
+\figc{244aa485}{Let the $\triangle_s ACB$ and $ADB$ have equal perimeters, and
+  let $AC$ and $CB$ be equal, and $AD$ and $DB$ be unequal.}
+
+\prove {$\triangle ACB > \triangle ADB$.}
+
+\step [\indent Proof.] {Produce $AC$ to $H$, making $CH = AC$; and draw $HB$.} {}
+
+\step {Produce $HB$, take $DP$ equal to $DB$, and draw $AP$.} {}
+
+\step {Draw $CE$ and $DF \perp$ to $AB$, and $CK$ and $DM \parallel$ to $AB$.} {}
+
+\step {The $\angle ABH$ is a right $\angle$, for it may be inscribed in the
+semicircle whose centre is $C$ and radius $CA$.} {§~290}
+
+$ADP$ is not a straight line, for then the $\angle_s DBA$ and $DAB$
+would be equal, being complements of the equal $\angle_s DBM$ and
+$DPM$, respectively; and $DA$ and $DB$ would be equal (§~147), which
+is contrary to the hypothesis.  Hence,
+
+\step {$AP < AD + DP$, $\therefore < AD+DB$, $\therefore < AC+CB$,
+$\therefore < AH$.} {}
+
+\eq {$\therefore BH $} {$> BP$.} {§~102}
+
+\eq {$\therefore CE(=\frac{1}{2} BH) $} {$> DF (=\frac{1}{2}BP)$.} {Ax.~7}
+
+\eq [\indent Therefore,] {$\triangle ACB $} {$> \triangle ADB$.} {§~405}
+
+\step {} {\qed}
+
+\end {proof}
+\scanpage{245.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of all polygons with sides all given but one, the
+  maximum can be inscribed in a semicircle which has the undetermined
+  side for its diameter.}
+
+\figc{245aa486}{Let $ABCDE$ be the maximum of polygons with sides $AB$, $BC$,
+  $CD$, $DE$, and the extremities $A$ and $E$ on the straight line
+  $MN$.}
+
+\prove{$ABCDE$ can be inscribed in a semicircle.}
+
+\textbf{Proof.} From \emph{any} vertex, as $C$, draw $CA$ and $CE$.
+
+The $\triangle ACE$ must be the maximum of all $\triangle_s$ having
+the
+sides $CA$ and $CE$, and the third side on $MN$; otherwise by
+increasing or diminishing the $\angle ACE$, keeping the lengths of
+the sides $CA$ and $CE$ unchanged, but sliding the extremities
+$A$ and $E$ along the line $MN$, we could increase the $\triangle ACE$,
+while the rest of the polygon would remain unchanged; and
+therefore increase the polygon.  But this is contrary to the hypothesis that the polygon is the maximum polygon.
+
+Hence, the $\triangle ACE$ is the maximum of $\triangle_s$ that have
+the sides $CA$ and $CE$.
+
+
+\step{Therefore, the $\angle ACE$ is a right angle.}{§~484}
+
+\step{Therefore, $C$ lies on the semicircumference.}{§~290}
+
+Hence, \emph{every} vertex lies on the circumference; that is, the
+maximum polygon can be inscribed in a semicircle having the
+undetermined side for a diameter.~\hfill\qed
+
+\end{proof}
+\scanpage{246.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of all polygons with given sides, that which can
+  be inscribed in a circle is the maximum.}
+
+\figc{246aa487}{Let $ABCDE$ be a polygon inscribed in a circle, and
+  $A'B'C'D'E'$ be a polygon, equilateral with respect to $ABCDE$,
+  which cannot be inscribed in a circle.}
+
+\prove{that $ABCDE > A'B'C'D'E'$.}
+
+\step [\indent Proof.] {Draw the diameter $AH$, and draw $CH$ and $DH$.} {}
+
+\step {Upon $C'D'$ construct the $\triangle C'H'D' = \triangle CHD$, and draw
+$A'H'$.} {}
+
+Since, by hypothesis, a $\odot$ cannot pass through \emph{all} the
+vertices of $A'B'C'D'E'$, \emph{one} or \emph{both} of the parts
+$ABCH$, $AEDH$ must be greater than the corresponding part of
+$A'B'C'H'D'E'$.\hfill§~486
+
+If either of these parts is \emph{not greater than} its corresponding
+part, it is equal to it,\hfill§~486
+
+\pnote{(for $ABCH$ and $AEDH$ are the maxima of polygons that have
+  sides equal to $AB$, $BC$, $CH$, and $AE$, $ED$, $DH$, respectively,
+  and the remaining side undetermined).}
+
+\eq {$\therefore ABCHDE $} {$> A'B'C'H'D'E'$.} {Ax.~4}
+
+\step {Take away from the two figures the equal $\triangle_s CHD$ and
+$C'H'D'$.}{}
+
+\eq [\indent Then] {$ABCDE $} {$> A'B'C'D'E'$.} {Ax.~5}
+
+\step {} {\qed}
+
+\end {proof}
+\scanpage{247.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of isoperimetric polygons of the same number of
+  sides, the maximum is equilateral.}
+
+\figc{247aa488}{Let $ABCD$ etc.\ be the maximum of isoperimetric polygons of
+  any given number of sides.}
+
+\prove{$AB$, $BC$, $CD$, etc., are equal.}
+
+\step[\indent\textbf{Proof.}]{Draw $AC$.}{}
+
+The $\triangle ABC$ must be the maximum of all the $\triangle_s$ which
+are formed upon $AC$ with a perimeter equal to that of $\triangle
+ABC$.
+
+Otherwise a greater $\triangle AKC$ could be substituted for
+$\triangle ABC$, without changing the perimeter of the polygon.
+
+But this is inconsistent with the hypothesis that the polygon $ABCD$
+etc.\ is the maximum polygon.
+
+
+\step{$\therefore$ the $\triangle ABC$ is isosceles.}{§~485}
+
+\step{$\therefore AB = BC$.}{}
+
+In like manner it may be proved that $BC=CD$, etc.~\hfill\qed
+
+\end{proof}
+
+\begin{point}%
+\cor{The maximum of isoperimetric
+  polygons of the same number of sides is a regular polygon.}
+
+For the maximum polygon is equilateral (§~488), and can be inscribed
+in a circle (§~487), and is, therefore, regular.~\hfill§~430
+\end{point}
+\scanpage{248.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of isoperimetric regular polygons, that which has
+  the greatest number of sides is the maximum.}
+
+\figc{248aa490}{Let $Q$ be a regular polygon of three sides, and $Q'$ a
+  regular polygon of four sides, and let the two polygons have equal
+  perimeters.}
+
+\prove{$Q'$ is greater than $Q$.}
+
+\textbf{Proof.} Draw $CD$ from $C$ to any point in $AB$.
+
+Invert the $\triangle CDA$ and place it in the position $DCE$, letting
+$D$ fall at $C$, $C$ at $D$, and $A$ at $E$.
+
+The polygon $DBCE$ is an irregular polygon of four sides, which by
+construction has the same perimeter as $Q'$, and the same area as $Q$.
+
+Then the irregular polygon $DBCE$ of four sides is less than the
+isoperimetric regular polygon $Q'$ of four sides.~\hfill§~489
+
+In like manner it may be shown that $Q'$ is less than an isoperimetric
+regular polygon of five sides, and so on.~\hfill\qed
+
+\end{proof}
+
+\ex{Of all equivalent parallelograms that have equal
+bases, the rectangle has the minimum perimeter.}
+
+\ex{Of all equivalent rectangles, the square has the
+minimum perimeter.}
+
+\ex{Of all triangles that have the same base and the
+same altitude, the isosceles has the minimum perimeter.}
+
+\ex{Of all triangles that can be inscribed in a given
+circle, the equilateral is the maximum and has the maximum perimeter.}
+\scanpage{249.png}%
+
+\proposition{Theorem.}
+
+\begin{proof}%
+\obs{Of regular polygons having a given area, that
+  which has the greatest number of sides has the least perimeter.}
+
+\figc{249aa491}{Let $Q$ and $Q'$ be regular polygons having the same area, and
+  let $Q'$ have the greater number of sides.}
+
+\proveq[\indent To prove ]{the perimeter of $Q$}{$>$ the perimeter of $Q'$.}
+
+\textbf{Proof.} Let $Q''$ be a regular polygon having the same
+perimeter as $Q'$, and the same number of sides as $Q$.
+
+\eq[\indent Then]{$Q'$}{$> Q''$}{§~490}
+
+\pnote{(of isoperimetric regular polygons, that which has the greatest
+number of sides is the maximum).}
+
+\eq[\indent But]{$Q$}{$\Bumpeq Q'$.}{Hyp.}
+
+\eq{$\therefore Q$}{$> Q''$.}{}
+
+\eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q''$.}{}
+
+\eq{But the perimeter of $Q'$}{$=$ the perimeter of $Q''$.}{Hyp.}
+
+\eq{$\therefore$ the perimeter of $Q$}{$>$ the perimeter of $Q'$.}{\qed}
+
+
+\end{proof}
+
+
+
+\ex{To inscribe in a semicircle the maximum rectangle.}
+
+\ex{Of all polygons of a given number of sides which may
+be inscribed in a given circle, that which is regular has the maximum
+area and the maximum perimeter.}
+
+\ex{Of all polygons of a given number of sides which may
+be circumscribed about a given circle, that which is regular has the
+minimum area and the minimum perimeter.
+}
+\scanpage{250.png}%
+
+
+\section[EXERCISES.]{THEOREMS.}
+
+\ex{Every equilateral polygon circumscribed about a circle is
+regular if it has an \emph{odd} number of sides.}
+
+\ex{Every equiangular polygon inscribed in a circle is regular if
+it has an \emph{odd} number of sides.}
+
+\ex{Every equiangular polygon circumscribed about a circle is
+regular.}
+
+\ex{The side of a circumscribed equilateral triangle is equal to
+twice the side of the similar inscribed triangle.}
+
+\ex{The apothem of an inscribed regular hexagon is equal to half
+the side of the inscribed equilateral triangle.}
+
+\ex{The area of an inscribed regular hexagon is three fourths of
+the area of the circumscribed regular hexagon.}
+
+\ex{The area of an inscribed regular hexagon is the mean proportional
+between the areas of the inscribed and the circumscribed equilateral
+triangles.}
+
+\ex{The square of the side of an inscribed equilateral triangle is
+equal to three times the square of a side of the inscribed regular hexagon.}
+
+\ex{The area of an inscribed equilateral triangle is equal to half
+the area of the inscribed regular hexagon.}
+
+\ex{The square of the side of an inscribed equilateral triangle is
+equal to the sum of the squares of the sides of the inscribed square and of
+the inscribed regular hexagon.}
+
+\ex{The square of the side of an inscribed regular pentagon is
+equal to the sum of the squares of the radius of the circle and the side of
+the inscribed regular decagon.}
+
+\exheader{If $R$ denotes the radius of a circle, and $a$ one side of an inscribed regular
+polygon, show that:}
+
+\ex{In a regular pentagon, $a = \frac{1}{2} R \sqrt{10 - 2 \sqrt{5}}$.}
+
+\ex{In a regular octagon, $a = R \sqrt{2 - \sqrt{2}}$.}
+
+\ex{In a regular dodecagon, $a = R \sqrt{2 - \sqrt{3}}$.}
+
+\ex{If two diagonals of a regular pentagon intersect, the longer
+segment of each is equal to a side of the pentagon.}
+\scanpage{251.png}%
+
+\ex{The apothem of an inscribed regular pentagon is equal to half
+the sum of the radius of the circle and the side of the inscribed regular
+decagon.}
+
+\ex{The side of an inscribed regular pentagon is equal to the
+hypotenuse of the right triangle which has for legs the radius of the
+circle and the side of the inscribed regular decagon.}
+
+\ex{The radius of an inscribed regular polygon is the mean proportional
+between its apothem and the radius of the similar circumscribed
+regular polygon.}
+
+\ex{If squares are constructed outwardly upon the six sides of a
+regular hexagon, the exterior vertices of these squares are the vertices of
+a regular dodecagon.}
+
+\ex{If the alternate vertices of a regular hexagon are joined by
+straight lines, show that another regular hexagon is thereby formed.
+Find the ratio of the areas of these two hexagons.}
+
+\ex{If on the legs of a right triangle as diameters semicircles are
+described external to the triangle, and from the whole figure a semicircle
+on the hypotenuse is subtracted, the remaining figure is equivalent to the
+given right triangle.}
+
+\ex{The star-shaped polygon, formed by producing the sides of a
+regular hexagon, is equivalent to twice the given hexagon.}
+
+\ex{The sum of the perpendiculars drawn to the sides of a regular
+polygon from any point within the polygon is equal to the apothem multiplied
+by the number of sides.}
+
+\ex{If two chords of a circle are perpendicular to each other, the
+sum of the four circles described on the four segments as diameters is
+equivalent to the given circle.}
+
+\ex{If the diameter of a circle is divided into any two segments,
+and upon these segments as diameters semicircumferences are described
+upon opposite sides of the diameter, these semicircumferences divide the
+circle into two parts which have the same ratio as the two segments of
+the diameter.}
+
+\ex{The diagonals that join any vertex of a regular polygon to
+all the vertices not adjacent divide the angle at that vertex into as many
+equal parts less two as the polygon has sides.}
+\scanpage{252.png}%
+
+
+\subsection{PROBLEMS OF CONSTRUCTION.}
+
+\ex{To circumscribe an equilateral triangle about a given circle.}
+
+\ex{To circumscribe a square about a given circle.}
+
+\ex{To circumscribe a regular hexagon about a given circle.}
+
+\ex{To circumscribe a regular octagon about a given circle.}
+
+\ex{To circumscribe a regular pentagon about a given circle.}
+
+\ex{To draw through a given point a line so as to divide a given
+circumference into two parts having the ratio $3:7$.}
+
+\ex{To construct a circumference equal to the sum of two given
+circumferences.}
+
+\ex{To construct a circumference equal to the difference of two
+given circumferences.}
+
+\ex{To construct a circle equivalent to the sum of two given
+circles.}
+
+\ex{To construct a circle equivalent to the difference of two given
+circles.}
+
+\ex{To construct a circle equivalent to three times a given circle.}
+
+\ex{To construct a circle equivalent to three fourths of a given
+circle.}
+
+\ex{To construct a circle whose ratio to a given circle shall be
+equal to the given ratio $m:n$.}
+
+\ex{To divide a given circle by a concentric circumference into
+two equivalent parts.}
+
+\ex{To divide a given circle by concentric circumferences into five
+equivalent parts.}
+
+\ex{To construct an angle of~$18°$; of~$36°$; of~$9°$.}
+
+\ex{To construct an angle of$12°$; of~$24°$; of~$6°$.}
+
+\exheader{To construct with a side of a given length:}
+
+\ex{An equilateral triangle.}
+
+\ex{A square.}
+
+\ex{A regular hexagon.}
+
+\ex{A regular octagon.}
+
+\ex{A regular pentagon.}
+
+\ex{A regular decagon.}
+
+\ex{A regular dodecagon.}
+
+\ex{A regular pentedecagon.}
+\scanpage{253.png}%
+
+
+\subsection{PROBLEMS OF COMPUTATION.}
+
+\ex{Find the area of a circle whose radius is $12$ inches.}
+
+\ex{Find the circumference and the area of a circle whose diameter
+is $8$ feet.}
+
+\ex{A regular pentagon is inscribed in a circle whose radius is $R$.
+If the length of a side is $a$, find the apothem.}
+
+\ex{A regular polygon is inscribed in a circle whose radius is $R$.
+If the length of a side is $a$, show that the apothem is $\frac{1}{2} \sqrt{R^2 - a^2}$.}
+
+\ex{Find the area of a regular decagon inscribed in a circle whose
+radius is $16$ inches.}
+
+\ex{Find the side of a regular dodecagon inscribed in a circle
+whose radius is $20$ inches.}
+
+\ex{Find the perimeter of a regular pentagon inscribed in a circle
+whose radius is $25$ feet.}
+
+\ex{The length of each side of a park in the shape of a regular
+decagon is $100$ yards. Find the area of the park.}
+
+\ex{Find the cost, at $\$2$ per yard, of building a wall around a
+cemetery in the shape of a regular hexagon, that contains $16,627.84$ square
+yards.}
+
+\ex{The side of an inscribed regular polygon of $n$ sides is $16$ feet.
+Find the side of an inscribed regular polygon of $2n$ sides.}
+
+\ex{If the radius of a circle is $R$, and the side of an inscribed
+regular polygon is $a$, show that the side of the similar circumscribed regular polygon is
+$\dfrac{2aR}{\sqrt{4R^2-a^2}}$.}
+
+\ex{What is the width of the circular ring between two concentric
+circumferences whose lengths are $650$ feet and $425$ feet?}
+
+\ex{Find the angle subtended at the centre by an arc $5$ feet $10$
+inches long, if the radius of the circle is $9$ feet $4$ inches.}
+
+\ex{The chord of a segment is $10$ feet, and the radius of the circle
+is $16$ feet. Find the area of the segment.}
+
+\ex{Find the area of a sector, if the angle at the centre is $20°$, and
+the radius of the circle is $20$ inches.}
+\scanpage{254.png}%
+
+\ex{The chord of half an arc is $12$~feet, and the radius of the
+circle is $18$~feet. Find the height of the segment subtended by the whole
+arc.}
+
+\ex{Find the side of a square which is equivalent to a circle whose
+diameter is $35$~feet.}
+
+\ex{The diameter of a circle is $15$~feet. Find the diameter of a
+circle twice as large. Three times as large.}
+
+\ex{Find the radii of the concentric circumferences that divide a
+circle $11$~inches in diameter into five equivalent parts.}
+
+\ex{The perimeter of a regular hexagon is $840$ feet, and that of a
+regular octagon is the same. By how many square feet is the octagon
+larger than the hexagon?}
+
+\ex{The diameter of a bicycle wheel is $28$~inches. How many
+revolutions does the wheel make in going $10$~miles?}
+
+\ex{Find the diameter of a carriage wheel that makes $264$~revolutions
+in going half a mile.}
+
+\ex{The sides of three regular octagons are $6$~feet, $7$~feet, $8$~feet,
+respectively. Find the side of a regular octagon equivalent to the sum of
+the three given octagons.}
+
+\ex{A circular pond $100$ yards in diameter is surrounded by a walk
+$10$~feet wide. Find the area of the walk.}
+
+\ex{The span (chord) of a bridge in the form of a circular arc is
+$120$~feet, and the highest point of the arch is $15$~feet above the piers. Find
+the radius of the arc.}
+
+\ex{Three equal circles are described each tangent to the other
+two. If the common radius is~$R$, find the area contained between the
+circles.}
+
+\ex{Given $p$, $P$, the perimeters of regular polygons of $n$~sides
+inscribed in and circumscribed about a given circle. Find $p'$, $P'$, the
+perimeters of regular polygons of $2n$~sides inscribed in and circumscribed
+about the given circle.}
+
+\ex{Given the radius $R$, and the apothem $r$ of an inscribed regular
+polygon of $n$ sides. Find the radius $R'$ and the apothem $r'$ of an isoperimetrical
+regular polygon of $2n$ sides.}
+\scanpage{255.png}%
+
+
+\subsection{MISCELLANEOUS EXERCISES.}
+
+\subsection{THEOREMS.}
+
+\ex{If two adjacent angles of a quadrilateral are right angles, the
+bisectors of the other two angles are perpendicular.}
+
+\ex{If two opposite angles of a quadrilateral are right angles, the
+bisectors of the other two angles are parallel.}
+
+\ex{The two lines that join the middle points of the opposite sides
+of a quadrilateral bisect each other.}
+
+\ex{The line that joins the feet of the perpendiculars dropped from
+the extremities of the base of an isosceles triangle to the opposite sides is
+parallel to the base.}
+
+\ex{If $AD$ bisects the angle $A$ of a triangle $ABC$, and $BD$ bisects
+the exterior angle $CBF$, then angle $ADB$ equals one half angle $ACB$.}
+
+\ex{The sum of the acute angles at the vertices of a pentagram\label{pentagram}
+(five-pointed star) is equal to two right angles.}
+
+\begin{proofex}%
+The altitudes $AD$, $BE$, $CF$ of the triangle $ABC$ bisect the
+angles of the triangle $DEF$.
+
+Circles with $AB$, $BC$, $AC$ as diameters will pass through $E$ and $D$, $E$
+and $F$, $D$ and $F$, respectively.
+
+\end{proofex}
+
+\ex{The segments of any straight line intercepted between the
+circumferences of two concentric circles are equal.}
+
+\ex{If a circle is circumscribed about any triangle, the feet of the
+perpendiculars dropped from any point in the circumference to the sides
+of the triangle lie in one straight line.}
+
+\ex{Two circles are tangent internally at $P$, and a chord $AB$ of
+the larger circle touches the smaller circle at $C$. Prove that $PC$ bisects
+the angle $APB$.}
+
+\ex{The diagonals of a trapezoid divide each other into segments
+which are proportional.}
+
+\ex{If through a point $P$ in the circumference of a circle two
+chords are drawn, the chords and the segments between $P$ and a chord
+parallel to the tangent at $P$ are reciprocally proportional.}
+\scanpage{256.png}%
+
+\ex{The perpendiculars from two vertices of a triangle upon the
+opposite sides divide each other into segments reciprocally proportional.}
+
+\ex{The perpendicular from any point of a circumference upon a
+chord is the mean proportional between the perpendiculars from the same
+point upon the tangents drawn at the extremities of the chord.}
+
+\ex{In an isosceles right triangle either leg is the mean proportional
+between the hypotenuse and the perpendicular upon it from the
+vertex of the right angle.}
+
+\ex{If two circles intersect in the points $A$ and $B$, and through $A$
+any secant $CAD$ is drawn limited by the circumferences at $C$ and $D$, the
+straight lines $BC$, $BD$ are to each other as the diameters of the circles.}
+
+\ex{The area of a triangle is equal to half the product of its perimeter
+by the radius of the inscribed circle.}
+
+\ex{The perimeter of a triangle is to one side as the perpendicular
+from the opposite vertex is to the radius of the inscribed circle.}
+
+\begin{proofex}%
+If three straight lines $AA'$, $BB'$, $CC'$, drawn from the vertices
+of a triangle $ABC$ to the opposite sides, pass through a common point $O$
+within the triangle, then
+
+\step{\( \dfrac{OA'}{AA'} + \dfrac{OB'}{BB'} + \dfrac{OC'}{CC'} = 1 \).}{}
+
+\end{proofex}
+
+\ex{$ABC$ is a triangle, $M$ the middle point of $AB$, $P$ any point
+in $AB$ between $A$ and $M$. If $MD$ is drawn parallel to $PC$, meeting $BC$
+at $D$, the triangle $BPD$ is equivalent to half the triangle $ABC$.}
+
+\ex{Two diagonals of a regular pentagon, not drawn from a common
+vertex, divide each other in extreme and mean ratio.}
+
+\ex{If all the diagonals of a regular pentagon are drawn, another
+regular pentagon is thereby formed.}
+
+\ex{The area of an inscribed regular dodecagon is equal to three
+times the square of the radius.}
+
+\ex{The area of a square inscribed in a semicircle is equal to two
+fifths the area of the square inscribed in the circle.}
+
+\ex{The area of a circle is greater than the area of any polygon
+of equal perimeter.}
+
+\ex{The circumference of a circle is less than the perimeter of any
+polygon of equal area.}
+\scanpage{257.png}%
+
+
+\subsection{PROBLEMS OF LOCI.}
+
+\ex{Find the locus of the centre of the circle inscribed in a triangle
+that has a given base and a given angle at the vertex.}
+
+\ex{Find the locus of the intersection of the altitudes of a triangle
+that has a given base and a given angle at the vertex.}
+
+\ex{Find the locus of the extremity of a tangent to a given circle,
+if the length of the tangent is equal to a given line.}
+
+\ex{Find the locus of a point, tangents drawn from which to a
+given circle form a given angle.}
+
+\ex{Find the locus of the middle point of a line drawn from a
+given point to a given straight line.}
+
+\ex{Find the locus of the vertex of a triangle that has a given
+base and a given altitude.}
+
+\ex{Find the locus of a point the sum of whose distances from
+two given parallel lines is equal to a given length.}
+
+\ex{Find the locus of a point the difference of whose distances
+from two given parallel lines is equal to a given length.}
+
+\ex{Find the locus of a point the sum of whose distances from two
+given intersecting lines is equal to a given length.}
+
+\ex{Find the locus of a point the difference of whose distances
+from two given intersecting lines is equal to a given length.}
+
+\ex{Find the locus of a point whose distances from two given
+points are in the given ratio $m:n$.}
+
+\ex{Find the locus of a point whose distances from two given
+parallel lines are in the given ratio $m:n$.}
+
+\ex{Find the locus of a point whose distances from two given
+intersecting lines are in the given ratio $m:n$.}
+
+\ex{Find the locus of a point the sum of the squares of whose
+distances from two given points is constant.}
+
+\ex{Find the locus of a point the difference of the squares of whose
+distances from two given points is constant.}
+
+\ex{Find the locus of the vertex of a triangle that has a given base
+and the other two sides in the given ratio $m:n$.
+}
+\scanpage{258.png}%
+
+
+\subsection{PROBLEMS OF CONSTRUCTION.}
+
+\ex{To divide a given trapezoid into two equivalent parts by a
+line parallel to the bases.}
+
+\ex{To divide a given trapezoid into two equivalent parts by a
+line through a given point in one of the bases.}
+
+\ex{To construct a regular pentagon, given one of the diagonals.}
+
+\ex{To divide a given straight line into two segments such that
+their product shall be the maximum.}
+
+\ex{To find a point in a semicircumference such that the sum of
+its distances from the extremities of the diameter shall be the maximum.}
+
+\ex{To draw a common secant to two given circles exterior to
+each other such that the intercepted chords shall have the given lengths
+$a$, $b$.}
+
+\ex{To draw through one of the points of intersection of two
+intersecting circles a common secant which shall have a given length.}
+
+\ex{To construct an isosceles triangle, given the altitude and one
+of the equal base angles.}
+
+\ex{To construct an equilateral triangle, given the altitude.}
+
+\ex{To construct a right triangle, given the radius of the inscribed
+circle and the difference of the acute angles.}
+
+\ex{To construct an equilateral triangle so that its vertices shall
+lie in three given parallel lines.}
+
+\ex{To draw a line from a given point to a given straight line
+which shall be to the perpendicular from the given point as $m : n$.}
+
+\ex{To find a point within a given triangle such that the perpendiculars
+from the point to the three sides shall be as the numbers $m$, $n$, $p$.}
+
+\ex{To draw a straight line equidistant from three given points.}
+
+\ex{To draw a tangent to a given circle such that the segment
+intercepted between the point of contact and a given straight line shall
+have a given length.}
+
+\ex{To inscribe a straight line of a given length between two given
+circumferences and parallel to a given straight line.
+}
+\scanpage{259.png}%
+
+\ex{To draw through a given point a straight line so that its distances
+from two other given points shall be in a given ratio.}
+
+\ex{To construct a square equivalent to the sum of a given triangle
+and a given parallelogram.}
+
+\ex{To construct a rectangle having the difference of its base and
+altitude equal to a given line, and its area equivalent to the sum of a given
+triangle and a given pentagon.}
+
+\ex{To construct a pentagon similar to a given pentagon and
+equivalent to a given trapezoid.}
+
+\ex{To find a point whose distances from three given straight
+lines shall be as the numbers $m$, $n$, $p$.}
+
+\ex{Given an angle and two points $P$ and $P'$ between the sides of
+the angle. To find the shortest path from $P$ to $P'$ that shall touch both
+sides of the angle.}
+
+\ex{To construct a triangle, given its angles and its area.}
+
+\ex{To transform a given triangle into a triangle similar to
+another given triangle.}
+
+\ex{Given three points $A$, $B$, $C$. To find a fourth point $P$ such
+that the areas of the triangles $APB$, $APC$, $BPC$ shall be equal.}
+
+\ex{To construct a triangle, given its base, the ratio of the other
+sides, and the angle included by them.}
+
+\ex{To divide a given circle into $n$ equivalent parts by concentric
+circumferences.}
+
+\ex{In a given equilateral triangle to inscribe three equal circles
+tangent to each other, each circle tangent to two sides of the triangle.}
+
+\ex{Given an angle and a point $P$ between the sides of the angle.
+To draw through $P$ a straight line that shall form with the sides of the
+angle a triangle with the perimeter equal to a given length $a$.}
+
+\ex{In a given square to inscribe four equal circles, so that each
+circle shall be tangent to two of the others and also tangent to two sides
+of the square.}
+
+\ex{In a given square to inscribe four equal circles, so that each
+circle shall be tangent to two of the others and also tangent to one side
+of the square.}
+\scanpage{260.png}%
+
+
+\chapter{TABLE OF FORMULAS.}
+\markboth{\Headings{PLANE GEOMETRY.}}{\Headings{TABLE OF FORMULAS.}}%
+
+\subsection{PLANE FIGURES.}
+
+\subsection{NOTATION.}
+
+\begin{tabular}{r@{~}c@{~}l}
+$P$ &=& perimeter. \\
+$h$ &=& altitude. \\
+$b$ &=& lower base. \\
+$b'$ &=& upper base. \\
+$R$ &=& radius of circle. \\
+$D$ &=& diameter of circle. \\
+$C$ &=& circumference of circle. \\
+$r$ &=& apothem of regular polygon. \\
+$a$, $b$, $c$ &=& sides of triangle. \\
+$s$ &=& \( \frac{1}{2}(a+b+c) \). \\
+$p$ &=& perpendicular of triangle. \\
+$m,n$ &=& segments of third side of triangle adjacent to \\
+&& sides $b$ and $a$, respectively. \\
+$S$ &=& area. \\
+$\pi$ &=& 3.1416.
+\end{tabular}
+
+
+\newpage
+\subsection{FORMULAS.}
+
+\noindent\begin{supertabular}{lr@{~}c@{~}l@{\qquad}r}
+
+\multicolumn{5}{l}{\hspace{-2ex}\textbf{Line Values.}} \\
+
+\multicolumn{5}{r}{\tiny PAGE}\\
+Right triangle, &
+  $b^2$ &=& $c × m$; $a^2 = c × n$ & \pageref{160} \\
+& $p^2$ &=& $m × n$ & \pageref{160} \\
+& $b^2:a^2$ &=& $m:n$ & \pageref{161} \\
+& $b^2:c^2$ &::& $m:c$ & \pageref{161} \\
+& $a^2+b^2$ &=& $c^2$ & \pageref{162} \\
+\scanpage{261.png}%
+
+Any triangle, &
+$a^2$ &=& $b^2+c^2 \pm 2c × m$ & \llap{\pageref{163},}\pageref{164} \\
+
+\multicolumn{4}{l}{Altitude of triangle on side $a$,} \\
+& $h$ &=& \( \displaystyle \frac{2}{a}
+             \sqrt{s(s-a)(s-b)(s-c)} \) & \pageref{formtrialtitude} \\
+
+\multicolumn{4}{l}{Median of triangle on side $a$,} \\
+& $m$ &=& \( \frac{1}{2} \sqrt{2(b^2+c^2) - a^2} \) & \pageref{formtrimedian} \\
+
+\multicolumn{4}{l}{Bisector of triangle on side $a$,} \\
+& $t$ &=& \( \displaystyle \frac{2}{b+c}
+             \sqrt{bcs(s-a)} \) & \pageref{formtribisector} \\
+
+\multicolumn{4}{l}{Radius of circumscribed circle,} \\
+& $R$ &=& \( \displaystyle \frac{abc}{4\sqrt{s(s-a)(s-b)(s-c)}} \) & \pageref{formradcircum} \\
+
+Circumference of circle, & $C$ &=& $2\pi R$ & \pageref{formcircum} \\
+\qquad\DittoMark\qquad\qquad\DittoMark & $C$ &=& $\pi D$ & \pageref{formcircum} \\
+
+\multicolumn{5}{l}{\hspace{-2ex}\textbf{Areas.}} \\
+
+Rectangle,     & $S$ &=& $b × h$ & \pageref{formarearect} \\
+Square,        & $S$ &=& $b^2$ & \pageref{formarearect} \\
+Parallelogram, & $S$ &=& $b × h$ & \pageref{formareapar} \\
+Triangle,      & $S$ &=& $\frac{1}{2}b × h$ & \pageref{formareatri} \\
+\qquad\DittoMark       & $S$ &=& $\sqrt{s(s-a)(s-b)(s-c)}$ & \pageref{formareatri2} \\
+\qquad\DittoMark       & $S$ &=& \( \displaystyle \frac{abc}{4R} \) & \pageref{formareatri3} \\
+Equilateral triangle, & $S$ &=&
+   \( \displaystyle \frac{a^2}{4}\sqrt{3} \) & \pageref{formareaequitri} \\
+Trapezoid,     & $S$ &=& $\frac{1}{2}h(b+b')$ & \pageref{formareatrap} \\
+Regular polygon, & $S$ &=& $\frac{1}{2}r × P$ & \pageref{formareapoly} \\
+Circle,        & $S$ &=& $\frac{1}{2}R × C$ & \pageref{formareacircle} \\
+\qquad\DittoMark       & $S$ &=& $\pi R^2$ & \pageref{formareacircle} \\
+Sector,        & $S$ &=& $\frac{1}{2}R × \arc$ & \pageref{formareasector} \\
+\end{supertabular}
+\scanpage{262.png}%
+
+
+\twocolumn
+\chapter{INDEX.}
+\markboth{INDEX.}{INDEX.}
+
+\small
+\noindent\begin{supertabular}{lr}
+
+\multicolumn{2}{r}{\tiny PAGE} \\
+\tablehead{\multicolumn{2}{r}{\tiny PAGE} \\}%
+
+Abbreviations & \pageref{abbr} \\
+Alternation & \pageref{alternation} \\
+Altitude of parallelogram & \pageref{altpar} \\
+\Ditto of trapezoid & \pageref{alttrap} \\
+\Ditto of triangle & \pageref{alttri} \\
+Analysis & \pageref{analysis} \\
+Angle & \pageref{angle} \\
+\Ditto acute & \pageref{acute} \\
+\multicolumn{2}{l}{\Ditto at centre of}\\
+\quad\quad\quad regular polygon & \pageref{anglecentreregpoly} \\
+\Ditto central & \pageref{central} \\
+\Ditto exterior of triangle & \pageref{exteriortri} \\
+\Ditto inscribed in circle & \pageref{inscribedcirc} \\
+\Ditto inscribed in segment & \pageref{inscribedseg} \\
+\Ditto oblique & \pageref{oblique} \\
+\Ditto obtuse & \pageref{obtuse} \\
+\Ditto reflex & \pageref{reflex} \\
+\Ditto right & \pageref{right} \\
+\Ditto salient & \pageref{salient} \\
+\Ditto straight & \pageref{straight} \\
+\Ditto vertical & \pageref{vertical angle} \\
+Angles, adjacent & \pageref{adjacent1},\pageref{adjacent2} \\
+\Ditto alternate-exterior & \pageref{altext} \\
+\Ditto alternate-interior & \pageref{altint} \\
+\Ditto complementary & \pageref{complementary} \\
+\Ditto conjugate & \pageref{conjugate angles} \\
+\Ditto exterior & \pageref{exterior} \\
+\Ditto exterior-interior & \pageref{extint} \\
+\Ditto interior & \pageref{interior} \\
+\Ditto supplementary & \pageref{supplementary} \\
+\Ditto supplementary-adjacent & \pageref{suppladj} \\
+\Ditto vertical & \pageref{vertical angles} \\
+Antecedents & \pageref{antecedents} \\
+Apothem & \pageref{apothem} \\
+Arc & \pageref{arc} \\
+Area & \pageref{area} \\
+Axiom & \pageref{axiom} \\
+\Ditto of parallel lines & \pageref{axiomparallel} \\
+Axioms of straight lines & \pageref{axiomstraight} \\
+\Ditto general & \pageref{generalaxioms} \\
+Axis of symmetry & \pageref{axissym} \\
+\\
+\textbf{B}ase of isosceles triangle & \pageref{baseiso} \\
+\Ditto of parallelogram & \pageref{basepar} \\
+\Ditto of triangle & \pageref{basetri} \\
+Bases of trapezoid & \pageref{basetrap} \\
+Bisector & \pageref{bisector} \\
+\\
+\textbf{C}entre of circle & \pageref{centrecirc} \\
+\Ditto of regular polygon & \pageref{centrepoly} \\
+\Ditto of symmetry & \pageref{centresym} \\
+Chord & \pageref{chord} \\
+Circle & \pageref{circle} \\
+\Ditto circumscribed & \pageref{circcircumscribed} \\
+\Ditto inscribed & \pageref{circinscribed} \\
+Circles, concentric & \pageref{concentric} \\
+\Ditto escribed & \pageref{escribed} \\
+Circum-centre of triangle & \pageref{circum-centre} \\
+Circumference & \pageref{circumference} \\
+Commensurable & \pageref{commensurable} \\
+Complement & \pageref{complement} \\
+Composition & \pageref{composition} \\
+Conclusion & \pageref{conclusion} \\
+Concurrent lines & \pageref{concurrent} \\
+Congruent figures & \pageref{congruent} \\
+Consequents & \pageref{consequents} \\
+Constant & \pageref{constant} \\
+Construction & \pageref{construction} \\
+Continued proportion & \pageref{continuedprop} \\
+Continuity, Principle of & \pageref{princcont} \\
+Contradictory of a theorem & \pageref{contradictory} \\
+Converse of a theorem & \pageref{converse1},\pageref{converse2} \\
+Convex curve & \pageref{convexcurve} \\
+Curved surface & \pageref{curvedsurf} \\
+\\
+\textbf{D}ecagon & \pageref{decagon} \\
+Diagonal & \pageref{diagonal1},\pageref{diagonal2} \\
+Diameter & \pageref{diameter} \\
+Dimensions & \pageref{dimensions} \\
+Distance & \pageref{distance1},\pageref{distance2} \\
+Division & \pageref{division} \\
+Dodecagon & \pageref{dodecagon} \\
+Duality, Principle of & \pageref{princduality} \\
+\\
+\textbf{E}qual figures & \pageref{equal} \\
+Equimultiples & \pageref{equimultiples} \\
+Equivalent figures & \pageref{equivalent1},\pageref{equivalent2} \\
+Ex-centres of triangle & \pageref{ex-centres} \\
+Extreme and mean ratio & \pageref{extrememean} \\
+Extremes & \pageref{extremes} \\
+\\
+\textbf{F}igure, curvilinear & \pageref{curvilinear} \\
+\Ditto geometrical & \pageref{geometrical figure} \\
+\Ditto plane & \pageref{plane figure} \\
+\Ditto rectilinear & \pageref{rectilinear} \\
+Foot of perpendicular & \pageref{foot} \\
+Fourth proportional & \pageref{fourth} \\
+\\
+\textbf{G}eometrical solid & \pageref{geometrical1},\pageref{geometrical2} \\
+Geometry & \pageref{Geometry} \\
+Geometry, Plane & \pageref{Plane Geometry} \\
+\Ditto Solid & \pageref{Solid Geometry} \\
+\\
+\textbf{H}armonic division & \pageref{divided harmonically} \\
+Heptagon & \pageref{heptagon} \\
+Hexagon & \pageref{hexagon} \\
+Homologous angles & \pageref{homologous angles},\pageref{homangles} \\
+\Ditto lines & \pageref{Homologous lines} \\
+\Ditto sides & \pageref{homologous sides},\pageref{homsides} \\
+Hypotenuse & \pageref{hypotenuse} \\
+Hypothesis & \pageref{hypothesis} \\
+\\
+\textbf{I}n-centre of triangle & \pageref{in-centre} \\
+Incommensurable ratio & \pageref{incommensurable ratio} \\
+Intersection & \pageref{intersection} \\
+Inversion & \pageref{inversion} \\
+Isoperimetric figures & \pageref{Isoperimetric} \\
+\\
+\textbf{L}egs of right triangle & \pageref{legs} \\
+\Ditto of trapezoid & \pageref{legstrap} \\
+Limit & \pageref{limit} \\
+Line & \pageref{line},\pageref{line2},\pageref{line3} \\
+\Ditto curved & \pageref{curved line} \\
+\Ditto of centres & \pageref{line of centres} \\
+\Ditto straight & \pageref{straight line} \\
+Lines, oblique & \pageref{oblique lines} \\
+\Ditto parallel & \pageref{parallel lines} \\
+\Ditto perpendicular & \pageref{perpendicular} \\
+\\
+\textbf{M}ajor arc & \pageref{major} \\
+Maximum & \pageref{maximum} \\
+Mean proportional & \pageref{mean proportional} \\
+Means & \pageref{means} \\
+Median of trapezoid & \pageref{mediantrap} \\
+Minimum & \pageref{minimum} \\
+Minor arc & \pageref{minor} \\
+\\
+\textbf{N}egative quantities & \pageref{negative} \\
+Numerical measure & \pageref{numerical measure} \\
+\\
+\textbf{O}ctagon & \pageref{octagon} \\
+Opposite of a theorem & \pageref{opposite} \\
+Origin & \pageref{origin} \\
+\\
+\textbf{P}arallel lines & \pageref{parallel lines} \\
+Parallelogram & \pageref{parallelogram} \\
+Pentagon & \pageref{pentagon} \\
+Pentagram & \pageref{pentagram} \\
+Perigon & \pageref{perigon} \\
+Perimeter & \pageref{perimeter},\pageref{perimeter2} \\
+Perpendicular bisector & \pageref{perpbisector} \\
+Perpendicular lines & \pageref{perpendicular} \\
+Pi ($\pi$) & \pageref{pi} \\
+Plane & \pageref{planes},\pageref{plane} \\
+Point & \pageref{point},\pageref{point2} \\
+\Ditto of contact & \pageref{point of contact} \\
+\Ditto of tangency & \pageref{point of tangency} \\
+Polygon & \pageref{polygon} \\
+\Ditto angles of & \pageref{polyangles} \\
+\Ditto circumscribed & \pageref{polycircumscribed} \\
+\Ditto concave & \pageref{concave polygon} \\
+\Ditto convex & \pageref{convex polygon} \\
+\Ditto equiangular & \pageref{equiangular polygon} \\
+\Ditto equilateral & \pageref{equilateral polygon} \\
+\Ditto inscribed & \pageref{polyinscribed} \\
+\Ditto regular & \pageref{regular polygon} \\
+Polygons mut.\ equiangular & \pageref{mutually equiangular} \\
+\Ditto mutually equilateral & \pageref{mutually equilateral} \\
+Positive quantities & \pageref{positive} \\
+Postulate & \pageref{postulate} \\
+Projection & \pageref{projection} \\
+Proof & \pageref{proof} \\
+Proportion & \pageref{proportion} \\
+Proposition & \pageref{proposition} \\
+\\
+\textbf{Q}uadrant & \pageref{quadrant} \\
+Quadrilateral & \pageref{quadrilateral},\pageref{quadrilateral2} \\
+Radius of regular polygon & \pageref{polyradius} \\
+Ratio & \pageref{ratio} \\
+Ratio of similitude & \pageref{ratio of similitude} \\
+Reciprocity, Principle of & \pageref{princreciprocity} \\
+Rectangle & \pageref{rectangle} \\
+Rhomboid & \pageref{rhomboid} \\
+Rhombus & \pageref{rhombus} \\
+\\
+\textbf{S}cholium & \pageref{scholium} \\
+Secant & \pageref{secant},\pageref{secant2} \\
+Sector & \pageref{sector} \\
+Segment of circle & \pageref{segment} \\
+\Ditto of line & \pageref{lineseg} \\
+Semicircle & \pageref{semicircle} \\
+Semicircumference & \pageref{semicircumference} \\
+Sides of an angle & \pageref{anglesides} \\
+\Ditto of polygon & \pageref{polysides} \\
+\Ditto of triangle & \pageref{trisides} \\
+Similar arcs & \pageref{similar arcs} \\
+\Ditto figures & \pageref{similar} \\
+\Ditto polygons & \pageref{Similar polygons} \\
+\Ditto sectors & \pageref{similar sectors} \\
+\Ditto segments & \pageref{similar segments} \\
+\Ditto triangles & \pageref{similar triangles} \\% not sure about this ref
+Square & \pageref{square} \\
+Superposition & \pageref{superposition} \\
+Supplement & \pageref{supplement} \\
+Surface & \pageref{surface},\pageref{surface2},\pageref{surface3} \\
+Symbols & \pageref{symbols} \\
+Symmetry & \pageref{symmetry} \\
+\\
+\textbf{T}angent & \pageref{tangent},\pageref{tangent2} \\
+\Ditto common external & \pageref{common external tangent} \\
+\Ditto common internal & \pageref{common internal tangent} \\
+Terms of a proportion & \pageref{terms} \\
+Theorem & \pageref{theorem} \\
+Third proportional & \pageref{third} \\
+Transversal & \pageref{transversal} \\
+Trapezium & \pageref{trapezium} \\
+Trapezoid & \pageref{trapezoid} \\
+\Ditto isosceles & \pageref{isosceles trapezoid} \\
+Triangle & \pageref{triangle},\pageref{triangle2} \\
+\Ditto equiangular & \pageref{equiangular triangle} \\
+\Ditto equilateral & \pageref{equilateral triangle} \\
+\Ditto isosceles & \pageref{isosceles triangle} \\
+\Ditto obtuse & \pageref{obtuse triangle} \\
+\Ditto right & \pageref{right triangle} \\
+\Ditto scalene & \pageref{scalene triangle} \\
+\Ditto altitudes of & \pageref{alttri} \\
+\Ditto angles of & \pageref{anglestri} \\
+\Ditto bisectors of & \pageref{tribisectors} \\
+\Ditto medians of & \pageref{trimedians} \\
+\Ditto vertices of & \pageref{trivertices} \\
+\\
+\textbf{V}ariable & \pageref{variable} \\
+Vertex of angle & \pageref{vertex} \\
+\Ditto of triangle & \pageref{trivertex} \\
+Vertices of polygon & \pageref{polyvertices} \\
+
+\end{supertabular}
+
+\onecolumn
+
+%%%%%%%%%%%%%%%%%%%%%%%%% GUTENBERG LICENSE %%%%%%%%%%%%%%%%%%%%%%%%%%
+
+\cleardoublepage
+
+\backmatter
+\phantomsection
+\pdfbookmark[-1]{BACK MATTER.}{BACK MATTER}
+\phantomsection
+\pdfbookmark[0]{PG LICENSE.}{LICENSE}
+\fancyhead[C]{\Headings{LICENSE.}}
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+% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %
+%                                                                         %
+% End of Project Gutenberg's Plane Geometry, by George Albert Wentworth   %
+%                                                                         %
+% *** END OF THIS PROJECT GUTENBERG EBOOK PLANE GEOMETRY ***              %
+%                                                                         %
+% ***** This file should be named 33063-t.tex or 33063-t.zip *****        %
+% This and all associated files of various formats will be found in:      %
+%         http://www.gutenberg.org/3/3/0/6/33063/                         %
+%                                                                         %
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diff --git a/33063-t/images/245aa486.pdf b/33063-t/images/245aa486.pdf
new file mode 100644
index 0000000..6eacb39
--- /dev/null
+++ b/33063-t/images/245aa486.pdf
diff --git a/33063-t/images/246aa487.pdf b/33063-t/images/246aa487.pdf
new file mode 100644
index 0000000..2cb35ee
--- /dev/null
+++ b/33063-t/images/246aa487.pdf
diff --git a/33063-t/images/247aa488.pdf b/33063-t/images/247aa488.pdf
new file mode 100644
index 0000000..d2b6b58
--- /dev/null
+++ b/33063-t/images/247aa488.pdf
diff --git a/33063-t/images/248aa490.pdf b/33063-t/images/248aa490.pdf
new file mode 100644
index 0000000..4eff8cd
--- /dev/null
+++ b/33063-t/images/248aa490.pdf
diff --git a/33063-t/images/249aa491.pdf b/33063-t/images/249aa491.pdf
new file mode 100644
index 0000000..db3d3d7
--- /dev/null
+++ b/33063-t/images/249aa491.pdf
diff --git a/33063-t/images/sources/012aaZ10.eepic b/33063-t/images/sources/012aaZ10.eepic
new file mode 100644
index 0000000..34b3ef3
--- /dev/null
+++ b/33063-t/images/sources/012aaZ10.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 012aaZ10.tex
+% Pt.10. Fig.2. (12, 7 )
+\PGset[0.8em]
+\begin{picture} (12, 7 )
+\drawline       ( 1  , 2  )( 3  , 6  )(11  , 6  )( 9  , 2  )( 1  , 2  )  %BCDFB
+\put(  5.5 ,  3.7 ){$\scriptstyle A$}   %A
+\put(  0.2 ,  1.2 ){$\scriptstyle B$}   %B
+\put(  2.7 ,  6.2 ){$\scriptstyle C$}   %C
+\put( 11.0 ,  6.2 ){$\scriptstyle D$}   %D
+\put(  8.8 ,  1.2 ){$\scriptstyle F$}   %F
+\put(  4   ,  0.2 ){{\footnotesize \textsc{Fig. 2.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/012bbZ14.eepic b/33063-t/images/sources/012bbZ14.eepic
new file mode 100644
index 0000000..5d24cc8
--- /dev/null
+++ b/33063-t/images/sources/012bbZ14.eepic
@@ -0,0 +1,23 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 012bbZ14.tex
+% Pt.14. Fig.3. (14,10 )
+\PGset[0.8em]
+\begin{picture} (14,10 )
+\drawline       ( 1  , 7  )( 1  , 2  )( 2  , 4 )( 2  , 9  )( 1  , 7  )  %ABCDA
+\drawline       (12  , 7  )(12  , 2  )(13  , 4 )(13  , 9  )(12  , 7  )  %EFGHE
+\dashline[80]{0.2}  ( 1  , 7  )(12  , 7  )  %AE
+\dashline[80]{0.2}  ( 1  , 2  )(12  , 2  )  %BF
+\dashline[80]{0.2}  ( 2  , 4  )(13  , 4  )  %CG
+\dashline[80]{0.2}  ( 2  , 9  )(13  , 9  )  %DH
+\put(  0   ,  6.6 ){$\scriptstyle A$}   %A
+\put(  0   ,  1.6 ){$\scriptstyle B$}   %B
+\put(  2.2 ,  4.2 ){$\scriptstyle C$}   %C
+\put(  1.5 ,  9.2 ){$\scriptstyle D$}   %D
+\put( 11.2 ,  7.2 ){$\scriptstyle E$}   %E
+\put( 12.2 ,  1.6 ){$\scriptstyle F$}   %F
+\put( 13.2 ,  3.5 ){$\scriptstyle G$}   %G
+\put( 12.5 ,  9.2 ){$\scriptstyle H$}   %H
+\put(  5   ,  0.2 ){{\footnotesize \textsc{Fig. 3.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/016aaZ37.eepic b/33063-t/images/sources/016aaZ37.eepic
new file mode 100644
index 0000000..e8896f5
--- /dev/null
+++ b/33063-t/images/sources/016aaZ37.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 016aaZ37.tex
+% Pt.37. Fig.4. (12, 9 )
+\PGset[0.8em]
+\begin{picture} (12, 9 )
+\drawline       ( 1  , 8   )             (11  , 8   )  %AB
+\qbezier        ( 1  , 4.5 )( 6  , 9.5  )(11  , 4.5 )  %arc CD
+\drawline       ( 1  , 2   )( 6  , 4    )(11  , 2   )  %E()F
+\put( 0.1   ,  8   ){$\scriptstyle A$}   %A
+\put(11.1   ,  8   ){$\scriptstyle B$}   %B
+\put( 0.1   ,  4.3 ){$\scriptstyle C$}   %C
+\put(  11   ,  4.5 ){$\scriptstyle D$}   %D
+\put( 0.1   ,  1.8 ){$\scriptstyle E$}   %E
+\put(11.1   ,  1.8 ){$\scriptstyle F$}   %F
+\put(   4.5 ,  0.2 ){{\footnotesize \textsc{Fig. 4.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/018aaZ55.eepic b/33063-t/images/sources/018aaZ55.eepic
new file mode 100644
index 0000000..c19199b
--- /dev/null
+++ b/33063-t/images/sources/018aaZ55.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 018aaZ55.tex
+% Pt.55. Fig.5. (19, 4 )
+\PGset[0.8em]
+\begin{picture} (18, 4 )
+\drawline(1,3)(17,3)
+\drawline(9,2.7)(9,3.3)
+\drawline(2.5,2.375)(1,3)(2.5,3.625)
+\drawline(15.5,2.375)(17,3)(15.5,3.625)
+\put( 0.1   ,  2.7 ){$\scriptstyle A$}   %A
+\put(17.1   ,  2.7 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  1.6 ){$\scriptstyle C$}   %C
+\put( 7.5   ,  0.2 ){{\footnotesize \textsc{Fig. 5.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/018bbZ56.eepic b/33063-t/images/sources/018bbZ56.eepic
new file mode 100644
index 0000000..ea793df
--- /dev/null
+++ b/33063-t/images/sources/018bbZ56.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 018bbZ56.tex
+% Pt.56. Fig.6. (22, 4 )
+\PGset[0.7em]
+\begin{picture} (22, 4 )
+\put( 1     ,  2   ){\line  ( 1, 0 ){20  }}  %A(BCD)E
+\put( 1     ,  1.7 ){\line  ( 0, 1 ){0.6 }}  %tick A
+\put( 6     ,  1.7 ){\line  ( 0, 1 ){0.6 }}  %tick B
+\put(11     ,  1.7 ){\line  ( 0, 1 ){0.6 }}  %tick C
+\put(16     ,  1.7 ){\line  ( 0, 1 ){0.6 }}  %tick D
+\put(21     ,  1.7 ){\line  ( 0, 1 ){0.6 }}  %tick E
+\put( 0.6   ,  2.5 ){$\scriptstyle A$}   %A
+\put( 5.6   ,  2.5 ){$\scriptstyle B$}   %B
+\put(10.6   ,  2.5 ){$\scriptstyle C$}   %C
+\put(15.6   ,  2.5 ){$\scriptstyle D$}   %D
+\put(20.6   ,  2.5 ){$\scriptstyle E$}   %E
+\put( 9.5   ,  0.2 ){{\footnotesize \textsc{Fig. 6.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/018ccZ57.eepic b/33063-t/images/sources/018ccZ57.eepic
new file mode 100644
index 0000000..6816c2b
--- /dev/null
+++ b/33063-t/images/sources/018ccZ57.eepic
@@ -0,0 +1,13 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 018ccZ57.tex
+% Pt.57. Fig.7. ( 9, 9 )
+\PGset[0.8em]
+\begin{picture} ( 9, 9 )
+\put( 1     ,  2   ){\line  ( 1, 0 ){7   }}  %ED
+\put( 1     ,  2   ){\line  ( 1, 2 ){3   }}  %EF
+\put( 8.1   ,  1.7 ){$\scriptstyle D$}   %D
+\put( 0.1   ,  1.7 ){$\scriptstyle E$}   %E
+\put( 4.1   ,  7.7 ){$\scriptstyle F$}   %F
+\put( 3     ,  0.2 ){{\footnotesize \textsc{Fig. 7.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/019aaZ58.eepic b/33063-t/images/sources/019aaZ58.eepic
new file mode 100644
index 0000000..db3db32
--- /dev/null
+++ b/33063-t/images/sources/019aaZ58.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 019aaZ58.tex
+% Pt.58. Fig.8. ( 8, 9 )
+\PGset[0.8em]
+\begin{picture} ( 8, 9 )
+\drawline( 2,  7   )( 4, 3 )( 6, 7 )         %CAB
+\drawline(1,3)(7,3) % EF
+\drawline(4,3)(4,7) % AD
+\put( 3.6   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 5.7   ,  7.1 ){$\scriptstyle B$}   %B
+\put( 1.6   ,  7.1 ){$\scriptstyle C$}   %C
+\put( 3.6   ,  7.1 ){$\scriptstyle D$}   %D
+\put( 0.6   ,  2.2 ){$\scriptstyle E$}   %E
+\put( 6.6   ,  2.2 ){$\scriptstyle F$}   %F
+\put( 2.5   ,  1   ){{\footnotesize \textsc{Fig. 8.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/019abZ58.eepic b/33063-t/images/sources/019abZ58.eepic
new file mode 100644
index 0000000..1d33179
--- /dev/null
+++ b/33063-t/images/sources/019abZ58.eepic
@@ -0,0 +1,31 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 019aaZ58.tex
+% Pt.58. Fig.8. ( 8, 9 )
+\PGset[0.6em]
+\begin{picture} ( 8, 8 )
+\drawline( 2,  7   )( 4, 3 )( 6, 7 )         %CAB
+\put( 1     ,  3   ){\line  ( 1, 0 ){6   }}  %E(A)F
+\put( 4     ,  3   ){\line  ( 0, 1 ){4   }}  %AD
+\put( 3.6   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 5.7   ,  7.1 ){$\scriptstyle B$}   %B
+\put( 1.6   ,  7.1 ){$\scriptstyle C$}   %C
+\put( 3.6   ,  7.1 ){$\scriptstyle D$}   %D
+\put( 0.6   ,  2.2 ){$\scriptstyle E$}   %E
+\put( 6.6   ,  2.2 ){$\scriptstyle F$}   %F
+\put( 2.5   ,  1   ){{\footnotesize \textsc{Fig. 8.}}}
+\end{picture}
+\newline
+\begin{picture} (11, 8 )%
+\drawline( 2.5,  6 )( 5.5, 3 )( 8.5, 6 )     %vee shape
+\put( 1     ,  3   ){\line  ( 1, 0 ){9   }}  %AB
+\put( 5.5   ,  3   ){\line  ( 0, 1 ){4   }}  %vertical
+\put( 0.8   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 9.0   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 6.5   ,  3.2 ){\emph{a}}   %a
+\put( 5.8   ,  4.2 ){\emph{b}}   %b
+\put( 4.5   ,  4.2 ){\emph{c}}   %c
+\put( 3.5   ,  3.2 ){\emph{d}}   %d
+\put( 4     ,  1   ){{\footnotesize \textsc{Fig. 9.}}}
+\end{picture}%
+\PGrestore
diff --git a/33063-t/images/sources/019bbZ58.eepic b/33063-t/images/sources/019bbZ58.eepic
new file mode 100644
index 0000000..98de3c8
--- /dev/null
+++ b/33063-t/images/sources/019bbZ58.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 019bbZ58.tex
+% Pt.58. Fig.9. (11, 8 )
+\PGset[0.8em]%
+\begin{picture} (11, 9 )%
+\drawline( 2.5,  6 )( 5.5, 3 )( 8.5, 6 )     %vee shape
+\drawline(1,3)(10,3) % AB
+\drawline(5.5,3)(5.5,7) % vertical
+\put( 0.8   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 9.0   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 6.5   ,  3.2 ){\emph{a}}   %a
+\put( 5.8   ,  4.2 ){\emph{b}}   %b
+\put( 4.5   ,  4.2 ){\emph{c}}   %c
+\put( 3.5   ,  3.2 ){\emph{d}}   %d
+\put( 4     ,  1   ){{\footnotesize \textsc{Fig. 9.}}}
+\end{picture}%
+\PGrestore%
diff --git a/33063-t/images/sources/019ccZ62.eepic b/33063-t/images/sources/019ccZ62.eepic
new file mode 100644
index 0000000..69646b9
--- /dev/null
+++ b/33063-t/images/sources/019ccZ62.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 019ccZ62.tex
+% Pt.62. Fig.10. (11, 8 )
+\PGset[0.8em]
+\begin{picture}  (11, 8 )
+\put( 1     ,  3   ){\line  ( 1, 0 ){9   }}  %A(O)B
+\put( 5.5   ,  3   ){\line  ( 5, 6 ){3.5 }}  %OD
+\put( 0.8   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 9.5   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 8.8   ,  7.4 ){$\scriptstyle D$}   %D
+\put( 5.1   ,  2.2 ){$\scriptstyle O$}   %O
+\put( 3.5   ,  1   ){{\footnotesize \textsc{Fig. 10.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/020aaZ63.eepic b/33063-t/images/sources/020aaZ63.eepic
new file mode 100644
index 0000000..db3dc38
--- /dev/null
+++ b/33063-t/images/sources/020aaZ63.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 020aaZ63.tex
+% Pt.63. Fig.11. (10, 9 )
+\PGset[0.8em]
+\begin{picture}  (10, 9 )
+\put( 1     ,  3   ){\line  ( 1, 0 ){8   }}  %A(C)B
+\put( 5     ,  3   ){\line  ( 0, 1 ){5   }}  %CD
+\put( 0.8   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 8.5   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 4.6   ,  2.2 ){$\scriptstyle C$}   %C
+\put( 4.6   ,  8.2 ){$\scriptstyle D$}   %D
+\put( 3     ,  1   ){{\footnotesize \textsc{Fig. 11.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/020bbZ66.eepic b/33063-t/images/sources/020bbZ66.eepic
new file mode 100644
index 0000000..a0a0d8b
--- /dev/null
+++ b/33063-t/images/sources/020bbZ66.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 018aaZ55.tex
+% Pt.55. Fig.5. (19, 4 )
+\PGset[0.8em]
+\begin{picture} (18, 4 )
+\drawline(1,3)(17,3)
+\drawline(9,2.7)(9,3.3)
+\drawline(2.5,2.375)(1,3)(2.5,3.625)
+\drawline(15.5,2.375)(17,3)(15.5,3.625)
+\put( 0.1   ,  2.7 ){$\scriptstyle A$}   %A
+\put(17.1   ,  2.7 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  1.6 ){$\scriptstyle C$}   %C
+\put( 7.5   ,  0.2 ){{\footnotesize \textsc{Fig. 12.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/020ccZ68.eepic b/33063-t/images/sources/020ccZ68.eepic
new file mode 100644
index 0000000..f388f50
--- /dev/null
+++ b/33063-t/images/sources/020ccZ68.eepic
@@ -0,0 +1,12 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 020ccZ68.tex
+% Pt.68. Fig.13. ( 9, 9 )
+\vspace{-2ex}
+\PGset[0.8em]
+\begin{picture}  ( 9, 6 )
+\drawline( 8 , 5.5 )( 1, 2 )( 8, 2 )  %ur A lr
+\put( 0.1   ,  1.7 ){$\scriptstyle A$}   %A
+\put( 2.5   ,  0.2 ){{\footnotesize \textsc{Fig. 13.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/020ddZ69.eepic b/33063-t/images/sources/020ddZ69.eepic
new file mode 100644
index 0000000..e22fe6f
--- /dev/null
+++ b/33063-t/images/sources/020ddZ69.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 020ddZ69.tex
+% Pt.68. Fig.14. (12, 6 )
+\PGset[0.8em]
+\begin{picture}  (12, 6 )
+\drawline( 1 , 5 )( 3, 3 )(11, 3 )  %DOA
+% begin rough solution with  epic
+\qbezier    (2.3, 3.7 )( 3.8, 5 )(4, 3 )        %obtuse arc
+\qbezier[20](1.6, 4.4 )( 1.3,-1 )(5.1, 3 )      %reflex arc
+\drawline( 2.3 , 4.2 )( 2.3, 3.7 )( 2.8, 3.7 )  %obtuse arrow
+\drawline( 4.4 , 2.7 )( 5  , 3   )( 5  , 2.5 )  %reflex arrow
+% end   rough solution with  epic
+\put(11     ,  2.7 ){$\scriptstyle A$}   %A
+\put( 0.5   ,  5.2 ){$\scriptstyle D$}   %D
+\put( 2.6   ,  2.2 ){$\scriptstyle O$}   %O
+\put( 5     ,  0.2 ){{\footnotesize \textsc{Fig. 14.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/021aaZ72.eepic b/33063-t/images/sources/021aaZ72.eepic
new file mode 100644
index 0000000..781b1cf
--- /dev/null
+++ b/33063-t/images/sources/021aaZ72.eepic
@@ -0,0 +1,30 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 021aaZ72.tex
+% Pt.72. Fig.15. (11,13 )
+\PGset[0.8em]
+\begin{picture}  (11,13 )
+\drawline( 2  , 11   )( 5.5, 7.5 )( 9.5, 9.5 )  %DOC
+\drawline(1,7.5)(10,7.5) % A'A
+\drawline(5.5,3)(5.5,12) % B'B
+
+% Ellipse:  u = 5.5  v = 7.5  a = 2.5  b = 2.5  phi = 0.0 Grad
+\qbezier[15](8.0, 7.5)(8.0, 8.5355)(7.2678, 9.2678)
+\qbezier[15](7.2678, 9.2678)(6.5355, 10.0)(5.5, 10.0)
+\qbezier[15](5.5, 10.0)(4.4645, 10.0)(3.7322, 9.2678)
+\qbezier[15](3.7322, 9.2678)(3.0, 8.5355)(3.0, 7.5)
+\qbezier[15](3.0, 7.5)(3.0, 6.4645)(3.7322, 5.7322)
+\qbezier[15](3.7322, 5.7322)(4.4645, 5.0)(5.5, 5.0)
+
+\drawline( 5   , 4.7 )( 5.5, 5   )( 5  , 5.4 )   %reflex arrow in Q3
+% end   rough solution
+\put(10.2   ,  7.4 ){$\scriptstyle A$}   %A
+\put( 5.1   , 12.2 ){$\scriptstyle B$}   %B
+\put( 9.6   ,  9.3 ){$\scriptstyle C$}   %C
+\put( 1.4   , 11.1 ){$\scriptstyle D$}   %D
+\put( 5.6   ,  6.7 ){$\scriptstyle O$}   %O
+\put( 0.1   ,  7.4 ){$\scriptstyle A'$}  %A'
+\put( 5     ,  2.2 ){$\scriptstyle B'$}  %B'
+\put( 3.5   ,  1   ){{\footnotesize \textsc{Fig. 15.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/022aaZ75.eepic b/33063-t/images/sources/022aaZ75.eepic
new file mode 100644
index 0000000..f23d38b
--- /dev/null
+++ b/33063-t/images/sources/022aaZ75.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 022aaZ75.tex
+% Pt.75. Fig.16. ( 8, 5 )
+\PGset[0.8em]
+\begin{picture}  ( 8, 8 )
+\put( 1     ,  3.5 ){\line  ( 2, 1 ){6   }}  %ll-ur
+\put( 1     ,  6.5 ){\line  ( 2,-1 ){6   }}  %ul-lr
+\put( 2.2   ,  4.7 ){\emph{a}}   %a
+\put( 5     ,  4.55 ){\emph{b}}   %b
+\put( 3.7   ,  5.4 ){\emph{c}}   %c
+\put( 3.5   ,  3.8 ){\emph{d}}   %d
+\put( 2.2   ,  1.0 ){{\footnotesize \textsc{Fig. 16.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/022bbZ76.eepic b/33063-t/images/sources/022bbZ76.eepic
new file mode 100644
index 0000000..cdb196c
--- /dev/null
+++ b/33063-t/images/sources/022bbZ76.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 022bbZ76.tex
+% Pt.76. Fig.17. ( 6, 8 )
+\PGset[0.8em]
+\begin{picture}  ( 6, 8 )
+\drawline( 1  ,  7 )( 1  , 3 )( 5  , 3 )     %COB
+\drawline(1,3)(3,6.4) % OD
+\put( 4.8   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 0.2   ,  7   ){$\scriptstyle C$}   %C
+\put( 3     ,  6.8 ){$\scriptstyle D$}   %D
+\put( 0.5   ,  2.2 ){$\scriptstyle O$}   %O
+\put( 1     ,  1   ){{\footnotesize \textsc{Fig. 17.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/022ccZ77.eepic b/33063-t/images/sources/022ccZ77.eepic
new file mode 100644
index 0000000..99a784b
--- /dev/null
+++ b/33063-t/images/sources/022ccZ77.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 022ccZ77.tex
+% Pt.77. Fig.18. (11, 8 )
+% *** Same as Pt.62. Fig.10. ***
+\PGset[0.8em]
+\begin{picture}  (11, 8 )
+\put( 1     ,  3   ){\line  ( 1, 0 ){9   }}  %A(O)B
+\put( 5.5   ,  3   ){\line  ( 5, 6 ){3.5 }}  %OD
+\put( 0.8   ,  2.2 ){$\scriptstyle A$}   %A
+\put( 9.5   ,  2.2 ){$\scriptstyle B$}   %B
+\put( 8.8   ,  7.4 ){$\scriptstyle D$}   %D
+\put( 5.1   ,  2.2 ){$\scriptstyle O$}   %O
+\put( 3.5   ,  1   ){{\footnotesize \textsc{Fig. 18.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/023aaZ79.eepic b/33063-t/images/sources/023aaZ79.eepic
new file mode 100644
index 0000000..57e7cdb
--- /dev/null
+++ b/33063-t/images/sources/023aaZ79.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 023aaZ79.tex
+% Pt.79. Fig.19. (20, 9 )
+\PGset[0.8em]
+\begin{picture}  (20, 9 )
+\drawline( 7,  6   )( 1, 2 )( 8  , 2 )         %FED
+\drawline(18,  6   )(12, 2 )(19  , 2 )         %CBA
+\dashline[80]{0.2}(15.8,  7.7   )(12, 2 )(18.2, 5.1 )         %HBG
+\put(19     ,  1.7 ){$\scriptstyle A$}   %A
+\put(11.1   ,  1.7 ){$\scriptstyle B$}   %B
+\put(18.1   ,  5.9 ){$\scriptstyle C$}   %C
+\put( 8     ,  1.7 ){$\scriptstyle D$}   %D
+\put( 0.2   ,  1.7 ){$\scriptstyle E$}   %E
+\put( 7     ,  5.7 ){$\scriptstyle F$}   %F
+\put(18.5   ,  4.9 ){$\scriptstyle G$}   %G
+\put(15.5   ,  7.9 ){$\scriptstyle H$}   %H
+\put( 8.1   ,  0.2 ){{\footnotesize \textsc{Fig. 19.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/023bcZ80.eepic b/33063-t/images/sources/023bcZ80.eepic
new file mode 100644
index 0000000..272aae5
--- /dev/null
+++ b/33063-t/images/sources/023bcZ80.eepic
@@ -0,0 +1,38 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 023bbZ80.tex
+% Pt.80. Fig.20. (32,10 )
+\PGset[0.8em]
+\begin{picture}  (32,10 )
+\drawline( 5,  9   )( 1, 2 )(10  , 2 )         %FED left
+\drawline(22,  9   )(22, 2 )(31  , 2 )         %CBA
+\drawline(14,  6   )(22, 2 )(25.8, 8.65 )         %HBF right
+\put(31     ,  1.8 ){$\scriptstyle A$}   %A
+\put(21.6   ,  1.2 ){$\scriptstyle B$}   %B
+\put(21.7   ,  9.2 ){$\scriptstyle C$}   %C
+\put(10.1   ,  1.9 ){$\scriptstyle D$}   %D
+\put( 0.2   ,  1.7 ){$\scriptstyle E$}   %E
+\put( 5     ,  8.8 ){$\scriptstyle F$}   %F left
+\put(25.7   ,  8.8 ){$\scriptstyle F$}   %F right
+\put(13.2   ,  6   ){$\scriptstyle H$}   %H
+\put(14     ,  0.2 ){{\footnotesize \textsc{Fig. 20.}}}
+\end{picture}
+\PGrestore
+%
+% ***  Uses  epic  package   ***
+% 023ccZ80.tex
+% Pt.80. Fig.21. ( 8, 9 )
+\PGset[0.8em]
+\begin{picture}  ( 8, 9 )
+\drawline( 1,  7.5 )( 1, 2 )( 7  , 2 )         %DBA
+\drawline( 3,  7   )( 1, 2 )( 6  , 4 )         %PBM
+\drawline(1,2)(5,6) % BC
+\put( 6.6   ,  1.2 ){$\scriptstyle A$}   %A
+\put( 0.6   ,  1.2 ){$\scriptstyle B$}   %B
+\put( 5.1   ,  6   ){$\scriptstyle C$}   %C
+\put( 0.6   ,  7.6 ){$\scriptstyle D$}   %D
+\put( 6     ,  3.8 ){$\scriptstyle M$}   %M
+\put( 2.8   ,  7.2 ){$\scriptstyle P$}   %P
+\put( 2     ,  0.2 ){{\footnotesize \textsc{Fig. 21.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/024aaZ81.eepic b/33063-t/images/sources/024aaZ81.eepic
new file mode 100644
index 0000000..ec5afe5
--- /dev/null
+++ b/33063-t/images/sources/024aaZ81.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 024aaZ81.tex
+% Pt.81. Fig.a. (18, 5 )
+\PGset[0.8em]
+\begin{picture} (18, 5 )
+\put( 1     ,  1   ){\line  ( 1, 0 ){16  }}    %D(E)F
+\put( 1     ,  4   ){\line  ( 1, 0 ){16  }}    %A(C)B
+\put( 0.2   ,  3.6 ){$\scriptstyle A$}   %A
+\put(17.1   ,  3.6 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  4.2 ){$\scriptstyle C$}   %C
+\put( 0.1   ,  0.7 ){$\scriptstyle D$}   %D
+\put( 8.5   ,  1.2 ){$\scriptstyle E$}   %E
+\put(17.1   ,  0.7 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/024bbZ83.eepic b/33063-t/images/sources/024bbZ83.eepic
new file mode 100644
index 0000000..622ee4f
--- /dev/null
+++ b/33063-t/images/sources/024bbZ83.eepic
@@ -0,0 +1,10 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 024bbZ83.tex
+% Pt.83. Fig.b. (10,11 )
+\PGset[0.8em]
+\begin{picture} (10,9 )
+\put( 1     ,  1   ){\line  ( 1, 0 ){ 8  }}    %horiz
+\put( 5     ,  1   ){\line  ( 0, 1 ){ 7  }}    %vert
+\put( 5     ,  1   ){\line  ( 1, 3 ){ 2  }}    %diag
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/025aaZ86.eepic b/33063-t/images/sources/025aaZ86.eepic
new file mode 100644
index 0000000..3ee750e
--- /dev/null
+++ b/33063-t/images/sources/025aaZ86.eepic
@@ -0,0 +1,13 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 025aaZ86.tex
+% Pt.86. Fig.a. (20, 9 )
+\PGset[0.8em]
+\begin{picture} (20, 9 )
+\put( 1     ,  1   ){\line  ( 1, 0 ){18   }}  %A(O)B
+\put(10     ,  1   ){\line  ( 5, 6 ){ 6   }}  %OD
+\put( 0.2   ,  1   ){$\scriptstyle A$}   %A
+\put(19.1   ,  1   ){$\scriptstyle B$}   %B
+\put(16     ,  7.9 ){$\scriptstyle D$}   %D
+\put( 9.6   ,  0.2 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/026aaZ90.eepic b/33063-t/images/sources/026aaZ90.eepic
new file mode 100644
index 0000000..7d90e83
--- /dev/null
+++ b/33063-t/images/sources/026aaZ90.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 026aaZ90.tex
+% Pt.90. Fig.a. (22,11 )
+\PGset[0.8em]
+\begin{picture} (22,9 )
+\drawline(1,1)(21,1) % ACB
+\drawline(11,1)(14,8.5) % CO
+\dashline[80]{0.2}(11, 1 )(21, 3)  %CF
+\put( 0.6   ,  0.2 ){$\scriptstyle A$}   %A
+\put(20.5   ,  0.2 ){$\scriptstyle B$}   %B
+\put(10.6   ,  0.2 ){$\scriptstyle C$}   %C
+\put(21     ,  2.7 ){$\scriptstyle F$}   %F
+\put(14     ,  8.1 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/027aaZ93.eepic b/33063-t/images/sources/027aaZ93.eepic
new file mode 100644
index 0000000..589e4f3
--- /dev/null
+++ b/33063-t/images/sources/027aaZ93.eepic
@@ -0,0 +1,14 @@
+%figshell.tex  REMOVE this line for actual diagram file
+% 027aaZ93.tex
+% Pt.93. Fig.a. (18,10 )
+\PGset[0.8em]
+\begin{picture} (18,10 )
+\put( 1     ,  1   ){\line  ( 2, 1 ){16   }}  %ACB
+\put( 1     ,  9   ){\line  ( 2,-1 ){16   }}  %OCP
+\put( 0.2   ,  0.8 ){$\scriptstyle A$}   %A
+\put(17     ,  8.7 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  4   ){$\scriptstyle C$}   %C
+\put(17.1   ,  0.6 ){$\scriptstyle P$}   %P
+\put( 0.2   ,  8.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/028aaZ95.eepic b/33063-t/images/sources/028aaZ95.eepic
new file mode 100644
index 0000000..85611ef
--- /dev/null
+++ b/33063-t/images/sources/028aaZ95.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 028aaZ95.tex
+% Pt.95. Fig.a.  (20,14 )
+\PGset[0.8em]
+\begin{picture}  (20,14 )
+\drawline(1,1)(19,1) % AB
+\drawline(10,1)(10,13) % FC
+\drawline( 4,  1   )(10,13 )(16  , 1 )         %ECK
+\put( 0.2   ,  0.7 ){$\scriptstyle A$}   %A
+\put(19.1   ,  0.6 ){$\scriptstyle B$}   %B
+\put( 9.7   , 13.2 ){$\scriptstyle C$}   %C
+\put( 3.5   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 9.5   ,  0.2 ){$\scriptstyle F$}   %F
+\put(15.5   ,  0.2 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/029aaZ96.eepic b/33063-t/images/sources/029aaZ96.eepic
new file mode 100644
index 0000000..97c6d1b
--- /dev/null
+++ b/33063-t/images/sources/029aaZ96.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 029aaZ96.tex
+% Pt.96. Fig.a.  (10,14 )
+\PGset[0.8em]
+\begin{picture}  (10,14 )
+\drawline(1,7)(9,7) % AB
+\drawline     ( 2,  7   )( 6,13 )( 6  , 7 )    %DPC
+\dashline[80]{0.2}( 2,  7   )( 6, 1 )( 6  , 7 )    %DP'C
+\put( 0.2   ,  7   ){$\scriptstyle A$}   %A
+\put( 9.1   ,  6.5 ){$\scriptstyle B$}   %B
+\put( 6.1   ,  6.2 ){$\scriptstyle C$}   %C
+\put( 1.2   ,  6.2 ){$\scriptstyle D$}   %D
+\put( 5.6   , 13.1 ){$\scriptstyle P$}   %P
+\put( 5.5   ,  0.2 ){$\scriptstyle P'$}  %P'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/030aaZ97.eepic b/33063-t/images/sources/030aaZ97.eepic
new file mode 100644
index 0000000..699ec8d
--- /dev/null
+++ b/33063-t/images/sources/030aaZ97.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 030aaZ97.tex
+% Pt.97. Fig.a.  (10,14 )
+\PGset[0.8em]
+\begin{picture}  (10,14 )
+\drawline(1,7)(9,7) % AB
+\drawline     ( 2,  7   )( 6,13 )( 6  , 7 )    %DPC
+\dashline[80]{0.2}( 2,  7   )( 6, 1 )( 6  , 7 )    %DP'C
+\put( 0.2   ,  7   ){$\scriptstyle A$}   %A
+\put( 9.1   ,  6.5 ){$\scriptstyle B$}   %B
+\put( 6.1   ,  6.2 ){$\scriptstyle C$}   %C
+\put( 1.2   ,  6.2 ){$\scriptstyle D$}   %D
+\put( 5.6   , 13.1 ){$\scriptstyle P$}   %P
+\put( 5.5   ,  0.2 ){$\scriptstyle P'$}  %P'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/031aa100.eepic b/33063-t/images/sources/031aa100.eepic
new file mode 100644
index 0000000..531c8c8
--- /dev/null
+++ b/33063-t/images/sources/031aa100.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 031aa100.tex
+% Pt.100. Fig.a.  (22,12 )
+\PGset[0.8em]
+\begin{picture}   (22,12 )
+\drawline( 1,  1   )(21, 1 )(11  ,11 )( 1  , 1 )  %ABCA
+\drawline( 1,  1   )(16, 6 )                      %A(O)E
+\drawline(21,  1   )(11, 4.333 )                      %BO
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put(20.5   ,  0.2 ){$\scriptstyle B$}   %B
+\put(10.7   , 11.1 ){$\scriptstyle C$}   %C
+\put(16.1   ,  5.9 ){$\scriptstyle E$}   %E
+\put(10.4   ,  4.45 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/032aa101.eepic b/33063-t/images/sources/032aa101.eepic
new file mode 100644
index 0000000..b7d25cc
--- /dev/null
+++ b/33063-t/images/sources/032aa101.eepic
@@ -0,0 +1,21 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 032aa101.tex
+% Pt.101. Fig.a.  (16,14 )
+\PGset[0.8em]
+\begin{picture}   (16,14 )
+\put( 1     ,  7   ){\line  ( 1, 0 ){14   }}      %A(EFCG)B
+\put( 9     ,  7   ){\line  ( 0, 1 ){ 6   }}      %CO
+\drawline     ( 3,  7   )( 9,13 )(12  , 7 )       %EOG
+\dashline[80]{0.2}( 3,  7   )( 9, 1 )( 9  , 7 )       %EDC
+\dashline[80]{0.2}( 9,  1   )( 6, 7 )( 9  ,13 )       %DFO
+\put( 0.2   ,  6.8 ){$\scriptstyle A$}   %A
+\put(15     ,  6.6 ){$\scriptstyle B$}   %B
+\put( 9.1   ,  6.2 ){$\scriptstyle C$}   %C
+\put( 8.6   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 2.2   ,  6.2 ){$\scriptstyle E$}   %E
+\put( 5.2   ,  6.2 ){$\scriptstyle F$}   %F
+\put(11.6   ,  6.2 ){$\scriptstyle G$}   %G
+\put( 8.7   , 13.2 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/033aa104.eepic b/33063-t/images/sources/033aa104.eepic
new file mode 100644
index 0000000..b775227
--- /dev/null
+++ b/33063-t/images/sources/033aa104.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 033aa104.tex
+% Pt.104. Fig.a.  (16,14 )
+\PGset[0.8em]
+\begin{picture}   (19,12 )
+\put( 1     ,  8   ){\line  ( 1, 0 ){17   }}      %AB
+\put( 1     ,  3   ){\line  ( 1, 0 ){17   }}      %CD
+\put( 1     ,  0.1 ){\line  ( 0, 1 ){10.8 }}      %vert w/CA
+\put( 0.1   ,  7.7 ){$\scriptstyle A$}   %A
+\put(18.1   ,  7.7 ){$\scriptstyle B$}   %B
+\put( 0.05   ,  2.7 ){$\scriptstyle C$}   %C
+\put(18.1   ,  2.7 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/034aa107.eepic b/33063-t/images/sources/034aa107.eepic
new file mode 100644
index 0000000..030e042
--- /dev/null
+++ b/33063-t/images/sources/034aa107.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 034aa107.tex
+% Pt.107. Fig.a.  (22,12 )
+\PGset[0.8em]
+\begin{picture}   (22,12 )
+\put( 1     ,  8   ){\line  ( 1, 0 ){20   }}      %A(O)B
+\put( 2     ,  4   ){\line  ( 1, 0 ){19   }}      %E(C)F
+\put(11     ,  1   ){\line  ( 0, 1 ){10   }}      %KCOH
+\dashline[80]{0.4}( 1,  6   )(21  , 2 )               %M(C)N
+\put( 0.1   ,  7.8 ){$\scriptstyle A$}   %A
+\put(21     ,  7.8 ){$\scriptstyle B$}   %B
+\put(11.2   ,  4.1 ){$\scriptstyle C$}   %C
+\put( 1     ,  3.7 ){$\scriptstyle E$}   %E
+\put(21.1   ,  3.7 ){$\scriptstyle F$}   %F
+\put(10.6   , 11.2 ){$\scriptstyle H$}   %H
+\put(10.5   ,  0.2 ){$\scriptstyle K$}   %K
+\put( 1     ,  5.6 ){$\scriptstyle M$}   %M
+\put(20     ,  1.2 ){$\scriptstyle N$}   %N
+\put(11.2   ,  8.2 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/034bb109.eepic b/33063-t/images/sources/034bb109.eepic
new file mode 100644
index 0000000..a6f5dfd
--- /dev/null
+++ b/33063-t/images/sources/034bb109.eepic
@@ -0,0 +1,24 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 034bb109.tex
+% Pt.109. Fig.b.  (18,14 )
+\PGset[0.8em]
+\begin{picture}   (18,14 )
+\put( 1     , 11   ){\line  ( 1, 0 ){16   }}      %AB
+\put( 1     ,  3   ){\line  ( 4, 1 ){16   }}      %CD
+\put( 7     ,  1   ){\line  ( 1, 3 ){ 4   }}      %FE
+\put( 0.1   , 10.7 ){$\scriptstyle A$}   %A
+\put(17     , 10.7 ){$\scriptstyle B$}   %B
+\put( 0.2   ,  2.6 ){$\scriptstyle C$}   %C
+\put(17     ,  6.7 ){$\scriptstyle D$}   %D
+\put(10.6   , 13.1 ){$\scriptstyle E$}   %E
+\put( 6.5   ,  0.1 ){$\scriptstyle F$}   %F
+\put( 9.1   , 10.1 ){\emph{a}}   %a
+\put( 9.6   , 11.2 ){\emph{b}}   %b
+\put(10.7   , 11.2 ){\emph{c}}   %c
+\put(10.2   ,  9.9 ){\emph{d}}   %d
+\put( 7.2   ,  3.8 ){\emph{e}}   %e
+\put( 7.7   ,  5.2 ){\emph{f}}   %f
+\put( 8.7   ,  5.4 ){\emph{g}}   %g
+\put( 8.2   ,  3.7 ){\emph{h}}   %h
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/035aa110.eepic b/33063-t/images/sources/035aa110.eepic
new file mode 100644
index 0000000..e878426
--- /dev/null
+++ b/33063-t/images/sources/035aa110.eepic
@@ -0,0 +1,21 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 035aa110.tex
+% Pt.110. Fig.a.  (28,10 )
+\PGset[0.8em]
+\begin{picture}   (28,10 )
+\put( 1     ,  1   ){\line  ( 1, 0 ){26   }}      %G(CD)H
+\put( 1     ,  9   ){\line  ( 1, 0 ){26   }}      %E(AB)F
+\put( 8     ,  1   ){\line  ( 3, 2 ){12   }}      %C(O)B
+\dashline[80]{0.2}(14,  1   )(14  , 9 )               %D(O)A
+\put(13.7   ,  9.2 ){$\scriptstyle A$}   %A
+\put(19.7   ,  9.2 ){$\scriptstyle B$}   %B
+\put( 7.6   ,  0.2 ){$\scriptstyle C$}   %C
+\put(13.6   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 0.1   ,  8.7 ){$\scriptstyle E$}   %E
+\put(27.1   ,  8.7 ){$\scriptstyle F$}   %F
+\put( 0.1   ,  0.7 ){$\scriptstyle G$}   %G
+\put(27.1   ,  0.7 ){$\scriptstyle H$}   %H
+\put(13.2   ,  4.9 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/036aa111.eepic b/33063-t/images/sources/036aa111.eepic
new file mode 100644
index 0000000..abedd35
--- /dev/null
+++ b/33063-t/images/sources/036aa111.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 036aa111.tex
+% Pt.111. Fig.a.  (24,15 )
+\PGset[0.8em]
+\begin{picture}   (24,15 )
+\put( 1     ,  5   ){\line  ( 1, 0 ){22   }}      %C(K)D
+\put( 1     , 10   ){\line  ( 1, 0 ){22   }}      %A(H)B
+\put( 6.5   ,  1   ){\line  ( 4, 5 ){10.4 }}      %F(KH)E
+\dashline[80]{0.2}( 1, 13   )(23  , 7.9 )             %M(H)N
+\put( 0.1   ,  9.7 ){$\scriptstyle A$}   %A
+\put(23     ,  9.7 ){$\scriptstyle B$}   %B
+\put( 0.1   ,  4.6 ){$\scriptstyle C$}   %C
+\put(23     ,  4.6 ){$\scriptstyle D$}   %D
+\put(16.9   , 13.5 ){$\scriptstyle E$}   %E
+\put( 5.7   ,  0.7 ){$\scriptstyle F$}   %F
+\put(13.2   , 10.4 ){$\scriptstyle H$}   %H
+\put( 9.6   ,  4.1 ){$\scriptstyle K$}   %K
+\put( 0.01   , 12.7 ){$\scriptstyle M$}   %M
+\put(23.1   ,  7.6 ){$\scriptstyle N$}   %N
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/037aa112.eepic b/33063-t/images/sources/037aa112.eepic
new file mode 100644
index 0000000..a0c31c2
--- /dev/null
+++ b/33063-t/images/sources/037aa112.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 037aa112.tex
+% Pt.112. Fig.a.  (25,17 )
+\PGset[0.8em]
+\begin{picture}   (25,17 )
+\put( 1     ,  6   ){\line  ( 1, 0 ){23   }}      %C(K)D
+\put( 1     , 11   ){\line  ( 1, 0 ){23   }}      %A(H)B
+\put( 5     ,  1   ){\line  ( 1, 1 ){15   }}      %F(KH)E
+\put( 0.1   , 10.7 ){$\scriptstyle A$}   %A
+\put(24     , 10.7 ){$\scriptstyle B$}   %B
+\put( 0.1   ,  5.7 ){$\scriptstyle C$}   %C
+\put(24     ,  5.7 ){$\scriptstyle D$}   %D
+\put(19.9   , 16.1 ){$\scriptstyle E$}   %E
+\put( 4.6   ,  0.1 ){$\scriptstyle F$}   %F
+\put(14.2   , 11.2 ){$\scriptstyle H$}   %H
+\put( 9.8   ,  5.1 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/038aa115.eepic b/33063-t/images/sources/038aa115.eepic
new file mode 100644
index 0000000..146d5f1
--- /dev/null
+++ b/33063-t/images/sources/038aa115.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 038aa115.tex
+% Pt.115. Fig.a.  (25,17 )
+% *** Same as Pt.112. Fig.a. ***
+\PGset[0.8em]
+\begin{picture}   (25,17 )
+\put( 1     ,  6   ){\line  ( 1, 0 ){23   }}      %C(K)D
+\put( 1     , 11   ){\line  ( 1, 0 ){23   }}      %A(H)B
+\put( 5     ,  1   ){\line  ( 1, 1 ){15   }}      %F(KH)E
+\put( 0.1   , 10.7 ){$\scriptstyle A$}   %A
+\put(24     , 10.7 ){$\scriptstyle B$}   %B
+\put( 0.1   ,  5.7 ){$\scriptstyle C$}   %C
+\put(24     ,  5.7 ){$\scriptstyle D$}   %D
+\put(19.9   , 16.1 ){$\scriptstyle E$}   %E
+\put( 4.6   ,  0.1 ){$\scriptstyle F$}   %F
+\put(14.2   , 11.2 ){$\scriptstyle H$}   %H
+\put( 9.8   ,  5.1 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/039aa117.eepic b/33063-t/images/sources/039aa117.eepic
new file mode 100644
index 0000000..624dfd2
--- /dev/null
+++ b/33063-t/images/sources/039aa117.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 039aa117.tex
+% Pt.117. Fig.1.  (11, 9 )
+\PGset[0.8em]
+\begin{picture}   (11, 9 )
+\drawline     ( 4, 8 )( 1, 2 )(10, 2 )( 4, 8 )    %AB(D)CA
+\drawline(4,2)(4,8) % DA
+\put( 3.6   ,  8.2 ){$\scriptstyle A$}   %A
+\put( 0.5   ,  1.2 ){$\scriptstyle B$}   %B
+\put( 9.5   ,  1.2 ){$\scriptstyle C$}   %C
+\put( 3.6   ,  1.2 ){$\scriptstyle D$}   %D
+\put( 4    ,   0.2 ){{\footnotesize \textsc{Fig. 1.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/039bb118.eepic b/33063-t/images/sources/039bb118.eepic
new file mode 100644
index 0000000..a0452ad
--- /dev/null
+++ b/33063-t/images/sources/039bb118.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 039bb118.tex
+% Pt.118. Fig.2.  (16,15 )
+\PGset[0.8em]
+\begin{picture}   (16,9 )
+\drawline     ( 4, 8 )( 1, 2 )(10, 2 )( 4, 8 )    %ABCA
+\drawline(10,2)(14,2) % CD
+\put( 3.6   ,  8.2 ){$\scriptstyle A$}   %A
+\put( 0.5   ,  1.2 ){$\scriptstyle B$}   %B
+\put( 9.5   ,  1.2 ){$\scriptstyle C$}   %C
+\put(13.5   ,  1.2 ){$\scriptstyle D$}   %D
+\put( 4    ,   0.2 ){{\footnotesize \textsc{Fig. 2.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/039ce119.eepic b/33063-t/images/sources/039ce119.eepic
new file mode 100644
index 0000000..6545fad
--- /dev/null
+++ b/33063-t/images/sources/039ce119.eepic
@@ -0,0 +1,29 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 039cc119.tex
+% Pt.119. Scalene.  (13, 8 )
+\PGset[0.8em]
+\begin{picture}     (13, 8 )
+\drawline  ( 1, 2 )(12, 2 )( 4, 7 )( 1, 2 )    %Scalene
+\put       ( 4, 0.2 ){{\footnotesize Scalene.}}
+\end{picture}
+\PGrestore
+%
+% 039dd119.tex
+% Pt.119. Isosceles.  ( 7,10 )
+\PGset[0.8em]
+\begin{picture}       ( 7,10 )
+%\drawline  ( 1, 2 )( 7, 2 )( 4, 9 )( 1, 2 )    %Isosceles
+\drawline  ( 1, 2 )( 5.8, 2 )( 3.4, 9.2 )( 1, 2 )    %Isosceles
+\put       ( 1.5, 0.2 ){{\footnotesize Isosceles.}}
+\end{picture}
+\PGrestore
+%
+% 039ee119.tex
+% Pt.119. Equilateral.  (10,10 )
+\PGset[0.8em]
+\begin{picture}         (10,10 )
+\drawline  ( 1, 2 )( 9, 2 )( 5, 8.93 )( 1, 2 )    %Equilateral
+\put       ( 2.5, 0.2 ){{\footnotesize Equilateral.}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/039fi120.eepic b/33063-t/images/sources/039fi120.eepic
new file mode 100644
index 0000000..7fa1599
--- /dev/null
+++ b/33063-t/images/sources/039fi120.eepic
@@ -0,0 +1,39 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 039ff120.tex
+% Pt.120. Right.  ( 9, 9 )
+\PGset[0.8em]
+\begin{picture}   ( 9, 9 )
+\drawline  ( 1, 2 )( 8.5, 2 )( 8.5, 8 )( 1, 2 )    %Right
+\put       ( 3.5, 0.2 ){{\footnotesize Right.}}
+\end{picture}
+\PGrestore
+%
+% 039gg120.tex
+% Pt.120. Obtuse.  (13, 6 )
+\PGset[0.8em]
+\begin{picture}    (13, 6 )
+\drawline  ( 1, 2 )(12.2, 2 )( 5.2, 5.5 )( 1, 2 )    %Obtuse
+\put       ( 4.7, 0.2 ){{\footnotesize Obtuse.}}
+\end{picture}
+\PGrestore
+%
+% 039hh120.tex
+% Pt.120. Acute.   ( 8,10 )
+\PGset[0.8em]
+\begin{picture}    ( 8,10 )
+%\drawline  ( 1, 2 )( 8, 2 )( 5, 9 )( 1, 2 )    %Acute
+\drawline  ( 1, 2 )( 7.5, 2 )( 4.9, 9.8 )( 1, 2 )    %Acute
+\put       ( 3, 0.2 ){{\footnotesize Acute.}}
+\end{picture}
+\PGrestore
+%
+% 039ii120.tex
+% Pt.120. Equiangular.  (10,10 )
+% *** Same as Pt.119. Fig.e. ***
+\PGset[0.8em]
+\begin{picture}         (10,10 )
+\drawline  ( 1, 2 )( 9, 2 )( 5, 8.93 )( 1, 2 )    %Equiangular
+\put       ( 2.3, 0.2 ){{\footnotesize Equiangular.}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/041aa129.eepic b/33063-t/images/sources/041aa129.eepic
new file mode 100644
index 0000000..a2c5f81
--- /dev/null
+++ b/33063-t/images/sources/041aa129.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 041aa129.tex
+% Pt.129. Fig.a.  (23,11 )
+\PGset[0.8em]
+\begin{picture}   (23,11 )
+\drawline     ( 1,  1   )( 7,10 )(16  , 1 )( 1,  1   )       %ABCA
+\dashline[80]{0.2}(22, 10   )(16, 1 )(22  , 1 )       %ECF
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 6.7   , 10.2 ){$\scriptstyle B$}   %B
+\put(15.7   ,  0.2 ){$\scriptstyle C$}   %C
+\put(21.8   , 10.2 ){$\scriptstyle E$}   %E
+\put(21.5   ,  0.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/042aa138.eepic b/33063-t/images/sources/042aa138.eepic
new file mode 100644
index 0000000..b350ddf
--- /dev/null
+++ b/33063-t/images/sources/042aa138.eepic
@@ -0,0 +1,12 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 042aa138.tex
+% Pt.138. Fig.a.  (18, 7 )
+\PGset[0.8em]
+\begin{picture}   (18, 7 )
+\drawline     ( 1, 1 )( 5, 6 )(17.5, 1 )( 1, 1 )    %ABCA
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.6   ,  6.2 ){$\scriptstyle B$}   %B
+\put(17.1   ,  0.2 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/043ab139.eepic b/33063-t/images/sources/043ab139.eepic
new file mode 100644
index 0000000..bec91b1
--- /dev/null
+++ b/33063-t/images/sources/043ab139.eepic
@@ -0,0 +1,25 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 043aa139.tex
+% Pt.139. Fig.a.  (17,10 )
+\PGset[0.8em]
+\begin{picture}   (17,10 )
+\drawline     ( 1, 1 )(16, 1 )( 6, 8.5 )( 1, 1 )    %ABCA
+\put( 0.6   ,  0.2 ){$\scriptstyle A$}   %A
+\put(15.5   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 5.7   ,  8.7 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
+%
+\quad
+% ***  Uses  epic  package   ***
+% 043bb139.tex
+% Pt.139. Fig.b.  (17,10 )
+\PGset[0.8em]
+\begin{picture}   (17,10 )
+\drawline     ( 1, 1 )(16, 1 )( 6, 8.5 )( 1, 1 )    %DEFD
+\put( 0.6   ,  0.2 ){$\scriptstyle D$}   %D
+\put(15.5   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 5.7   ,  8.7 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/044ab143.eepic b/33063-t/images/sources/044ab143.eepic
new file mode 100644
index 0000000..2165e43
--- /dev/null
+++ b/33063-t/images/sources/044ab143.eepic
@@ -0,0 +1,27 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 044aa143.tex
+% Pt.143. Fig.a.  (17,10 )
+% *** Same as Pt.139. Fig.a. ***
+\PGset[0.8em]
+\begin{picture}   (17,10 )
+\drawline     ( 1, 1 )(16, 1 )( 6, 8.5 )( 1, 1 )    %ABCA
+\put( 0.6   ,  0.2 ){$\scriptstyle A$}   %A
+\put(15.5   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 5.7   ,  8.7 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
+%
+\quad
+% ***  Uses  epic  package   ***
+% 044bb143.tex
+% Pt.143. Fig.b.  (17,10 )
+% *** Same as Pt.139. Fig.b. ***
+\PGset[0.8em]
+\begin{picture}   (17,10 )
+\drawline     ( 1, 1 )(16, 1 )( 6, 8.5 )( 1, 1 )    %DEFD
+\put( 0.6   ,  0.2 ){$\scriptstyle D$}   %D
+\put(15.5   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 5.7   ,  8.7 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/045aa145.eepic b/33063-t/images/sources/045aa145.eepic
new file mode 100644
index 0000000..1bedb5f
--- /dev/null
+++ b/33063-t/images/sources/045aa145.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 045aa145.tex
+% Pt.145. Fig.a.  (18,12 )
+\PGset[0.8em]
+\begin{picture}   (18,12 )
+\drawline     ( 9,11 )( 1, 1 )(17, 1 )( 9,11 )    %AB(D)CA
+\dashline[80]{0.2}( 9, 1 )( 9,11 )                    %DA
+\put( 8.7   , 11.2 ){$\scriptstyle A$}   %A
+\put( 0.7   ,  0.2 ){$\scriptstyle B$}   %B
+\put(16.3   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 8.5   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/046aa147.eepic b/33063-t/images/sources/046aa147.eepic
new file mode 100644
index 0000000..d67eda5
--- /dev/null
+++ b/33063-t/images/sources/046aa147.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 046aa147.tex
+% Pt.147. Fig.a.  (18,12 )
+% *** Same as Pt.145. Fig.a. ***
+\PGset[0.8em]
+\begin{picture}   (18,12 )
+\drawline     ( 9,11 )( 1, 1 )(17, 1 )( 9,11 )    %AB(D)CA
+\dashline[80]{0.2}( 9, 1 )( 9,11 )                    %DA
+\put( 8.7   , 11.2 ){$\scriptstyle A$}   %A
+\put( 0.7   ,  0.2 ){$\scriptstyle B$}   %B
+\put(16.3   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 8.5   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/047ab150.eepic b/33063-t/images/sources/047ab150.eepic
new file mode 100644
index 0000000..4e6716e
--- /dev/null
+++ b/33063-t/images/sources/047ab150.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 047aa150.tex
+% Pt.150. Fig.a.  (16,16 )
+\PGset[0.8em]
+\begin{picture}   (16,16 )
+\drawline     ( 1  , 8 )( 4.5,15 )(15, 8 )( 1, 8 )    %ABCA
+\dashline[80]{0.2}( 1  , 8 )( 4.5, 1 )(15, 8)             %AB'C
+\dashline[80]{0.2}( 4.5, 1 )( 4.5,15 )                    %B'B
+\put( 0.1   ,  7.6 ){$\scriptstyle A$}   %A
+\put( 4.2   , 15.2 ){$\scriptstyle B$}   %B
+\put(15.2   ,  7.6 ){$\scriptstyle C$}   %C
+\put( 4.2   ,  0.2 ){$\scriptstyle B'$}   %B'
+\end{picture}
+\PGrestore
+%
+\quad
+% ***  Uses  epic  package   ***
+% 047bb150.tex
+% Pt.150. Fig.b.  (16,16 )
+\PGset[0.8em]
+\begin{picture}   (16,16 )
+\drawline     ( 1  , 8 )( 4.5,15 )(15, 8 )( 1, 8 )    %A'B'C'A'
+\put( 0.1   ,  8   ){$\scriptstyle A'$}   %A'
+\put( 4.2   , 15.2 ){$\scriptstyle B'$}   %B'
+\put(15     ,  8   ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/048ac151.eepic b/33063-t/images/sources/048ac151.eepic
new file mode 100644
index 0000000..6bc81b7
--- /dev/null
+++ b/33063-t/images/sources/048ac151.eepic
@@ -0,0 +1,38 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 048aa151.tex
+% Pt.151. Fig.a.  ( 8,12 )
+\PGset[0.8em]
+\begin{picture}   ( 8,12 )
+\drawline     ( 7  ,11 )( 1  , 1 )( 7, 1 )( 7,11 )    %ACBA
+\put( 6.7   , 11.3 ){$\scriptstyle A$}   %A
+\put( 6.7   ,  0.1 ){$\scriptstyle B$}   %B
+\put( 0.5   ,  0.1 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
+%
+\qquad
+% 048bb151.tex
+% Pt.151. Fig.b.  (14,12 )
+\PGset[0.8em]
+\begin{picture}   (14,12 )
+\dashline[80]{0.2}( 7  ,11 )( 1  , 1 )( 7, 1 )( 7,11 )    %A'CB'A'
+\dashline[80]{0.2}( 7  ,11 )(13  , 1 )( 7, 1 )            %A'C'B'
+\put( 0.5   ,  0.1 ){$\scriptstyle C$}   %C
+\put( 6.7   , 11.3 ){$\scriptstyle A'$}   %A'
+\put( 6.6   ,  0.1 ){$\scriptstyle B'$}   %B'
+\put(12.6   ,  0.1 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
+%
+\qquad
+% 048cc151.tex
+% Pt.151. Fig.c.  ( 8,12 )
+\PGset[0.8em]
+\begin{picture}   ( 8,12 )
+\drawline     ( 1  ,11 )( 1  , 1 )( 7, 1 )( 1,11 )    %A'B'C'A'
+\put( 0.7   , 11.3 ){$\scriptstyle A'$}   %A'
+\put( 0.6   ,  0.1 ){$\scriptstyle B'$}   %B'
+\put( 6.5   ,  0.1 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/049aa152.eepic b/33063-t/images/sources/049aa152.eepic
new file mode 100644
index 0000000..3da5e1b
--- /dev/null
+++ b/33063-t/images/sources/049aa152.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 049aa152.tex
+% Pt.152. Fig.a.  (23, 9 )
+\PGset[0.8em]
+\begin{picture}   (23, 9 )
+\drawline     ( 8  , 8 )( 1  , 1 )(22, 1 )( 8, 8 )    %ACB(E)A
+\dashline[80]{0.2}( 1  , 1 )(17  , 3.5 )                    %CE
+\put( 7.7   ,  8.2 ){$\scriptstyle A$}   %A
+\put(21.4   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 0.5   ,  0.2 ){$\scriptstyle C$}   %C
+\put(17.1   ,  3.5 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/050aa153.eepic b/33063-t/images/sources/050aa153.eepic
new file mode 100644
index 0000000..ccae4a2
--- /dev/null
+++ b/33063-t/images/sources/050aa153.eepic
@@ -0,0 +1,12 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 050aa153.tex
+% Pt.153. Fig.a.  (18,10 )
+\PGset[0.8em]
+\begin{picture}   (18,10 )
+\drawline     ( 5  , 9 )( 1  , 1 )(17, 1 )( 5, 9 )    %ACBA
+\put( 4.7   ,  9.2 ){$\scriptstyle A$}   %A
+\put(16.5   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 0.5   ,  0.2 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/051ac154.eepic b/33063-t/images/sources/051ac154.eepic
new file mode 100644
index 0000000..045ae9b
--- /dev/null
+++ b/33063-t/images/sources/051ac154.eepic
@@ -0,0 +1,40 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 051aa154.tex
+% Pt.154. Fig.a.  ( 9,14 )
+\PGset[0.8em]
+\begin{picture}   ( 9,14 )
+\drawline     ( 1  , 8 )( 8  , 1 )( 5,13 )( 1 , 8 )            %AEBA
+\put( 0.1   ,  7.7 ){$\scriptstyle A$}   %A
+\put( 4.6   , 13.2 ){$\scriptstyle B$}   %B
+\put( 7.7   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
+%
+\hspace{-1.5em}
+% 051bb154.tex
+% Pt.154. Fig.b.  (17,14 )
+\PGset[0.8em]
+\begin{picture}   (17,14 )
+\drawline     ( 1  , 8 )(15  , 8 )( 5,13 )( 1, 8 )    %A(F)CBA
+\drawline     ( 1  , 8 )( 8  , 1 )( 5,13 )            %AEB
+\dashline[80]{0.2}( 5  ,13 )( 9  , 8 )( 8, 1 )            %BFE
+\put( 0.1   ,  7.7 ){$\scriptstyle A$}   %A
+\put( 4.6   , 13.2 ){$\scriptstyle B$}   %B
+\put(15.2   ,  7.6 ){$\scriptstyle C$}   %C
+\put( 7.7   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 9     ,  7.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
+%
+\hspace{-1.5em}
+% 051cc154.tex
+% Pt.154. Fig.c.  (16,14 )
+\PGset[0.8em]
+\begin{picture}   (16,14 )
+\drawline     ( 1  , 8 )(15  , 8 )( 5,13 )( 1, 8 )    %ACBA
+\put( 0.8   ,  7.2 ){$\scriptstyle A$}   %A
+\put( 4.6   , 13.2 ){$\scriptstyle B$}   %B
+\put(14.5   ,  7.2 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/052ab155.eepic b/33063-t/images/sources/052ab155.eepic
new file mode 100644
index 0000000..066b476
--- /dev/null
+++ b/33063-t/images/sources/052ab155.eepic
@@ -0,0 +1,24 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 052aa155.tex
+% Pt.155. Fig.a.  (18,11 )
+\PGset[0.8em]
+\begin{picture}   (18,11 )
+\drawline     ( 5.5,10 )( 1  , 1 )(16.75, 1 )( 5.5,10 )    %ABCA
+\put( 5.2   , 10.2 ){$\scriptstyle A$}   %A
+\put( 0.7   ,  0.2 ){$\scriptstyle B$}   %B
+\put(16.2   ,  0.2 ){$\scriptstyle C$}   %C
+\end{picture}
+\PGrestore
+%
+\qquad
+% 052bb155.tex
+% Pt.155. Fig.b.  (14,12 )
+\PGset[0.8em]
+\begin{picture}   (14,12 )
+\drawline     ( 3  ,11 )( 1  , 1 )(13  , 1 )( 3  ,11 )    %DEFD
+\put( 2.6   , 11.2 ){$\scriptstyle D$}   %D
+\put( 0.7   ,  0.2 ){$\scriptstyle E$}   %E
+\put(12.5   ,  0.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/054aa160.eepic b/33063-t/images/sources/054aa160.eepic
new file mode 100644
index 0000000..73a94f5
--- /dev/null
+++ b/33063-t/images/sources/054aa160.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 054aa160.tex
+% Pt.160. Fig.a.  (14,12 )
+\PGset[0.8em]
+\begin{picture}   (14,12 )
+\drawline(1,1)(13,1) % AB
+\drawline(7,1)(7,11) % PR
+\dashline[80]{0.4}( 1, 1    )( 7   , 4 )(13, 1  )  %AOB
+\dashline[80]{0.4}( 1, 1    )( 8.125 ,10.5 )(13, 1  )  %A(D)CB
+\dashline[80]{0.4}(13, 1    )( 7   , 9 )           %BD
+\put( 0.1   ,  0.9 ){$\scriptstyle A$}   %A
+\put(13.1   ,  0.9 ){$\scriptstyle B$}   %B
+\put( 8     , 10.7 ){$\scriptstyle C$}   %C
+\put( 5.9   ,  8.9 ){$\scriptstyle D$}   %D
+\put( 6     ,  4   ){$\scriptstyle O$}   %O
+\put( 6.5   ,  0.2 ){$\scriptstyle P$}   %P
+\put( 6.6   , 11.2 ){$\scriptstyle R$}   %R
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/055aa162.eepic b/33063-t/images/sources/055aa162.eepic
new file mode 100644
index 0000000..822783f
--- /dev/null
+++ b/33063-t/images/sources/055aa162.eepic
@@ -0,0 +1,21 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 055aa162.tex
+% Pt.162. Fig.a.  (13,11 )
+\PGset[0.8em]
+\begin{picture}   (13,11 )
+\drawline(1,1)(13,7) % AD
+\drawline     ( 7.6, 9.8  )( 1   , 1    )(12   , 1  )  %P(BF)A(GC)Q
+\dashline[80]{0.2}( 7  , 9    )(11   , 6    )(11   , 1  )  %BDC
+\dashline[80]{0.2}( 5, 6.33 )( 7.67, 4.33 )( 7.67, 1  )  %FOG
+\put( 0.5   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 6.1   ,  9   ){$\scriptstyle B$}   %B
+\put(10.5   ,  0.2 ){$\scriptstyle C$}   %C
+\put(11.1   ,  5.2 ){$\scriptstyle D$}   %D
+\put( 4.1   ,  6.2 ){$\scriptstyle F$}   %F
+\put( 7.2   ,  0.2 ){$\scriptstyle G$}   %G
+\put( 7.8   ,  3.5 ){$\scriptstyle O$}   %O
+\put( 7.5   , 10   ){$\scriptstyle P$}   %P
+\put(12.2   ,  0.6 ){$\scriptstyle Q$}   %Q
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/056ac166.eepic b/33063-t/images/sources/056ac166.eepic
new file mode 100644
index 0000000..418806f
--- /dev/null
+++ b/33063-t/images/sources/056ac166.eepic
@@ -0,0 +1,30 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 056aa166.tex
+% Pt.166. Trapezium.  (11, 7.5 )
+\PGset[0.8em]
+\begin{picture}       (11, 7.5 )
+\drawline  ( 0, 1.5 )(11, 1.5 )( 7.5, 6.5 )( 1.5 , 7.5 )( 0, 1.5 )    %Trapezium
+\put       ( 3, 0.2 ){{\footnotesize Trapezium.}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 056bb166.tex
+% Pt.166. Trapezoid.  (10, 6.5 )
+\PGset[0.8em]
+\begin{picture}       (10, 6.5 )
+\drawline  ( 0, 1.5 )(10, 1.5 )( 6.5, 6.5 )( 1.25 , 6.5 )( 0, 1.5 )    %Trapezoid
+\put       ( 2.5, 0.2 ){{\footnotesize Trapezoid.}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 056cc166.tex
+% Pt.166. Parallelogram.  (12, 6.5 )
+\PGset[0.8em]
+\begin{picture}           (12, 6.5 )
+\drawline  ( 0, 1.5 )( 8, 1.5 )(12, 6.5 )( 4, 6.5 )( 0, 1.5 )    %Parallelogram
+\put       ( 3, 0.2 ){{\footnotesize Parallelogram.}}
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/056dg170.eepic b/33063-t/images/sources/056dg170.eepic
new file mode 100644
index 0000000..cd0f53a
--- /dev/null
+++ b/33063-t/images/sources/056dg170.eepic
@@ -0,0 +1,40 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 056dd170.tex
+% Pt.170. Square.  ( 5, 6.5 )
+\PGset[0.8em]
+\begin{picture}    ( 5, 6.5 )
+\drawline  ( 0, 1.5 )( 5, 1.5 )( 5, 6.5 )( 0, 6.5 )( 0, 1.5 )    %Square
+\put       ( 1, 0.2 ){{\footnotesize Square.}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 056ee170.tex
+% Pt.170. Rectangle.  (10, 6.5 )
+\PGset[0.8em]
+\begin{picture}       (10, 6.5 )
+\drawline  ( 0, 1.5 )(10, 1.5 )(10, 6.5 )( 0, 6.5 )( 0, 1.5 )    %Rectangle
+\put       ( 2.5, 0.2 ){{\footnotesize Rectangle.}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 056ff170.tex
+% Pt.166. Rhombus.  (10, 6.5 )
+\PGset[0.8em]
+\begin{picture}     (10, 6.5 )
+\drawline  ( 0, 1.5 )( 6, 1.5 )(10, 6.5 )( 4, 6.5 )( 0, 1.5 )    %Rhombus
+\put       ( 1, 0.2 ){{\footnotesize Rhombus.}}
+\end{picture}
+\PGrestore
+%
+\hspace{-2.5em}
+% 056gg170.tex
+% Pt.166. Rhomboid.  (13, 6.5 )
+\PGset[0.8em]
+\begin{picture}      (13, 6.5 )
+\drawline  ( 0, 1.5 )( 9, 1.5 )(13, 6.5 )( 4, 6.5 )( 0, 1.5 )    %Rhomboid
+\put       ( 4, 0.2 ){{\footnotesize Rhomboid.}}
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/057aa174.eepic b/33063-t/images/sources/057aa174.eepic
new file mode 100644
index 0000000..7189c46
--- /dev/null
+++ b/33063-t/images/sources/057aa174.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 057aa174.tex
+% Pt.174. Fig.a.  (14.5, 7 )
+\PGset[0.8em]
+\begin{picture}   (14.5, 7 )
+\drawline  ( 1, 1 )( 9.5, 1 )(13.5, 6 )( 5, 6 )( 1, 1 )  %A(Q)EC(P)BA
+\drawline  ( 1, 1 )(13.5, 6 )                            %AC
+\drawline  ( 6, 1 )(6, 6)
+\put( 0.6   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.2   ,  6.1 ){$\scriptstyle B$}   %B
+\put(13.6   ,  6   ){$\scriptstyle C$}   %C
+\put( 9.2   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 5.8   ,  6.2 ){$\scriptstyle P$}   %P
+\put( 5.7   ,  0.2 ){$\scriptstyle Q$}   %Q
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/057bb176.eepic b/33063-t/images/sources/057bb176.eepic
new file mode 100644
index 0000000..4d4ec69
--- /dev/null
+++ b/33063-t/images/sources/057bb176.eepic
@@ -0,0 +1,25 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 057bb176.tex
+% Pt.176. Fig.b.  (20, 9 )
+\PGset[0.8em]
+\begin{picture}   (20, 9 )
+\drawline     ( 6, 8.5 )( 1, 6 )( 6.5, 6 )  %ABC
+\drawline     ( 1, 2.5 )(19, 2.5 )          %MN
+\drawline     ( 2, 0   )(11, 4.5 )          %HD
+\dashline[80]{0.2}( 6.5, 6 )(19, 6   )          %extend BC->
+\dashline[80]{0.2}(11, 4.5 )(18, 8   )          %extend HD/
+\put( 4.4   ,  8.1 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  5.7 ){$\scriptstyle B$}   %B
+\put( 6.3   ,  5.2 ){$\scriptstyle C$}   %C
+\put( 9.8   ,  4.5 ){$\scriptstyle D$}   %D
+\put( 3.7   ,  0   ){$\scriptstyle H$}   %H
+\put( 0     ,  2.2 ){$\scriptstyle M$}   %M
+\put(19.1   ,  2.2 ){$\scriptstyle N$}   %N
+\put( 2.5   ,  6.2 ){$a$}%\emph{a}}   %a
+\put( 5     ,  1.8 ){$c$}%\emph{c}}   %c
+\put(16     ,  6.2 ){$x$}%\emph{x}}   %x
+\put( 8.7   ,  2.7 ){$a'$}%\emph{a'}}   %a'
+\put( 6.6   ,  2.8 ){$c'$}%\emph{c'}}   %c'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/058aa178.eepic b/33063-t/images/sources/058aa178.eepic
new file mode 100644
index 0000000..dc691f7
--- /dev/null
+++ b/33063-t/images/sources/058aa178.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 058aa178.tex
+% Pt.178. Fig.a.  (24,11 )
+\PGset[0.8em]
+\begin{picture}   (24,11 )
+\drawline     ( 1, 1 )(19.9, 1   )(23.5,10 )( 4.6,10 )( 1, 1 )  %AECBA
+\dashline[80]{0.2}( 1, 1 )(23.5,10 )                            %AC
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.5   , 10.2 ){$\scriptstyle B$}   %B
+\put(23.2   , 10.2 ){$\scriptstyle C$}   %C
+\put(19.2   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/058bb181.eepic b/33063-t/images/sources/058bb181.eepic
new file mode 100644
index 0000000..16f8768
--- /dev/null
+++ b/33063-t/images/sources/058bb181.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 058bb181.tex
+% Pt.181. Fig.b.  (20, 7 )
+\PGset[0.8em]
+\begin{picture}   (20, 7 )
+\put( 0    ,  1   ){\line  ( 1, 0 ){20 }}                %lower horiz
+\put( 0    ,  6   ){\line  ( 1, 0 ){20 }}                %upper horiz
+\put( 5    ,  1   ){\line  ( 0, 1 ){ 5 }}                %DA
+\put(15    ,  1   ){\line  ( 0, 1 ){ 5 }}                %CB
+\put( 4.6   ,  6.2 ){$\scriptstyle A$}   %A
+\put(14.7   ,  6.2 ){$\scriptstyle B$}   %B
+\put(14.6   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 4.6   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/059aa182.eepic b/33063-t/images/sources/059aa182.eepic
new file mode 100644
index 0000000..e2d83be
--- /dev/null
+++ b/33063-t/images/sources/059aa182.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 059aa182.tex
+% Pt.182. Fig.a.  (24,11 )
+% *** Same as Pt.178. Fig.a. ***
+\PGset[0.8em]
+\begin{picture}   (24,11 )
+\drawline     ( 1, 1 )(19.9, 1   )(23.5,10 )( 4.6,10 )( 1, 1 )  %AECBA
+\dashline[80]{0.2}( 1, 1 )(23.5,10 )                            %AC
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.5   , 10.2 ){$\scriptstyle B$}   %B
+\put(23.2   , 10.2 ){$\scriptstyle C$}   %C
+\put(19.2   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/060aa183.eepic b/33063-t/images/sources/060aa183.eepic
new file mode 100644
index 0000000..ece4115
--- /dev/null
+++ b/33063-t/images/sources/060aa183.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 060aa183.tex
+% Pt.183. Fig.a.  (24,11 )
+% *** Same as Pt.178. Fig.a. ***
+\PGset[0.8em]
+\begin{picture}   (24,11 )
+\drawline     ( 1, 1 )(19.9, 1   )(23.5,10 )( 4.6,10 )( 1, 1 )  %AECBA
+\dashline[80]{0.2}( 1, 1 )(23.5,10 )                            %AC
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.5   , 10.2 ){$\scriptstyle B$}   %B
+\put(23.2   , 10.2 ){$\scriptstyle C$}   %C
+\put(19.2   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/061aa184.eepic b/33063-t/images/sources/061aa184.eepic
new file mode 100644
index 0000000..26e0f86
--- /dev/null
+++ b/33063-t/images/sources/061aa184.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 061aa184.tex
+% Pt.184. Fig.a.  (24,11 )
+\PGset[0.8em]
+\begin{picture}   (24,11 )
+\drawline     ( 1  , 1 )(19.9, 1   )(23.5,10 )( 4.6,10 )( 1, 1 )  %AECBA
+\drawline     ( 1  , 1 )(23.5,10 )                            %A(O)C
+\drawline     ( 4.6,10 )(19.9, 1 )                            %B(O)E
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.5   , 10.2 ){$\scriptstyle B$}   %B
+\put(23.2   , 10.2 ){$\scriptstyle C$}   %C
+\put(19.2   ,  0.2 ){$\scriptstyle E$}   %E
+\put(11.95  ,  5.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/062ab185.eepic b/33063-t/images/sources/062ab185.eepic
new file mode 100644
index 0000000..c1f9df5
--- /dev/null
+++ b/33063-t/images/sources/062ab185.eepic
@@ -0,0 +1,26 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 062aa185.tex
+% Pt.185. Fig.a.  (17, 8 )
+\PGset[0.8em]
+\begin{picture}   (17, 8 )
+\drawline     ( 1  , 1 )(14, 1   )(16.4, 7 )( 3.4, 7 )( 1, 1 )  %ADCBA
+\put( 0.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 3.3   ,  7.2 ){$\scriptstyle B$}   %B
+\put(16.1   ,  7.2 ){$\scriptstyle C$}   %C
+\put(13.5   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
+%
+\quad
+% 062bb185.tex
+% Pt.185. Fig.b.  (17, 8 )
+\PGset[0.8em]
+\begin{picture}   (17, 8 )
+\drawline     ( 1  , 1 )(14, 1   )(16.4, 7 )( 3.4, 7 )( 1, 1 )  %ADCBA
+\put( 0.7   ,  0.2 ){$\scriptstyle A'$}   %A'
+\put( 3.3   ,  7.2 ){$\scriptstyle B'$}   %B'
+\put(16.1   ,  7.2 ){$\scriptstyle C'$}   %C'
+\put(13.5   ,  0.2 ){$\scriptstyle D'$}   %D'
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/063aa187.eepic b/33063-t/images/sources/063aa187.eepic
new file mode 100644
index 0000000..3308f6f
--- /dev/null
+++ b/33063-t/images/sources/063aa187.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 063aa187.tex
+% Pt.187. Fig.a.  (24.5,17.5 )
+\PGset[0.8em]
+\begin{picture}   (24.5,17.5 )
+\drawline(0, 4.5)(24.5, 4.5) % DGP
+\drawline(3.5,7.5)(24.5,7.5) % CFM
+\drawline(5,10.5)(23.7,10.5) % BEK
+\drawline(8.2,13.5)(22.7,13.5) % AH
+\drawline(1,1.5)(15.5,16) % DCBA
+\drawline(22.9,0)(19.4,17.5) % PMKH
+\dashline[80]{0.2}( 7, 7.5 )( 7.6, 4.5 )               %CG
+\dashline[80]{0.2}(10,10.5 )(10.6, 7.5 )               %BF
+\dashline[80]{0.2}(13,13.5 )(13.6,10.5 )               %AE
+\put(12.3   , 13.7 ){$\scriptstyle A$}   %A
+\put( 9.3   , 10.7 ){$\scriptstyle B$}   %B
+\put( 6.3   ,  7.7 ){$\scriptstyle C$}   %C
+\put( 3.3   ,  4.7 ){$\scriptstyle D$}   %D
+\put(13.3   ,  9.7 ){$\scriptstyle E$}   %E
+\put(10.3   ,  6.7 ){$\scriptstyle F$}   %F
+\put( 7.3   ,  3.7 ){$\scriptstyle G$}   %G
+\put(20.3   , 13.7 ){$\scriptstyle H$}   %H
+\put(20.9   , 10.7 ){$\scriptstyle K$}   %K
+\put(21.5   ,  7.7 ){$\scriptstyle M$}   %M
+\put(22.1   ,  4.7 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/064aa188.eepic b/33063-t/images/sources/064aa188.eepic
new file mode 100644
index 0000000..3083030
--- /dev/null
+++ b/33063-t/images/sources/064aa188.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 064aa188.tex
+% Pt.188. Fig.a.  (13,10 )
+\PGset[0.8em]
+\begin{picture}   (13,10 )
+\drawline     ( 4.5, 9 )( 0.5, 1 )(12.5, 1 )( 4.5, 9 ) %A(D)B(F)C(E)A
+\drawline(2.5,5)(8.5,5) % DE
+\dashline[80]{0.2}( 6.5, 1   )( 8.5, 5   )                 %FE
+\dashline[80]{0.2}( 0.5, 9   )(10.5, 9   )                 %horiz (A)
+\put( 4.1   ,  9.2 ){$\scriptstyle A$}   %A
+\put( 0     ,  0.2 ){$\scriptstyle B$}   %B
+\put(12     ,  0.2 ){$\scriptstyle C$}   %C
+\put( 1.6   ,  4.9 ){$\scriptstyle D$}   %D
+\put( 8.6   ,  5   ){$\scriptstyle E$}   %E
+\put( 6     ,  0.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/064bb190.eepic b/33063-t/images/sources/064bb190.eepic
new file mode 100644
index 0000000..bee8db8
--- /dev/null
+++ b/33063-t/images/sources/064bb190.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 064bb190.tex
+% Pt.190. Fig.b.  (17.3, 9.2 )
+\PGset[0.8em]
+\begin{picture}   (17.3, 9.2 )
+\drawline     ( 0.5, 1 )(16.9, 1 )( 9.7, 8.2 )( 2.9, 8.2 )( 0.5, 1 ) %AB(G)CD(E)A
+\drawline(1.7,4.6)(13.3,4.6) % EG
+\dashline[80]{0.2}( 2.9, 8.2 )(16.9  , 1   )                             %D(F)B
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put(16.4   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 9.5   ,  8.4 ){$\scriptstyle C$}   %C
+\put( 2.7   ,  8.4 ){$\scriptstyle D$}   %D
+\put( 0.9   ,  4.5 ){$\scriptstyle E$}   %E
+\put( 9     ,  3.8 ){$\scriptstyle F$}   %F
+\put(13.4   ,  4.5 ){$\scriptstyle G$}   %G
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/065ac192.eepic b/33063-t/images/sources/065ac192.eepic
new file mode 100644
index 0000000..5de3c44
--- /dev/null
+++ b/33063-t/images/sources/065ac192.eepic
@@ -0,0 +1,62 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 065aa192.tex
+% Pt.192. Fig.a.  (13,12 )
+\PGset[0.8em]
+\begin{picture}   (13,12 )
+\drawline     ( 1, 6 )( 4,11 )( 9,10 )(12, 6 )( 6, 2 )( 1, 6 ) %ABCDEA
+\drawline     ( 1, 6 )( 9,10 )                                 %AC
+\put( 0.2   ,  5.6 ){$\scriptstyle A$}   %A
+\put( 3.8   , 11.2 ){$\scriptstyle B$}   %B
+\put( 9.1   ,  9.8 ){$\scriptstyle C$}   %C
+\put(12.1   ,  5.6 ){$\scriptstyle D$}   %D
+\put( 5.7   ,  1.1 ){$\scriptstyle E$}   %E
+\put( 5     ,  0.2 ){{\footnotesize \textsc{Fig. 1.}}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 065bb192.tex
+% Pt.192. Fig.b.  (13,12 )
+\PGset[0.8em]
+\begin{picture}   (13,12 )
+\drawline     ( 1, 6 )( 4,11 )( 9,10 )(12, 6 )( 6, 2 )( 1, 6 ) %ABCDEA
+\put( 0.2   ,  5.6 ){$\scriptstyle A$}   %A
+\put( 3.8   , 11.2 ){$\scriptstyle B$}   %B
+\put( 9.1   ,  9.8 ){$\scriptstyle C$}   %C
+\put(12.1   ,  5.6 ){$\scriptstyle D$}   %D
+\put( 5.7   ,  1.1 ){$\scriptstyle E$}   %E
+\put( 4.5   ,  0.2 ){{\footnotesize \textsc{Fig. 2.}}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 065cc192.tex
+% Pt.192. Fig.b.  (11.1, 9.6 )
+\PGset[0.8em]
+\begin{picture}   (11.1, 9.6 )
+\drawline     ( 5.1, 1.5 )( 0.1, 3.5 )( 0.1, 7.5 )( 5.1, 9.5 )
+(10.1, 7.571)( 5.1, 5.5 )(10.1, 3.429 )( 5.1, 1.5 ) %bot-ll-ul-top-EDF-bot
+% Ellipse:  u = 5.1  v = 5.5  a = 1.43  b = 1.43  phi = 0.0 Grad
+
+\qbezier[5](6.4211, 6.0472)(6.3206, 6.29)(6.1112, 6.5112)
+\qbezier[5](6.1112, 6.5112)(5.9253, 6.697)(5.6472, 6.8211)
+\qbezier[5](5.6472, 6.8211)(5.4044, 6.9217)(5.1, 6.93)
+\qbezier[5](5.1, 6.93)(4.8372, 6.93)(4.5528, 6.8211)
+\qbezier[5](4.5528, 6.8211)(4.31, 6.7206)(4.0888, 6.5112)
+\qbezier[5](4.0888, 6.5112)(3.903, 6.3253)(3.7789, 6.0472)
+\qbezier[5](3.7789, 6.0472)(3.6783, 5.8044)(3.67, 5.5)
+\qbezier[5](3.67, 5.5)(3.67, 5.2372)(3.7789, 4.9528)
+\qbezier[5](3.7789, 4.9528)(3.8794, 4.71)(4.0888, 4.4888)
+\qbezier[5](4.0888, 4.4888)(4.2747, 4.303)(4.5528, 4.1789)
+\qbezier[5](4.5528, 4.1789)(4.7956, 4.0783)(5.1, 4.07)
+\qbezier[5](5.1, 4.07)(5.3628, 4.07)(5.6472, 4.1789)
+\qbezier[5](5.6472, 4.1789)(5.89, 4.2794)(6.1112, 4.4888)
+\qbezier[5](6.1112, 4.4888)(6.297, 4.6747)(6.4211, 4.9528)
+
+\put( 4.1   ,  5.2 ){$\scriptstyle D$}   %D
+\put(10.2   ,  7.1 ){$\scriptstyle E$}   %E
+\put(10.2   ,  3.1 ){$\scriptstyle F$}   %F
+\put( 4     ,  0.2 ){{\footnotesize \textsc{Fig. 3.}}}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/066ad203.eepic b/33063-t/images/sources/066ad203.eepic
new file mode 100644
index 0000000..40fe740
--- /dev/null
+++ b/33063-t/images/sources/066ad203.eepic
@@ -0,0 +1,40 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 066aa203.tex
+% Pt.203. Fig.a.  ( 8, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 8 )
+\drawline     ( 0, 1.5 )( 0.5, 8 )( 5, 7.5 )( 8, 1.5 )( 0, 1.5 ) %ll-ul-ur-lr-ll
+\put( 2.5   ,  0.2 ){{\footnotesize \textsc{Fig. 4.}}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 066bb203.tex
+% Pt.203. Fig.b.  ( 8, 5 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 5 )
+\drawline     ( 0, 1.5 )( 0.38, 5 )( 6.60, 4.31 )( 8, 1.5 )( 0, 1.5 ) %ll-ul-ur-lr-ll
+\put( 2.5   ,  0.2 ){{\footnotesize \textsc{Fig. 5.}}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 066cc203.tex
+% Pt.203. Fig.c.  ( 8, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 7 )
+\drawline     ( 0, 1.5 )( 3, 7 )( 8, 6.8 )( 7, 1.5 )( 0, 1.5 ) %ll-ul-ur-lr-ll
+\put( 2     ,  0.2 ){{\footnotesize \textsc{Fig. 6.}}}
+\end{picture}
+\PGrestore
+%
+\quad
+% 066dd203.tex
+% Pt.203. Fig.d.  ( 7, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 8 )
+\drawline     ( 0, 1.5 )( 1, 7.7 )( 6, 6.8 )( 7, 1.5 )( 0, 1.5 ) %ll-ul-ur-lr-ll
+\put( 2     ,  0.2 ){{\footnotesize \textsc{Fig. 7.}}}
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/067aa205.eepic b/33063-t/images/sources/067aa205.eepic
new file mode 100644
index 0000000..d997a8b
--- /dev/null
+++ b/33063-t/images/sources/067aa205.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 067aa205.tex
+% Pt.205. Fig.a.  (20,17 )
+\PGset[0.8em]
+\begin{picture}   (20,17 )
+\drawline     ( 5.5, 1 )(14.5, 1   )(19, 9.1 )(14.5,16 )( 5.5,16 )( 1, 8.5 )( 5.5, 1 ) %ABCDEFA
+\dashline[80]{0.2}( 5.5, 1 )(19  , 9.1 )                             %AC
+\dashline[80]{0.2}( 5.5, 1 )(14.5,16   )                             %AD
+\dashline[80]{0.2}( 5.5, 1 )( 5.5,16   )                             %AE
+\put( 5.1   ,  0.2 ){$\scriptstyle A$}   %A
+\put(14     ,  0.2 ){$\scriptstyle B$}   %B
+\put(19.1   ,  8.7 ){$\scriptstyle C$}   %C
+\put(14     , 16.2 ){$\scriptstyle D$}   %D
+\put( 5.1   , 16.2 ){$\scriptstyle E$}   %E
+\put( 0.2   ,  8.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/068aa207.eepic b/33063-t/images/sources/068aa207.eepic
new file mode 100644
index 0000000..4889a76
--- /dev/null
+++ b/33063-t/images/sources/068aa207.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 068aa207.tex
+% Pt.207. Fig.a.  (17,18 )
+\PGset[0.8em]
+\begin{picture}   (17,18 )
+\put( 4     ,  2.5  ){\line  ( 1, 0 ){12    }}        %ABext
+\put(13.93  ,  2.5  ){\line  ( 1, 5 ){ 2.3  }}        %BCext
+\put(15.65  , 11.11 ){\line  (-4, 3 ){ 9    }}        %CDext
+\put( 9     , 16.1  ){\line  (-5,-4 ){ 9    }}        %DEext
+\put( 2.34  , 10.78 ){\line  ( 1,-5 ){ 2.15 }}        %EAext
+\put( 4.1   ,  2.7 ){$\scriptstyle A$}   %A
+\put(13     ,  2.7 ){$\scriptstyle B$}   %B
+\put(14.5   , 10.5 ){$\scriptstyle C$}   %C
+\put( 8.5   , 15   ){$\scriptstyle D$}   %D
+\put( 2.8   , 10.3 ){$\scriptstyle E$}   %E
+\put( 4.3   ,  1.7 ){\emph{a}}   %a
+\put(14.2   ,  2.7 ){\emph{b}}   %b
+\put(15.0   , 11.7 ){\emph{c}}   %c
+\put( 7.5   , 15.5 ){\emph{d}}   %d
+\put( 1.7   ,  9.6 ){\emph{e}}   %e
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/069ac208.eepic b/33063-t/images/sources/069ac208.eepic
new file mode 100644
index 0000000..92c5563
--- /dev/null
+++ b/33063-t/images/sources/069ac208.eepic
@@ -0,0 +1,52 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 069aa208.tex
+% Pt.208. Fig.a.  (13,13 )
+\PGset[0.8em]
+\begin{picture}   (13,13 )
+\drawline(1,6.5)(12,6.5) % X'X
+\dashline[80]{0.2}( 6.5, 1 )( 6.5,12 )          %P'(O)P
+\put( 5.6   ,  5.6 ){$\scriptstyle O$}   %O
+\put( 6.2   , 12.2 ){$\scriptstyle P$}   %P
+\put( 6.1   ,  0.1 ){$\scriptstyle P'$}   %P'
+\put(12     ,  6.2 ){$\scriptstyle X$}   %X
+\put( 0.1   ,  6.2 ){$\scriptstyle X'$}   %X'
+\end{picture}
+\PGrestore
+%
+% 069bb208.tex
+% Pt.208. Fig.b.  (11,11 )
+\PGset[0.8em]
+\begin{picture}   (11,11 )
+\drawline     ( 1, 6.5 )( 3, 9.5 )( 8, 9.5 )(10, 6.5)( 8, 3.5 )( 3, 3.5 )( 1, 6.5)%A'BCAB'C'A'
+\dashline[80]{0.2}( 1, 6.5 )(10, 6.5 )          %A'A
+\dashline[80]{0.2}( 3, 9.5 )( 8, 3.5 )          %BB'
+\dashline[80]{0.2}( 3, 3.5 )( 8, 9.5 )          %C'C
+\put(10.1   ,  6.1 ){$\scriptstyle A$}   %A
+\put( 2.5   ,  9.7 ){$\scriptstyle B$}   %B
+\put( 8     ,  9.6 ){$\scriptstyle C$}   %C
+\put( 0.0   ,  6.2 ){$\scriptstyle A'$}   %A'
+\put( 7.8   ,  2.7 ){$\scriptstyle B'$}   %B'
+\put( 2.4   ,  2.6 ){$\scriptstyle C'$}   %C'
+\put( 5.2   ,  7.1 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+% 069cc208.tex
+% Pt.208. Fig.c.  (11,11 )
+\PGset[0.8em]
+\begin{picture}   (14,11 )
+\drawline     ( 3, 6.5 )( 5, 8.5 )(10, 9.5 )(11.5, 6.5)(10, 3.5 )( 5, 4.5 )( 3, 6.5)%ABCDC'B'A
+\drawline(1,6.5)(13.5,6.5) % X'X
+\dashline[80]{0.2}( 5, 4.5 )( 5, 8.5 )          %B'B
+\dashline[80]{0.2}(10, 3.5 )(10, 9.5 )          %C'C
+\put( 2.2   ,  6.7 ){$\scriptstyle A$}   %A
+\put( 4.5   ,  8.7 ){$\scriptstyle B$}   %B
+\put( 9.6   ,  9.6 ){$\scriptstyle C$}   %C
+\put(11.5   ,  6.7 ){$\scriptstyle D$}   %D
+\put( 4.5   ,  3.5 ){$\scriptstyle B'$}   %B'
+\put( 9.5   ,  2.6 ){$\scriptstyle C'$}   %C'
+\put(13.6   ,  6.1 ){$\scriptstyle X$}   %X
+\put( 0.1   ,  6.2 ){$\scriptstyle X'$}   %X'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/069dd211.eepic b/33063-t/images/sources/069dd211.eepic
new file mode 100644
index 0000000..fd475e4
--- /dev/null
+++ b/33063-t/images/sources/069dd211.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 069dd211.tex
+% Pt.211. Fig.d.  (13,12 )
+\PGset[0.8em]
+\begin{picture}   (13,12 )
+\drawline     ( 3, 9 )( 6,11 )(10, 8 )( 3, 9 )   %ABCA
+\drawline     ( 3, 3 )( 6, 1 )(10, 4 )( 3, 3 )   %A'B'C'A'
+\drawline(1,6)(12,6) % X'X
+\dashline[80]{0.2}( 3, 3 )( 3, 9 )        %A'A
+\dashline[80]{0.2}( 6, 1 )( 6,11 )        %B'B
+\dashline[80]{0.2}(10, 4 )(10, 8 )        %C'C
+\put( 2.2   ,  8.9 ){$\scriptstyle A$}   %A
+\put( 5.7   , 11.2 ){$\scriptstyle B$}   %B
+\put(10.2   ,  7.7 ){$\scriptstyle C$}   %C
+\put( 1.8   ,  2.8 ){$\scriptstyle A'$}   %A'
+\put( 5.6   ,  0.1 ){$\scriptstyle B'$}   %B'
+\put(10.2   ,  3.6 ){$\scriptstyle C'$}   %C'
+\put(12.1   ,  5.7 ){$\scriptstyle X$}   %X
+\put( 0.1   ,  5.8 ){$\scriptstyle X'$}   %X'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/070aa212.eepic b/33063-t/images/sources/070aa212.eepic
new file mode 100644
index 0000000..62fdd51
--- /dev/null
+++ b/33063-t/images/sources/070aa212.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 070aa212.tex
+% Pt.212. Fig.a.  (23,16 )
+\PGset[0.8em]
+\begin{picture}   (23,16 )
+\drawline     ( 1, 8 )( 8,15 )(22, 8 )( 8, 1 )( 1, 8 )   %ABCDA
+\drawline(1,8)(22,8) % AC
+\drawline(8,1)(8,15) % DB
+\put( 0.1   ,  7.6 ){$\scriptstyle A$}   %A
+\put( 7.6   , 15.2 ){$\scriptstyle B$}   %B
+\put(22.2   ,  7.6 ){$\scriptstyle C$}   %C
+\put( 7.6   ,  0.1 ){$\scriptstyle D$}   %D
+\put( 8.2   ,  8.2 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/071aa213.eepic b/33063-t/images/sources/071aa213.eepic
new file mode 100644
index 0000000..b04535f
--- /dev/null
+++ b/33063-t/images/sources/071aa213.eepic
@@ -0,0 +1,32 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 071aa213.tex
+% Pt.213. Fig.a.  (24,16 )
+\PGset[0.8em]
+\begin{picture}   (24,16 )
+\drawline     ( 4,11 )( 9,13 )(15,13 )(20,11 )(20, 5 )(15, 3 )( 9, 3 )( 4, 5 )( 4,11 )   %A(I)BC(N)DEFG(L)HA
+\drawline(1,8)(23,8) % X'X
+\drawline(12,1)(12,15) % Y'Y
+\dashline[80]{0.2}( 6.5, 4 )( 6.5,12 )(17.5,12 )( 6.5, 4 )   %L(K)I(M)N(O)L
+\dashline[80]{0.2}( 6.5, 8 )(12  ,12 )                       %KM
+
+\put( 3.2   , 11.2 ){$\scriptstyle A$}   %A
+\put( 8.5   , 13.2 ){$\scriptstyle B$}   %B
+\put(14.6   , 13.2 ){$\scriptstyle C$}   %C
+\put(20     , 11.2 ){$\scriptstyle D$}   %D
+\put(19.9   ,  4.1 ){$\scriptstyle E$}   %E
+\put(14.5   ,  2.1 ){$\scriptstyle F$}   %F
+\put( 8.5   ,  2.1 ){$\scriptstyle G$}   %G
+\put( 3.1   ,  4.1 ){$\scriptstyle H$}   %H
+\put( 6.2   , 12.2 ){$\scriptstyle I$}   %I
+\put( 5.5   ,  7.1 ){$\scriptstyle K$}   %K
+\put( 6     ,  3.1 ){$\scriptstyle L$}   %L
+\put(10.9   , 12.1 ){$\scriptstyle M$}   %M
+\put(17.5   , 12.2 ){$\scriptstyle N$}   %N
+\put(12.2   ,  7.1 ){$\scriptstyle O$}   %O
+\put(23.1   ,  7.7 ){$\scriptstyle X$}   %X
+\put( 0.1   ,  7.8 ){$\scriptstyle X'$}   %X'
+\put(11.7   , 15.2 ){$\scriptstyle Y$}   %Y
+\put(11.6   ,  0.1 ){$\scriptstyle Y'$}   %Y'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/074aaZ19.eepic b/33063-t/images/sources/074aaZ19.eepic
new file mode 100644
index 0000000..fa26f12
--- /dev/null
+++ b/33063-t/images/sources/074aaZ19.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 074aaZ19.tex
+% Ex.19. Fig.a.  (10,10 )
+\PGset[0.8em]
+\begin{picture}  (10,10 )
+\drawline     ( 7, 9 )( 1, 5 )( 9, 5 )( 7, 9 )   %AB(D)CA
+\drawline     ( 5, 5 )( 7, 9 )                   %DA
+\dashline[80]{0.2}( 1, 5 )( 3, 1 )( 5, 5 )       %BED
+\put( 6.5   ,  9.2 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  4.4 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  4.2 ){$\scriptstyle C$}   %C
+\put( 5     ,  4.1 ){$\scriptstyle D$}   %D
+\put( 2.7   ,  0.1 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/074bbZ20.eepic b/33063-t/images/sources/074bbZ20.eepic
new file mode 100644
index 0000000..8991e45
--- /dev/null
+++ b/33063-t/images/sources/074bbZ20.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 074bbZ20.tex
+% Ex.20. Fig.b.  ( 8, 8 )
+\PGset[0.8em]
+\begin{picture}  ( 8, 8 )
+\drawline     ( 2.5, 7 )( 0.5, 1 )( 6.5, 1 )( 2.5, 7 )   %A(D)BC(E)A
+\drawline     ( 1.5, 4 )( 4.5, 4 )                   %DE
+\dashline[80]{0.2}( 6.5, 1 )( 7.5, 4 )( 4.5, 4 )         %CGE
+\put( 2.2   ,  7.2 ){$\scriptstyle A$}   %A
+\put( 0     ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6.3   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 0.6   ,  3.9 ){$\scriptstyle D$}   %D
+\put( 4.5   ,  4.2 ){$\scriptstyle E$}   %E
+\put( 7.1   ,  4.2 ){$\scriptstyle G$}   %G
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/075aaZ21.eepic b/33063-t/images/sources/075aaZ21.eepic
new file mode 100644
index 0000000..52278bd
--- /dev/null
+++ b/33063-t/images/sources/075aaZ21.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 075aaZ21.tex
+% Ex.21. Fig.a.  (12, 8 )
+\PGset[0.8em]
+\begin{picture}  (12, 8 )
+\drawline     ( 1, 7 )(11, 1 )( 1, 1 )( 1, 7 )   %A(D)BC(E)A
+\dashline[80]{0.2}( 1, 1 )( 6, 4 )( 6, 1 )           %CDE
+\put( 0.6   ,  7.2 ){$\scriptstyle A$}   %A
+\put(11.1   ,  0.5 ){$\scriptstyle B$}   %B
+\put( 0.2   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 6     ,  4.2 ){$\scriptstyle D$}   %D
+\put( 5.6   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/075bbZ22.eepic b/33063-t/images/sources/075bbZ22.eepic
new file mode 100644
index 0000000..ffcd6d6
--- /dev/null
+++ b/33063-t/images/sources/075bbZ22.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 075bbZ22.tex
+% Ex.22. Fig.b.  (12, 8 )
+\PGset[0.8em]
+\begin{picture}  (12, 8 )
+\drawline     ( 1, 7 )(11, 1 )( 1, 1 )( 1, 7 )   %A(D)BC(E)A
+\drawline     ( 1, 1 )( 6, 4 )                   %CD
+\dashline[80]{0.2}( 6, 4 )( 6, 1 )                   %DE
+\put( 0.6   ,  7.2 ){$\scriptstyle A$}   %A
+\put(11.1   ,  0.5 ){$\scriptstyle B$}   %B
+\put( 0.2   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 6     ,  4.2 ){$\scriptstyle D$}   %D
+\put( 5.6   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 9.2   ,  1.1 ){\emph{a}}   %a
+\put( 2.7   ,  1.1 ){\emph{b}}   %b
+\put( 1.1   ,  1.6 ){\emph{c}}   %c
+\put( 1.1   ,  5.5 ){2\emph{a}}   %2a
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/075ceZ23.eepic b/33063-t/images/sources/075ceZ23.eepic
new file mode 100644
index 0000000..43f46d1
--- /dev/null
+++ b/33063-t/images/sources/075ceZ23.eepic
@@ -0,0 +1,38 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 075ccZ23.tex
+% Ex.23. Fig.c.  ( 9, 6 )
+\PGset[0.8em]
+\begin{picture}  ( 9, 6 )
+\drawline     ( 1, 1 )( 8, 1 )( 3, 5 )( 1, 1 )   %A(D)BCA
+\dashline[80]{0.2}( 3, 5 )( 5, 1 )           %CD
+\put( 0.6   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 7.8   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 2.7   ,  5.2 ){$\scriptstyle C$}   %C
+\put( 4.6   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
+%
+\quad
+% 075ddZ23.tex
+% Ex.23. Fig.d.  ( 9, 6 )
+\PGset[0.8em]
+\begin{picture}  ( 9, 6 )
+\drawline     ( 1, 1 )( 8, 1 )( 3, 5 )( 1, 1 )   %A'B'C'A'
+\put( 0.6   ,  0.15 ){$\scriptstyle A'$}   %A'
+\put( 7.8   ,  0.2 ){$\scriptstyle B'$}   %B'
+\put( 2.7   ,  5.1 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
+%
+\quad
+% 075eeZ23.tex
+% Ex.23. Fig.e.  ( 7, 6 )
+\PGset[0.8em]
+\begin{picture}  ( 7, 6 )
+\drawline     ( 3, 1 )( 6, 1 )( 1, 5 )( 3, 1 )   %A'B'C'A'
+\put( 2.6   ,  0.15 ){$\scriptstyle A'$}   %A'
+\put( 5.8   ,  0.2 ){$\scriptstyle B'$}   %B'
+\put( 0.7   ,  5.1 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/076aaZ24.eepic b/33063-t/images/sources/076aaZ24.eepic
new file mode 100644
index 0000000..37e7be0
--- /dev/null
+++ b/33063-t/images/sources/076aaZ24.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 076aaZ24.tex
+% Ex.24. Fig.a.  (10, 8 )
+\PGset[0.8em]
+\begin{picture}  (10, 8 )
+\drawline     ( 1, 1 )( 9, 1 )( 3, 7 )( 1, 1 )   %ABCA
+\drawline     ( 1, 1 )( 4.75, 3.5  )               %AD
+\drawline     ( 9, 1 )( 3.2 , 3.32 )               %BE
+\drawline     ( 3, 7 )( 4.2 , 2.2  )               %CF
+\put( 0.1   ,  0.7 ){$\scriptstyle A$}   %A
+\put( 9.1   ,  0.7 ){$\scriptstyle B$}   %B
+\put( 2.6   ,  7.2 ){$\scriptstyle C$}   %C
+\put( 4.8   ,  3.3 ){$\scriptstyle D$}   %D
+\put( 2.4   ,  3.2 ){$\scriptstyle E$}   %E
+\put( 4.1   ,  1.6 ){$\scriptstyle F$}   %F
+\put( 3.3   ,  1.9 ){$\scriptstyle O$}   %O 4,3
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/076bbZ25.eepic b/33063-t/images/sources/076bbZ25.eepic
new file mode 100644
index 0000000..2a4bfce
--- /dev/null
+++ b/33063-t/images/sources/076bbZ25.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 076bbZ25.tex
+% Ex.25. Fig.b.  ( 9, 7 )
+\PGset[0.8em]
+\begin{picture}  ( 9, 7 )
+\drawline     ( 3, 6 )( 1, 1 )( 8, 1 )( 3, 6 )   %A(D)B(F)C(E)A
+\drawline     ( 2  , 3.5 )( 5.25, 2.2  )               %DD'
+\drawline     ( 5.5, 3.5 )( 4.2 , 2.2  )               %EE'
+\drawline     ( 4.5, 1   )( 4.5 , 2.9  )               %FF'
+\put( 2.6   ,  6.2 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.7 ){$\scriptstyle B$}   %B
+\put( 8.2   ,  0.7 ){$\scriptstyle C$}   %C
+\put( 1.1   ,  3.3 ){$\scriptstyle D$}   %D
+\put( 5.6   ,  3.6 ){$\scriptstyle E$}   %E
+\put( 4.1   ,  0.1 ){$\scriptstyle F$}   %F
+\put( 5.2   ,  1.7 ){$\scriptstyle D'$}   %D'
+\put( 3.2   ,  1.3 ){$\scriptstyle E'$}   %E'
+\put( 4.1   ,  3   ){$\scriptstyle F'$}   %F'
+\put( 5     ,  2.4 ){$\scriptstyle O$}   %O 4.5, 2.5
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/076ccZ26.eepic b/33063-t/images/sources/076ccZ26.eepic
new file mode 100644
index 0000000..cc5b01c
--- /dev/null
+++ b/33063-t/images/sources/076ccZ26.eepic
@@ -0,0 +1,23 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 076ccZ26.tex
+% Ex.26. Fig.c.  (13.3, 9 )
+\PGset[0.8em]
+\begin{picture}  (13.3, 9 )
+\dashline[80]{0.2}( 1  , 8 )(12    , 8     )( 8.333, 0.667 )( 1  , 8 )   %C'(A)B'(C)A'(B)C'
+\drawline     ( 6.5, 8 )( 4.667, 4.333 )(10.167, 4.333 )( 6.5, 8 )   %A(K)B(H)C(P)A
+\drawline     ( 6.5  , 8     )( 6.5   , 4.333   )               %AH
+\drawline     ( 4.667, 4.333 )( 7.417 , 7.083   )               %BP
+\drawline     (10.167, 4.333 )( 5.767 , 6.533   )               %CK
+\put( 6.1   ,  8.2 ){$\scriptstyle A$}   %A
+\put( 3.8   ,  3.5 ){$\scriptstyle B$}   %B
+\put(10.2   ,  3.6 ){$\scriptstyle C$}   %C
+\put( 6     ,  3.5 ){$\scriptstyle H$}   %H
+\put( 4.9   ,  6.3 ){$\scriptstyle K$}   %K
+\put( 6.6   ,  5   ){$\scriptstyle O$}   %O 6.5, 6.166...
+\put( 7.6   ,  7   ){$\scriptstyle P$}   %P
+\put( 8.3   ,  0.1 ){$\scriptstyle A'$}   %A'
+\put(12.1   ,  7.7 ){$\scriptstyle B'$}   %B'
+\put( 0     ,  7.8 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/076ddZ27.eepic b/33063-t/images/sources/076ddZ27.eepic
new file mode 100644
index 0000000..eee9ace
--- /dev/null
+++ b/33063-t/images/sources/076ddZ27.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 076ddZ27.tex
+% Ex.27. Fig.d.  (10, 9 )
+\PGset[0.8em]
+\begin{picture}  (10, 9 )
+\drawline     ( 0.5, 1   )( 9.5, 1   )( 7.5, 8 )( 0.5, 1 )   %A(E)B(D)C(H)A
+\drawline     ( 0.5, 1   )( 8.5, 4.5 )                      %AD
+\drawline     ( 9.5, 1   )( 4  , 4.5 )                      %BH
+\drawline     ( 7.5, 8   )( 5  , 1   )                      %CE
+\dashline[80]{0.2}( 8.5, 4.5 )( 5  , 1   )( 3.167, 2.167 )( 6.667, 5.667 )( 8.5, 4.5 )   %DEFGD
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 9     ,  0.2 ){$\scriptstyle B$}   %B
+\put( 7.4   ,  8.1 ){$\scriptstyle C$}   %C
+\put( 8.6   ,  4.5 ){$\scriptstyle D$}   %D
+\put( 4.6   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 2.6   ,  2.2 ){$\scriptstyle F$}   %F
+\put( 5.8   ,  5.6 ){$\scriptstyle G$}   %G
+\put( 3.2   ,  4.5 ){$\scriptstyle H$}   %H
+\put( 5.65  ,  2.28 ){$\scriptstyle O$}   %O 5.833..., 3.333...
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/077adZ28.eepic b/33063-t/images/sources/077adZ28.eepic
new file mode 100644
index 0000000..4d62ea3
--- /dev/null
+++ b/33063-t/images/sources/077adZ28.eepic
@@ -0,0 +1,72 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 077aaZ28.tex
+% Ex.28. Fig.a.   ( 7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 7 )
+\put( 0.5   ,  1   ){\line  ( 1, 0 ){ 6 }}             %E(A)F
+\put( 3.5   ,  1   ){\line  (-2, 5 ){ 2 }}             %AC
+\put( 3.5   ,  1   ){\line  ( 0, 1 ){ 5 }}             %AD
+\put( 3.5   ,  1   ){\line  ( 2, 5 ){ 2 }}             %AB
+\put( 3.1   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 5.3   ,  6.1 ){$\scriptstyle B$}   %B
+\put( 1.1   ,  6.1 ){$\scriptstyle C$}   %C
+\put( 3.1   ,  6.1 ){$\scriptstyle D$}   %D
+\put( 0.1   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 6     ,  0.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
+%
+\quad
+% 077bbZ28.tex
+% Ex.28. Fig.b.   ( 8.5, 5.6 )
+\PGset[0.8em]
+\begin{picture}   ( 8.5, 5.6 )
+\put( 0.5   ,  1   ){\line  ( 1, 0 ){ 7   }}             %D(A)B
+\put( 4     ,  1   ){\line  (-5, 6 ){ 2.5 }}             %AF
+\put( 4     ,  1   ){\line  ( 1, 2 ){ 2   }}             %AC
+\put( 4     ,  1   ){\line  ( 6, 5 ){ 3.5 }}             %AE
+\put( 3.7   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 7.1   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 5.2   ,  4.8 ){$\scriptstyle C$}   %C
+\put( 0     ,  0.2 ){$\scriptstyle D$}   %D
+\put( 7.5   ,  3.8 ){$\scriptstyle E$}   %E
+\put( 1.1   ,  4   ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
+%
+\quad
+% 077ccZ28.tex
+% Ex.28. Fig.c.   ( 7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 7 )
+\put( 0.5   ,  3.5 ){\line  ( 1, 0 ){ 6   }}             %D(A)B
+\put( 1     ,  2   ){\line  ( 5, 3 ){ 5   }}             %G(A)F
+\put( 2     ,  1   ){\line  ( 3, 5 ){ 3   }}             %E(A)C
+\put( 3.3   ,  2.7 ){$\scriptstyle A$}   %A 3.5, 3.5
+\put( 6.1   ,  2.7 ){$\scriptstyle B$}   %B
+\put( 4.8   ,  6.1 ){$\scriptstyle C$}   %C
+\put( 0.2   ,  2.7 ){$\scriptstyle D$}   %D
+\put( 1.5   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 6.1   ,  4.9 ){$\scriptstyle F$}   %F
+\put( 0.2   ,  1.3 ){$\scriptstyle G$}   %G
+\end{picture}
+\PGrestore
+%
+\quad
+% 077ddZ28.tex
+% Ex.28. Fig.d.   ( 7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 7 )
+\put( 0.5   ,  3.5 ){\line  ( 1, 0 ){ 6   }}             %D(A)B
+\put( 1     ,  2   ){\line  ( 5, 3 ){ 5   }}             %G(A)F
+\put( 2     ,  1   ){\line  ( 3, 5 ){ 3   }}             %E(A)C
+\put( 2     ,  6   ){\line  ( 3,-5 ){ 3   }}             %ul-lr, perp to G(A)F
+\put( 3.1   ,  2.4 ){$\scriptstyle A$}   %A 3.5, 3.5
+\put( 6.1   ,  2.7 ){$\scriptstyle B$}   %B
+\put( 4.8   ,  6.1 ){$\scriptstyle C$}   %C
+\put( 0.2   ,  2.7 ){$\scriptstyle D$}   %D
+\put( 1.5   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 6.1   ,  4.9 ){$\scriptstyle F$}   %F
+\put( 0.2   ,  1.3 ){$\scriptstyle G$}   %G
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/077eeZ34.eepic b/33063-t/images/sources/077eeZ34.eepic
new file mode 100644
index 0000000..b8cd8f4
--- /dev/null
+++ b/33063-t/images/sources/077eeZ34.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 077eeZ34.tex
+% Ex.34. Fig.e.   ( 5.8, 6 )
+\PGset[0.8em]
+\begin{picture}   ( 5.8, 6 )
+\drawline    ( 0.5, 1 )( 5.3, 1 )( 2.9, 5 )( 0.5, 1 )  %A(D)BCA
+\drawline(2.9,1)(2.9,5) % DC
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 4.9   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 2.7   ,  5.2 ){$\scriptstyle C$}   %C
+\put( 2.6   ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/077ffZ35.eepic b/33063-t/images/sources/077ffZ35.eepic
new file mode 100644
index 0000000..cbc6d4e
--- /dev/null
+++ b/33063-t/images/sources/077ffZ35.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 077ffZ35.tex
+% Ex.35. Fig.f.   ( 7.7, 9 )
+\PGset[0.8em]
+\begin{picture}   ( 7.7, 9 )
+\drawline    ( 0.567, 1 )( 7.233, 1 )( 3.9, 6 )( 0.567, 1 )  %A(D)BCA
+\drawline(3.9,1)(3.9,8) % DE
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 6.8   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 3.0   ,  5.9 ){$\scriptstyle C$}   %C
+\put( 3.4   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 3.6   ,  8.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/078aaZ41.eepic b/33063-t/images/sources/078aaZ41.eepic
new file mode 100644
index 0000000..7bfeea3
--- /dev/null
+++ b/33063-t/images/sources/078aaZ41.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 078aaZ41.tex
+% Ex.41. Fig.a.   ( 9, 8.5 )
+\PGset[0.8em]
+\begin{picture}   ( 9, 8.5 )
+\drawline    ( 4.4, 7.5 )( 0.5, 1 )( 6.5, 1 )( 3.5, 6 )( 9, 6 )  %D(A)BCAE
+\put( 2.6   ,  5.9 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6.1   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 4.1   ,  7.7 ){$\scriptstyle D$}   %D
+\put( 8.2   ,  6.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/078bbZ43.eepic b/33063-t/images/sources/078bbZ43.eepic
new file mode 100644
index 0000000..acde9c6
--- /dev/null
+++ b/33063-t/images/sources/078bbZ43.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 078bbZ43.tex
+% Ex.43. Fig.b.   ( 5.9, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 5.9, 8 )
+\drawline    ( 0.5, 1 )( 5.3, 1 )( 5.3, 7 )( 0.5, 1 )  %ABD(C)A
+\drawline(2.9,4)(5.3,1) % CB
+\put( 0.1   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 5.1   ,  0.3 ){$\scriptstyle B$}   %B
+\put( 2.1   ,  3.9 ){$\scriptstyle C$}   %C
+\put( 4.9   ,  7.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/078ccZ46.eepic b/33063-t/images/sources/078ccZ46.eepic
new file mode 100644
index 0000000..b171fc0
--- /dev/null
+++ b/33063-t/images/sources/078ccZ46.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 078ccZ46.tex
+% Ex.46. Fig.c.   (10, 7 )
+\PGset[0.8em]
+\begin{picture}   (10, 7 )
+\put( 1   ,  1   ){\line  ( 3, 4 ){ 3.75 }}         %B(H)A
+\put( 1   ,  1   ){\line  ( 2, 1 ){ 8    }}         %B(P)D
+\put( 1   ,  1   ){\line  ( 1, 0 ){ 8    }}         %BC
+\put( 3.25,  4   ){\line  ( 1, 0 ){ 5.75 }}         %H(P)K
+\put( 4.4   ,  6.2 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.3 ){$\scriptstyle B$}   %B
+\put( 8.2   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 8.6   ,  5.2 ){$\scriptstyle D$}   %D
+\put( 2.4   ,  3.9 ){$\scriptstyle H$}   %H
+\put( 9.0   ,  3.7 ){$\scriptstyle K$}   %K
+\put( 6.8   ,  3.2 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/078ddZ49.eepic b/33063-t/images/sources/078ddZ49.eepic
new file mode 100644
index 0000000..b22b7e6
--- /dev/null
+++ b/33063-t/images/sources/078ddZ49.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 078ddZ49.tex
+% Ex.49. Fig.d.   ( 7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 7 )
+\drawline    ( 1  , 1 )( 6  , 1   )( 5, 6   )( 1  , 1 )  %ABCA
+\drawline    ( 3.5, 1 )( 5.5, 3.5 )( 3, 3.5 )( 3.5, 1 )  %DEFD
+\put( 0.4   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 5.9   ,  0.3 ){$\scriptstyle B$}   %B
+\put( 4.6   ,  6.1 ){$\scriptstyle C$}   %C
+\put( 3.1   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 5.65  ,  3.3 ){$\scriptstyle E$}   %E
+\put( 2.2   ,  3.3 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/079adZ51.eepic b/33063-t/images/sources/079adZ51.eepic
new file mode 100644
index 0000000..2ba16b3
--- /dev/null
+++ b/33063-t/images/sources/079adZ51.eepic
@@ -0,0 +1,70 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 079aaZ51.tex
+% Ex.51. Fig.a.   ( 7, 9.5 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 9.5 )
+\drawline    ( 3.5, 8.5 )( 0.5  , 1   )( 6.5, 1   )( 3.5, 8.5 )  %A(E)BC(D)A
+\put( 0.5 ,  1   ){\line  ( 5, 2 ){ 5.172  }}         %BD
+\put( 6.5 ,  1   ){\line  (-5, 2 ){ 5.172  }}         %CE
+\put( 3.2   ,  8.7 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6     ,  0.2 ){$\scriptstyle C$}   %C
+\put( 5.7   ,  3.2 ){$\scriptstyle D$}   %D
+\put( 0.6   ,  3.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
+%
+\qquad
+% 079bbZ51.tex
+% Ex.51. Fig.b.   ( 7, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 8 )
+\drawline    ( 3.5, 7 )( 0.5  , 1   )( 6.5, 1   )( 3.5, 7 )  %A(E)BC(D)A
+\drawline              ( 0.5  , 1   )( 5  , 4   )            %BD
+\drawline                            ( 6.5, 1   )( 2  , 4 )  %CE
+\put( 3.2   ,  7.2 ){$\scriptstyle A$}   %A
+\put( 0.2   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6     ,  0.2 ){$\scriptstyle C$}   %C
+\put( 5.1   ,  4   ){$\scriptstyle D$}   %D
+\put( 1.2   ,  4   ){$\scriptstyle E$}   %E
+\put( 3.14  ,  3.3 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+\qquad
+% 079ccZ51.tex
+% Ex.51. Fig.c.   ( 7, 9.5 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 9.5 )
+\drawline    ( 3.5, 8.5 )( 0.5  , 1   )( 6.5, 1   )( 3.5, 8.5 )  %A(E)BC(D)A
+\put( 0.5 ,  1   ){\line  ( 3, 2 ){ 4.737  }}         %BD
+\put( 6.5 ,  1   ){\line  (-3, 2 ){ 4.737  }}         %CE
+\put( 3.2   ,  8.7 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6     ,  0.2 ){$\scriptstyle C$}   %C
+\put( 5.3   ,  4.1 ){$\scriptstyle D$}   %D
+\put( 1     ,  4.1 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
+%
+\qquad
+% 079ddZ51.tex
+% Ex.51. Fig.d.   (11.5,10 )
+\PGset[0.8em]
+\begin{picture}   (11.5,10 )
+\drawline     ( 3, 9 )( 1, 1   )(11, 1   )( 3, 9 )  %A(D)BC(FE)A
+\put( 1 ,  1   ){\line  ( 5, 4 ){ 5.556  }}         %B(KH)E
+\put(11 ,  1   ){\line  (-3, 1 ){ 9.231  }}         %C(GK)D
+\dashline[80]{0.2}( 8.805, 3.195 )( 7.514, 2.162 )( 5.729, 4.783 )  %FGH
+\put( 2.5   ,  9.2 ){$\scriptstyle A$}   %A
+\put( 0.3   ,  0.2 ){$\scriptstyle B$}   %B
+\put(10.6   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 0.8   ,  3.9 ){$\scriptstyle D$}   %D
+\put( 6.6   ,  5.5 ){$\scriptstyle E$}   %E
+\put( 8.9   ,  3.3 ){$\scriptstyle F$}   %F
+\put( 6.8   ,  1.4 ){$\scriptstyle G$}   %G
+\put( 4.8   ,  4.9 ){$\scriptstyle H$}   %H
+\put( 3.6   ,  3.7 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/079eeZ57.eepic b/33063-t/images/sources/079eeZ57.eepic
new file mode 100644
index 0000000..fce414b
--- /dev/null
+++ b/33063-t/images/sources/079eeZ57.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 079eeZ57.tex
+% Ex.57. Fig.e.   ( 6, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 6, 7 )
+\drawline    ( 3, 6 )( 1, 1 )( 5, 1 )( 3, 6 )  %A(E)B(D)CA
+\put( 3 ,  1   ){\line  (-5, 2 ){ 1.724 }}     %DE
+\put( 3 ,  1   ){\line  ( 5, 2 ){ 1.724 }}     %DF
+\put( 2.55  ,  6.2 ){$\scriptstyle A$}   %A
+\put( 0.3   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 4.9   ,  0.3 ){$\scriptstyle C$}   %C
+\put( 2.55   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 0.5   ,  1.7 ){$\scriptstyle E$}   %E
+\put( 4.8   ,  1.7 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/080aaZ59.eepic b/33063-t/images/sources/080aaZ59.eepic
new file mode 100644
index 0000000..39c4812
--- /dev/null
+++ b/33063-t/images/sources/080aaZ59.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080aaZ59.tex
+% Ex.59. Fig.a.   (11.5, 9 )
+\PGset[0.8em]
+\begin{picture}   (11.5, 9 )
+\drawline     ( 7.45, 0.1 )( 3.5, 8   )( 1  , 3   )(11.4, 3 )  %lr(FC)A(E)B(CD)farright
+\drawline     ( 6.6 , 1.8 )( 9  , 3   )( 2.6, 6.2 )            %FD(G)E
+\dashline[80]{0.2}( 6   , 3   )( 8.2, 7.4 )                        %C(G)H
+\put( 3.05  ,  8.2 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  2.5 ){$\scriptstyle B$}   %B
+\put( 5.2   ,  2.2 ){$\scriptstyle C$}   %C
+\put( 8.9   ,  3.2 ){$\scriptstyle D$}   %D
+\put( 1.8   ,  6.1 ){$\scriptstyle E$}   %E
+\put( 5.8   ,  1.05 ){$\scriptstyle F$}   %F
+\put( 7     ,  4.2 ){$\scriptstyle G$}   %G
+\put( 8.1   ,  7.6 ){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/080bbZ60.eepic b/33063-t/images/sources/080bbZ60.eepic
new file mode 100644
index 0000000..2911b90
--- /dev/null
+++ b/33063-t/images/sources/080bbZ60.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080bbZ60.tex
+% Ex.60. Fig.b.   (10,10 )
+\PGset[0.8em]
+\begin{picture}   (10,10 )
+\drawline     ( 1, 1 )( 9, 1   )( 5  , 9   )( 1, 1 )  %A(P)B(D)C(FE)A
+\drawline     ( 1.6, 2.2 )( 4  , 1   )( 8, 3 )            %EPD
+\put( 9 ,  1   ){\line  (-2, 1 ){ 6.4 }}                     %B(G)F
+\dashline[80]{0.2}( 4   , 1   )( 5, 3 )                        %PG
+\put( 0.3   ,  0.3 ){$\scriptstyle A$}   %A
+\put( 9     ,  0.5 ){$\scriptstyle B$}   %B
+\put( 4.6   ,  9.1 ){$\scriptstyle C$}   %C
+\put( 8.15  ,  2.8 ){$\scriptstyle D$}   %D
+\put( 0.8   ,  2.2 ){$\scriptstyle E$}   %E
+\put( 1.9   ,  4.1 ){$\scriptstyle F$}   %F
+\put( 5     ,  3.1 ){$\scriptstyle G$}   %G
+\put( 3.6   ,  0.2 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/080ccZ61.eepic b/33063-t/images/sources/080ccZ61.eepic
new file mode 100644
index 0000000..91d3c5d
--- /dev/null
+++ b/33063-t/images/sources/080ccZ61.eepic
@@ -0,0 +1,24 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080ccZ61.tex
+% Ex.61. Fig.c.   ( 9.2, 9 )
+\PGset[0.8em]
+\begin{picture}   ( 9.2, 9 )
+\drawline     ( 4.5  , 7.928 )( 0.5, 1   )( 8.5  , 1     )( 4.5, 7.928 )  %A(GH)B(ED)C(KF)A
+\drawline     ( 2.296, 4.111 )( 3.7, 3.3 )( 6.304, 4.803 )               %GPF
+\drawline(4.5,1)(4.5,7.928) % DA
+\drawline(3.7,3.3)(3.7,1) % PE
+\dashline[80]{0.2}( 1.828, 3.3   )( 7.172, 3.3 )                   %H(PM)K
+\put( 4.1   ,  8.1 ){$\scriptstyle A$}   %A
+\put( 0.1   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 8.3   ,  0.2 ){$\scriptstyle C$}   %C
+\put( 4.2   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 3.2   ,  0.2 ){$\scriptstyle E$}   %E
+\put( 6.3   ,  4.9 ){$\scriptstyle F$}   %F
+\put( 1.5   ,  4.1 ){$\scriptstyle G$}   %G
+\put( 0.85   ,  3   ){$\scriptstyle H$}   %H
+\put( 7.2   ,  3.3 ){$\scriptstyle K$}   %K
+\put( 4.6   ,  2.5 ){$\scriptstyle M$}   %M 4.5 3.3
+\put( 2.8   ,  2.5 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/080ddZ62.eepic b/33063-t/images/sources/080ddZ62.eepic
new file mode 100644
index 0000000..648cf2d
--- /dev/null
+++ b/33063-t/images/sources/080ddZ62.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080ddZ62.tex
+% Ex.62. Fig.d.   ( 7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 7, 7 )
+\drawline     ( 1.5, 1 )( 6.5, 1 )( 5.5, 5 )( 1.5, 1 )  %ABCA
+\drawline     ( 1.5, 1 )( 0.5, 6 )( 6.5, 1 )            %ADB
+\put( 1.1   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 6     ,  0.2 ){$\scriptstyle B$}   %B
+\put( 5.2   ,  5.2 ){$\scriptstyle C$}   %C
+\put( 0.1   ,  6.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/080eeZ63.eepic b/33063-t/images/sources/080eeZ63.eepic
new file mode 100644
index 0000000..ddecbb7
--- /dev/null
+++ b/33063-t/images/sources/080eeZ63.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080eeZ63.tex
+% Ex.63. Fig.e.   ( 9, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 9, 8 )
+\drawline     ( 0.5, 1 )( 8.5, 1 )( 6.5, 7 )( 0.5, 1 )  %ABCA
+\put( 5.5   ,  3   ){\line  (-5,-2 ){ 5  }}             %OA
+\put( 5.5   ,  3   ){\line  ( 3,-2 ){ 3  }}             %OB
+\put( 5.5   ,  3   ){\line  ( 1, 4 ){ 1  }}             %OC
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 8     ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6.3   ,  7.2 ){$\scriptstyle C$}   %C
+\put( 4.95   ,  2.0 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/080ffZ64.eepic b/33063-t/images/sources/080ffZ64.eepic
new file mode 100644
index 0000000..889e3b7
--- /dev/null
+++ b/33063-t/images/sources/080ffZ64.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 080ffZ64.tex
+% Ex.64. Fig.f.   ( 6, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 6, 7 )
+\drawline    ( 3, 6 )( 1, 1 )( 5, 1 )( 3, 6 )  %A(E)B(D)CA
+\put( 2.4 ,  1   ){\line  (-2, 5 ){ 0.7 }}     %DE
+\put( 2.4 ,  1   ){\line  ( 2, 5 ){ 1.3 }}     %DF
+\put( 2.6   ,  6.2 ){$\scriptstyle A$}   %A
+\put( 0.3   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 4.9   ,  0.3 ){$\scriptstyle C$}   %C
+\put( 2.0   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 0.9   ,  2.6 ){$\scriptstyle E$}   %E
+\put( 3.8   ,  4.1 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/081aaZ65.eepic b/33063-t/images/sources/081aaZ65.eepic
new file mode 100644
index 0000000..00799b4
--- /dev/null
+++ b/33063-t/images/sources/081aaZ65.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081aaZ65.tex
+% Ex.65. Fig.a.   ( 8.5, 9.5 )
+\PGset[0.8em]
+\begin{picture}   ( 8.5, 9.5 )
+\drawline    ( 1  , 0.5 )( 4.2, 8.5 )( 7.2, 1.0 )               %M(B)A(C)N
+\drawline    ( 3.4, 0.5 )( 5.8, 4.5 )( 2.6, 4.5 )( 4.88, 0.7 )  %F(D)CB(D)E
+\drawline(4.2,8.5)(4.2,0.6) % A(D)
+\put( 3.8   ,  8.7 ){$\scriptstyle A$}   %A
+\put( 1.6   ,  4.2 ){$\scriptstyle B$}   %B
+\put( 5.95  ,  4.2 ){$\scriptstyle C$}   %C
+\put( 3.0   ,  1.4 ){$\scriptstyle D$}   %D
+\put( 4.9   ,  0.35 ){$\scriptstyle E$}   %E
+\put( 2.5   ,  0.1 ){$\scriptstyle F$}   %F
+\put( 0.1   ,  0.6 ){$\scriptstyle M$}   %M
+\put( 7.25  ,  1.1 ){$\scriptstyle N$}   %N
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/081bbZ66.eepic b/33063-t/images/sources/081bbZ66.eepic
new file mode 100644
index 0000000..6b4a676
--- /dev/null
+++ b/33063-t/images/sources/081bbZ66.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081bbZ66.tex
+% Ex.66. Fig.b.   ( 8, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 7 )
+\drawline    ( 6, 6 )( 1, 1 )( 7, 1 )( 6, 6 )  %A(E)BC(D)A
+\put( 1     ,  1   ){\line  ( 5, 2 ){ 5.2 }}   %B(O)ur
+\put( 7     ,  1   ){\line  (-5, 4 ){ 3.0 }}   %C(O)ul
+\put( 6.6   ,  2.6 ){\line  (-1, 0 ){ 4.35 }}   %D(OE)left
+\put( 5.6   ,  6.2 ){$\scriptstyle A$}   %A
+\put( 0.4   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 6.9   ,  0.3 ){$\scriptstyle C$}   %C
+\put( 6.7   ,  2.3 ){$\scriptstyle D$}   %D
+\put( 1.3   ,  2.3 ){$\scriptstyle E$}   %E
+\put( 4.8   ,  2.9 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/081ccZ67.eepic b/33063-t/images/sources/081ccZ67.eepic
new file mode 100644
index 0000000..fabed17
--- /dev/null
+++ b/33063-t/images/sources/081ccZ67.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081ccZ67.tex
+% Ex.67. Fig.c.   ( 9, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 9, 7 )
+\drawline    ( 1  , 1 )( 8  , 1   )( 6.3, 6.1 )( 1, 1 )  %A(DE)BCA
+\drawline    ( 4.6, 1 )( 6.3, 6.1 )( 6.3, 1   )          %DCE
+\put( 0.4   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 7.8   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 5.9   ,  6.3 ){$\scriptstyle C$}   %C
+\put( 4.1   ,  0.2 ){$\scriptstyle D$}   %D
+\put( 5.9   ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/081ddZ68.eepic b/33063-t/images/sources/081ddZ68.eepic
new file mode 100644
index 0000000..88da017
--- /dev/null
+++ b/33063-t/images/sources/081ddZ68.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081ddZ68.tex
+% Ex.68. Fig.d.   ( 9.4, 6.2 )
+\PGset[0.8em]
+\begin{picture}   ( 9.4, 6.2 )
+\drawline    ( 0.5, 1 )( 7.5, 1   )( 8.9, 5.2 )( 1.9, 5.2 )( 0.5, 1 )  %ABCDA
+\drawline    ( 0.5, 1 )( 8.9, 5.2 )                      %AC
+\drawline    ( 7.5, 1 )( 1.9, 5.2 )                      %BD
+\put( 0.2   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 6.95   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 8.5   ,  5.4 ){$\scriptstyle C$}   %C
+\put( 1.6   ,  5.4 ){$\scriptstyle D$}   %D
+\put( 4.15  ,  2.1 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/081eeZ69.eepic b/33063-t/images/sources/081eeZ69.eepic
new file mode 100644
index 0000000..6dd96d7
--- /dev/null
+++ b/33063-t/images/sources/081eeZ69.eepic
@@ -0,0 +1,15 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081eeZ69.tex
+% Ex.69. Fig.e.   (10, 5 )
+\PGset[0.8em]
+\begin{picture}   (10, 5 )
+\drawline    ( 1, 4.5 )( 9, 4.5 )( 9, 0.5 )( 1, 0.5 )( 1, 4.5 )  %ABCDA
+\drawline    ( 1, 4.5 )( 9, 0.5 )                                %AC
+\drawline    ( 9, 4.5 )( 1, 0.5 )                                %BD
+\put( 0.1   ,  4.2 ){$\scriptstyle A$}   %A
+\put( 9.1   ,  4.2 ){$\scriptstyle B$}   %B
+\put( 9.1   ,  0.1 ){$\scriptstyle C$}   %C
+\put( 0.05  ,  0.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/081ffZ73.eepic b/33063-t/images/sources/081ffZ73.eepic
new file mode 100644
index 0000000..71bfffb
--- /dev/null
+++ b/33063-t/images/sources/081ffZ73.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 081ffZ73.tex
+% Ex.73. Fig.f.   (10.4, 6.2 )
+\PGset[0.8em]
+\begin{picture}   (10.4, 6.2 )
+\drawline     ( 1   , 1   )( 7.3 , 1   )( 9.4, 5.2 )( 3.1, 5.2 )( 1, 1 )  %AB(F)CD(E)A
+\drawline     ( 1   , 1   )( 9.4 , 5.2 )                   %AC
+\dashline[80]{0.2}( 7.3 , 1   )( 2.05, 3.1 )                   %B(M)E
+\dashline[80]{0.2}( 8.35, 3.1 )( 3.1 , 5.2 )                   %F(N)D
+\put( 0.4   ,  0.2 ){$\scriptstyle A$}   %A
+\put( 7.2   ,  0.2 ){$\scriptstyle B$}   %B
+\put( 9.1   ,  5.3 ){$\scriptstyle C$}   %C
+\put( 2.3   ,  5.2 ){$\scriptstyle D$}   %D
+\put( 1.1   ,  2.9 ){$\scriptstyle E$}   %E
+\put( 8.45   ,  2.6 ){$\scriptstyle F$}   %F
+\put( 3.3   ,  1.35 ){$\scriptstyle M$}   %M
+\put( 6.1   ,  2.9 ){$\scriptstyle N$}   %N
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/082adZ74.eepic b/33063-t/images/sources/082adZ74.eepic
new file mode 100644
index 0000000..debe0e8
--- /dev/null
+++ b/33063-t/images/sources/082adZ74.eepic
@@ -0,0 +1,78 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 082aaZ74.tex
+% Ex.74. Fig.a.   (10, 8 )
+\PGset[0.8em]
+\begin{picture}   (10, 8 )
+\drawline(1  ,  1)(9  ,  1)(8  ,7  )(3  ,6  )(1  ,1)  %ABCDA
+\drawline(5  ,  1)(8.5,  4)(5.5,6.5)(2  ,3.5)(5  ,1)  %EFGHE
+\dashline[80]{0.2}    (  1,  1)( 8,7)   %AC
+\dashline[80]{0.2}    (  9,  1)( 3,6)   %BD
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  9   ,  0.2 ){$\scriptstyle B$}   %B
+\put(  8.1 ,  7.1 ){$\scriptstyle C$}   %C
+\put(  2   ,  6   ){$\scriptstyle D$}   %D
+\put(  4.5 ,  0.2 ){$\scriptstyle E$}   %E
+\put(  8.6 ,  3.7 ){$\scriptstyle F$}   %F
+\put(  5.05,  6.6 ){$\scriptstyle G$}   %G
+\put(  1   ,  3.15){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
+%
+% 082bbZ74.tex
+% Ex.74. Fig.b.   ( 8,10 )
+\PGset[0.8em]
+\begin{picture}   ( 8,10 )
+\drawline( 4  , 1  )( 7  , 5  )( 4  , 9  )( 1  , 5  )( 4  , 1  )  %ABCDA
+\drawline( 5.5, 3  )( 5.5, 7  )( 2.5, 7  )( 2.5, 3  )( 5.5, 3  )  %EFGHE
+\dashline[80]{0.2}      ( 4  , 1  )( 4  , 9  )   %AC
+\dashline[80]{0.2}      ( 7  , 5  )( 1  , 5  )   %BD
+\put(  3.6 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  7.1 ,  4.7 ){$\scriptstyle B$}   %B
+\put(  3.6 ,  9.2 ){$\scriptstyle C$}   %C
+\put(  0   ,  4.7 ){$\scriptstyle D$}   %D
+\put(  5.5 ,  2.2 ){$\scriptstyle E$}   %E
+\put(  5.7 ,  7.0 ){$\scriptstyle F$}   %F
+\put(  1.6 ,  7.0 ){$\scriptstyle G$}   %G
+\put(  1.5 ,  2.2 ){$\scriptstyle H$}   %H
+\put(  4.1 ,  4.1 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+% 082ccZ74.tex
+% Ex.74. Fig.c.   (12, 7 )
+\PGset[0.8em]
+\begin{picture}   (12, 7 )
+\drawline( 1  , 1  )(11  , 1  )(11  , 6  )( 1  , 6  )( 1  , 1  )  %ABCDA
+\drawline( 6  , 1  )(11  , 3.5)( 6  , 6  )( 1  , 3.5)( 6  , 1  )  %EFGHE
+%\dashline{0.5}      ( 4  , 1  )( 4  , 9  )   %AC
+%\dashline{0.5}      ( 7  , 5  )( 1  , 5  )   %BD
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put( 10.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put( 10.7 ,  6.2 ){$\scriptstyle C$}   %C
+\put(  0.3 ,  6.2 ){$\scriptstyle D$}   %D
+\put(  5.5 ,  0.2 ){$\scriptstyle E$}   %E
+\put( 11.1 ,  3.2 ){$\scriptstyle F$}   %F
+\put(  5.6 ,  6.2 ){$\scriptstyle G$}   %G
+\put(  0   ,  3.2 ){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
+%
+% 082ddZ74.tex
+% Ex.74. Fig.d.   ( 8, 8 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 8 )
+\drawline( 1  , 1  )( 7  , 1  )( 7  , 7  )( 1  , 7  )( 1  , 1  )  %ABCDA
+\drawline( 4  , 1  )( 7  , 4  )( 4  , 7  )( 1  , 4  )( 4  , 1  )  %EFGHE
+%\dashline{0.5}      ( 4  , 1  )( 4  , 9  )   %AC
+%\dashline{0.5}      ( 7  , 5  )( 1  , 5  )   %BD
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  6.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  6.7 ,  7.2 ){$\scriptstyle C$}   %C
+\put(  0.3 ,  7.2 ){$\scriptstyle D$}   %D
+\put(  3.5 ,  0.2 ){$\scriptstyle E$}   %E
+\put(  7.1 ,  3.7 ){$\scriptstyle F$}   %F
+\put(  3.6 ,  7.2 ){$\scriptstyle G$}   %G
+\put(  0   ,  3.7 ){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/082ehZ78.eepic b/33063-t/images/sources/082ehZ78.eepic
new file mode 100644
index 0000000..c09cbbe
--- /dev/null
+++ b/33063-t/images/sources/082ehZ78.eepic
@@ -0,0 +1,78 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 082eeZ78.tex
+% Ex.78. Fig.e.   (10,10 )
+\PGset[0.8em]
+\begin{picture}   (10,10 )
+\drawline( 1  , 1  )( 9  , 1  )( 7  , 9  )( 3  , 9  )( 1  , 1  )  %ABCDA
+\drawline( 5  , 1  )( 8  , 5  )( 5  , 9  )( 2  , 5  )( 5  , 1  )  %EFGHE
+\dashline[80]{0.2}      ( 8  , 1  )( 8  , 9  )   %PQ
+\dashline[80]{0.2}      ( 2  , 1  )( 2  , 9  )   %RS
+\dashline[80]{0.2}      ( 7  , 9  )( 8  , 9  )   %CQ
+\dashline[80]{0.2}      ( 3  , 9  )( 2  , 9  )   %DS
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  8.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  6.5 ,  9.2 ){$\scriptstyle C$}   %C
+\put(  2.8 ,  9.2 ){$\scriptstyle D$}   %D
+\put(  4.5 ,  0.2 ){$\scriptstyle E$}   %E
+\put(  8.1 ,  4.7 ){$\scriptstyle F$}   %F
+\put(  4.6 ,  9.2 ){$\scriptstyle G$}   %G
+\put(  1   ,  4.7 ){$\scriptstyle H$}   %H
+\put(  7.5 ,  0.2 ){$\scriptstyle P$}   %P
+\put(  7.7 ,  9.2 ){$\scriptstyle Q$}   %Q
+\put(  1.5 ,  0.2 ){$\scriptstyle R$}   %R
+\put(  1.5 ,  9.2 ){$\scriptstyle S$}   %S
+\end{picture}
+\PGrestore
+%
+% 082ffZ78.tex
+% Ex.78. Fig.f.   ( 9, 6 )
+\PGset[0.8em]
+\begin{picture}   ( 9, 6 )
+\drawline( 2  , 1  )( 8  , 1  )( 7  , 5  )( 1  , 5  )( 2  , 1  )  %ABCDA
+\drawline( 2  , 1  )( 4.3  , 3.9  )( 8  , 1  )  %ALB
+\drawline( 1  , 5  )( 4.7  , 2.1  )( 7  , 5  )  %DFC
+\put(  1.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  7.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  6.7 ,  5.2 ){$\scriptstyle C$}   %C
+\put(  0.6 ,  5.2 ){$\scriptstyle D$}   %D
+\put(  3.1 ,  1.9 ){$\scriptstyle E$}   %E
+\put(  4.5 ,  1.2 ){$\scriptstyle F$}   %F
+\put(  3.8 ,  4   ){$\scriptstyle L$}   %L
+\put(  5.7 ,  2.8 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
+%
+% 082ggZ78.tex
+% Ex.78. Fig.g.   (11, 7 )
+\PGset[0.8em]
+\begin{picture}   (11, 7 )
+\drawline( 1  , 1  )(10  , 1  )(10  , 6  )( 1  , 6  )( 1  , 1  )  %ABCDA
+\drawline( 1  , 1  )( 5.5  , 5.5  )(10  , 1  )  %ALB
+\drawline( 1  , 6  )( 5.5  , 1.5  )(10  , 6  )  %DFC
+\dashline[80]{0.2}      ( 5.5 ,  1.5  )( 6  , 1  )  %F-lr
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  9.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  9.7 ,  6.2 ){$\scriptstyle C$}   %C
+\put(  0.6 ,  6.2 ){$\scriptstyle D$}   %D
+\put(  2.1 ,  3.2 ){$\scriptstyle E$}   %E
+\put(  4.1 ,  1.2 ){$\scriptstyle F$}   %F
+\put(  4.4 ,  5.1 ){$\scriptstyle L$}   %L
+\put(  8   ,  3.3 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
+%
+% 082hhZ78.tex
+% Ex.78. Fig.h.   ( 8, 9 )
+\PGset[0.8em]
+\begin{picture}   ( 8, 9 )
+\drawline( 1   , 4    )( 2.5  , 2  )( 6.5  , 5  )( 5  , 7  )( 1  , 4  )  %ABCDA
+\drawline( 0   , 2    )( 8     , 2    )  %lower horiz
+\drawline( 0   , 7    )( 8     , 7    )  %upper horiz
+\drawline( 1.75, 0.5  )( 5.75  , 8.5  )  %ll-ur diag
+\put(  0.2 ,  3.8 ){$\scriptstyle A$}   %A
+\put(  2.3 ,  1.2 ){$\scriptstyle B$}   %B
+\put(  6.6 ,  4.7 ){$\scriptstyle C$}   %C
+\put(  4.3 ,  7.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/082iiZ82.eepic b/33063-t/images/sources/082iiZ82.eepic
new file mode 100644
index 0000000..d4629b6
--- /dev/null
+++ b/33063-t/images/sources/082iiZ82.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 082iiZ82.tex
+% Ex.82. Fig.i.   ( 9.7, 9 )
+\PGset[0.8em]
+\begin{picture}   ( 9.7, 9 )
+\drawline( 1  , 1  )( 9  , 1  )( 7  , 8  )( 2  , 8  )( 1  , 1  )  %ABCDA
+\drawline( 1  , 1  )( 7  , 8  )  %AC
+\drawline( 9  , 1  )( 2  , 8  )  %BD
+\drawline( 1.5, 4.5)( 8  , 4.5)  %EF
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  8.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  6.7 ,  8.2 ){$\scriptstyle C$}   %C
+\put(  1.6 ,  8.2 ){$\scriptstyle D$}   %D
+\put(  0.5 ,  4.2 ){$\scriptstyle E$}   %E
+\put(  8.2 ,  4.2 ){$\scriptstyle F$}   %F
+\put(  3.2 ,  4.6 ){$\scriptstyle G$}   %G
+\put(  5.5 ,  4.6 ){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
\ No newline at end of file
diff --git a/33063-t/images/sources/083adZ83.eepic b/33063-t/images/sources/083adZ83.eepic
new file mode 100644
index 0000000..0153346
--- /dev/null
+++ b/33063-t/images/sources/083adZ83.eepic
@@ -0,0 +1,66 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 083aaZ83.tex
+% Ex.83. Fig.a.   (10, 9 )
+\PGset[0.8em]
+\begin{picture}   (10, 9 )
+\drawline       ( 1  , 1  )( 9  , 1  )( 8  , 8  )( 3  , 8  )( 1  , 1  )  %ABCDA
+\drawline       ( 1  , 1  )( 8  , 8  )  %AC
+\drawline       ( 9  , 1  )( 3  , 8  )  %BD
+\drawline       ( 4.5, 4.5)( 6  , 4.5)  %EF
+\dashline[80]{0.2}  ( 4  , 1  )( 8  , 8  )  %GC
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  8.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  7.7 ,  8.2 ){$\scriptstyle C$}   %C
+\put(  2.6 ,  8.2 ){$\scriptstyle D$}   %D
+\put(  3.5 ,  4.2 ){$\scriptstyle E$}   %E
+\put(  6.2 ,  4.2 ){$\scriptstyle F$}   %F
+\put(  3.5 ,  0.2 ){$\scriptstyle G$}   %G
+\end{picture}
+\PGrestore
+%
+\hspace{-0.5em} %keeps these diagrams within margins
+%
+% 083bbZ83.tex
+% Ex.83. Fig.b.   (10, 6 )
+\PGset[0.8em]
+\begin{picture}   (10, 6 )
+\drawline       ( 1  , 1  )(9  , 1  )( 8  , 5  )( 2  , 5  )( 1  , 1  )  %ABDCA
+\dashline[80]{0.2}  ( 2  , 5  )( 3  , 1  )  %CE
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  8.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  1.7 ,  5.2 ){$\scriptstyle C$}   %C
+\put(  7.6 ,  5.2 ){$\scriptstyle D$}   %D
+\put(  2.5 ,  0.2 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
+%
+\hspace{-0.5em} %keeps these diagrams within margins
+%
+% 083ccZ83.tex
+% Ex.83. Fig.c.   (12, 7 )
+\PGset[0.8em]
+\begin{picture}   (12, 7 )
+\drawline       ( 1  , 1  )(11  , 1  )( 9  , 6  )( 3  , 6  )( 1  , 1  )  %ABCDA
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put( 10.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  8.7 ,  6.2 ){$\scriptstyle C$}   %C
+\put(  2.6 ,  6.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
+%
+\hspace{-0.5em} %keeps these diagrams within margins
+%
+% 083ddZ83.tex
+% Ex.83. Fig.d.   (11, 6 )
+\PGset[0.8em]
+\begin{picture}   (11, 6 )
+\drawline       ( 1  , 1  )(10  , 1  )( 8  , 5  )( 3  , 5  )( 1  , 1  )  %ABCDA
+\drawline       ( 1  , 1  )( 8  , 5  )  %AC
+\drawline       (10  , 1  )( 3  , 5  )  %BD
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  9.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  7.7 ,  5.2 ){$\scriptstyle C$}   %C
+\put(  2.6 ,  5.2 ){$\scriptstyle D$}   %D
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/083eeZ88.eepic b/33063-t/images/sources/083eeZ88.eepic
new file mode 100644
index 0000000..5a45e5b
--- /dev/null
+++ b/33063-t/images/sources/083eeZ88.eepic
@@ -0,0 +1,20 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 083eeZ88.tex
+% Ex.88. Fig.e.   (11.7,10 )
+\PGset[0.8em]
+\begin{picture}   (11.7,10 )
+\drawline       ( 1  , 1  )(11  , 1  )( 9  , 9 )( 3  , 9  )( 1  , 1  )  %ABDCA
+\drawline       ( 1  , 1  )( 9  , 9  )  %AD
+\drawline       (11  , 1  )( 3  , 9  )  %BC
+\dashline[80]{0.2}  ( 3  , 9  )( 3  , 1  )  %CE
+\dashline[80]{0.2}  ( 9  , 9  )( 9  , 1  )  %DF
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put( 10.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  2.7 ,  9.2 ){$\scriptstyle C$}   %C
+\put(  8.6 ,  9.2 ){$\scriptstyle D$}   %D
+\put(  2.5 ,  0.2 ){$\scriptstyle E$}   %E
+\put(  8.5 ,  0.2 ){$\scriptstyle F$}   %F
+\put(  5.5 ,  4.9 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/083ffZ89.eepic b/33063-t/images/sources/083ffZ89.eepic
new file mode 100644
index 0000000..dbce314
--- /dev/null
+++ b/33063-t/images/sources/083ffZ89.eepic
@@ -0,0 +1,18 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 083ffZ89.tex
+% Ex.89. Fig.f.   ( 6.7, 7 )
+\PGset[0.8em]
+\begin{picture}   ( 6.7, 7 )
+\drawline       ( 1     , 1     )( 6  , 1  )( 6  , 6  )( 1  , 6  )( 1  , 1  )  %ABCDA
+\drawline       ( 6     , 1     )( 1  , 6  )  %BD
+\drawline       ( 2.5   , 4.5   )( 4  , 6  )  %EF
+\dashline[80]{0.2}  ( 6     , 1     )( 4  , 6  )  %BF
+\put(  0.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  5.8 ,  0.2 ){$\scriptstyle B$}   %B
+\put(  5.7 ,  6.2 ){$\scriptstyle C$}   %C
+\put(  0.6 ,  6.2 ){$\scriptstyle D$}   %D
+\put(  1.5 ,  3.7 ){$\scriptstyle E$}   %E
+\put(  3.3 ,  6.2 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/084aa220.eepic b/33063-t/images/sources/084aa220.eepic
new file mode 100644
index 0000000..b88e880
--- /dev/null
+++ b/33063-t/images/sources/084aa220.eepic
@@ -0,0 +1,26 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 084aa220.tex
+% Pt.220. Fig.1.   (13,12.1 )
+\PGset[0.8em]
+\begin{picture}    (13,12.1 )
+\drawline( 1 ,11)(12.1 , 7.633)  %AD
+\drawline( 1 , 2)(11.8 , 2    )  %BC
+
+% Ellipse:  u = 6.0  v = 7.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 7.0)(11.0, 9.0711)(9.5355, 10.5355)
+\qbezier(9.5355, 10.5355)(8.0711, 12.0)(6.0, 12.0)
+\qbezier(6.0, 12.0)(3.9289, 12.0)(2.4645, 10.5355)
+\qbezier(2.4645, 10.5355)(1.0, 9.0711)(1.0, 7.0)
+\qbezier(1.0, 7.0)(1.0, 4.9289)(2.4645, 3.4645)
+\qbezier(2.4645, 3.4645)(3.9289, 2.0)(6.0, 2.0)
+\qbezier(6.0, 2.0)(8.0711, 2.0)(9.5355, 3.4645)
+\qbezier(9.5355, 3.4645)(11.0, 4.9289)(11.0, 7.0)
+
+\put(  0.2 , 10.8 ){$\scriptstyle A$}   %A
+\put(  0.1 ,  1.7 ){$\scriptstyle B$}   %B
+\put( 11.8 ,  1.6 ){$\scriptstyle C$}   %C
+\put( 12.1 ,  7.2 ){$\scriptstyle D$}   %D
+\put(  4.5 ,  0.2 ){$\scriptstyle \textsc{Fig.~1.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/085ac224.eepic b/33063-t/images/sources/085ac224.eepic
new file mode 100644
index 0000000..422ae48
--- /dev/null
+++ b/33063-t/images/sources/085ac224.eepic
@@ -0,0 +1,114 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 085aa224.tex
+% Pt.224. Fig.2.   (13,13.1 )
+\PGset[0.8em]
+\begin{picture}    (13,13.1 )
+
+% Ellipse:  u = 6.5  v = 7.5  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier(12.0, 7.5)(12.0, 9.7782)(10.3891, 11.3891)
+\qbezier(10.3891, 11.3891)(8.7782, 13.0)(6.5, 13.0)
+\qbezier(6.5, 13.0)(4.2218, 13.0)(2.6109, 11.3891)
+\qbezier(2.6109, 11.3891)(1.0, 9.7782)(1.0, 7.5)
+\qbezier(1.0, 7.5)(1.0, 5.2218)(2.6109, 3.6109)
+\qbezier(2.6109, 3.6109)(4.2218, 2.0)(6.5, 2.0)
+\qbezier(6.5, 2.0)(8.7782, 2.0)(10.3891, 3.6109)
+\qbezier(10.3891, 3.6109)(12.0, 5.2218)(12.0, 7.5)
+
+\drawline( 3.064,11.795)( 9.936,11.795)( 6.5, 7.5 )( 3.064,11.795 )    %ABOA
+\drawline(1,7.5)(12,7.5) % DC
+\drawline(6.5,7.5)(6.5,2) % ON
+
+\drawline(3.643,12.2)(9.356,12.2)
+\drawline(4.441,12.6)(8.559,12.6)
+\drawline( 6.5  , 7.5   )(11.916, 8.455)  %sector shading long
+\drawline( 9.319, 8.526 )(11.668, 9.381)  %sector shading med
+\drawline( 9.098, 9.000 )(11.263,10.250)  %sector shading med
+\drawline( 6.5  , 7.5   )(10.713,11.035)  %sector shading long
+\drawline(10.983, 7.892 )(11.979, 7.979)  %sector shading short
+\drawline(10.847, 8.665 )(11.813, 8.924)  %sector shading short
+\drawline(11.032, 9.613 )(11.485, 9.824)  %sector shading shortest
+\drawline(10.186,10.081 )(11.005,10.655)  %sector shading short
+\drawline( 9.682,10.682 )(10.389,11.389)  %sector shading short
+\put(  2.2 , 11.7 ){$\scriptstyle A$}   %A
+\put( 10.1 , 11.7 ){$\scriptstyle B$}   %B
+\put( 12.1 ,  7.1 ){$\scriptstyle C$}   %C
+\put(  0.1 ,  7.1 ){$\scriptstyle D$}   %D
+\put(  6.1 ,  1.1 ){$\scriptstyle N$}   %N
+\put(  6.7 ,  6.6 ){$\scriptstyle O$}   %O
+\put(  4.1 ,  4   ){{\tiny SEMICIRCLE}}
+\put(  4.4 , 11.95){{\tiny SEGMENT}}
+%\put(  7.9 ,  8.3 ){{\tiny SECTOR}}
+\put(  7.7 ,  7.9 ){{\tiny S}}
+\put(  8.1 ,  8.2 ){{\tiny E}}
+\put(  8.7 ,  8.5 ){{\tiny C}}
+\put(  9.3 ,  8.8 ){{\tiny T}}
+\put(  9.75,  8.95 ){{\tiny O}}
+\put( 10.3 ,  9.2 ){{\tiny R}}
+%\put(  1.4 ,  6   ){{\tiny QUADRANT}}
+\put(  2.3 ,  4.3 ){{\tiny Q}}
+\put(  2.8 ,  4.5 ){{\tiny U}}
+\put(  3.3 ,  4.9 ){{\tiny A}}
+\put(  3.8 ,  5.2 ){{\tiny D}}
+\put(  4.3 ,  5.6 ){{\tiny R}}
+\put(  4.8 ,  5.9 ){{\tiny A}}
+\put(  5.3 ,  6.2 ){{\tiny N}}
+\put(  5.8 ,  6.6 ){{\tiny T}}
+\put(  5   ,  0.2 ){$\scriptstyle \textsc{Fig.~2.}$}
+\end{picture}
+\PGrestore
+%
+\quad
+% 085bb224.tex
+% Pt.224. Fig.3.   (10.5,12.1 )
+\PGset[0.8em]
+\begin{picture}    (10.5,12.1 )
+
+% Ellipse:  u = 5.1  v = 7.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(10.1, 7.0)(10.1, 9.0711)(8.6355, 10.5355)
+\qbezier(8.6355, 10.5355)(7.1711, 12.0)(5.1, 12.0)
+\qbezier(5.1, 12.0)(3.0289, 12.0)(1.5645, 10.5355)
+\qbezier(1.5645, 10.5355)(0.1, 9.0711)(0.1, 7.0)
+\qbezier(0.1, 7.0)(0.1, 4.9289)(1.5645, 3.4645)
+\qbezier(1.5645, 3.4645)(3.0289, 2.0)(5.1, 2.0)
+\qbezier(5.1, 2.0)(7.1711, 2.0)(8.6355, 3.4645)
+\qbezier(8.6355, 3.4645)(10.1, 4.9289)(10.1, 7.0)
+
+\drawline( 0.904, 4.333)( 1.890,10.833)( 8.310,10.833 )( 9.515, 4.667)( 6.895, 2.333)( 0.904, 4.333 )    %ABCDEA
+\put(  0.1 ,  3.6 ){$\scriptstyle A$}   %A
+\put(  1   , 10.8 ){$\scriptstyle B$}   %B
+\put(  8.4 , 10.8 ){$\scriptstyle C$}   %C
+\put(  9.6 ,  4.1 ){$\scriptstyle D$}   %D
+\put(  6.6 ,  1.4 ){$\scriptstyle E$}   %E
+\put(  3.8 ,  0.2 ){$\scriptstyle \textsc{Fig.~3.}$}
+\end{picture}
+\PGrestore
+%
+\quad
+% 085cc224.tex
+% Pt.224. Fig.4.   (11.9,13.5 )
+\PGset[0.8em]
+\begin{picture}    (11.9,13.5 )
+
+% Ellipse:  u = 6.0  v = 7.2  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(10.6, 7.2)(10.6, 9.1054)(9.2527, 10.4527)
+\qbezier(9.2527, 10.4527)(7.9054, 11.8)(6.0, 11.8)
+\qbezier(6.0, 11.8)(4.0946, 11.8)(2.7473, 10.4527)
+\qbezier(2.7473, 10.4527)(1.4, 9.1054)(1.4, 7.2)
+\qbezier(1.4, 7.2)(1.4, 5.2946)(2.7473, 3.9473)
+\qbezier(2.7473, 3.9473)(4.0946, 2.6)(6.0, 2.6)
+\qbezier(6.0, 2.6)(7.9054, 2.6)(9.2527, 3.9473)
+\qbezier(9.2527, 3.9473)(10.6, 5.2946)(10.6, 7.2)
+
+\drawline( 1.021, 7.2)( 2.479,10.721)( 6,12.179 )( 9.521,10.721)(10.979, 7.2)( 9.521, 3.679)( 6, 2.221)( 2.479, 3.679)( 1.021, 7.2 )    %ABCDEFGHA
+\put(  0.1 ,  6.9 ){$\scriptstyle A$}   %A
+\put(  1.7 , 10.8 ){$\scriptstyle B$}   %B
+\put(  5.7 , 12.5 ){$\scriptstyle C$}   %C
+\put(  9.6 , 10.8 ){$\scriptstyle D$}   %D
+\put( 11.1 ,  6.9 ){$\scriptstyle E$}   %E
+\put(  9.6 ,  3.1 ){$\scriptstyle F$}   %F
+\put(  5.5 ,  1.2 ){$\scriptstyle G$}   %G
+\put(  1.5 ,  3.1 ){$\scriptstyle H$}   %H
+\put(  5.1 ,  0.2 ){$\scriptstyle \textsc{Fig.~4.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/086aa235.eepic b/33063-t/images/sources/086aa235.eepic
new file mode 100644
index 0000000..9ace1df
--- /dev/null
+++ b/33063-t/images/sources/086aa235.eepic
@@ -0,0 +1,27 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 086aa235.tex
+% Pt.235. Fig.a.   (12,13 )
+\PGset[0.8em]
+\begin{picture}    (12,13 )
+
+% Ellipse:  u = 6.0  v = 6.5  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier(11.5, 6.5)(11.5, 8.7782)(9.8891, 10.3891)
+\qbezier(9.8891, 10.3891)(8.2782, 12.0)(6.0, 12.0)
+\qbezier(6.0, 12.0)(3.7218, 12.0)(2.1109, 10.3891)
+\qbezier(2.1109, 10.3891)(0.5, 8.7782)(0.5, 6.5)
+\qbezier(0.5, 6.5)(0.5, 4.2218)(2.1109, 2.6109)
+\qbezier(2.1109, 2.6109)(3.7218, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.2782, 1.0)(9.8891, 2.6109)
+\qbezier(9.8891, 2.6109)(11.5, 4.2218)(11.5, 6.5)
+
+\drawline     ( 1.75 , 3.009 )(10.25 , 3.009 )                  %HK
+\dashline[80]{0.2}( 6    , 6.5   )( 6    , 1     )                  %OP
+\dashline[80]{0.2}( 1.75 , 3.009 )( 6    , 6.5   )(10.25 , 3.009 )  %HOP
+\put(  0.7 ,  2.4 ){$\scriptstyle H$}   %H
+\put( 10.4 ,  2.4 ){$\scriptstyle K$}   %K
+\put(  5.5 , 12.2 ){$\scriptstyle M$}   %M
+\put(  5.7 ,  6.7 ){$\scriptstyle O$}   %O
+\put(  5.5 ,  0.1 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/087ab236.eepic b/33063-t/images/sources/087ab236.eepic
new file mode 100644
index 0000000..473fcaa
--- /dev/null
+++ b/33063-t/images/sources/087ab236.eepic
@@ -0,0 +1,50 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 087aa236.tex
+% Pt.236. Fig.a.   (12.5,11.5 )
+\PGset[0.8em]
+\begin{picture}    (12.5,11.5 )
+
+% Ellipse:  u = 6.0  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 5.5)(11.0, 7.5711)(9.5355, 9.0355)
+\qbezier(9.5355, 9.0355)(8.0711, 10.5)(6.0, 10.5)
+\qbezier(6.0, 10.5)(3.9289, 10.5)(2.4645, 9.0355)
+\qbezier(2.4645, 9.0355)(1.0, 7.5711)(1.0, 5.5)
+\qbezier(1.0, 5.5)(1.0, 3.4289)(2.4645, 1.9645)
+\qbezier(2.4645, 1.9645)(3.9289, 0.5)(6.0, 0.5)
+\qbezier(6.0, 0.5)(8.0711, 0.5)(9.5355, 1.9645)
+\qbezier(9.5355, 1.9645)(11.0, 3.4289)(11.0, 5.5)
+
+\drawline(11, 5.5 )( 6, 5.5 )( 8.9, 9.573 )     %AOB
+\drawline          ( 6, 5.5 )( 7  ,10.399 )     %OC
+\put( 11.1 ,  5.2 ){$\scriptstyle A$}   %A
+\put(  8.9 ,  9.7 ){$\scriptstyle B$}   %B
+\put(  6.7 , 10.6 ){$\scriptstyle C$}   %C
+\put(  5.6 ,  4.7 ){$\scriptstyle O$}   %O
+\put(  0.2 ,  3.7 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
+%
+\qquad
+% 087bb236.tex
+% Pt.236. Fig.b.   (12.5,11.5 )
+\PGset[0.8em]
+\begin{picture}    (12.5,11.5 )
+
+% Ellipse:  u = 6.0  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 5.5)(11.0, 7.5711)(9.5355, 9.0355)
+\qbezier(9.5355, 9.0355)(8.0711, 10.5)(6.0, 10.5)
+\qbezier(6.0, 10.5)(3.9289, 10.5)(2.4645, 9.0355)
+\qbezier(2.4645, 9.0355)(1.0, 7.5711)(1.0, 5.5)
+\qbezier(1.0, 5.5)(1.0, 3.4289)(2.4645, 1.9645)
+\qbezier(2.4645, 1.9645)(3.9289, 0.5)(6.0, 0.5)
+\qbezier(6.0, 0.5)(8.0711, 0.5)(9.5355, 1.9645)
+\qbezier(9.5355, 1.9645)(11.0, 3.4289)(11.0, 5.5)
+
+\drawline(11, 5.5 )( 6, 5.5 )( 8.9, 9.573 )     %A'O'B'
+\put( 11.1 ,  5.2 ){$\scriptstyle A'$}   %A'
+\put(  8.9 ,  9.7 ){$\scriptstyle B'$}   %B'
+\put(  5.6 ,  4.5 ){$\scriptstyle O'$}   %O'
+\put(  0.2 ,  2.7 ){$\scriptstyle P'$}   %P'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/089ab241.eepic b/33063-t/images/sources/089ab241.eepic
new file mode 100644
index 0000000..e8e7601
--- /dev/null
+++ b/33063-t/images/sources/089ab241.eepic
@@ -0,0 +1,50 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 089aa241.tex
+% Pt.241. Fig.a.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 5.5  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(10.5, 5.5)(10.5, 7.5711)(9.0355, 9.0355)
+\qbezier(9.0355, 9.0355)(7.5711, 10.5)(5.5, 10.5)
+\qbezier(5.5, 10.5)(3.4289, 10.5)(1.9645, 9.0355)
+\qbezier(1.9645, 9.0355)(0.5, 7.5711)(0.5, 5.5)
+\qbezier(0.5, 5.5)(0.5, 3.4289)(1.9645, 1.9645)
+\qbezier(1.9645, 1.9645)(3.4289, 0.5)(5.5, 0.5)
+\qbezier(5.5, 0.5)(7.5711, 0.5)(9.0355, 1.9645)
+\qbezier(9.0355, 1.9645)(10.5, 3.4289)(10.5, 5.5)
+
+\dashline[80]{0.4}(10.5, 5.5 )( 5.5, 5.5 )( 8.5, 9.5 )     %AOB
+\dashline[80]{0.4}            ( 5.5, 5.5 )( 2.5, 9.5 )     %OF
+\drawline     ( 8.5, 9.5 )(10.5, 5.5 )( 2.5, 9.5 )     %BAF
+\put( 10.6 ,  5.2 ){$\scriptstyle A$}   %A
+\put(  8.5 ,  9.6 ){$\scriptstyle B$}   %B
+\put(  2.1 ,  9.6 ){$\scriptstyle F$}   %F
+\put(  5.1 ,  4.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+\qquad
+% 089bb241.tex
+% Pt.241. Fig.b.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 5.5  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(10.5, 5.5)(10.5, 7.5711)(9.0355, 9.0355)
+\qbezier(9.0355, 9.0355)(7.5711, 10.5)(5.5, 10.5)
+\qbezier(5.5, 10.5)(3.4289, 10.5)(1.9645, 9.0355)
+\qbezier(1.9645, 9.0355)(0.5, 7.5711)(0.5, 5.5)
+\qbezier(0.5, 5.5)(0.5, 3.4289)(1.9645, 1.9645)
+\qbezier(1.9645, 1.9645)(3.4289, 0.5)(5.5, 0.5)
+\qbezier(5.5, 0.5)(7.5711, 0.5)(9.0355, 1.9645)
+\qbezier(9.0355, 1.9645)(10.5, 3.4289)(10.5, 5.5)
+
+\dashline[80]{0.4}(10.5, 5.5 )( 5.5, 5.5 )( 8.5, 9.5 )     %AOB
+\drawline     ( 8.5, 9.5 )(10.5, 5.5 )                 %B'A'
+\put( 10.6 ,  5.2 ){$\scriptstyle A'$}   %A'
+\put(  8.5 ,  9.6 ){$\scriptstyle B'$}   %B'
+\put(  4.9 ,  4.6 ){$\scriptstyle O'$}   %O'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/090ab243.eepic b/33063-t/images/sources/090ab243.eepic
new file mode 100644
index 0000000..e8e7601
--- /dev/null
+++ b/33063-t/images/sources/090ab243.eepic
@@ -0,0 +1,50 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 089aa241.tex
+% Pt.241. Fig.a.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 5.5  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(10.5, 5.5)(10.5, 7.5711)(9.0355, 9.0355)
+\qbezier(9.0355, 9.0355)(7.5711, 10.5)(5.5, 10.5)
+\qbezier(5.5, 10.5)(3.4289, 10.5)(1.9645, 9.0355)
+\qbezier(1.9645, 9.0355)(0.5, 7.5711)(0.5, 5.5)
+\qbezier(0.5, 5.5)(0.5, 3.4289)(1.9645, 1.9645)
+\qbezier(1.9645, 1.9645)(3.4289, 0.5)(5.5, 0.5)
+\qbezier(5.5, 0.5)(7.5711, 0.5)(9.0355, 1.9645)
+\qbezier(9.0355, 1.9645)(10.5, 3.4289)(10.5, 5.5)
+
+\dashline[80]{0.4}(10.5, 5.5 )( 5.5, 5.5 )( 8.5, 9.5 )     %AOB
+\dashline[80]{0.4}            ( 5.5, 5.5 )( 2.5, 9.5 )     %OF
+\drawline     ( 8.5, 9.5 )(10.5, 5.5 )( 2.5, 9.5 )     %BAF
+\put( 10.6 ,  5.2 ){$\scriptstyle A$}   %A
+\put(  8.5 ,  9.6 ){$\scriptstyle B$}   %B
+\put(  2.1 ,  9.6 ){$\scriptstyle F$}   %F
+\put(  5.1 ,  4.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+\qquad
+% 089bb241.tex
+% Pt.241. Fig.b.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 5.5  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(10.5, 5.5)(10.5, 7.5711)(9.0355, 9.0355)
+\qbezier(9.0355, 9.0355)(7.5711, 10.5)(5.5, 10.5)
+\qbezier(5.5, 10.5)(3.4289, 10.5)(1.9645, 9.0355)
+\qbezier(1.9645, 9.0355)(0.5, 7.5711)(0.5, 5.5)
+\qbezier(0.5, 5.5)(0.5, 3.4289)(1.9645, 1.9645)
+\qbezier(1.9645, 1.9645)(3.4289, 0.5)(5.5, 0.5)
+\qbezier(5.5, 0.5)(7.5711, 0.5)(9.0355, 1.9645)
+\qbezier(9.0355, 1.9645)(10.5, 3.4289)(10.5, 5.5)
+
+\dashline[80]{0.4}(10.5, 5.5 )( 5.5, 5.5 )( 8.5, 9.5 )     %AOB
+\drawline     ( 8.5, 9.5 )(10.5, 5.5 )                 %B'A'
+\put( 10.6 ,  5.2 ){$\scriptstyle A'$}   %A'
+\put(  8.5 ,  9.6 ){$\scriptstyle B'$}   %B'
+\put(  4.9 ,  4.6 ){$\scriptstyle O'$}   %O'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/091aa245.eepic b/33063-t/images/sources/091aa245.eepic
new file mode 100644
index 0000000..d5d5a86
--- /dev/null
+++ b/33063-t/images/sources/091aa245.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 091aa245.tex
+% Pt.245. Fig.a.   (12,13 )
+\PGset[0.8em]
+\begin{picture}    (12,13 )
+
+% Ellipse:  u = 6.0  v = 6.5  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier(11.5, 6.5)(11.5, 8.7782)(9.8891, 10.3891)
+\qbezier(9.8891, 10.3891)(8.2782, 12.0)(6.0, 12.0)
+\qbezier(6.0, 12.0)(3.7218, 12.0)(2.1109, 10.3891)
+\qbezier(2.1109, 10.3891)(0.5, 8.7782)(0.5, 6.5)
+\qbezier(0.5, 6.5)(0.5, 4.2218)(2.1109, 2.6109)
+\qbezier(2.1109, 2.6109)(3.7218, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.2782, 1.0)(9.8891, 2.6109)
+\qbezier(9.8891, 2.6109)(11.5, 4.2218)(11.5, 6.5)
+
+\drawline     ( 1.757, 3   )(10.243, 3   )                 %A(M)B
+\drawline     ( 6    , 1   )( 6    ,12   )                 %S(M)E
+\dashline[80]{0.2}( 1.757, 3   )( 6    , 6.5 )(10.243, 3   )   %AOB
+\put(  0.9 ,  2.4 ){$\scriptstyle A$}   %A
+\put( 10.3 ,  2.4 ){$\scriptstyle B$}   %B
+\put(  5.7 , 12.2 ){$\scriptstyle E$}   %E
+\put(  4.9 ,  2.1 ){$\scriptstyle M$}   %M
+\put(  6.2 ,  6.5 ){$\scriptstyle O$}   %O
+\put(  5.7 ,  0.2 ){$\scriptstyle S$}   %S
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/092aa249.eepic b/33063-t/images/sources/092aa249.eepic
new file mode 100644
index 0000000..a818361
--- /dev/null
+++ b/33063-t/images/sources/092aa249.eepic
@@ -0,0 +1,30 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 092aa249.tex
+% Pt.249. Fig.a.   (13.4,12.4 )
+\PGset[0.8em]
+\begin{picture}    (13.4,12.4 )
+
+% Ellipse:  u = 6.7  v = 6.2  a = 5.7  b = 5.7  phi = 0.0 Grad
+\qbezier(12.4, 6.2)(12.4, 8.561)(10.7305, 10.2305)
+\qbezier(10.7305, 10.2305)(9.061, 11.9)(6.7, 11.9)
+\qbezier(6.7, 11.9)(4.339, 11.9)(2.6695, 10.2305)
+\qbezier(2.6695, 10.2305)(1.0, 8.561)(1.0, 6.2)
+\qbezier(1.0, 6.2)(1.0, 3.839)(2.6695, 2.1695)
+\qbezier(2.6695, 2.1695)(4.339, 0.5)(6.7, 0.5)
+\qbezier(6.7, 0.5)(9.061, 0.5)(10.7305, 2.1695)
+\qbezier(10.7305, 2.1695)(12.4, 3.839)(12.4, 6.2)
+
+\drawline     ( 1.495, 8.524 )(10.610,10.347 )                 %A(P)B
+\drawline     ( 2.052, 2.9   )(11.348, 2.9   )                 %C(H)F
+\dashline[80]{0.2}( 1.495, 8.524 )( 6.7  , 6.2   )( 6.053, 9.436 ) %AOP
+\dashline[80]{0.2}( 2.052, 2.9   )( 6.7  , 6.2   )( 6.7  , 2.9   ) %COH
+\put(  0.7 ,  8.5 ){$\scriptstyle A$}   %A
+\put( 10.7 , 10.3 ){$\scriptstyle B$}   %B
+\put(  1   ,  2.2 ){$\scriptstyle C$}   %C
+\put( 11.5 ,  2.2 ){$\scriptstyle F$}   %F
+\put(  6.2 ,  2.0 ){$\scriptstyle H$}   %H
+\put(  6.9 ,  5.9 ){$\scriptstyle O$}   %O
+\put(  5.7 ,  9.6 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/093aa250.eepic b/33063-t/images/sources/093aa250.eepic
new file mode 100644
index 0000000..7a530b8
--- /dev/null
+++ b/33063-t/images/sources/093aa250.eepic
@@ -0,0 +1,33 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 093aa250.tex
+% Pt.250. Fig.a.   (10.7,10.7 )
+\PGset[0.8em]
+\begin{picture}    (10.7,10.7 )
+
+% Ellipse:  u = 5.1  v = 5.1  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(9.7, 5.1)(9.7, 7.0054)(8.3527, 8.3527)
+\qbezier(8.3527, 8.3527)(7.0054, 9.7)(5.1, 9.7)
+\qbezier(5.1, 9.7)(3.1946, 9.7)(1.8473, 8.3527)
+\qbezier(1.8473, 8.3527)(0.5, 7.0054)(0.5, 5.1)
+\qbezier(0.5, 5.1)(0.5, 3.1946)(1.8473, 1.8473)
+\qbezier(1.8473, 1.8473)(3.1946, 0.5)(5.1, 0.5)
+\qbezier(5.1, 0.5)(7.0054, 0.5)(8.3527, 1.8473)
+\qbezier(8.3527, 1.8473)(9.7, 3.1946)(9.7, 5.1)
+
+\drawline     ( 1.116, 2.8   )( 9.084, 2.8   )                 %A(E)B
+\drawline     ( 4.477, 9.658 )( 9.683, 5.493 )                 %C(F)D
+\drawline     ( 5.1  , 2.8   )( 5.1  , 5.1   )( 7.080, 7.576 ) %EOF
+\dashline[80]{0.2}( 1.116, 2.8   )( 3.003, 9.194 )                 %A(H)G
+\dashline[80]{0.2}( 5.1  , 2.8   )( 2.060, 5.997 )( 5.1  , 5.1   ) %EHO
+\put(  0.2 ,  2.4 ){$\scriptstyle A$}   %A
+\put(  9.2 ,  2.4 ){$\scriptstyle B$}   %B
+\put(  4.2 ,  9.9 ){$\scriptstyle C$}   %C
+\put(  9.8 ,  5   ){$\scriptstyle D$}   %D
+\put(  4.7 ,  1.9 ){$\scriptstyle E$}   %E
+\put(  7.2 ,  7.6 ){$\scriptstyle F$}   %F
+\put(  2.5 ,  9.4 ){$\scriptstyle G$}   %G
+\put(  1.1 ,  5.7 ){$\scriptstyle H$}   %H
+\put(  4.6 ,  5.5 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/094aa251.eepic b/33063-t/images/sources/094aa251.eepic
new file mode 100644
index 0000000..7a530b8
--- /dev/null
+++ b/33063-t/images/sources/094aa251.eepic
@@ -0,0 +1,33 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 093aa250.tex
+% Pt.250. Fig.a.   (10.7,10.7 )
+\PGset[0.8em]
+\begin{picture}    (10.7,10.7 )
+
+% Ellipse:  u = 5.1  v = 5.1  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(9.7, 5.1)(9.7, 7.0054)(8.3527, 8.3527)
+\qbezier(8.3527, 8.3527)(7.0054, 9.7)(5.1, 9.7)
+\qbezier(5.1, 9.7)(3.1946, 9.7)(1.8473, 8.3527)
+\qbezier(1.8473, 8.3527)(0.5, 7.0054)(0.5, 5.1)
+\qbezier(0.5, 5.1)(0.5, 3.1946)(1.8473, 1.8473)
+\qbezier(1.8473, 1.8473)(3.1946, 0.5)(5.1, 0.5)
+\qbezier(5.1, 0.5)(7.0054, 0.5)(8.3527, 1.8473)
+\qbezier(8.3527, 1.8473)(9.7, 3.1946)(9.7, 5.1)
+
+\drawline     ( 1.116, 2.8   )( 9.084, 2.8   )                 %A(E)B
+\drawline     ( 4.477, 9.658 )( 9.683, 5.493 )                 %C(F)D
+\drawline     ( 5.1  , 2.8   )( 5.1  , 5.1   )( 7.080, 7.576 ) %EOF
+\dashline[80]{0.2}( 1.116, 2.8   )( 3.003, 9.194 )                 %A(H)G
+\dashline[80]{0.2}( 5.1  , 2.8   )( 2.060, 5.997 )( 5.1  , 5.1   ) %EHO
+\put(  0.2 ,  2.4 ){$\scriptstyle A$}   %A
+\put(  9.2 ,  2.4 ){$\scriptstyle B$}   %B
+\put(  4.2 ,  9.9 ){$\scriptstyle C$}   %C
+\put(  9.8 ,  5   ){$\scriptstyle D$}   %D
+\put(  4.7 ,  1.9 ){$\scriptstyle E$}   %E
+\put(  7.2 ,  7.6 ){$\scriptstyle F$}   %F
+\put(  2.5 ,  9.4 ){$\scriptstyle G$}   %G
+\put(  1.1 ,  5.7 ){$\scriptstyle H$}   %H
+\put(  4.6 ,  5.5 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/095aa253.eepic b/33063-t/images/sources/095aa253.eepic
new file mode 100644
index 0000000..1fa46f4
--- /dev/null
+++ b/33063-t/images/sources/095aa253.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 095aa253.tex
+% Pt.253. Fig.a.   (18,12.5 )
+\PGset[0.8em]
+\begin{picture}    (18,12.5 )
+
+% Ellipse:  u = 10.0  v = 6.5  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier(15.5, 6.5)(15.5, 8.7782)(13.8891, 10.3891)
+\qbezier(13.8891, 10.3891)(12.2782, 12.0)(10.0, 12.0)
+\qbezier(10.0, 12.0)(7.7218, 12.0)(6.1109, 10.3891)
+\qbezier(6.1109, 10.3891)(4.5, 8.7782)(4.5, 6.5)
+\qbezier(4.5, 6.5)(4.5, 4.2218)(6.1109, 2.6109)
+\qbezier(6.1109, 2.6109)(7.7218, 1.0)(10.0, 1.0)
+\qbezier(10.0, 1.0)(12.2782, 1.0)(13.8891, 2.6109)
+\qbezier(13.8891, 2.6109)(15.5, 4.2218)(15.5, 6.5)
+
+\drawline     ( 1, 1   )(17, 1   )            %M(HA)B
+\drawline     (10, 1   )(10, 6.5 )            %AO
+\dashline[80]{0.2}( 4, 1   )(10, 6.5 )            %H(C)O
+\put(  9.6 ,  0.2 ){$\scriptstyle A$}   %A
+\put( 17.1 ,  0.6 ){$\scriptstyle B$}   %B
+\put(  5.7 ,  3.4 ){$\scriptstyle C$}   %C
+\put(  3.5 ,  0.2 ){$\scriptstyle H$}   %H
+\put(  0.1 ,  0.6 ){$\scriptstyle M$}   %M
+\put(  9.9 ,  6.8 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/096ac257.eepic b/33063-t/images/sources/096ac257.eepic
new file mode 100644
index 0000000..6280355
--- /dev/null
+++ b/33063-t/images/sources/096ac257.eepic
@@ -0,0 +1,95 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 096aa257.tex
+% Pt.257. Fig.1.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 6.5  v = 6.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(10.5, 6.0)(10.5, 7.6569)(9.3284, 8.8284)
+\qbezier(9.3284, 8.8284)(8.1569, 10.0)(6.5, 10.0)
+\qbezier(6.5, 10.0)(4.8431, 10.0)(3.6716, 8.8284)
+\qbezier(3.6716, 8.8284)(2.5, 7.6569)(2.5, 6.0)
+\qbezier(2.5, 6.0)(2.5, 4.3431)(3.6716, 3.1716)
+\qbezier(3.6716, 3.1716)(4.8431, 2.0)(6.5, 2.0)
+\qbezier(6.5, 2.0)(8.1569, 2.0)(9.3284, 3.1716)
+\qbezier(9.3284, 3.1716)(10.5, 4.3431)(10.5, 6.0)
+
+\drawline     ( 1  ,10   )(11  ,10   )          %A(F)B
+\drawline     ( 2  , 8.3 )(11  , 8.3 )          %left(CD)right
+\drawline     ( 1.2, 2   )(10.7, 2   )          %A'(F')B'
+\dashline[80]{0.2}( 6.5, 2   )( 6.5,10   )          %F'F
+\put(  0.2 ,  9.7 ){$\scriptstyle A$}   %A
+\put( 11.1 ,  9.7 ){$\scriptstyle B$}   %B
+\put(  2.5 ,  8.4 ){$\scriptstyle C$}   %C
+\put(  9.8 ,  8.4 ){$\scriptstyle D$}   %D
+\put(  6.3 , 10.2 ){$\scriptstyle F$}   %F
+\put(  0.1 ,  1.6 ){$\scriptstyle A'$}   %A'
+\put( 10.8 ,  1.6 ){$\scriptstyle B'$}   %B'
+\put(  6   ,  1.1 ){$\scriptstyle F'$}   %F'
+\put(  5.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~1.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+\qquad
+% 096bb257.tex
+% Pt.257. Fig.2.   (10,11 )
+\PGset[0.8em]
+\begin{picture}    (10,11 )
+
+% Ellipse:  u = 5.0  v = 6.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 6.0)(9.0, 7.6569)(7.8284, 8.8284)
+\qbezier(7.8284, 8.8284)(6.6569, 10.0)(5.0, 10.0)
+\qbezier(5.0, 10.0)(3.3431, 10.0)(2.1716, 8.8284)
+\qbezier(2.1716, 8.8284)(1.0, 7.6569)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 4.3431)(2.1716, 3.1716)
+\qbezier(2.1716, 3.1716)(3.3431, 2.0)(5.0, 2.0)
+\qbezier(5.0, 2.0)(6.6569, 2.0)(7.8284, 3.1716)
+\qbezier(7.8284, 3.1716)(9.0, 4.3431)(9.0, 6.0)
+
+\drawline     ( 1  , 4   )( 9  , 4   )          %AB
+\drawline     ( 1  , 8.3 )( 9  , 8.3 )          %CD
+\dashline[80]{0.2}( 1  ,10   )( 9  ,10   )          %E(M)F
+\put(  0.2 ,  3.8 ){$\scriptstyle A$}   %A
+\put(  9.1 ,  3.8 ){$\scriptstyle B$}   %B
+\put(  0.2 ,  7.9 ){$\scriptstyle C$}   %C
+\put(  9.1 ,  7.9 ){$\scriptstyle D$}   %D
+\put(  0.2 ,  9.7 ){$\scriptstyle E$}   %E
+\put(  9.1 ,  9.7 ){$\scriptstyle F$}   %F
+\put(  4.6 , 10.2 ){$\scriptstyle M$}   %M
+\put(  3.8 ,  0.2 ){$\scriptstyle \textsc{Fig.~2.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+\qquad
+% 096cc257.tex
+% Pt.257. Fig.3.   (10,11 )
+\PGset[0.8em]
+\begin{picture}    (10,11 )
+
+% Ellipse:  u = 5.0  v = 6.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 6.0)(9.0, 7.6569)(7.8284, 8.8284)
+\qbezier(7.8284, 8.8284)(6.6569, 10.0)(5.0, 10.0)
+\qbezier(5.0, 10.0)(3.3431, 10.0)(2.1716, 8.8284)
+\qbezier(2.1716, 8.8284)(1.0, 7.6569)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 4.3431)(2.1716, 3.1716)
+\qbezier(2.1716, 3.1716)(3.3431, 2.0)(5.0, 2.0)
+\qbezier(5.0, 2.0)(6.6569, 2.0)(7.8284, 3.1716)
+\qbezier(7.8284, 3.1716)(9.0, 4.3431)(9.0, 6.0)
+
+\drawline     ( 1  ,10   )( 9  ,10   )          %A(E)B
+\drawline     ( 1  , 2   )( 9  , 2   )          %C(F)D
+\dashline[80]{0.2}( 1  , 7.7 )( 9  , 7.7 )          %GH
+\put(  0.1 ,  9.7 ){$\scriptstyle A$}   %A
+\put(  9.1 ,  9.6 ){$\scriptstyle B$}   %B
+\put(  0.1 ,  1.7 ){$\scriptstyle C$}   %C
+\put(  9.1 ,  1.6 ){$\scriptstyle D$}   %D
+\put(  4.7 , 10.2 ){$\scriptstyle E$}   %E
+\put(  4.6 ,  1.1 ){$\scriptstyle F$}   %F
+\put(  0.7 ,  7.85 ){$\scriptstyle G$}   %G
+\put(  8.7 ,  7.85 ){$\scriptstyle H$}   %H
+\put(  3.8 ,  0.2 ){$\scriptstyle \textsc{Fig.~3.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/097aa258.eepic b/33063-t/images/sources/097aa258.eepic
new file mode 100644
index 0000000..081eaa2
--- /dev/null
+++ b/33063-t/images/sources/097aa258.eepic
@@ -0,0 +1,25 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 097aa258.tex
+% Pt.258. Fig.a.   (13.5,13 )
+\PGset[0.8em]
+\begin{picture}    (13.5,13 )
+
+% Ellipse:  u = 6.5  v = 6.5  a = 6.0  b = 6.0  phi = 0.0 Grad
+\qbezier(12.5, 6.5)(12.5, 8.9853)(10.7426, 10.7426)
+\qbezier(10.7426, 10.7426)(8.9853, 12.5)(6.5, 12.5)
+\qbezier(6.5, 12.5)(4.0147, 12.5)(2.2574, 10.7426)
+\qbezier(2.2574, 10.7426)(0.5, 8.9853)(0.5, 6.5)
+\qbezier(0.5, 6.5)(0.5, 4.0147)(2.2574, 2.2574)
+\qbezier(2.2574, 2.2574)(4.0147, 0.5)(6.5, 0.5)
+\qbezier(6.5, 0.5)(8.9853, 0.5)(10.7426, 2.2574)
+\qbezier(10.7426, 2.2574)(12.5, 4.0147)(12.5, 6.5)
+
+\dashline[80]{0.2}( 2.028, 2.5 )(10.972, 2.5 )(12.167, 8.472 ) %ABC
+\dashline[80]{0.2}( 6.5  , 2.5 )( 6.5  , 6.5 )(11.570, 5.486 ) %midAB-O-midBC
+\put(  1.3 ,  1.9 ){$\scriptstyle A$}   %A
+\put( 11.1 ,  1.95 ){$\scriptstyle B$}   %B
+\put( 12.2 ,  8.5 ){$\scriptstyle C$}   %C
+\put(  6.2 ,  6.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/098aa261.eepic b/33063-t/images/sources/098aa261.eepic
new file mode 100644
index 0000000..4905c01
--- /dev/null
+++ b/33063-t/images/sources/098aa261.eepic
@@ -0,0 +1,26 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 098aa261.tex
+% Pt.261. Fig.a.   (20.5,12 )
+\PGset[0.8em]
+\begin{picture}    (20.5,12 )
+
+% Ellipse:  u = 14.5  v = 6.0  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier(20.0, 6.0)(20.0, 8.2782)(18.3891, 9.8891)
+\qbezier(18.3891, 9.8891)(16.7782, 11.5)(14.5, 11.5)
+\qbezier(14.5, 11.5)(12.2218, 11.5)(10.6109, 9.8891)
+\qbezier(10.6109, 9.8891)(9.0, 8.2782)(9.0, 6.0)
+\qbezier(9.0, 6.0)(9.0, 3.7218)(10.6109, 2.1109)
+\qbezier(10.6109, 2.1109)(12.2218, 0.5)(14.5, 0.5)
+\qbezier(14.5, 0.5)(16.7782, 0.5)(18.3891, 2.1109)
+\qbezier(18.3891, 2.1109)(20.0, 3.7218)(20.0, 6.0)
+
+\drawline     (12.231,11.010 )( 1.167, 6 )(12.231, 0.990 ) %BAC
+\drawline     ( 1.167, 6     )(14.5  , 6 )                 %AO
+\dashline[80]{0.2}(12.231,11.010 )(14.5  , 6 )(12.231, 0.990 ) %BOC
+\put(  0.4 ,  6   ){$\scriptstyle A$}   %A
+\put( 11.5 , 11.1 ){$\scriptstyle B$}   %B
+\put( 11.5 ,  0.2 ){$\scriptstyle C$}   %C
+\put( 14.6 ,  5.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/099aa264.eepic b/33063-t/images/sources/099aa264.eepic
new file mode 100644
index 0000000..9141680
--- /dev/null
+++ b/33063-t/images/sources/099aa264.eepic
@@ -0,0 +1,36 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 099aa264.tex
+% Pt.264. Fig.a.   (16.2,10.4 )
+\PGset[0.8em]
+\begin{picture}    (16.2,10.4 )
+
+% Ellipse:  u = 5.2  v = 5.2  a = 4.7  b = 4.7  phi = 0.0 Grad
+\qbezier(9.9, 5.2)(9.9, 7.1468)(8.5234, 8.5234)
+\qbezier(8.5234, 8.5234)(7.1468, 9.9)(5.2, 9.9)
+\qbezier(5.2, 9.9)(3.2532, 9.9)(1.8766, 8.5234)
+\qbezier(1.8766, 8.5234)(0.5, 7.1468)(0.5, 5.2)
+\qbezier(0.5, 5.2)(0.5, 3.2532)(1.8766, 1.8766)
+\qbezier(1.8766, 1.8766)(3.2532, 0.5)(5.2, 0.5)
+\qbezier(5.2, 0.5)(7.1468, 0.5)(8.5234, 1.8766)
+\qbezier(8.5234, 1.8766)(9.9, 3.2532)(9.9, 5.2)
+
+% Ellipse:  u = 12.2  v = 5.2  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(15.7, 5.2)(15.7, 6.6497)(14.6749, 7.6749)
+\qbezier(14.6749, 7.6749)(13.6497, 8.7)(12.2, 8.7)
+\qbezier(12.2, 8.7)(10.7503, 8.7)(9.7251, 7.6749)
+\qbezier(9.7251, 7.6749)(8.7, 6.6497)(8.7, 5.2)
+\qbezier(8.7, 5.2)(8.7, 3.7503)(9.7251, 2.7251)
+\qbezier(9.7251, 2.7251)(10.7503, 1.7)(12.2, 1.7)
+\qbezier(12.2, 1.7)(13.6497, 1.7)(14.6749, 2.7251)
+\qbezier(14.6749, 2.7251)(15.7, 3.7503)(15.7, 5.2)
+
+\drawline     ( 5.2  , 5.2   )(12.2  , 5.2   )     %CC'
+\drawline     ( 9.403, 7.304 )( 9.403, 3.096 )     %AB
+\dashline[80]{0.2}( 5.2  , 5.2   )( 9.403, 3.096 )(12.2, 5.2 )( 9.403, 7.304 )( 5.2, 5.2 ) %CBC'AC
+\put(  9.1 ,  7.95 ){$\scriptstyle A$}   %A
+\put(  9   ,  1.95 ){$\scriptstyle B$}   %B
+\put(  4.2 ,  4.95 ){$\scriptstyle C$}   %C
+\put( 12.4 ,  4.9 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/100aa265.eepic b/33063-t/images/sources/100aa265.eepic
new file mode 100644
index 0000000..b2fefd4
--- /dev/null
+++ b/33063-t/images/sources/100aa265.eepic
@@ -0,0 +1,36 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 100aa265.tex
+% Pt.265. Fig.a.   (17.4,11.4 )
+\PGset[0.8em]
+\begin{picture}    (17.4,11.4 )
+
+% Ellipse:  u = 5.2  v = 6.0  a = 4.7  b = 4.7  phi = 0.0 Grad
+\qbezier(9.9, 6.0)(9.9, 7.9468)(8.5234, 9.3234)
+\qbezier(8.5234, 9.3234)(7.1468, 10.7)(5.2, 10.7)
+\qbezier(5.2, 10.7)(3.2532, 10.7)(1.8766, 9.3234)
+\qbezier(1.8766, 9.3234)(0.5, 7.9468)(0.5, 6.0)
+\qbezier(0.5, 6.0)(0.5, 4.0532)(1.8766, 2.6766)
+\qbezier(1.8766, 2.6766)(3.2532, 1.3)(5.2, 1.3)
+\qbezier(5.2, 1.3)(7.1468, 1.3)(8.5234, 2.6766)
+\qbezier(8.5234, 2.6766)(9.9, 4.0532)(9.9, 6.0)
+
+% Ellipse:  u = 13.4  v = 6.0  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(16.9, 6.0)(16.9, 7.4497)(15.8749, 8.4749)
+\qbezier(15.8749, 8.4749)(14.8497, 9.5)(13.4, 9.5)
+\qbezier(13.4, 9.5)(11.9503, 9.5)(10.9251, 8.4749)
+\qbezier(10.9251, 8.4749)(9.9, 7.4497)(9.9, 6.0)
+\qbezier(9.9, 6.0)(9.9, 4.5503)(10.9251, 3.5251)
+\qbezier(10.9251, 3.5251)(11.9503, 2.5)(13.4, 2.5)
+\qbezier(13.4, 2.5)(14.8497, 2.5)(15.8749, 3.5251)
+\qbezier(15.8749, 3.5251)(16.9, 4.5503)(16.9, 6.0)
+
+\drawline     ( 5.2  , 6     )(13.4  , 6     )   %CC'
+\drawline     ( 9.9  ,10.4   )( 9.9  , 1     )   %AB
+\put(  9.55, 10.6 ){$\scriptstyle A$}   %A
+\put(  9.45,  0.2 ){$\scriptstyle B$}   %B
+\put(  4.3 ,  5.75 ){$\scriptstyle C$}   %C
+\put( 10.1 ,  6.15 ){$\scriptstyle O$}   %O
+\put( 13.5 ,  5.65 ){$\scriptstyle C'$}   %C'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/103aa272.eepic b/33063-t/images/sources/103aa272.eepic
new file mode 100644
index 0000000..5848324
--- /dev/null
+++ b/33063-t/images/sources/103aa272.eepic
@@ -0,0 +1,16 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 103aa272.tex
+% Pt.272. Fig.a.   (28.2, 2.3 )
+\PGset[0.8em]
+\begin{picture}    (28.2, 2.3 )
+\drawline(0.1,1.1)(28.1,1.1) % AB
+\drawline(13.6,1.1)(13.6,0.1) % M
+\drawline(20.267,1.1)(20.267,0.1) % M'
+\drawline(24.767,1.1)(24.767,0.1) % M''
+\put(  0.1 ,  1.5 ){$\scriptstyle A$}   %A
+\put( 27.4 ,  1.5 ){$\scriptstyle B$}   %B
+\put( 13.2 ,  1.5 ){$\scriptstyle M$}   %M
+\put( 19.7 ,  1.5 ){$\scriptstyle M'$}   %M'
+\put( 24.1 ,  1.5 ){$\scriptstyle M''$}   %M''
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/103bb273.eepic b/33063-t/images/sources/103bb273.eepic
new file mode 100644
index 0000000..ff41448
--- /dev/null
+++ b/33063-t/images/sources/103bb273.eepic
@@ -0,0 +1,31 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 103bb273.tex
+% Pt.273. Fig.b.   (15.4,15.4 )
+\PGset[0.8em]
+\begin{picture}    (15.4,15.4 )
+
+% Ellipse:  u = 7.7  v = 7.7  a = 6.7  b = 6.7  phi = 0.0 Grad
+\qbezier(14.4, 7.7)(14.4, 10.4752)(12.4376, 12.4376)
+\qbezier(12.4376, 12.4376)(10.4752, 14.4)(7.7, 14.4)
+\qbezier(7.7, 14.4)(4.9248, 14.4)(2.9624, 12.4376)
+\qbezier(2.9624, 12.4376)(1.0, 10.4752)(1.0, 7.7)
+\qbezier(1.0, 7.7)(1.0, 4.9248)(2.9624, 2.9624)
+\qbezier(2.9624, 2.9624)(4.9248, 1.0)(7.7, 1.0)
+\qbezier(7.7, 1.0)(10.4752, 1.0)(12.4376, 2.9624)
+\qbezier(12.4376, 2.9624)(14.4, 4.9248)(14.4, 7.7)
+
+\drawline     ( 2.962,12.438 )(12.438,12.438 )(12.438, 2.962 )         %ABC
+\drawline     (12.438, 2.962 )( 2.962, 2.962 )( 2.962,12.438 )         %CDA
+\dashline[80]{0.2}( 2.962,12.438 )( 7.7,14.4 )(12.438,12.438 )(14.4 , 7.7 )(12.438, 2.962 ) %AEBFC
+\dashline[80]{0.2}(12.438, 2.962 )( 7.7, 1   )( 2.962, 2.962 )( 1   , 7.7 )( 2.962,12.438 ) %CHDKA
+\put(  2.1 , 12.45 ){$\scriptstyle A$}   %A
+\put( 12.6 , 12.45 ){$\scriptstyle B$}   %B
+\put( 12.6 ,  2.4 ){$\scriptstyle C$}   %C
+\put(  2.1 ,  2.3 ){$\scriptstyle D$}   %D
+\put(  7.3 , 14.6 ){$\scriptstyle E$}   %E
+\put( 14.6 ,  7.4 ){$\scriptstyle F$}   %F
+\put(  7.2 ,  0.2 ){$\scriptstyle H$}   %H
+\put(  0.2 ,  7.4 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/108aa286.eepic b/33063-t/images/sources/108aa286.eepic
new file mode 100644
index 0000000..d1ec72d
--- /dev/null
+++ b/33063-t/images/sources/108aa286.eepic
@@ -0,0 +1,26 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 108aa286.tex
+% Pt.286. Fig.a.   (27.1, 6.5 )
+\PGset[0.8em]
+\begin{picture}    (27.1, 6.5 )
+\drawline(0.9,4.7)(26.3,4.7) % AB
+\drawline(7.9,1.5)(19.2,1.5) 
+\drawline(11.9,4.7)(11.9,5.7)
+\drawline(23.2,4.7)(23.2,5.45) % E
+\drawline(25,4.7)(25,5.45) % H
+\drawline(7.9,1.5)(7.9,2.3) % C
+\drawline(11,1.5)(11,0.9)
+\drawline(14.1,1.5)(14.1,0.9)
+\drawline(17.2,1.5)(17.2,0.9) % F
+\drawline(18.5,1.5)(18.5,2.3) % K
+\drawline(19.2,1.5)(19.2,2.3) % D
+\put(  0.1 ,  4.3 ){$\scriptstyle A$}   %A
+\put( 26.2 ,  4.4 ){$\scriptstyle B$}   %B
+\put(  6.9 ,  1.2 ){$\scriptstyle C$}   %C
+\put( 19.5 ,  1.1 ){$\scriptstyle D$}   %D
+\put( 22.9 ,  5.7 ){$\scriptstyle E$}   %E
+\put( 16.8 ,  0.1 ){$\scriptstyle F$}   %F
+\put( 24.7 ,  5.7 ){$\scriptstyle H$}   %H
+\put( 18.3 ,  2.4 ){$\scriptstyle K$}   %K
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/109ac287.eepic b/33063-t/images/sources/109ac287.eepic
new file mode 100644
index 0000000..2e605e5
--- /dev/null
+++ b/33063-t/images/sources/109ac287.eepic
@@ -0,0 +1,89 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 109aa287.tex
+% Pt.287. Fig.1.   (13.5,11.7 )
+\PGset[0.8em]
+\begin{picture}    (13.5,11.7 )
+
+% Ellipse:  u = 5.1  v = 6.6  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(9.7, 6.6)(9.7, 8.5054)(8.3527, 9.8527)
+\qbezier(8.3527, 9.8527)(7.0054, 11.2)(5.1, 11.2)
+\qbezier(5.1, 11.2)(3.1946, 11.2)(1.8473, 9.8527)
+\qbezier(1.8473, 9.8527)(0.5, 8.5054)(0.5, 6.6)
+\qbezier(0.5, 6.6)(0.5, 4.6946)(1.8473, 3.3473)
+\qbezier(1.8473, 3.3473)(3.1946, 2.0)(5.1, 2.0)
+\qbezier(5.1, 2.0)(7.0054, 2.0)(8.3527, 3.3473)
+\qbezier(8.3527, 3.3473)(9.7, 4.6946)(9.7, 6.6)
+
+\drawline     ( 5.1  , 6.6   )( 2.800, 2.616 )  %C'A'
+\dashline[80]{0.2}( 5.1  , 6.6   )( 3.909, 2.157 )  %C'6:30
+\dashline[80]{0.2}( 5.1  , 6.6   )( 5.1  , 2     )  %C'6:00
+\dashline[80]{0.2}( 5.1  , 6.6   )( 6.291, 2.157 )  %C'5:30
+\drawline     ( 5.1  , 6.6   )( 7.400, 2.616 )  %C'B'
+\dashline[80]{0.2}(12.4  , 6.6   )(11.800, 2.039 )  %wedge6:30
+\dashline[80]{0.2}(12.4  , 6.6   )(13.000, 2.039 )  %wedge5:30
+\qbezier      (11.800, 2.039 )(12.4  , 1.8   )(13.000, 2.039 ) %wedge arc
+\put(  1.8 ,  1.7 ){$\scriptstyle A'$}   %A'
+\put(  7.3 ,  1.7 ){$\scriptstyle B'$}   %B'
+\put(  4.6 ,  6.9 ){$\scriptstyle C'$}   %C'
+\put( 11.8 ,  1   ){$m$}   %m
+\put(  4   ,  0.2 ){$\scriptstyle \textsc{Fig.~1.}$}
+\end{picture}
+\PGrestore
+%
+\quad
+% 109bb287.tex
+% Pt.287. Fig.2.   (10.2,11.7 )
+\PGset[0.8em]
+\begin{picture}    (10.2,11.7 )
+
+% Ellipse:  u = 5.1  v = 6.6  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(9.7, 6.6)(9.7, 8.5054)(8.3527, 9.8527)
+\qbezier(8.3527, 9.8527)(7.0054, 11.2)(5.1, 11.2)
+\qbezier(5.1, 11.2)(3.1946, 11.2)(1.8473, 9.8527)
+\qbezier(1.8473, 9.8527)(0.5, 8.5054)(0.5, 6.6)
+\qbezier(0.5, 6.6)(0.5, 4.6946)(1.8473, 3.3473)
+\qbezier(1.8473, 3.3473)(3.1946, 2.0)(5.1, 2.0)
+\qbezier(5.1, 2.0)(7.0054, 2.0)(8.3527, 3.3473)
+\qbezier(8.3527, 3.3473)(9.7, 4.6946)(9.7, 6.6)\drawline     ( 5.1  , 6.6   )( 1.451, 3.800 )  %CA
+
+\dashline[80]{0.2}( 5.1  , 6.6   )( 2.300, 2.951 )  %C 1
+\dashline[80]{0.2}( 5.1  , 6.6   )( 3.340, 2.350 )  %C 2
+\dashline[80]{0.2}( 5.1  , 6.6   )( 4.500, 2.039 )  %C 3
+\dashline[80]{0.2}( 5.1  , 6.6   )( 5.700, 2.039 )  %C 4
+\dashline[80]{0.2}( 5.1  , 6.6   )( 6.860, 2.350 )  %C 5
+\dashline[80]{0.2}( 5.1  , 6.6   )( 7.900, 2.951 )  %C 6
+\drawline     ( 5.1  , 6.6   )( 8.749, 3.800 )  %CB
+\put(  0.4 ,  3.4 ){$\scriptstyle A$}   %A
+\put(  8.8 ,  3.3 ){$\scriptstyle B$}   %B
+\put(  4.7 ,  6.9 ){$\scriptstyle C$}   %C
+\put(  4   ,  0.2 ){$\scriptstyle \textsc{Fig.~2.}$}
+\end{picture}
+\PGrestore
+%
+\quad
+% 109cc287.tex
+% Pt.287. Fig.3.   (10.2,11.7 )
+\PGset[0.8em]
+\begin{picture}    (10.2,11.7 )
+
+% Ellipse:  u = 5.1  v = 6.6  a = 4.6  b = 4.6  phi = 0.0 Grad
+\qbezier(9.7, 6.6)(9.7, 8.5054)(8.3527, 9.8527)
+\qbezier(8.3527, 9.8527)(7.0054, 11.2)(5.1, 11.2)
+\qbezier(5.1, 11.2)(3.1946, 11.2)(1.8473, 9.8527)
+\qbezier(1.8473, 9.8527)(0.5, 8.5054)(0.5, 6.6)
+\qbezier(0.5, 6.6)(0.5, 4.6946)(1.8473, 3.3473)
+\qbezier(1.8473, 3.3473)(3.1946, 2.0)(5.1, 2.0)
+\qbezier(5.1, 2.0)(7.0054, 2.0)(8.3527, 3.3473)
+\qbezier(8.3527, 3.3473)(9.7, 4.6946)(9.7, 6.6)
+
+\drawline     ( 5.1  , 6.6   )( 1.847, 3.347 )  %C'A'
+\dashline[80]{0.2}( 5.1  , 6.6   )( 7.900, 2.951 )  %C'D
+\drawline     ( 5.1  , 6.6   )( 8.749, 3.800 )  %C'B'
+\put(  0.6 ,  2.7 ){$\scriptstyle A'$}   %A'
+\put(  8.8 ,  3.3 ){$\scriptstyle B'$}   %B'
+\put(  4.6 ,  6.9 ){$\scriptstyle C'$}   %C'
+\put(  7.9 ,  2.2 ){$\scriptstyle D$}   %D
+\put(  4   ,  0.2 ){$\scriptstyle \textsc{Fig.~3.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/111ac289.eepic b/33063-t/images/sources/111ac289.eepic
new file mode 100644
index 0000000..a2e549a
--- /dev/null
+++ b/33063-t/images/sources/111ac289.eepic
@@ -0,0 +1,80 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 111aa289.tex
+% Pt.289. Fig.1.   (10.8,12.8 )
+\PGset[0.8em]
+\begin{picture}    (10.8,12.8 )
+
+% Ellipse:  u = 5.4  v = 6.9  a = 4.9  b = 4.9  phi = 0.0 Grad
+\qbezier(10.3, 6.9)(10.3, 8.9296)(8.8648, 10.3648)
+\qbezier(8.8648, 10.3648)(7.4296, 11.8)(5.4, 11.8)
+\qbezier(5.4, 11.8)(3.3704, 11.8)(1.9352, 10.3648)
+\qbezier(1.9352, 10.3648)(0.5, 8.9296)(0.5, 6.9)
+\qbezier(0.5, 6.9)(0.5, 4.8704)(1.9352, 3.4352)
+\qbezier(1.9352, 3.4352)(3.3704, 2.0)(5.4, 2.0)
+\qbezier(5.4, 2.0)(7.4296, 2.0)(8.8648, 3.4352)
+\qbezier(8.8648, 3.4352)(10.3, 4.8704)(10.3, 6.9)
+
+\drawline     ( 8.779, 3.352 )( 5.4  ,11.8   )( 5.4  , 2     )  %AB(C)P
+\dashline[80]{0.4}( 5.4  , 6.9   )( 8.779, 3.352 )  %CA
+\put(  8.7 ,  2.8 ){$\scriptstyle A$}   %A
+\put(  5   , 12   ){$\scriptstyle B$}   %B
+\put(  4.4 ,  6.65 ){$\scriptstyle C$}   %C
+\put(  5   ,  1.1 ){$\scriptstyle P$}   %P
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~1.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+% 111bb289.tex
+% Pt.289. Fig.2.   (10.8,12.8 )
+\PGset[0.8em]
+\begin{picture}    (10.8,12.8 )
+
+% Ellipse:  u = 5.4  v = 6.9  a = 4.9  b = 4.9  phi = 0.0 Grad
+\qbezier(10.3, 6.9)(10.3, 8.9296)(8.8648, 10.3648)
+\qbezier(8.8648, 10.3648)(7.4296, 11.8)(5.4, 11.8)
+\qbezier(5.4, 11.8)(3.3704, 11.8)(1.9352, 10.3648)
+\qbezier(1.9352, 10.3648)(0.5, 8.9296)(0.5, 6.9)
+\qbezier(0.5, 6.9)(0.5, 4.8704)(1.9352, 3.4352)
+\qbezier(1.9352, 3.4352)(3.3704, 2.0)(5.4, 2.0)
+\qbezier(5.4, 2.0)(7.4296, 2.0)(8.8648, 3.4352)
+\qbezier(8.8648, 3.4352)(10.3, 4.8704)(10.3, 6.9)
+
+\drawline     ( 9.320, 3.960 )( 5.4  ,11.8   )( 1.480, 3.960 )  %ABP
+\dashline[80]{0.4}( 5.4  ,11.8   )( 5.4  , 2     )  %B(C)E
+\put(  9.3 ,  3.4 ){$\scriptstyle A$}   %A
+\put(  5   , 12   ){$\scriptstyle B$}   %B
+\put(  5.6 ,  6.65 ){$\scriptstyle C$}   %C
+\put(  5   ,  1.1 ){$\scriptstyle E$}   %E
+\put(  0.6 ,  3.2 ){$\scriptstyle P$}   %P
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~2.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+% 111cc289.tex
+% Pt.289. Fig.3.   (11.3,12.8 )
+\PGset[0.8em]
+\begin{picture}    (11.3,12.8 )
+
+% Ellipse:  u = 5.4  v = 6.9  a = 4.9  b = 4.9  phi = 0.0 Grad
+\qbezier(10.3, 6.9)(10.3, 8.9296)(8.8648, 10.3648)
+\qbezier(8.8648, 10.3648)(7.4296, 11.8)(5.4, 11.8)
+\qbezier(5.4, 11.8)(3.3704, 11.8)(1.9352, 10.3648)
+\qbezier(1.9352, 10.3648)(0.5, 8.9296)(0.5, 6.9)
+\qbezier(0.5, 6.9)(0.5, 4.8704)(1.9352, 3.4352)
+\qbezier(1.9352, 3.4352)(3.3704, 2.0)(5.4, 2.0)
+\qbezier(5.4, 2.0)(7.4296, 2.0)(8.8648, 3.4352)
+\qbezier(8.8648, 3.4352)(10.3, 4.8704)(10.3, 6.9)
+
+\drawline     ( 9.858, 8.934 )( 5.4  ,11.8   )( 8.606, 3.194 )  %ABP
+\dashline[80]{0.4}( 5.4  ,11.8   )( 5.4  , 2     )  %B(C)F
+\put( 10.1 ,  8.7 ){$\scriptstyle A$}   %A
+\put(  5   , 12   ){$\scriptstyle B$}   %B
+\put(  4.4 ,  6.65 ){$\scriptstyle C$}   %C
+\put(  5   ,  1.1 ){$\scriptstyle F$}   %F
+\put(  8.55,  2.4 ){$\scriptstyle P$}   %P
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~3.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/112ac290.eepic b/33063-t/images/sources/112ac290.eepic
new file mode 100644
index 0000000..810bafa
--- /dev/null
+++ b/33063-t/images/sources/112ac290.eepic
@@ -0,0 +1,78 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 112aa290.tex
+% Pt.290. Fig.4.   (11,12 )
+\PGset[0.8em]
+\begin{picture}    (11,12 )
+
+% Ellipse:  u = 5.5  v = 6.5  a = 4.5  b = 4.5  phi = 0.0 Grad
+\qbezier(10.0, 6.5)(10.0, 8.364)(8.682, 9.682)
+\qbezier(8.682, 9.682)(7.364, 11.0)(5.5, 11.0)
+\qbezier(5.5, 11.0)(3.636, 11.0)(2.318, 9.682)
+\qbezier(2.318, 9.682)(1.0, 8.364)(1.0, 6.5)
+\qbezier(1.0, 6.5)(1.0, 4.636)(2.318, 3.318)
+\qbezier(2.318, 3.318)(3.636, 2.0)(5.5, 2.0)
+\qbezier(5.5, 2.0)(7.364, 2.0)(8.682, 3.318)
+\qbezier(8.682, 3.318)(10.0, 4.636)(10.0, 6.5)
+
+\drawline     ( 2.8  ,10.1   )( 1  , 6.5   )(10  , 6.5   )( 2.8,10.1 )  %ABCA
+\put(  1.9 , 10.1 ){$\scriptstyle A$}   %A
+\put(  0.1 ,  6.2 ){$\scriptstyle B$}   %B
+\put( 10.2 ,  6.6 ){$\scriptstyle C$}   %C
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~4.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+% 112bb290.tex
+% Pt.290. Fig.5.   (11,12 )
+\PGset[0.8em]
+\begin{picture}    (11,12 )
+
+% Ellipse:  u = 5.5  v = 6.5  a = 4.5  b = 4.5  phi = 0.0 Grad
+\qbezier(10.0, 6.5)(10.0, 8.364)(8.682, 9.682)
+\qbezier(8.682, 9.682)(7.364, 11.0)(5.5, 11.0)
+\qbezier(5.5, 11.0)(3.636, 11.0)(2.318, 9.682)
+\qbezier(2.318, 9.682)(1.0, 8.364)(1.0, 6.5)
+\qbezier(1.0, 6.5)(1.0, 4.636)(2.318, 3.318)
+\qbezier(2.318, 3.318)(3.636, 2.0)(5.5, 2.0)
+\qbezier(5.5, 2.0)(7.364, 2.0)(8.682, 3.318)
+\qbezier(8.682, 3.318)(10.0, 4.636)(10.0, 6.5)
+
+\drawline     ( 3.2  ,10.368 )( 1.632, 4.2   )( 9.368, 4.2   )( 3.2  ,10.368 )  %ACDA
+\drawline     ( 1.632, 4.2   )( 5.019, 2.026 )( 9.368, 4.2   )                   %CBD
+\put(  2.4 , 10.5 ){$\scriptstyle A$}   %A
+\put(  4.5 ,  1.2 ){$\scriptstyle B$}   %B
+\put(  0.6 ,  3.6 ){$\scriptstyle C$}   %C
+\put(  9.4 ,  3.6 ){$\scriptstyle D$}   %D
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~5.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+% 112cc290.tex
+% Pt.290. Fig.6.   (11,12 )
+\PGset[0.8em]
+\begin{picture}    (11,12 )
+
+% Ellipse:  u = 5.5  v = 6.5  a = 4.5  b = 4.5  phi = 0.0 Grad
+\qbezier(10.0, 6.5)(10.0, 8.364)(8.682, 9.682)
+\qbezier(8.682, 9.682)(7.364, 11.0)(5.5, 11.0)
+\qbezier(5.5, 11.0)(3.636, 11.0)(2.318, 9.682)
+\qbezier(2.318, 9.682)(1.0, 8.364)(1.0, 6.5)
+\qbezier(1.0, 6.5)(1.0, 4.636)(2.318, 3.318)
+\qbezier(2.318, 3.318)(3.636, 2.0)(5.5, 2.0)
+\qbezier(5.5, 2.0)(7.364, 2.0)(8.682, 3.318)
+\qbezier(8.682, 3.318)(10.0, 4.636)(10.0, 6.5)
+
+\drawline     ( 1    , 6.5   )( 1.469, 4.5   )( 9.531, 4.5   )( 1    , 6.5 )  %ADEA
+\drawline     ( 1.469, 4.5   )( 3.438,10.5   )( 9.531, 4.5   )                %DBE
+\drawline     ( 1.469, 4.5   )( 9.531, 8.5   )( 9.531, 4.5   )                %DCE
+\put(  0.05,  6.3 ){$\scriptstyle A$}   %A
+\put(  2.9 , 10.7 ){$\scriptstyle B$}   %B
+\put(  9.7 ,  8.4 ){$\scriptstyle C$}   %C
+\put(  0.45,  3.85){$\scriptstyle D$}   %D
+\put(  9.6 ,  3.85){$\scriptstyle E$}   %E
+\put(  4.3 ,  0.2 ){$\scriptstyle \textsc{Fig.~6.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/113aa294.eepic b/33063-t/images/sources/113aa294.eepic
new file mode 100644
index 0000000..448faba
--- /dev/null
+++ b/33063-t/images/sources/113aa294.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 113aa294.tex
+% Pt.294. Fig.a.   (12,11 )
+\PGset[0.8em]
+\begin{picture}    (12,11 )
+
+% Ellipse:  u = 6.0  v = 5.5  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 5.5)(11.0, 7.5711)(9.5355, 9.0355)
+\qbezier(9.5355, 9.0355)(8.0711, 10.5)(6.0, 10.5)
+\qbezier(6.0, 10.5)(3.9289, 10.5)(2.4645, 9.0355)
+\qbezier(2.4645, 9.0355)(1.0, 7.5711)(1.0, 5.5)
+\qbezier(1.0, 5.5)(1.0, 3.4289)(2.4645, 1.9645)
+\qbezier(2.4645, 1.9645)(3.9289, 0.5)(6.0, 0.5)
+\qbezier(6.0, 0.5)(8.0711, 0.5)(9.5355, 1.9645)
+\qbezier(9.5355, 1.9645)(11.0, 3.4289)(11.0, 5.5)
+
+\drawline     ( 2.379, 8.948 )(11    , 5.5   )  %C(O)A
+\drawline     ( 1    , 5.5   )( 9.621, 8.948 )  %D(O)B
+\dashline[80]{0.2}( 2.379, 2.052 )(11    , 5.5   )  %EA
+\put( 11.2 ,  5.2 ){$\scriptstyle A$}   %A
+\put(  9.7 ,  9   ){$\scriptstyle B$}   %B
+\put(  1.6 ,  8.9 ){$\scriptstyle C$}   %C
+\put(  0.1 ,  5.15){$\scriptstyle D$}   %D
+\put(  1.45,  1.35){$\scriptstyle E$}   %E
+\put(  5.7 ,  7.8 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/114aa295.eepic b/33063-t/images/sources/114aa295.eepic
new file mode 100644
index 0000000..eff539f
--- /dev/null
+++ b/33063-t/images/sources/114aa295.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 114aa295.tex
+% Pt.295. Fig.a.   (16,11.5 )
+\PGset[0.8em]
+\begin{picture}    (16,11.5 )
+
+% Ellipse:  u = 7.7  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(12.7, 6.0)(12.7, 8.0711)(11.2355, 9.5355)
+\qbezier(11.2355, 9.5355)(9.7711, 11.0)(7.7, 11.0)
+\qbezier(7.7, 11.0)(5.6289, 11.0)(4.1645, 9.5355)
+\qbezier(4.1645, 9.5355)(2.7, 8.0711)(2.7, 6.0)
+\qbezier(2.7, 6.0)(2.7, 3.9289)(4.1645, 2.4645)
+\qbezier(4.1645, 2.4645)(5.6289, 1.0)(7.7, 1.0)
+\qbezier(7.7, 1.0)(9.7711, 1.0)(11.2355, 2.4645)
+\qbezier(11.2355, 2.4645)(12.7, 3.9289)(12.7, 6.0)
+
+\drawline     ( 1    , 1     )(15    , 1     )  %M(A)O
+\drawline     ( 7.7  , 1     )( 4.7  ,10.000 )  %AH
+\dashline[80]{0.2}( 4.7  ,10.000 )(10.700,10.000 )  %HF
+\put(  7.3 ,  0.2 ){$\scriptstyle A$}   %A
+\put(  2.8 ,  2.2 ){$\scriptstyle E$}   %E
+\put( 11   ,  9.9 ){$\scriptstyle F$}   %F
+\put(  3.7 ,  9.8 ){$\scriptstyle H$}   %H
+\put(  0.05,  0.7 ){$\scriptstyle M$}   %M
+\put( 15.1 ,  0.6 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/115ac296.eepic b/33063-t/images/sources/115ac296.eepic
new file mode 100644
index 0000000..17fe3fa
--- /dev/null
+++ b/33063-t/images/sources/115ac296.eepic
@@ -0,0 +1,81 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 115aa296.tex
+% Pt.296. Fig.a.   ( 8.8,12.5 )
+\PGset[0.8em]
+\begin{picture}    ( 8.8,12.5 )
+
+% Ellipse:  u = 4.4  v = 4.9  a = 3.9  b = 3.9  phi = 0.0 Grad
+\qbezier(8.3, 4.9)(8.3, 6.5154)(7.1577, 7.6577)
+\qbezier(7.1577, 7.6577)(6.0154, 8.8)(4.4, 8.8)
+\qbezier(4.4, 8.8)(2.7846, 8.8)(1.6423, 7.6577)
+\qbezier(1.6423, 7.6577)(0.5, 6.5154)(0.5, 4.9)
+\qbezier(0.5, 4.9)(0.5, 3.2846)(1.6423, 2.1423)
+\qbezier(1.6423, 2.1423)(2.7846, 1.0)(4.4, 1.0)
+\qbezier(4.4, 1.0)(6.0154, 1.0)(7.1577, 2.1423)
+\qbezier(7.1577, 2.1423)(8.3, 3.2846)(8.3, 4.9)
+
+\drawline     ( 1.4  , 2.408 )( 4.4  ,11.5   )( 7.4  , 2.408 )  %E(C)O(B)A
+\dashline[80]{0.2}( 2.891, 1.304 )( 5.328, 8.688 )  %PB
+\put(  7.3 ,  1.7 ){$\scriptstyle A$}   %A
+\put(  5.3 ,  8.9 ){$\scriptstyle B$}   %B
+\put(  2.6 ,  8.9 ){$\scriptstyle C$}   %C
+\put(  0.7 ,  1.6 ){$\scriptstyle E$}   %E
+\put(  4.1 , 11.6 ){$\scriptstyle O$}   %O
+\put(  2.5 ,  0.4 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
+%
+\qquad
+% 115bb296.tex
+% Pt.296. Fig.b.   (11,12.5 )
+\PGset[0.8em]
+\begin{picture}    (11,12.5 )
+
+% Ellipse:  u = 4.4  v = 4.9  a = 3.9  b = 3.9  phi = 0.0 Grad
+\qbezier(8.3, 4.9)(8.3, 6.5154)(7.1577, 7.6577)
+\qbezier(7.1577, 7.6577)(6.0154, 8.8)(4.4, 8.8)
+\qbezier(4.4, 8.8)(2.7846, 8.8)(1.6423, 7.6577)
+\qbezier(1.6423, 7.6577)(0.5, 6.5154)(0.5, 4.9)
+\qbezier(0.5, 4.9)(0.5, 3.2846)(1.6423, 2.1423)
+\qbezier(1.6423, 2.1423)(2.7846, 1.0)(4.4, 1.0)
+\qbezier(4.4, 1.0)(6.0154, 1.0)(7.1577, 2.1423)
+\qbezier(7.1577, 2.1423)(8.3, 3.2846)(8.3, 4.9)
+
+\drawline     ( 1.254, 7.205 )( 4.4  ,11.5   )(10     , 3.855 )  %BO(A)E
+\dashline[80]{0.2}( 3.152, 1.205 )( 7.546, 7.205 )  %PA
+\put(  7.65,  7.2 ){$\scriptstyle A$}   %A
+\put(  0.4 ,  7.2 ){$\scriptstyle B$}   %B
+\put( 10   ,  3.1 ){$\scriptstyle E$}   %E
+\put(  6.7 ,  1.0 ){$\scriptstyle M$}   %M
+\put(  4.1 , 11.6 ){$\scriptstyle O$}   %O
+\put(  2.5 ,  0.4 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
+%
+\quad
+% 115cc296.tex
+% Pt.296. Fig.c.   (11.3,12.5 )
+\PGset[0.8em]
+\begin{picture}    (11.3,12.5 )
+
+% Ellipse:  u = 4.4  v = 4.9  a = 3.9  b = 3.9  phi = 0.0 Grad
+\qbezier(8.3, 4.9)(8.3, 6.5154)(7.1577, 7.6577)
+\qbezier(7.1577, 7.6577)(6.0154, 8.8)(4.4, 8.8)
+\qbezier(4.4, 8.8)(2.7846, 8.8)(1.6423, 7.6577)
+\qbezier(1.6423, 7.6577)(0.5, 6.5154)(0.5, 4.9)
+\qbezier(0.5, 4.9)(0.5, 3.2846)(1.6423, 2.1423)
+\qbezier(1.6423, 2.1423)(2.7846, 1.0)(4.4, 1.0)
+\qbezier(4.4, 1.0)(6.0154, 1.0)(7.1577, 2.1423)
+\qbezier(7.1577, 2.1423)(8.3, 3.2846)(8.3, 4.9)
+
+\drawline     ( 1.4  , 2.408 )( 4.4  ,11.5   )(10.3  , 3.445 )  %C(B)O(A)E
+\dashline[80]{0.2}( 5.557, 1.176 )( 7.546, 7.205 )  %PA
+\put(  7.65,  7.25){$\scriptstyle A$}   %A
+\put(  2.6 ,  8.9 ){$\scriptstyle B$}   %B
+\put(  0.7 ,  1.5 ){$\scriptstyle C$}   %C
+\put( 10.3 ,  2.8 ){$\scriptstyle E$}   %E
+\put(  4.1 , 11.6 ){$\scriptstyle O$}   %O
+\put(  5.4 ,  0.4 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/115dd297.eepic b/33063-t/images/sources/115dd297.eepic
new file mode 100644
index 0000000..294234d
--- /dev/null
+++ b/33063-t/images/sources/115dd297.eepic
@@ -0,0 +1,14 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% 115dd297.tex
+% Pt.297. Fig.d.   (10,10 )
+\PGset[0.8em]
+\begin{picture}    (10,10 )
+\drawline(1,5)(9,5)
+\drawline(5,1)(5,9)
+\put(  9.0 ,  4.8 ){$\scriptstyle A$}   %A
+\put(  4.6 ,  9.2 ){$\scriptstyle B$}   %B
+\put(  0.1 ,  4.7 ){$\scriptstyle C$}   %C
+\put(  4.6 ,  0.1 ){$\scriptstyle D$}   %D
+\put(  5.2 ,  4.15){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/115ee297.eepic b/33063-t/images/sources/115ee297.eepic
new file mode 100644
index 0000000..eaa0771
--- /dev/null
+++ b/33063-t/images/sources/115ee297.eepic
@@ -0,0 +1,22 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 115ee297.tex
+% Pt.297. Fig.e.   (11,12.6 )
+\PGset[0.8em]
+\begin{picture}    (11,12.6 )
+\drawline     ( 8.2 ,11.7  )( 1   , 6.3   )( 8.2 , 0.9  )   %BCB'
+\drawline(1,6.3)(10,6.3)
+\qbezier      ( 8.2 ,11.7  )(10   , 9.3   )(10   , 6.3  )   %B-control-A
+\qbezier      ( 8.2 , 0.9  )(10   , 3.3   )(10   , 6.3  )   %B'-control-A
+\qbezier [ 8] ( 3.88, 8.46 )( 4.6 , 7.5   )( 4.6 , 6.3  )   %upper dotted arc
+\qbezier [ 8] ( 4.28, 3.84 )( 5.1 , 4.933 )( 5.1 , 6.3  )   %lower dotted arc
+%\drawline     ( 4.3 , 4.3  )( 4.3, 3.9    )( 4.7  , 3.9 )   %lower arrow
+%\drawline     ( 3.9 , 7.9  )( 3.9, 8.4    )( 4.3  , 8.4 )   %upper arrow
+\put(  3.9 ,  8.5 ){\vector(-3, 4 ){0.001}}   %upper arrow
+\put(  4.3 ,  3.9 ){\vector(-3,-4 ){0.001}}   %lower arrow
+\put( 10.3 ,  6   ){$\scriptstyle A$}   %A
+\put(  8.2 , 11.8 ){$\scriptstyle B$}   %B
+\put(  0.1 ,  6   ){$\scriptstyle C$}   %C
+\put(  8.2 ,  0.2 ){$\scriptstyle B'$}   %B'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/116ag299.eepic b/33063-t/images/sources/116ag299.eepic
new file mode 100644
index 0000000..b2dddbe
--- /dev/null
+++ b/33063-t/images/sources/116ag299.eepic
@@ -0,0 +1,184 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 116aa299.tex
+% Pt.299. Fig.a.   ( 4.8, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 4.8, 8 )
+
+% Ellipse:  u = 2.4  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(4.7, 3.5)(4.7, 4.4527)(4.0263, 5.1263)
+\qbezier(4.0263, 5.1263)(3.3527, 5.8)(2.4, 5.8)
+\qbezier(2.4, 5.8)(1.4473, 5.8)(0.7737, 5.1263)
+\qbezier(0.7737, 5.1263)(0.1, 4.4527)(0.1, 3.5)
+\qbezier(0.1, 3.5)(0.1, 2.5473)(0.7737, 1.8737)
+\qbezier(0.7737, 1.8737)(1.4473, 1.2)(2.4, 1.2)
+\qbezier(2.4, 1.2)(3.3527, 1.2)(4.0263, 1.8737)
+\qbezier(4.0263, 1.8737)(4.7, 2.5473)(4.7, 3.5)
+
+\drawline     ( 0.963, 1.704 )( 3.837, 5.296 )   %c(o)a
+\drawline     ( 2.958, 1.269 )( 1.842, 5.731 )   %d(o)b
+\put( 3.8   ,  5.4 ){\emph{a}}   %a
+\put( 1.3   ,  5.95){\emph{b}}   %b
+\put( 0.4   ,  0.95){\emph{c}}   %c
+\put( 2.7   ,  0.1 ){\emph{d}}   %d
+\put( 1.8  ,  2.4 ){\emph{o}}   %o
+\end{picture}
+\PGrestore
+%
+% 116bb299.tex
+% Pt.299. Fig.b.   ( 4.8, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 4.8, 8 )
+
+% Ellipse:  u = 2.4  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(4.7, 3.5)(4.7, 4.4527)(4.0263, 5.1263)
+\qbezier(4.0263, 5.1263)(3.3527, 5.8)(2.4, 5.8)
+\qbezier(2.4, 5.8)(1.4473, 5.8)(0.7737, 5.1263)
+\qbezier(0.7737, 5.1263)(0.1, 4.4527)(0.1, 3.5)
+\qbezier(0.1, 3.5)(0.1, 2.5473)(0.7737, 1.8737)
+\qbezier(0.7737, 1.8737)(1.4473, 1.2)(2.4, 1.2)
+\qbezier(2.4, 1.2)(3.3527, 1.2)(4.0263, 1.8737)
+\qbezier(4.0263, 1.8737)(4.7, 2.5473)(4.7, 3.5)
+
+\drawline     ( 0.455, 2.273 )( 2.784, 5.768 )   %c(o)a
+\drawline     ( 4.328, 2.246 )( 1.146, 5.428 )   %d(o)b
+\put( 2.6   ,  6   ){\emph{a}}   %a
+\put( 0.6   ,  5.7 ){\emph{b}}   %b
+\put( 0     ,  1.4 ){\emph{c}}   %c
+\put( 4     ,  1.2 ){\emph{d}}   %d
+\put( 1.6   ,  3.5 ){\emph{o}}   %o
+\end{picture}
+\PGrestore
+%
+% 116cc299.tex
+% Pt.299. Fig.c.   ( 4.8, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 4.8, 8 )
+
+% Ellipse:  u = 2.4  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(4.7, 3.5)(4.7, 4.4527)(4.0263, 5.1263)
+\qbezier(4.0263, 5.1263)(3.3527, 5.8)(2.4, 5.8)
+\qbezier(2.4, 5.8)(1.4473, 5.8)(0.7737, 5.1263)
+\qbezier(0.7737, 5.1263)(0.1, 4.4527)(0.1, 3.5)
+\qbezier(0.1, 3.5)(0.1, 2.5473)(0.7737, 1.8737)
+\qbezier(0.7737, 1.8737)(1.4473, 1.2)(2.4, 1.2)
+\qbezier(2.4, 1.2)(3.3527, 1.2)(4.0263, 1.8737)
+\qbezier(4.0263, 1.8737)(4.7, 2.5473)(4.7, 3.5)
+
+\drawline     ( 0.434, 2.306 )( 2.192, 6.7   )   %ca-ext
+\drawline     ( 3.816, 1.688 )( 1.310, 6.7   )   %db-ext
+\put( 2     ,  6   ){\emph{a}}   %a
+\put( 0.8   ,  5.7 ){\emph{b}}   %b
+\put( 0     ,  1.4 ){\emph{c}}   %c
+\put( 3.6   ,  0.8 ){\emph{d}}   %d
+\put( 1.4   ,  4.4 ){\emph{o}}   %o
+\end{picture}
+\PGrestore
+%
+% 116dd299.tex
+% Pt.299. Fig.d.   ( 6.5, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 6.5, 8 )
+
+% Ellipse:  u = 4.1  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(6.4, 3.5)(6.4, 4.4527)(5.7263, 5.1263)
+\qbezier(5.7263, 5.1263)(5.0527, 5.8)(4.1, 5.8)
+\qbezier(4.1, 5.8)(3.1473, 5.8)(2.4737, 5.1263)
+\qbezier(2.4737, 5.1263)(1.8, 4.4527)(1.8, 3.5)
+\qbezier(1.8, 3.5)(1.8, 2.5473)(2.4737, 1.8737)
+\qbezier(2.4737, 1.8737)(3.1473, 1.2)(4.1, 1.2)
+\qbezier(4.1, 1.2)(5.0527, 1.2)(5.7263, 1.8737)
+\qbezier(5.7263, 1.8737)(6.4, 2.5473)(6.4, 3.5)
+
+\drawline     ( 0.6  , 1.643 )( 3.043, 6.529 )   %c(a)o-ext
+\drawline     ( 4.258, 1.205 )( 1.043, 6.029 )   %d(b)o-ext
+\put( 1.6   ,  5.2 ){\emph{a}}   %a
+\put( 1     ,  4.3 ){\emph{b}}   %b
+\put( 0.1   ,  1.05){\emph{c}}   %c
+\put( 3.8   ,  0.1 ){\emph{d}}   %d
+\put( 1.5   ,  0.8 ){\emph{m}}   %m
+\put( 1.8   ,  3.5 ){\emph{o}}   %o
+\end{picture}
+\PGrestore
+%
+% 116ee299.tex
+% Pt.299. Fig.e.   ( 6.5, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 6.5, 8 )
+
+% Ellipse:  u = 4.1  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(6.4, 3.5)(6.4, 4.4527)(5.7263, 5.1263)
+\qbezier(5.7263, 5.1263)(5.0527, 5.8)(4.1, 5.8)
+\qbezier(4.1, 5.8)(3.1473, 5.8)(2.4737, 5.1263)
+\qbezier(2.4737, 5.1263)(1.8, 4.4527)(1.8, 3.5)
+\qbezier(1.8, 3.5)(1.8, 2.5473)(2.4737, 1.8737)
+\qbezier(2.4737, 1.8737)(3.1473, 1.2)(4.1, 1.2)
+\qbezier(4.1, 1.2)(5.0527, 1.2)(5.7263, 1.8737)
+\qbezier(5.7263, 1.8737)(6.4, 2.5473)(6.4, 3.5)
+
+\drawline     ( 0.6  , 2.240 )( 3.920, 7.552 )   %c(a)o-ext
+\drawline     ( 5.789, 1.939 )( 3.066, 7.386 )   %d(b)o-ext
+\put( 1.4   ,  4.7 ){\emph{a}}   %a
+\put( 3.8   ,  6   ){\emph{b}}   %b
+\put( 0.1   ,  1.55){\emph{c}}   %c
+\put( 5.6   ,  1   ){\emph{d}}   %d
+\put( 2.3   ,  0.6 ){\emph{m}}   %m
+\put( 5.7   ,  5.2 ){\emph{n}}   %n
+% select ONE of the following for  o
+%\put( 2.9   ,  5.7 ){\emph{o}}   %o
+\put( 3.1   ,  5.9 ){\scriptsize o}   %scrptsz o
+\end{picture}
+\PGrestore
+%
+% 116ff299.tex
+% Pt.299. Fig.f.   ( 4.8, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 4.8, 8 )
+
+% Ellipse:  u = 2.4  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(4.7, 3.5)(4.7, 4.4527)(4.0263, 5.1263)
+\qbezier(4.0263, 5.1263)(3.3527, 5.8)(2.4, 5.8)
+\qbezier(2.4, 5.8)(1.4473, 5.8)(0.7737, 5.1263)
+\qbezier(0.7737, 5.1263)(0.1, 4.4527)(0.1, 3.5)
+\qbezier(0.1, 3.5)(0.1, 2.5473)(0.7737, 1.8737)
+\qbezier(0.7737, 1.8737)(1.4473, 1.2)(2.4, 1.2)
+\qbezier(2.4, 1.2)(3.3527, 1.2)(4.0263, 1.8737)
+\qbezier(4.0263, 1.8737)(4.7, 2.5473)(4.7, 3.5)
+
+\drawline     ( 0.445, 2.289 )( 3.000, 7.4   )   %c(o)a-ext
+\drawline     ( 3.162, 1.330 )( 2.408, 7.7   )   %d(o)b-ext
+\put( 1.5   ,  6   ){\emph{a}}   %a
+\put( 2.7   ,  5.9 ){\emph{b}}   %b
+\put( 0     ,  1.4 ){\emph{c}}   %c
+\put( 2.8   ,  0.3 ){\emph{d}}   %d
+\end{picture}
+\PGrestore
+%
+% 116gg299.tex
+% Pt.299. Fig.g.   ( 7.6, 8 )
+\PGset[0.8em]
+\begin{picture}    ( 7.6, 8 )
+
+% Ellipse:  u = 4.1  v = 3.5  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(6.4, 3.5)(6.4, 4.4527)(5.7263, 5.1263)
+\qbezier(5.7263, 5.1263)(5.0527, 5.8)(4.1, 5.8)
+\qbezier(4.1, 5.8)(3.1473, 5.8)(2.4737, 5.1263)
+\qbezier(2.4737, 5.1263)(1.8, 4.4527)(1.8, 3.5)
+\qbezier(1.8, 3.5)(1.8, 2.5473)(2.4737, 1.8737)
+\qbezier(2.4737, 1.8737)(3.1473, 1.2)(4.1, 1.2)
+\qbezier(4.1, 1.2)(5.0527, 1.2)(5.7263, 1.8737)
+\qbezier(5.7263, 1.8737)(6.4, 2.5473)(6.4, 3.5)
+
+\drawline     ( 0.6  , 3.120 )( 4.967, 7.8   )   %c(a)o-ext
+\drawline     ( 7.1  , 2.973 )( 4.109, 7.95  )   %d(b)o-ext
+\put( 1.7   ,  5.1 ){\emph{a}}   %a
+\put( 5.9   ,  5   ){\emph{b}}   %b
+\put( 0.1   ,  2.5 ){\emph{c}}   %c
+\put( 6.8   ,  2.1 ){\emph{d}}   %d
+\put( 3.5   ,  0.4 ){\emph{m}}   %m
+\put( 3.65  ,  5.9 ){\emph{n}}   %n
+% select ONE of the following for  o
+%\put( 4     ,  6.3 ){\emph{o}}   %o
+\put( 4.2   ,  6.5 ){\scriptsize o}   %scrptsz o
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/117aa104.eepic b/33063-t/images/sources/117aa104.eepic
new file mode 100644
index 0000000..5ed3dd8
--- /dev/null
+++ b/33063-t/images/sources/117aa104.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 117aa104.tex
+% Ex.104. Fig.a.   ( 7.1, 7 )
+\PGset[0.8em]
+\begin{picture}    ( 7.1, 7 )
+
+% Ellipse:  u = 3.6  v = 3.5  a = 2.7  b = 2.7  phi = 0.0 Grad
+\qbezier(6.3, 3.5)(6.3, 4.6184)(5.5092, 5.4092)
+\qbezier(5.5092, 5.4092)(4.7184, 6.2)(3.6, 6.2)
+\qbezier(3.6, 6.2)(2.4816, 6.2)(1.6908, 5.4092)
+\qbezier(1.6908, 5.4092)(0.9, 4.6184)(0.9, 3.5)
+\qbezier(0.9, 3.5)(0.9, 2.3816)(1.6908, 1.5908)
+\qbezier(1.6908, 1.5908)(2.4816, 0.8)(3.6, 0.8)
+\qbezier(3.6, 0.8)(4.7184, 0.8)(5.5092, 1.5908)
+\qbezier(5.5092, 1.5908)(6.3, 2.3816)(6.3, 3.5)
+
+\drawline     ( 2.506, 5.968 )( 5.315, 5.585 )( 6.095, 2.467 )( 1.105, 2.467 )( 2.506, 5.968 )  %ABCDA
+\dashline[80]{0.2}( 2.506, 5.968 )( 3.6  , 3.5   )( 5.315, 5.585 )  %AOB
+\dashline[80]{0.2}( 6.095, 2.467 )( 3.6  , 3.5   )( 1.105, 2.467 )  %COD
+\put(  1.9 ,  6.15){$\scriptstyle A$}   %A
+\put(  5.3 ,  5.8 ){$\scriptstyle B$}   %B
+\put(  6.25,  2.1 ){$\scriptstyle C$}   %C
+\put(  0.1 ,  1.9 ){$\scriptstyle D$}   %D
+\put(  2.5 ,  3.4 ){$\scriptstyle O$}   %O
+\put(  4.4 ,  0.1 ){$\scriptstyle M$}   %M
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/117bb105.eepic b/33063-t/images/sources/117bb105.eepic
new file mode 100644
index 0000000..37d970f
--- /dev/null
+++ b/33063-t/images/sources/117bb105.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 117bb105.tex
+% Ex.105. Fig.b.   ( 7.4, 5.6 )
+\PGset[0.8em]
+\begin{picture}    ( 7.4, 5.6 )
+
+% Ellipse:  u = 3.7  v = 2.8  a = 2.7  b = 2.7  phi = 0.0 Grad
+\qbezier(6.4, 2.8)(6.4, 3.9184)(5.6092, 4.7092)
+\qbezier(5.6092, 4.7092)(4.8184, 5.5)(3.7, 5.5)
+\qbezier(3.7, 5.5)(2.5816, 5.5)(1.7908, 4.7092)
+\qbezier(1.7908, 4.7092)(1.0, 3.9184)(1.0, 2.8)
+\qbezier(1.0, 2.8)(1.0, 1.6816)(1.7908, 0.8908)
+\qbezier(1.7908, 0.8908)(2.5816, 0.1)(3.7, 0.1)
+\qbezier(3.7, 0.1)(4.8184, 0.1)(5.6092, 0.8908)
+\qbezier(5.6092, 0.8908)(6.4, 1.6816)(6.4, 2.8)
+
+%\put          ( 3.7  , 2.75  ){\circle*{ 0.05}}     %circle dot center (part of dashed line)
+\drawline     ( 1    , 2.8   )( 6.4  , 2.8   )    %C(OA)B
+\drawline     ( 5.3  , 2.8   )( 6.208, 1.8   )    %AD
+\dashline[80]{0.2}( 3.7  , 2.8   )( 6.208, 1.8   )    %OD
+\put(  5   ,  3   ){$\scriptstyle A$}   %A
+\put(  6.5 ,  2.5 ){$\scriptstyle B$}   %B
+\put(  0.1 ,  2.5 ){$\scriptstyle C$}   %C
+\put(  6.2 ,  1.3 ){$\scriptstyle D$}   %D
+\put(  3.5 ,  3   ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/118aa107.eepic b/33063-t/images/sources/118aa107.eepic
new file mode 100644
index 0000000..b27e45f
--- /dev/null
+++ b/33063-t/images/sources/118aa107.eepic
@@ -0,0 +1,27 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 118aa107.tex
+% Ex.107. Fig.a.   (11.1, 6.6 )
+\PGset[0.8em]
+\begin{picture}    (11.1, 6.6 )
+
+% Ellipse:  u = 6.95  v = 3.3  a = 3.25  b = 3.25  phi = 0.0 Grad
+\qbezier(10.2, 3.3)(10.2, 4.6462)(9.2481, 5.5981)
+\qbezier(9.2481, 5.5981)(8.2962, 6.55)(6.95, 6.55)
+\qbezier(6.95, 6.55)(5.6038, 6.55)(4.6519, 5.5981)
+\qbezier(4.6519, 5.5981)(3.7, 4.6462)(3.7, 3.3)
+\qbezier(3.7, 3.3)(3.7, 1.9538)(4.6519, 1.0019)
+\qbezier(4.6519, 1.0019)(5.6038, 0.05)(6.95, 0.05)
+\qbezier(6.95, 0.05)(8.2962, 0.05)(9.2481, 1.0019)
+\qbezier(9.2481, 1.0019)(10.2, 1.9538)(10.2, 3.3)
+
+\drawline     ( 9.124, 5.716 )( 0.5  , 3.3   )(10.2  , 3.3   )    %D(C)P(AO)B
+\dashline[80]{0.2}( 3.837, 4.235 )( 6.95 , 3.35  )( 9.124, 5.716 )    %COD
+\put(  2.85,  2.5 ){$\scriptstyle A$}   %A
+\put( 10.2 ,  2.6 ){$\scriptstyle B$}   %B
+\put(  3.05,  4.35){$\scriptstyle C$}   %C
+\put(  9.1 ,  5.8 ){$\scriptstyle D$}   %D
+\put(  6.5 ,  2.5 ){$\scriptstyle O$}   %O
+\put(  0.1 ,  2.5 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/118be108.eepic b/33063-t/images/sources/118be108.eepic
new file mode 100644
index 0000000..8515aca
--- /dev/null
+++ b/33063-t/images/sources/118be108.eepic
@@ -0,0 +1,128 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 118bb108.tex
+% Ex.108. Fig.b.   (7.2, 9.7 )
+\PGset[0.8em]
+\begin{picture}    (7.2, 9.7 )
+
+% Ellipse:  u = 3.4  v = 3.8  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(6.2, 3.8)(6.2, 4.9598)(5.3799, 5.7799)
+\qbezier(5.3799, 5.7799)(4.5598, 6.6)(3.4, 6.6)
+\qbezier(3.4, 6.6)(2.2402, 6.6)(1.4201, 5.7799)
+\qbezier(1.4201, 5.7799)(0.6, 4.9598)(0.6, 3.8)
+\qbezier(0.6, 3.8)(0.6, 2.6402)(1.4201, 1.8201)
+\qbezier(1.4201, 1.8201)(2.2402, 1.0)(3.4, 1.0)
+\qbezier(3.4, 1.0)(4.5598, 1.0)(5.3799, 1.8201)
+\qbezier(5.3799, 1.8201)(6.2, 2.6402)(6.2, 3.8)
+
+\drawline     ( 3.4  , 1     )( 3.4  , 6.6   )    %bottom(OP)top
+\drawline     ( 1.175, 5.5   )( 5.625, 5.5   )    %A(P)B
+\drawline     ( 2.258, 6.357 )( 6.174, 3.419 )    %C(PQ)D
+\dashline[80]{0.2}( 3.4  , 3.8   )( 4.216, 4.888 )    %OQ
+\put(  0.3 ,  5.3 ){$\scriptstyle A$}   %A
+\put(  5.65,  5.5 ){$\scriptstyle B$}   %B
+\put(  1.7 ,  6.5 ){$\scriptstyle C$}   %C
+\put(  6.3 ,  3   ){$\scriptstyle D$}   %D
+\put(  2.5 ,  3.4 ){$\scriptstyle O$}   %O 3.4,3.8
+\put(  3.5 ,  5.6 ){$\scriptstyle P$}   %P 3.4,5.5
+\put(  4.4 ,  4.75){$\scriptstyle Q$}   %Q
+\end{picture}
+\PGrestore
+%
+\qquad
+% 118cc108.tex
+% Ex.108. Fig.c.   (7.6, 9.7 )
+\PGset[0.8em]
+\begin{picture}    (7.6, 9.7 )
+
+% Ellipse:  u = 3.8  v = 3.8  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(6.6, 3.8)(6.6, 4.9598)(5.7799, 5.7799)
+\qbezier(5.7799, 5.7799)(4.9598, 6.6)(3.8, 6.6)
+\qbezier(3.8, 6.6)(2.6402, 6.6)(1.8201, 5.7799)
+\qbezier(1.8201, 5.7799)(1.0, 4.9598)(1.0, 3.8)
+\qbezier(1.0, 3.8)(1.0, 2.6402)(1.8201, 1.8201)
+\qbezier(1.8201, 1.8201)(2.6402, 1.0)(3.8, 1.0)
+\qbezier(3.8, 1.0)(4.9598, 1.0)(5.7799, 1.8201)
+\qbezier(5.7799, 1.8201)(6.6, 2.6402)(6.6, 3.8)
+
+\drawline     ( 1.296, 2.548 )( 6.304, 5.052 )    %H(OC)K
+\drawline     ( 1.680, 5.630 )( 6.531, 4.417 )    %A(MC)B
+\drawline     ( 5.932, 5.615 )( 3.992, 1.007 )    %D(CN)E
+\dashline[80]{0.2}( 4.106, 5.023 )( 3.8  , 3.8   )( 4.962, 3.311 )    %MON
+\put(  0.7 ,  5.6 ){$\scriptstyle A$}   %A
+\put(  6.7 ,  3.9 ){$\scriptstyle B$}   %B
+\put(  4.9 ,  4.9 ){$\scriptstyle C$}   %C 5.533, 4.667
+\put(  5.8 ,  5.9 ){$\scriptstyle D$}   %D
+\put(  3.6 ,  0.2 ){$\scriptstyle E$}   %E
+\put(  0.2 ,  1.9 ){$\scriptstyle H$}   %H
+\put(  6.5 ,  4.9 ){$\scriptstyle K$}   %K
+\put(  3.6 ,  5.2 ){$\scriptstyle M$}   %M
+\put(  5.1 ,  2.9 ){$\scriptstyle N$}   %N
+\put(  2.9 ,  3.8 ){$\scriptstyle O$}   %O 
+\end{picture}
+\PGrestore
+%
+\qquad
+% 118dd108.tex
+% Ex.108. Fig.d.   (9.2, 9.7 )
+\PGset[0.8em]
+\begin{picture}    (9.2, 9.7 )
+
+% Ellipse:  u = 3.8  v = 4.6  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(6.6, 4.6)(6.6, 5.7598)(5.7799, 6.5799)
+\qbezier(5.7799, 6.5799)(4.9598, 7.4)(3.8, 7.4)
+\qbezier(3.8, 7.4)(2.6402, 7.4)(1.8201, 6.5799)
+\qbezier(1.8201, 6.5799)(1.0, 5.7598)(1.0, 4.6)
+\qbezier(1.0, 4.6)(1.0, 3.4402)(1.8201, 2.6201)
+\qbezier(1.8201, 2.6201)(2.6402, 1.8)(3.8, 1.8)
+\qbezier(3.8, 1.8)(4.9598, 1.8)(5.7799, 2.6201)
+\qbezier(5.7799, 2.6201)(6.6, 3.4402)(6.6, 4.6)
+
+\drawline     ( 1    , 7.7   )( 1    , 1     )( 3.8  , 4.6   )  %A(L)DO
+\drawline     ( 1    , 1     )( 8.325, 2.860 )( 3.8  , 4.6   )  %D(K)CO
+\drawline     ( 8.325, 2.860 )( 5.397, 7.252 )( 3.8  , 4.6   )  %C(H)BO
+\drawline     ( 5.397, 7.252 )( 1    , 7.7   )( 3.8  , 4.6   )  %B(G)AO
+\dashline[80]{0.2}( 3.8  , 4.6   )( 1    , 4.35  )                  %OL
+\dashline[80]{0.2}( 3.8  , 4.6   )( 4.489, 1.886 )                  %OK
+\dashline[80]{0.2}( 3.8  , 4.6   )( 6.130, 6.153 )                  %OH
+\dashline[80]{0.2}( 3.8  , 4.6   )( 4.084, 7.386 )                  %OG
+\put(  0.1 ,  7.6 ){$\scriptstyle A$}   %A
+\put(  5.5 ,  7.2 ){$\scriptstyle B$}   %B
+\put(  8.4 ,  2.5 ){$\scriptstyle C$}   %C
+\put(  0.5 ,  0.2 ){$\scriptstyle D$}   %D
+\put(  3.8 ,  7.6 ){$\scriptstyle G$}   %G
+\put(  6.2 ,  6.1 ){$\scriptstyle H$}   %H
+\put(  4.2 ,  1   ){$\scriptstyle K$}   %K
+\put(  0.2 ,  4   ){$\scriptstyle L$}   %L
+\put(  4.2 ,  4.6 ){$\scriptstyle O$}   %O 
+\end{picture}
+\PGrestore
+%
+\qquad
+% 118ee108.tex
+% Ex.108. Fig.e.   (9.8, 9.7 )
+\PGset[0.8em]
+\begin{picture}    (9.8, 9.7 )
+
+% Ellipse:  u = 5.034  v = 3.56  a = 2.56  b = 2.56  phi = 0.0 Grad
+\qbezier(7.594, 3.56)(7.594, 4.6204)(6.8442, 5.3702)
+\qbezier(6.8442, 5.3702)(6.0944, 6.12)(5.034, 6.12)
+\qbezier(5.034, 6.12)(3.9736, 6.12)(3.2238, 5.3702)
+\qbezier(3.2238, 5.3702)(2.474, 4.6204)(2.474, 3.56)
+\qbezier(2.474, 3.56)(2.474, 2.4996)(3.2238, 1.7498)
+\qbezier(3.2238, 1.7498)(3.9736, 1.0)(5.034, 1.0)
+\qbezier(5.034, 1.0)(6.0944, 1.0)(6.8442, 1.7498)
+\qbezier(6.8442, 1.7498)(7.594, 2.4996)(7.594, 3.56)
+
+\drawline     ( 5.034, 8.68  )( 0.6  , 1     )( 9.468, 1     )( 5.034, 8.68  )  %A(E)B(D)CA
+\drawline     ( 5.034, 1     )( 5.034, 8.68  )                                  %D(OF)A
+\dashline[80]{0.2}( 5.034, 6.12  )( 2.817, 4.84  )( 5.034, 3.56  )                  %FEO
+\put(  4.7 ,  8.9 ){$\scriptstyle A$}   %A
+\put(  0.1 ,  0.1 ){$\scriptstyle B$}   %B
+\put(  8.9 ,  0.1 ){$\scriptstyle C$}   %C
+\put(  4.5 ,  0.1 ){$\scriptstyle D$}   %D
+\put(  1.9 ,  4.8 ){$\scriptstyle E$}   %E
+\put(  5.1 ,  6.3 ){$\scriptstyle F$}   %F
+\put(  5.2 ,  3.1 ){$\scriptstyle O$}   %O 
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119aa120.eepic b/33063-t/images/sources/119aa120.eepic
new file mode 100644
index 0000000..c12702a
--- /dev/null
+++ b/33063-t/images/sources/119aa120.eepic
@@ -0,0 +1,39 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119aa120.tex
+% Ex.120. Fig.a.   (11.2, 7.5 )
+\PGset[0.8em]
+\begin{picture}    (11.2, 7.5 )
+
+% Ellipse:  u = 3.5  v = 3.6  a = 3.2  b = 3.2  phi = 0.0 Grad
+\qbezier(6.7, 3.6)(6.7, 4.9255)(5.7627, 5.8627)
+\qbezier(5.7627, 5.8627)(4.8255, 6.8)(3.5, 6.8)
+\qbezier(3.5, 6.8)(2.1745, 6.8)(1.2373, 5.8627)
+\qbezier(1.2373, 5.8627)(0.3, 4.9255)(0.3, 3.6)
+\qbezier(0.3, 3.6)(0.3, 2.2745)(1.2373, 1.3373)
+\qbezier(1.2373, 1.3373)(2.1745, 0.4)(3.5, 0.4)
+\qbezier(3.5, 0.4)(4.8255, 0.4)(5.7627, 1.3373)
+\qbezier(5.7627, 1.3373)(6.7, 2.2745)(6.7, 3.6)
+
+% Ellipse:  u = 8.8  v = 3.6  a = 2.1  b = 2.1  phi = 0.0 Grad
+\qbezier(10.9, 3.6)(10.9, 4.4698)(10.2849, 5.0849)
+\qbezier(10.2849, 5.0849)(9.6698, 5.7)(8.8, 5.7)
+\qbezier(8.8, 5.7)(7.9302, 5.7)(7.3151, 5.0849)
+\qbezier(7.3151, 5.0849)(6.7, 4.4698)(6.7, 3.6)
+\qbezier(6.7, 3.6)(6.7, 2.7302)(7.3151, 2.1151)
+\qbezier(7.3151, 2.1151)(7.9302, 1.5)(8.8, 1.5)
+\qbezier(8.8, 1.5)(9.6698, 1.5)(10.2849, 2.1151)
+\qbezier(10.2849, 2.1151)(10.9, 2.7302)(10.9, 3.6)
+
+\drawline     ( 0.94 , 5.52  )(10.48 , 2.34  )(10.06 , 5.28  )  %A(P)BD
+\drawline     ( 0.94 , 5.52  )( 1.58 , 1.04  )(10.06 , 5.28  )  %AC(P)D
+\dashline[80]{0.2}( 6.7  , 0.1   )( 6.7  , 7.4   )                  %M(P)N 
+\put(  0.1 ,  5.5 ){$\scriptstyle A$}   %A
+\put( 10.4 ,  1.6 ){$\scriptstyle B$}   %B
+\put(  1   ,  0.1 ){$\scriptstyle C$}   %C
+\put( 10   ,  5.5 ){$\scriptstyle D$}   %D
+\put(  6.8 ,  0.1 ){$\scriptstyle M$}   %M
+\put(  5.8 ,  6.8 ){$\scriptstyle N$}   %N
+\put(  5.55,  4.2 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119bb121.eepic b/33063-t/images/sources/119bb121.eepic
new file mode 100644
index 0000000..a5d7b88
--- /dev/null
+++ b/33063-t/images/sources/119bb121.eepic
@@ -0,0 +1,47 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119bb121.tex
+%*****************************************************************************
+% Steeper lines than original in order to make smaller jaggies
+% See below for original diagram if needed
+% Ex.121. Fig.b.   ( 7, 5 )
+%*****************************************************************************
+%
+% For original diagram:
+%  1. Delete -- or otherwise inactivate -- all above between star lines ( %************ )
+%  2. Uncomment all double percent lines ( %% ) below i.e remove %%
+% 119bb121.tex
+% Ex.121. Fig.b.   ( 7.2, 5 )
+\PGset[0.8em]
+\begin{picture}    ( 7.2, 5 )
+
+% Ellipse:  u = 2.3  v = 2.5  a = 1.7  b = 1.7  phi = 0.0 Grad
+\qbezier(4.0, 2.5)(4.0, 3.2042)(3.5021, 3.7021)
+\qbezier(3.5021, 3.7021)(3.0042, 4.2)(2.3, 4.2)
+\qbezier(2.3, 4.2)(1.5958, 4.2)(1.0979, 3.7021)
+\qbezier(1.0979, 3.7021)(0.6, 3.2042)(0.6, 2.5)
+\qbezier(0.6, 2.5)(0.6, 1.7958)(1.0979, 1.2979)
+\qbezier(1.0979, 1.2979)(1.5958, 0.8)(2.3, 0.8)
+\qbezier(2.3, 0.8)(3.0042, 0.8)(3.5021, 1.2979)
+\qbezier(3.5021, 1.2979)(4.0, 1.7958)(4.0, 2.5)
+
+% Ellipse:  u = 4.9  v = 2.5  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier(6.9, 2.5)(6.9, 3.3284)(6.3142, 3.9142)
+\qbezier(6.3142, 3.9142)(5.7284, 4.5)(4.9, 4.5)
+\qbezier(4.9, 4.5)(4.0716, 4.5)(3.4858, 3.9142)
+\qbezier(3.4858, 3.9142)(2.9, 3.3284)(2.9, 2.5)
+\qbezier(2.9, 2.5)(2.9, 1.6716)(3.4858, 1.0858)
+\qbezier(3.4858, 1.0858)(4.0716, 0.5)(4.9, 0.5)
+\qbezier(4.9, 0.5)(5.7284, 0.5)(6.3142, 1.0858)
+\qbezier(6.3142, 1.0858)(6.9, 1.6716)(6.9, 2.5)
+
+\drawline     ( 1 , 1.405 )( 1 , 3.595 )( 6.159, 4.054 )( 6.159, 0.946 )( 1 , 1.405 )  %EC(A)DF(B)E
+\dashline[80]{0.2}( 3.387, 1.193 )( 3.387, 3.807 )     %BA
+\put(  3   ,  4.3 ){$\scriptstyle A$}   %A
+\put(  2.9 ,  0.1 ){$\scriptstyle B$}   %B
+\put(  0.3 ,  3.8 ){$\scriptstyle C$}   %C
+\put(  6.3 ,  4.15){$\scriptstyle D$}   %D
+\put(  0.1 ,  0.6 ){$\scriptstyle E$}   %E
+\put(  6.2 ,  0.1 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119cc122.eepic b/33063-t/images/sources/119cc122.eepic
new file mode 100644
index 0000000..57a9c8a
--- /dev/null
+++ b/33063-t/images/sources/119cc122.eepic
@@ -0,0 +1,39 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119cc122.tex
+% Ex.122. Fig.c.   ( 7.8, 5.2 )
+\PGset[0.8em]
+\begin{picture}    ( 7.8, 5.2 )
+
+% Ellipse:  u = 2.8  v = 2.6  a = 2.3  b = 2.3  phi = 0.0 Grad
+\qbezier(5.1, 2.6)(5.1, 3.5527)(4.4263, 4.2263)
+\qbezier(4.4263, 4.2263)(3.7527, 4.9)(2.8, 4.9)
+\qbezier(2.8, 4.9)(1.8473, 4.9)(1.1737, 4.2263)
+\qbezier(1.1737, 4.2263)(0.5, 3.5527)(0.5, 2.6)
+\qbezier(0.5, 2.6)(0.5, 1.6473)(1.1737, 0.9737)
+\qbezier(1.1737, 0.9737)(1.8473, 0.3)(2.8, 0.3)
+\qbezier(2.8, 0.3)(3.7527, 0.3)(4.4263, 0.9737)
+\qbezier(4.4263, 0.9737)(5.1, 1.6473)(5.1, 2.6)
+
+% Ellipse:  u = 5.9  v = 2.6  a = 1.8  b = 1.8  phi = 0.0 Grad
+\qbezier(7.7, 2.6)(7.7, 3.3456)(7.1728, 3.8728)
+\qbezier(7.1728, 3.8728)(6.6456, 4.4)(5.9, 4.4)
+\qbezier(5.9, 4.4)(5.1544, 4.4)(4.6272, 3.8728)
+\qbezier(4.6272, 3.8728)(4.1, 3.3456)(4.1, 2.6)
+\qbezier(4.1, 2.6)(4.1, 1.8544)(4.6272, 1.3272)
+\qbezier(4.6272, 1.3272)(5.1544, 0.8)(5.9, 0.8)
+\qbezier(5.9, 0.8)(6.6456, 0.8)(7.1728, 1.3272)
+\qbezier(7.1728, 1.3272)(7.7, 1.8544)(7.7, 2.6)
+
+\put          ( 2.8  , 2.6   ){\circle*{ 0.2}}     %circle left  center dot
+\put          ( 5.9  , 2.6   ){\circle*{ 0.2}}     %circle right center dot
+\drawline     ( 0.919, 1.276 )( 7.119, 1.276 )( 4.681, 3.924 )( 0.919, 1.276 )  %C(B)D(O')A(O)C
+\dashline[80]{0.2}( 4.681, 1.276 )( 4.681, 3.924 )     %BA
+\put(  4.4 ,  4.5){$\scriptstyle A$}   %A
+\put(  4.3 ,  0.1 ){$\scriptstyle B$}   %B
+\put(  0.1 ,  0.5 ){$\scriptstyle C$}   %C
+\put(  6.9 ,  0.5 ){$\scriptstyle D$}   %D
+\put(  2   ,  2.7 ){$\scriptstyle O$}   %O
+\put(  6   ,  2.7 ){$\scriptstyle O'$}   %O'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119dd123.eepic b/33063-t/images/sources/119dd123.eepic
new file mode 100644
index 0000000..3b392e6
--- /dev/null
+++ b/33063-t/images/sources/119dd123.eepic
@@ -0,0 +1,30 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119dd123.tex
+% Ex.123. Fig.d.   (10.9, 8.2 )
+\PGset[0.8em]
+\begin{picture}    (10.9, 8.2 )
+
+% Ellipse:  u = 7.906  v = 3.194  a = 2.194  b = 2.194  phi = 0.0 Grad
+\qbezier(10.1, 3.194)(10.1, 4.1028)(9.4574, 4.7454)
+\qbezier(9.4574, 4.7454)(8.8148, 5.388)(7.906, 5.388)
+\qbezier(7.906, 5.388)(6.9972, 5.388)(6.3546, 4.7454)
+\qbezier(6.3546, 4.7454)(5.712, 4.1028)(5.712, 3.194)
+\qbezier(5.712, 3.194)(5.712, 2.2852)(6.3546, 1.6426)
+\qbezier(6.3546, 1.6426)(6.9972, 1.0)(7.906, 1.0)
+\qbezier(7.906, 1.0)(8.8148, 1.0)(9.4574, 1.6426)
+\qbezier(9.4574, 1.6426)(10.1, 2.2852)(10.1, 3.194)
+
+\drawline     ( 0.4  , 1     )(10.1  , 1     )(10.1  , 7.2   )( 0.4  , 1     )  %A(D)B(E)C(F)A
+\dashline[80]{0.2}( 7.906, 3.194 )( 7.906, 1     )     %OD
+\dashline[80]{0.2}( 7.906, 3.194 )(10.1  , 3.194 )     %OE
+\dashline[80]{0.2}( 7.906, 3.194 )( 6.725, 5.042 )     %OF
+\put(  0.1 ,  0.1 ){$\scriptstyle A$}   %A
+\put(  9.6 ,  0.1 ){$\scriptstyle B$}   %B
+\put(  9.7 ,  7.4 ){$\scriptstyle C$}   %C
+\put(  7.4 ,  0.1 ){$\scriptstyle D$}   %D
+\put( 10.2 ,  2.85){$\scriptstyle E$}   %E
+\put(  6.3 ,  5.3 ){$\scriptstyle F$}   %F
+\put(  7   ,  2.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119ee125.eepic b/33063-t/images/sources/119ee125.eepic
new file mode 100644
index 0000000..af05a6a
--- /dev/null
+++ b/33063-t/images/sources/119ee125.eepic
@@ -0,0 +1,29 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119ee125.tex
+% Ex.125. Fig.e.   ( 7.6, 7.6 )
+\PGset[0.8em]
+\begin{picture}    ( 7.6, 7.6 )
+
+% Ellipse:  u = 2.9  v = 3.8  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(5.7, 3.8)(5.7, 4.9598)(4.8799, 5.7799)
+\qbezier(4.8799, 5.7799)(4.0598, 6.6)(2.9, 6.6)
+\qbezier(2.9, 6.6)(1.7402, 6.6)(0.9201, 5.7799)
+\qbezier(0.9201, 5.7799)(0.1, 4.9598)(0.1, 3.8)
+\qbezier(0.1, 3.8)(0.1, 2.6402)(0.9201, 1.8201)
+\qbezier(0.9201, 1.8201)(1.7402, 1.0)(2.9, 1.0)
+\qbezier(2.9, 1.0)(4.0598, 1.0)(4.8799, 1.8201)
+\qbezier(4.8799, 1.8201)(5.7, 2.6402)(5.7, 3.8)
+
+\drawline     ( 2.9  , 6.6   )( 2.9  , 1     )( 6.9  , 1     )( 2.9  , 6.6   )  %A(O)B(E)C(D)A
+\drawline     ( 4.591, 0.1   )( 6.2  , 4.792 )                     %low(ED)high
+\dashline[80]{0.2}( 5.549, 2.892 )( 2.9  , 3.8   )( 4.9  , 1     )     %DOE
+\drawline     ( 4.9  , 1     )( 4.7  , 1.28 )   %on OE ensures dash near E
+\put(  2.6 ,  6.8 ){$\scriptstyle A$}   %A
+\put(  2.5 ,  0.1 ){$\scriptstyle B$}   %B
+\put(  6.8 ,  0.3 ){$\scriptstyle C$}   %C
+\put(  5.7 ,  2.8 ){$\scriptstyle D$}   %D
+\put(  4.9 ,  0.1 ){$\scriptstyle E$}   %E 4.9, 1
+\put(  2.1 ,  3.6 ){$\scriptstyle O$}   %O 2.9, 3.8
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/119ff126.eepic b/33063-t/images/sources/119ff126.eepic
new file mode 100644
index 0000000..b8b9d6b
--- /dev/null
+++ b/33063-t/images/sources/119ff126.eepic
@@ -0,0 +1,29 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 119ff126.tex
+% Ex.126. Fig.f.   ( 9.5, 6.6 )
+\PGset[0.8em]
+\begin{picture}    ( 9.5, 6.6 )
+
+% Ellipse:  u = 2.9  v = 2.9  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(5.7, 2.9)(5.7, 4.0598)(4.8799, 4.8799)
+\qbezier(4.8799, 4.8799)(4.0598, 5.7)(2.9, 5.7)
+\qbezier(2.9, 5.7)(1.7402, 5.7)(0.9201, 4.8799)
+\qbezier(0.9201, 4.8799)(0.1, 4.0598)(0.1, 2.9)
+\qbezier(0.1, 2.9)(0.1, 1.7402)(0.9201, 0.9201)
+\qbezier(0.9201, 0.9201)(1.7402, 0.1)(2.9, 0.1)
+\qbezier(2.9, 0.1)(4.0598, 0.1)(4.8799, 0.9201)
+\qbezier(4.8799, 0.9201)(5.7, 1.7402)(5.7, 2.9)
+
+\drawline     ( 0.1  , 5.7   )( 8.6  , 5.7   )                  %left(AK)C
+\drawline     ( 2.9  , 5.7   )( 5.277, 1.421 )                  %A(H)B
+\drawline     ( 4.089, 3.560 )( 5.348, 4.260 )( 5.348, 5.7   )  %HMK
+\dashline[80]{0.2}( 2.9  , 5.7   )( 5.348, 4.260 )                  %AM
+\put(  2.3 ,  5.8 ){$\scriptstyle A$}   %A
+\put(  5.2 ,  0.7 ){$\scriptstyle B$}   %B
+\put(  8.7 ,  5.4 ){$\scriptstyle C$}   %C
+\put(  2.9 ,  3   ){$\scriptstyle H$}   %H
+\put(  5   ,  5.9 ){$\scriptstyle K$}   %K
+\put(  5.5 ,  4.3 ){$\scriptstyle M$}   %M
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/120aa130.eepic b/33063-t/images/sources/120aa130.eepic
new file mode 100644
index 0000000..f7494c7
--- /dev/null
+++ b/33063-t/images/sources/120aa130.eepic
@@ -0,0 +1,27 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 120aa130.tex
+% Ex.130. Fig.a.   ( 8.4, 8.4 )
+\PGset[0.8em]
+\begin{picture}    ( 8.4, 8.4 )
+\drawline     ( 1.2 , 7.3  )( 7.5 , 7.3  )( 7.5 , 1    )( 1.2 , 1    )( 1.2 , 7.3  )  %A(F')B(E)C(F)D(E')A
+\drawline     ( 1.2 , 7.3  )( 7.5 , 1    )  %A(PO)C
+\drawline     ( 1.2 , 1    )( 7.5 , 7.3  )  %D(O)B
+\drawline     ( 1.2 , 5.6  )( 7.5 , 5.6  )  %E'(P)E
+\drawline     ( 2.9 , 1    )( 2.9 , 7.3  )  %F(P)F'
+\dashline[80]{0.2}( 4.35, 4.15 )( 7.5 , 5.6  )     %OE
+\dashline[80]{0.2}( 4.35, 4.15 )( 2.9 , 1    )     %OF
+\dashline[80]{0.2}( 4.35, 4.15 )( 1.2 , 5.6  )     %OE'
+\dashline[80]{0.2}( 4.35, 4.15 )( 2.9 , 7.3  )     %OF'
+\put(  0.7 ,  7.4 ){$\scriptstyle A$}   %A
+\put(  7.4 ,  7.4 ){$\scriptstyle B$}   %B
+\put(  7.2 ,  0.2 ){$\scriptstyle C$}   %C
+\put(  0.5 ,  0.1 ){$\scriptstyle D$}   %D
+\put(  7.6 ,  5.3 ){$\scriptstyle E$}   %E
+\put(  2.5 ,  0.1 ){$\scriptstyle F$}   %F
+\put(  4.9 ,  3.7 ){$\scriptstyle O$}   %O
+\put(  3.1 ,  5.9 ){$\scriptstyle P$}   %P
+\put(  0.1 ,  5.4 ){$\scriptstyle E'$}   %E'
+\put(  2.6 ,  7.5 ){$\scriptstyle F'$}   %F'
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/120bb131.eepic b/33063-t/images/sources/120bb131.eepic
new file mode 100644
index 0000000..4b7b4f3
--- /dev/null
+++ b/33063-t/images/sources/120bb131.eepic
@@ -0,0 +1,28 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 120bb131.tex
+% Ex.131. Fig.b.   ( 6.6, 7.6 )
+\PGset[0.8em]
+\begin{picture}    ( 6.6, 7.6 )
+
+% Ellipse:  u = 3.3  v = 3.8  a = 2.8  b = 2.8  phi = 0.0 Grad
+\qbezier(6.1, 3.8)(6.1, 4.9598)(5.2799, 5.7799)
+\qbezier(5.2799, 5.7799)(4.4598, 6.6)(3.3, 6.6)
+\qbezier(3.3, 6.6)(2.1402, 6.6)(1.3201, 5.7799)
+\qbezier(1.3201, 5.7799)(0.5, 4.9598)(0.5, 3.8)
+\qbezier(0.5, 3.8)(0.5, 2.6402)(1.3201, 1.8201)
+\qbezier(1.3201, 1.8201)(2.1402, 1.0)(3.3, 1.0)
+\qbezier(3.3, 1.0)(4.4598, 1.0)(5.2799, 1.8201)
+\qbezier(5.2799, 1.8201)(6.1, 2.6402)(6.1, 3.8)
+
+\drawline     ( 3.3  , 6.6   )( 0.875, 2.4   )( 5.725, 2.4   )( 3.3  , 6.6   ) %ABCA
+\drawline     ( 0.875, 2.4   )( 3.8  , 1.045 )( 5.725, 2.4   )                 %BPC
+\drawline     ( 3.3  , 6.6   )( 3.8  , 1.045 )                                 %A(M)P
+\dashline[80]{0.2}( 0.875, 2.4   )( 3.511, 4.256 )                                 %BM
+\put(  3   ,  6.8 ){$\scriptstyle A$}   %A
+\put(  0.0 ,  1.7 ){$\scriptstyle B$}   %B
+\put(  5.7 ,  1.6 ){$\scriptstyle C$}   %C
+\put(  2.4 ,  4.25){$\scriptstyle M$}   %M
+\put(  3.4 ,  0.1 ){$\scriptstyle P$}   %P
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/120cd132.eepic b/33063-t/images/sources/120cd132.eepic
new file mode 100644
index 0000000..1766ad8
--- /dev/null
+++ b/33063-t/images/sources/120cd132.eepic
@@ -0,0 +1,67 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 120cc132.tex
+% Ex.132. Fig.c.   (12  , 9.4 )
+\PGset[0.8em]
+\begin{picture}    (12  , 9.4 )
+
+% Ellipse:  u = 6.0  v = 4.7  a = 2.95  b = 2.95  phi = 0.0 Grad
+\qbezier(8.95, 4.7)(8.95, 5.9219)(8.086, 6.786)
+\qbezier(8.086, 6.786)(7.2219, 7.65)(6.0, 7.65)
+\qbezier(6.0, 7.65)(4.7781, 7.65)(3.914, 6.786)
+\qbezier(3.914, 6.786)(3.05, 5.9219)(3.05, 4.7)
+\qbezier(3.05, 4.7)(3.05, 3.4781)(3.914, 2.614)
+\qbezier(3.914, 2.614)(4.7781, 1.75)(6.0, 1.75)
+\qbezier(6.0, 1.75)(7.2219, 1.75)(8.086, 2.614)
+\qbezier(8.086, 2.614)(8.95, 3.4781)(8.95, 4.7)
+
+\drawline     ( 4.25 , 2.325 )( 7.75 , 2.325 )( 7.75 , 7.075 )( 4.25 , 7.075 )( 4.25 , 2.325 ) %ABCDA
+\drawline     ( 6    , 1.036 )(10.973, 4.7   )( 6    , 8.364 )( 1.027, 4.7   )( 6    , 1.036 ) %E(B)F(C)G(D)H(A)E
+\dashline[80]{0.2}( 4.25 , 2.325 )( 7.75 , 7.075 )                                 %AC
+\dashline[80]{0.2}( 4.25 , 7.075 )( 7.75 , 2.325 )                                 %DB
+\put(  3.5 ,  1.6 ){$\scriptstyle A$}   %A
+\put(  7.7 ,  1.6 ){$\scriptstyle B$}   %B
+\put(  7.8 ,  7.2 ){$\scriptstyle C$}   %C
+\put(  3.4 ,  7.3 ){$\scriptstyle D$}   %D
+\put(  5.7 ,  0.1 ){$\scriptstyle E$}   %E
+\put( 11.1 ,  4.4 ){$\scriptstyle F$}   %F
+\put(  5.6 ,  8.6 ){$\scriptstyle G$}   %G
+\put(  0.1 ,  4.5 ){$\scriptstyle H$}   %H
+\put(  6.4 ,  4.4 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
+%
+\qquad
+\quad
+% 120dd132.tex
+% Ex.132. Fig.c.   (16  ,13.7 )
+\PGset[0.8em]
+\begin{picture}    (16  ,13.7 )
+
+% Ellipse:  u = 4.4  v = 4.4  a = 3.4  b = 3.4  phi = 0.0 Grad
+\qbezier(7.8, 4.4)(7.8, 5.8083)(6.8042, 6.8042)
+\qbezier(6.8042, 6.8042)(5.8083, 7.8)(4.4, 7.8)
+\qbezier(4.4, 7.8)(2.9917, 7.8)(1.9958, 6.8042)
+\qbezier(1.9958, 6.8042)(1.0, 5.8083)(1.0, 4.4)
+\qbezier(1.0, 4.4)(1.0, 2.9917)(1.9958, 1.9958)
+\qbezier(1.9958, 1.9958)(2.9917, 1.0)(4.4, 1.0)
+\qbezier(4.4, 1.0)(5.8083, 1.0)(6.8042, 1.9958)
+\qbezier(6.8042, 1.9958)(7.8, 2.9917)(7.8, 4.4)
+
+\drawline     ( 2.577, 1.530 )(15.029, 4.844 )( 4.005, 7.777 ) %A(FD)G(CM)B
+\drawline     ( 1.029, 4.844 )(15.029, 4.844 )                 %H(IN)G
+\drawline     ( 2.577, 1.530 )( 5.140,12.738 )( 7.408, 2.816 ) %A(B)E(C)D
+\drawline     ( 5.140, 1.081 )( 5.140,12.738 )                 %F(IM)E
+\put(  1.9 ,  0.7 ){$\scriptstyle A$}   %A
+\put(  3.1 ,  7.9 ){$\scriptstyle B$}   %B
+\put(  6.6 ,  7.3 ){$\scriptstyle C$}   %C
+\put(  7.3 ,  2   ){$\scriptstyle D$}   %D
+\put(  4.9 , 12.9 ){$\scriptstyle E$}   %E
+\put(  4.8 ,  0.2 ){$\scriptstyle F$}   %F
+\put( 15.1 ,  4.5 ){$\scriptstyle G$}   %G
+\put(  0.1 ,  4.6 ){$\scriptstyle H$}   %H
+\put(  4.5 ,  4   ){$\scriptstyle I$}   %I
+\put(  5.15,  7.9 ){$\scriptstyle M$}   %M
+\put(  7.9 ,  5.1 ){$\scriptstyle N$}   %N
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/121aa300.eepic b/33063-t/images/sources/121aa300.eepic
new file mode 100644
index 0000000..2b0958f
--- /dev/null
+++ b/33063-t/images/sources/121aa300.eepic
@@ -0,0 +1,23 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 121aa300.tex
+% Pt.300. Fig.a.   (21.7,12.8 )
+\PGset[0.8em]
+\begin{picture}    (21.7,12.8 )
+\drawline     ( 0.9  , 1.75  )(20.9  , 1.75  ) %A(HMK)B
+\dashline[80]{0.4}(10.9  , 1.75  )(10.9  ,12.1   ) %M(O)C
+\qbezier[40]  ( 2.369, 3.569 )( 5.903, 0.035 )(10.9  , 0.035 ) %left  lower arc
+\qbezier[40]  (10.9  , 0.035 )(15.897, 0.035 )(19.431, 3.569 ) %right lower arc
+%\dashline{0.2}( 9.615,10     )(10.9  , 9.083 )(12.31 , 7.607 ) %H arc @ O
+%\dashline{0.2}( 9.567, 7.706 )(10.9  , 9.083 )(12.631,10.25  ) %K arc @ O
+\dottedline[.]{0.4}( 9.615,10     )(10.9  , 9.083 )(12.31 , 7.607 ) %H arc @ O
+\dottedline[.]{0.4}( 9.567, 7.706 )(10.9  , 9.083 )(12.631,10.25  ) %K arc @ O
+\put(  0.1 ,  1.4 ){$\scriptstyle A$}   %A
+\put( 20.9 ,  1.4 ){$\scriptstyle B$}   %B
+\put( 10.6 , 12.15){$\scriptstyle C$}   %C
+\put(  4   ,  0.8 ){$\scriptstyle H$}   %H
+\put( 16.9 ,  0.8 ){$\scriptstyle K$}   %K
+\put( 10.4 ,  0.8 ){$\scriptstyle M$}   %M
+\put( 11.4 ,  8.7 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/122ab301.eepic b/33063-t/images/sources/122ab301.eepic
new file mode 100644
index 0000000..14dba55
--- /dev/null
+++ b/33063-t/images/sources/122ab301.eepic
@@ -0,0 +1,54 @@
+%figshell.tex  or  0.tex  REMOVE this line for actual diagram file
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% ***  Uses  epic  &  eepic  packages   ***
+% 122aa301.tex
+% Pt.301. Fig.a.   (16.9,10.7 )
+\PGset[0.8em]
+\begin{picture}    (16.9,10.7 )
+\drawline     ( 0.9  , 2     )(15.9  , 2     ) %A(HOB)C
+\dashline[80]{0.4}( 8.4  , 2     )( 8.4  ,10.6   ) %O(R)top
+\qbezier[8]  ( 5.1  , 2     )( 5.1  , 3.367 )( 6.067, 4.333 ) %left  lower arc 4th octant
+\qbezier[8]  ( 6.067, 4.333 )( 7.033, 5.3   )( 8.4  , 5.3   ) %right lower arc 3rd octant
+\qbezier[8]  ( 8.4  , 5.3   )( 9.767, 5.3   )(10.733, 4.333 ) %left  lower arc 2nd octant
+\qbezier[8]  (10.733, 4.333 )(11.7  , 3.367 )(11.7  , 2     ) %right lower arc 1st octant
+%\dottedline[.]{0.4}( 6.734, 9.288 )( 8.4  , 8.7   )(10.042, 7.6   ) %H arc @ R
+%\dottedline[.]{0.4}( 6.734, 7.758 )( 8.4  , 8.7   )(10.122, 9.3   ) %K arc @ R
+\dottedline[.]{0.4}( 6.734, 9.288 )( 7.567, 9.049 )( 8.4  , 8.7   ) %H left  arc @ R
+\dottedline[.]{0.4}( 8.4  , 8.7   )( 9.338, 8.15  )(10.042, 7.6   ) %H right arc @ R
+\dottedline[.]{0.4}( 6.734, 7.758 )( 7.567, 8.221 )( 8.4  , 8.7   ) %K left  arc @ R
+\dottedline[.]{0.4}( 8.4  , 8.7   )( 9.096, 9     )(10.122, 9.3   ) %K right arc @ R
+\put(  0.1 ,  1.8 ){$\scriptstyle A$}   %A
+\put( 11.2 ,  1.2 ){$\scriptstyle B$}   %B
+\put( 16   ,  1.8 ){$\scriptstyle C$}   %C
+\put(  4.6 ,  1.2 ){$\scriptstyle H$}   %H
+\put(  8   ,  1.2 ){$\scriptstyle O$}   %O
+\put(  7.5 ,  7   ){$\scriptstyle R$}   %R
+\put(  7   ,  0.1 ){$\scriptstyle \textsc{Fig.~1.}$}
+\end{picture}
+\PGrestore
+%
+\qquad
+\qquad
+\quad
+% 122bb301.tex
+% Pt.301. Fig.b.   (11.9,13.4 )
+\PGset[0.8em]
+\begin{picture}    (11.9,13.4 )
+\drawline     ( 1    , 2.7   )(10    , 2.7   ) %A(E)B
+\dashline[80]{0.4}(10    , 2.7   )(10    ,13.3   ) %B(D)top
+\qbezier[10]       (11.842, 6.6   )(11.842, 8.692 )(10.363,10.171 ) %Octant 1
+\qbezier[10]       (10.363,10.171 )( 8.884,11.65  )( 6.792,11.65  ) %Octant 2
+\qbezier[10]       ( 1.742, 6.6   )( 1.742, 4.508 )( 3.221, 3.029 ) %Octant 5
+\qbezier[10]       ( 3.221, 3.029 )( 4.700, 1.55  )( 6.792, 1.55  ) %Octant 6
+\qbezier[10]       ( 6.792, 1.55  )( 8.884, 1.55  )(10.363, 3.029 ) %Octant 7
+\qbezier[10]       (10.363, 3.029 )(11.842, 4.508 )(11.842, 6.6   ) %Octant 8
+\dashline[80]{0.2}( 3.584, 2.7   )(10    ,10.5   )                 %E(C)D
+\put(  0.1 ,  2.6 ){$\scriptstyle A$}   %A
+\put( 10.2 ,  2   ){$\scriptstyle B$}   %B
+\put(  6   ,  6.7 ){$\scriptstyle C$}   %C
+\put( 10.2 , 10.5 ){$\scriptstyle D$}   %D
+\put(  2.7 ,  1.8 ){$\scriptstyle E$}   %E
+\put(  5   ,  0.1 ){$\scriptstyle \textsc{Fig.~2.}$}
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/123aa302.eepic b/33063-t/images/sources/123aa302.eepic
new file mode 100644
index 0000000..758728f
--- /dev/null
+++ b/33063-t/images/sources/123aa302.eepic
@@ -0,0 +1,27 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 123aa302.tex
+% Pt.302. Fig.a.   ( 16.8 ,15 )
+\PGset[0.8em]
+\begin{picture}    ( 16.8 ,15 )
+\drawline     ( 1    , 7.4   )(16    , 7.4   ) %A(D)B
+\dashline[80]{0.2}( 8.5  , 0.7   )( 8.5  ,14.1   ) %E(D)C
+%\put( 1,7.4){\arc{19.209}{5.504}{5.673}} %arc A @ C
+%\put( 1,7.4){\arc{19.209}{0.548}{0.779}} %arc A @ E
+%\put(16,7.4){\arc{19.209}{3.709}{3.909}} %arc B @ C
+%\put(16,7.4){\arc{19.209}{2.370}{2.614}} %arc B @ E
+\spline( 7.834,14.149 )( 8.5  ,13.4   )( 8.874,12.9   ) %arc A @ C
+\spline( 7.834, 0.651 )( 8.5  , 1.4   )( 9.201, 2.4   ) %arc A @ E
+\spline( 7.9  ,12.561 )( 8.5  ,13.4   )( 9.086,14.067 ) %arc B @ C
+\spline( 7.7  , 2.567 )( 8.5  , 1.4   )( 9.118, 0.7   ) %arc B @ E
+%\drawline( 7.834,14.149 )( 8.5  ,13.4   )( 8.874,12.9   ) %arc A @ C
+%\drawline( 7.834, 0.651 )( 8.5  , 1.4   )( 9.201, 2.4   ) %arc A @ E
+%\drawline( 7.9  ,12.561 )( 8.5  ,13.4   )( 9.086,14.067 ) %arc B @ C
+%\drawline( 7.7  , 2.567 )( 8.5  , 1.4   )( 9.118, 0.7   ) %arc B @ E
+\put(  0.1 ,  7.1 ){$\scriptstyle A$}   %A
+\put( 16   ,  7.2 ){$\scriptstyle B$}   %B
+\put(  8.1 , 14.3 ){$\scriptstyle C$}   %C
+\put(  7.5 ,  7.6 ){$\scriptstyle D$}   %D
+\put(  8.1 ,  0.1 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/123bb303.eepic b/33063-t/images/sources/123bb303.eepic
new file mode 100644
index 0000000..e55ee3d
--- /dev/null
+++ b/33063-t/images/sources/123bb303.eepic
@@ -0,0 +1,29 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  &  eepic  packages   ***
+% 123bb303.tex
+% Pt.303. Fig.b.   ( 15.8 ,13.8 )
+\PGset[0.8em]
+\begin{picture}    ( 15.8 ,13.8 )
+\put( 8,2.128){\arc{16.943}{3.740}{5.685}}     %arc ACB
+%\spline( 1    , 6.9   )( 2    , 8.109 )( 4    , 9.596 )( 8    ,10.6   )(12    , 9.596 )(14    , 8.109 )(15    , 6.9   ) %arc ACB
+\dashline[80]{0.2}( 1    , 6.9   )(15    , 6.9   ) %AB
+\dashline[80]{0.2}( 8    , 1.4   )( 8    ,12.4   ) %E(C)D
+\put( 1,6.9){\arc{17.804}{5.531}{5.753}} %arc A @ D
+\put( 1,6.9){\arc{17.804}{0.556}{0.779}} %arc A @ E
+\put(15,6.9){\arc{17.804}{3.689}{3.907}} %arc B @ D
+\put(15,6.9){\arc{17.804}{2.376}{2.600}} %arc B @ E
+%\spline( 7.5  ,12.983 )( 8    ,12.4   )( 8.681,11.4   ) %arc A @ D
+%\spline( 8.560, 2.2   )( 8    , 1.4   )( 7.334, 0.645 ) %arc A @ E
+%\spline( 7.4  ,11.536 )( 8    ,12.4   )( 8.580,13.067 ) %arc B @ D
+%\spline( 8.579, 0.734 )( 8    , 1.4   )( 7.37 , 2.314 ) %arc B @ E
+%\drawline( 7.5  ,12.983 )( 8    ,12.4   )( 8.681,11.4   ) %arc A @ D
+%\drawline( 8.560, 2.2   )( 8    , 1.4   )( 7.334, 0.645 ) %arc A @ E
+%\drawline( 7.4  ,11.536 )( 8    ,12.4   )( 8.580,13.067 ) %arc B @ D
+%\drawline( 8.579, 0.734 )( 8    , 1.4   )( 7.37 , 2.314 ) %arc B @ E
+\put(  0.1 ,  6.6 ){$\scriptstyle A$}   %A
+\put( 15   ,  6.9 ){$\scriptstyle B$}   %B
+\put(  8.2 ,  9.6 ){$\scriptstyle C$}   %C
+\put(  7.65, 13.1 ){$\scriptstyle D$}   %D
+\put(  7.5 ,  0.1 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/124aa304.eepic b/33063-t/images/sources/124aa304.eepic
new file mode 100644
index 0000000..ec8085e
--- /dev/null
+++ b/33063-t/images/sources/124aa304.eepic
@@ -0,0 +1,23 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 124aa304.tex
+% Pt.304. Fig.a.   ( 16 ,18.7 )
+\PGset[0.8em]
+\begin{picture}    ( 16 ,18.7 )
+\qbezier[40]  ( 8    ,12.5   )( 3.858,12.5   )( 0.929, 9.571 ) %arc AC 3rd octant
+\qbezier[40]  (15.071, 9.571 )(12.142,12.5   )( 8    ,12.5   ) %arc CB 2nd octant
+\drawline     ( 0.9  , 9.6   )( 8    , 2.5   )(15.1  , 9.6   ) %upperleft(A)E(B)upperright
+\dashline[80]{0.2}( 7.2  , 1.7   )( 8    , 2.5   )                 %lowleft -E(B)
+\dashline[80]{0.2}( 8.831, 1.67  )( 8    , 2.5   )                 %lowright-E(A)
+\dashline[80]{0.2}( 8    , 0.1   )( 8    ,18.3   )                 %bottom(ECD)top
+\dottedline[.]{0.4}( 6.6  ,17.943 )( 7.3  ,17.424 )( 8    ,16.8   ) %A left  arc @ D
+\dottedline[.]{0.4}( 8    ,16.8   )( 8.522,16.25  )( 8.972,15.7   ) %A right arc @ D
+\dottedline[.]{0.4}( 6.7  ,15.244 )( 7.35 ,16.101 )( 8    ,16.8   ) %B left  arc @ D
+\dottedline[.]{0.4}( 8    ,16.8   )( 8.550,17.3   )( 9.194,17.8   ) %B right arc @ D
+\put(  0.1 ,  9.6 ){$\scriptstyle A$}   %A
+\put( 15.2 ,  9.6 ){$\scriptstyle B$}   %B
+\put(  8.2 , 12.9 ){$\scriptstyle C$}   %C
+\put(  8.3 , 18   ){$\scriptstyle D$}   %D
+\put(  8.6 ,  2.3 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/125ab305.eepic b/33063-t/images/sources/125ab305.eepic
new file mode 100644
index 0000000..d51d471
--- /dev/null
+++ b/33063-t/images/sources/125ab305.eepic
@@ -0,0 +1,36 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 125aa305.tex
+% Pt.305. Fig.a.   ( 11.2, 9.4 )
+\PGset[0.8em]
+\begin{picture}    ( 11.2, 9.4 )
+\drawline     (10    , 8.4   )( 1    , 4.8   )(10    , 1.2   ) %upperright(F)A(E)lowerright
+\dashline[80]{0.4}( 8    , 2     )( 8    , 7.6   )                 %EF
+\qbezier[24]  ( 8    , 2     )( 9.12 , 4.8   )( 8    , 7.6   ) %arc EF
+\put(  0.1 ,  4.5 ){$\scriptstyle A$}   %A
+\put(  7.5 ,  1   ){$\scriptstyle E$}   %E
+\put(  7.5 ,  7.7 ){$\scriptstyle F$}   %F
+\end{picture}
+\PGrestore
+%
+\qquad
+\qquad
+\qquad
+% 125bb305.tex
+% Pt.305. Fig.b.   ( 11.2, 9.4 )
+\PGset[0.8em]
+\begin{picture}    ( 11.2, 9.4 )
+\drawline     ( 1    , 4.3   )(10.3  , 0.58  )                 %C(H)M
+%\dashline{0.4}( 1    , 4.3   )(10.4  , 8.06  )                 %C(O)upperright
+\dashline[80]{0.4}( 1    , 4.3   )(10.4  , 8.06  )                 %C(O)upperright
+%\qbezier[26]  ( 1    , 4.3   )( 5.7  , 6.18  )(10.4  , 8.06  )                 %C(O)upperright
+\qbezier[28]  ( 8    , 1.5   )( 9.12 , 4.3   )( 8    , 7.1   ) %arc HO
+\qbezier[12]  ( 8    , 7.1   )( 7.587, 8.131 )( 6.895, 9     ) %arc OG
+\qbezier[18]  ( 6.7  , 6.947 )( 8.558, 7.391 )(10.3  , 6.606 ) %arc H @ O
+\put(  0.1 ,  4   ){$\scriptstyle C$}   %C
+\put(  6   ,  8.7 ){$\scriptstyle G$}   %G
+\put(  7.2 ,  0.6 ){$\scriptstyle H$}   %H
+\put( 10.2 ,  0.1 ){$\scriptstyle M$}   %M
+\put(  8   ,  7.5 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/125cc306.eepic b/33063-t/images/sources/125cc306.eepic
new file mode 100644
index 0000000..03817c6
--- /dev/null
+++ b/33063-t/images/sources/125cc306.eepic
@@ -0,0 +1,24 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 125cc306.tex
+% Pt.306. Fig.c.   ( 22,13.3 )
+\PGset[0.8em]
+\begin{picture}    ( 22,13.3 )
+\drawline          ( 1    , 1.4   )(20    , 1.4   )                 %A(D)B
+\dashline[80]{0.2}     ( 3.4  , 7.4   )(21    , 7.4   )                 %H(C)F
+\dashline[80]{0.2}     ( 9    , 1.4   )(12.032,12.1   )                 %D(C)E
+%\dottedline[.]{0.4}( 9    , 1.4   )(12.032,12.1   )                 %D(C)E
+\qbezier[12]  ( 7.7  , 4.864 )( 9.931, 5.701 )(11.616, 4.016 ) %arc from D top   half
+\qbezier[12]  (11.616, 4.016 )(13.301, 2.331 )(12.464, 0.1   ) %arc from D right half
+\qbezier[12]  ( 9.9  ,11.012 )(12.050,11.489 )(13.494, 9.825 ) %arc from C top   half
+\qbezier[12]  (13.494, 9.825 )(14.938, 8.162 )(14.164, 6.1   ) %arc from C right half
+\qbezier[12]  (13.5  , 6.872 )(14.436, 7.283 )(15.1  , 8.059 ) %small arc on CF
+\put(  0.2 ,  1.35){$\scriptstyle A$}   %A
+\put( 20   ,  1.35){$\scriptstyle B$}   %B
+\put(  9.8 ,  7.7 ){$\scriptstyle C$}   %C
+\put(  8.5 ,  0.5 ){$\scriptstyle D$}   %D
+\put( 11.8 , 12.5 ){$\scriptstyle E$}   %E
+\put( 21.1 ,  7   ){$\scriptstyle F$}   %F
+\put(  2.5 ,  7.1 ){$\scriptstyle H$}   %H
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/126aa307.eepic b/33063-t/images/sources/126aa307.eepic
new file mode 100644
index 0000000..048d7af
--- /dev/null
+++ b/33063-t/images/sources/126aa307.eepic
@@ -0,0 +1,17 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 126aa307.tex
+% Pt.307. Fig.a.   ( 28.8,11.8 )
+\PGset[0.8em]
+\begin{picture}    ( 28.8,11.8 )
+\drawline          ( 1    ,11.4   )(28    ,11.4   )                 %AB
+\dashline[80]{0.2}     ( 1    ,11.4   )(28    , 0.6   )                 %A(C)O
+\dashline[80]{0.2}     (10    ,11.4   )( 8.5  , 8.4   )                 %first  slant
+\dashline[80]{0.2}     (19    ,11.4   )(16    , 5.4   )                 %second slant
+\dashline[80]{0.2}     (28    ,11.4   )(23.5  , 2.4   )                 %BC     slant
+\put(  0.1 , 11.1 ){$\scriptstyle A$}   %A
+\put( 28   , 11   ){$\scriptstyle B$}   %B
+\put( 22.9 ,  1.6 ){$\scriptstyle C$}   %C
+\put( 28   ,  0.1 ){$\scriptstyle O$}   %O
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/127ab308.eepic b/33063-t/images/sources/127ab308.eepic
new file mode 100644
index 0000000..0fc8ab0
--- /dev/null
+++ b/33063-t/images/sources/127ab308.eepic
@@ -0,0 +1,41 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 127aa308.tex
+% Pt.308. Fig.a.   ( 8.2,12.8 )
+\PGset[0.8em]
+\begin{picture}    ( 8.2,12.8 )
+\drawline     ( 6.5  ,12.1   )( 1    , 9.1   )( 7.7  , 9.1   ) %ur-A-lr
+\drawline     ( 4.6  , 5.8   )( 1    , 1.4   )( 8.1  , 1.4   ) %ur-B-lr
+\qbezier[12]  ( 3.530,12.7   )( 4.849,11.773 )( 5.257,10.214 ) %arc from A top   half
+\qbezier[12]  ( 5.257,10.214 )( 5.665, 8.654 )( 4.969, 7.2   ) %arc from A right half
+\qbezier[12]  ( 2.3  , 5.604 )( 4.041, 5.065 )( 4.892, 3.453 ) %arc from B top   half
+\qbezier[12]  ( 4.892, 3.453 )( 5.742, 1.841 )( 5.204, 0.1   ) %arc from B right half
+\put(  0.1 ,  9.1 ){$\scriptstyle A$}   %A
+\put(  0.1 ,  1.4 ){$\scriptstyle B$}   %B
+\end{picture}
+\PGrestore
+%
+\qquad
+% 127bb308.tex
+% Pt.308. Fig.b.   (14.4,12.8 )
+\PGset[0.8em]
+\begin{picture}    (14.4,12.8 )
+\drawline          ( 1    , 1.9   )(13.6  , 1.9   ) %E(H)F
+\dashline[80]{0.4}     ( 6.1  , 1.9   )( 7.552, 9.6   ) %HR
+\dashline[80]{0.2}( 6.1  , 1.9   )(12.7  , 5.5   ) %H-ur
+%\qbezier[24]  ( 6.1  , 1.9   )( 9.4  , 3.7   )(12.7  , 5.5   ) %H-ur
+\qbezier[12]  ( 3.719, 5.6   )( 5.193, 6.549 )( 6.915, 6.224 ) %arc from H left   part
+\qbezier[12]  ( 6.915, 6.224 )( 8.502, 5.925 )( 9.519, 4.670 ) %arc from H middle part
+\qbezier[12]  ( 9.519, 4.670 )(10.536, 3.415 )(10.499, 1.8   ) %arc from H right  part
+%\qbezier[12]  ( 8.993, 3.467 )( 9.834, 4.276 )(10.973, 4.022 ) %arc for  a (geom. accurate, but uninstructive)
+\qbezier[ 8]  ( 8.993, 3.767 )( 9.834, 4.376 )(10.973, 4.022 ) %arc for  a (helpful - matches inaccurate diagram)
+\qbezier[12]  ( 6.302, 4.9   )( 6.664, 6.387 )( 7.961, 7.2   ) %arc for  b
+\put(  0.2 ,  1.5 ){$\scriptstyle E$}   %E
+\put( 13.6 ,  1.5 ){$\scriptstyle F$}   %F
+\put(  5.6 ,  1.1 ){$\scriptstyle H$}   %H
+\put(  7.1 ,  9.8 ){$\scriptstyle R$}   %R
+\put(  7.5 ,  2   ){\emph{a}}           %a
+\put(  6.4 ,  2.6 ){\emph{b}}           %b
+\put(  5.3 ,  2.1 ){\emph{c}}           %c
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/127cd309.eepic b/33063-t/images/sources/127cd309.eepic
new file mode 100644
index 0000000..d6df60f
--- /dev/null
+++ b/33063-t/images/sources/127cd309.eepic
@@ -0,0 +1,38 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 127cc309.tex
+% Pt.309. Fig.c.   (15.7,13.9 )
+\PGset[0.8em]
+\begin{picture}    (15.7,13.9 )
+\drawline     ( 6.8  , 6.95  )( 0.5  , 1.7   )( 9.5  , 1.7   ) %ur-E-lr
+\drawline     ( 1.5  , 9.2   )(15.6  , 9.2   )                 %left(c)right
+\drawline     ( 2.3  ,12.1   )(14.6  ,12.1   )                 %left(b)right
+\qbezier[12]  ( 3.661, 6.8   )( 5.286, 5.792 )( 6.029, 4.030 ) %arc from E top    half
+\qbezier[12]  ( 6.029, 4.030 )( 6.772, 2.267 )( 6.357, 0.4   ) %arc from E bottom half
+\put(  0.1 ,  0.9 ){$\scriptstyle E$}   %E
+\put(  8.3 , 12.2 ){\emph{b}}           %b
+\put(  8.3 ,  9.3 ){\emph{c}}           %c
+\end{picture}
+\PGrestore
+%
+% 127dd309.tex
+% Pt.309. Fig.d.   (21,13.9 )
+\PGset[0.8em]
+\begin{picture}    (21,13.9 )
+\dashline[80]{0.4}     ( 0.5  , 1.7   )(14.6  , 1.7   ) %A(c)B
+\dashline[80]{0.2}(14.6  , 1.7   )(20.9  , 1.7   ) %B-lr
+\dashline[80]{0.4}     ( 0.5  , 1.7   )( 9.949, 9.574 ) %A(b)C *** ROUGH C ***
+\dashline[80]{0.2}( 9.949, 9.574 )(14.66 ,13.5   ) %CD    *** ROUGH C ***
+\dashline[80]{0.4}     ( 9.949, 9.574 )(14.6  , 1.7   ) %CB    *** ROUGH C ***
+\qbezier[12]  ( 3.661, 6.8   )( 5.359, 5.747 )( 6.087, 3.887 ) %arc from A for angle E - top    part
+\qbezier[12]  ( 6.087, 3.887 )( 6.816, 2.026 )( 6.283, 0.1   ) %arc from A for angle E - bottom part
+\qbezier[10]  ( 3.840, 4.8   )( 5.363, 6.108 )( 7.330, 5.7   ) %short arc  for angle E
+\qbezier[20]  ( 7.662,11.7   )( 9.926,10.079 )(11.271, 7.64  ) %arc from A @ C for length b
+\put(  0.1 ,  0.8 ){$\scriptstyle A$}   %A
+\put( 14.2 ,  0.8 ){$\scriptstyle B$}   %B
+\put( 10.4 ,  9.2 ){$\scriptstyle C$}   %C
+\put( 14.7 , 13.1 ){$\scriptstyle D$}   %D
+\put(  6.3 ,  7.3 ){\emph{b}}           %b
+\put(  8   ,  0.9 ){\emph{c}}           %c
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/128ab310.eepic b/33063-t/images/sources/128ab310.eepic
new file mode 100644
index 0000000..bba96d6
--- /dev/null
+++ b/33063-t/images/sources/128ab310.eepic
@@ -0,0 +1,38 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 128aa310.tex
+% Pt.310. Fig.a.   (20,10.2 )
+\PGset[0.8em]
+\begin{picture}    (20,10.2 )
+\drawline     ( 7.7  ,10.1   )( 1    , 3.4   )( 8.3  , 3.4   ) %ur-A-lr
+\drawline     (11    , 7.4   )(19    , 3.4   )(11.5  , 3.4   ) %ul-B-ll
+\drawline     ( 3.7  , 1     )(14.7  , 1     )                 %left(c)right
+\qbezier[12]  ( 3.63 , 8.457 )( 5.467, 7.501 )( 6.263, 5.589 ) %arc from A top    half
+\qbezier[12]  ( 6.263, 5.589 )( 7.058, 3.677 )( 6.441, 1.7   ) %arc from A bottom half
+\qbezier[10]  (16.421, 7.13  )(15.118, 6.226 )(14.661, 4.701 ) %arc from B top    half
+\qbezier[10]  (14.661, 4.701 )(14.204, 3.176 )(14.801, 1.7   ) %arc from B bottom half
+\put(  0.1 ,  3.4 ){$\scriptstyle A$}   %A
+\put( 19.1 ,  3.4 ){$\scriptstyle B$}   %B
+\put(  9.8 ,  0.2 ){\emph{c}}           %c
+\end{picture}
+\PGrestore
+%
+\qquad
+\quad
+% 128bb310.tex
+% Pt.310. Fig.b.   (15.4,10.2 )
+\PGset[0.8em]
+\begin{picture}    (15.4,10.2 )
+\dashline[80]{0.4}( 7.03 , 8.93  )( 1    , 2.9   )(14.3  , 2.9   )( 4.3  , 8.9   ) %ur(OH)EC(KO)ul
+\qbezier[12]  ( 3.263, 8.457 )( 5.449, 7.567 )( 6.443, 5.426 ) %arc from E top    half
+\qbezier[12]  ( 6.443, 5.426 )( 7.436, 3.284 )( 6.704, 1.04  ) %arc from E bottom half
+\qbezier[ 8]  (11.025, 6.03  )(10.081, 5.043 )( 9.841, 3.698 ) %arc from C top    half
+\qbezier[ 8]  ( 9.841, 3.698 )( 9.600, 2.353 )(10.143, 1.1   ) %arc from C bottom half
+\put( 14.6 ,  2.7 ){$\scriptstyle C$}   %C
+\put(  0.1 ,  2.7 ){$\scriptstyle E$}   %E
+\put(  3.8 ,  6.8 ){$\scriptstyle H$}   %H
+\put( 11.3 ,  5.1 ){$\scriptstyle K$}   %K
+\put(  5.55,  8.4 ){$\scriptstyle O$}   %O
+\put(  7.9 ,  3.2 ){\emph{c}}           %c
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/129ab311.eepic b/33063-t/images/sources/129ab311.eepic
new file mode 100644
index 0000000..d62925c
--- /dev/null
+++ b/33063-t/images/sources/129ab311.eepic
@@ -0,0 +1,41 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 129aa311.tex
+% Pt.311. Fig.a.   ( 9,11.7 )
+\PGset[0.8em]
+\begin{picture}    ( 9,11.7 )
+\drawline     ( 8    ,11.6   )( 1.9  , 6.9   )( 8.2  , 6.9   ) %ur-A-lr
+\drawline     ( 0.1  , 4.3   )( 8.3  , 4.3   )                 %left-b
+\drawline     ( 2.1  , 1.6   )( 8.3  , 1.6   )                 %left-a
+\put(  1.1 ,  6.8 ){$\scriptstyle A$}   %A
+\put(  8.3 ,  1.4 ){\emph{a}}           %a
+\put(  8.3 ,  4.1 ){\emph{b}}           %b
+\end{picture}
+\PGrestore
+%
+\qquad
+\qquad
+\quad
+% 129bb311.tex
+% Pt.311. Fig.b.   (16,11.7 )
+\PGset[0.8em]
+\begin{picture}    (16,11.7 )
+\drawline          ( 0.1  , 1.3   )(15    , 1.3   ) %left(AC')E
+\dashline[80]{0.4}     ( 4.4  , 1.3   )(11.07 , 7.046 ) %AB
+\dashline[80]{0.2}(11.07 , 7.046 )(15.1  ,10.517 ) %BD
+\dashline[80]{0.4}     (13.4  , 1.3   )(11.07 , 7.046 ) %CB
+\dashline[80]{0.4}     ( 8.74 , 1.3   )(11.07 , 7.046 ) %C'B
+\qbezier[10]  (15.9  , 3.158 )(14.891, 1.905 )(13.4  , 1.3   ) %arc from B - right  part
+\qbezier[12]  (13.4  , 1.3   )(11.07 , 0.355 )( 8.74 , 1.3   ) %arc from B - middle part
+\qbezier[ 4]  ( 8.74 , 1.3   )( 8.217, 1.512 )( 7.74 , 1.816 ) %arc from B - left   part (short)
+\put(  4   ,  0.4 ){$\scriptstyle A$}   %A
+\put( 10.4 ,  7.2 ){$\scriptstyle B$}   %B
+\put( 12.8 ,  0.3 ){$\scriptstyle C$}   %C
+\put(  8   ,  0.3 ){$\scriptstyle C'$}   %C'
+\put( 14.6 , 10.7 ){$\scriptstyle D$}   %D
+\put( 15.1 ,  1   ){$\scriptstyle E$}   %E
+\put( 12.4 ,  3.9 ){\emph{a}}           %a on BC
+\put(  9.6 ,  3   ){\emph{a}}           %a on BC'
+\put(  6.9 ,  4.1 ){\emph{b}}           %b
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/129cc311.eepic b/33063-t/images/sources/129cc311.eepic
new file mode 100644
index 0000000..d5a9abe
--- /dev/null
+++ b/33063-t/images/sources/129cc311.eepic
@@ -0,0 +1,21 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 129cc311.tex
+% Pt.311. Fig.c.   (15.4, 9.1 )
+\PGset[0.8em]
+\begin{picture}    (15.4, 9.1 )
+\drawline          ( 0.1  , 1     )(14.7  , 1     ) %left(AH)E
+\dashline[80]{0.4}     ( 3    , 1     )( 9.8  , 6.33  ) %AB
+\dashline[80]{0.2}( 9.8  , 6.33  )(11.8  , 7.898 ) %BD
+\dashline[80]{0.4}     ( 9.8  , 1     )( 9.8  , 6.33  ) %HB
+\qbezier[10]  (13.288, 2.3   )(11.786, 1     )( 9.8  , 1     ) %arc from B - right  part
+\qbezier[10]  ( 9.8  , 1     )( 7.814, 1     )( 6.312, 2.3   ) %arc from B - left   part
+\put(  2.5 ,  0.1 ){$\scriptstyle A$}   %A
+\put(  9.1 ,  6.5 ){$\scriptstyle B$}   %B
+\put( 11.8 ,  8.3 ){$\scriptstyle D$}   %D
+\put( 14.7 ,  0.6 ){$\scriptstyle E$}   %E
+\put(  9.3 ,  0.1 ){$\scriptstyle H$}   %H
+\put( 10   ,  3.1 ){$\scriptstyle a$}           %a
+\put(  5.8 ,  3.8 ){$\scriptstyle b$}           %b
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/129dd311.eepic b/33063-t/images/sources/129dd311.eepic
new file mode 100644
index 0000000..344dec3
--- /dev/null
+++ b/33063-t/images/sources/129dd311.eepic
@@ -0,0 +1,19 @@
+% Diagram by nilrem for PGDP project: Plane Geometry by G.A. Wentworth
+% ***  Uses  epic  package   ***
+% 129dd311.tex
+% Pt.311. Fig.d.   (16.3, 9.5 )
+\PGset[0.8em]
+\begin{picture}    (16.3, 9.5 )
+\drawline          ( 0.1  , 1     )(15.4  , 1     )                 %left(A)E
+\dashline[80]{0.4}     ( 2.3  , 1     )( 8.8  , 6.6   )                 %AB
+\dashline[80]{0.2}( 8.8  , 6.6   )(11.238, 8.7   )                 %BD
+\dashline[80]{0.4}     ( 8.8  , 6.6   )(10.049, 2.8   )                 %B-arc
+\qbezier[24]       (11.751, 3.9   )(10.007, 1.993 )( 7.551, 2.8   ) %arc from B
+\put(  1.8 ,  0.1 ){$\scriptstyle A$}   %A
+\put(  7.9 ,  6.6 ){$\scriptstyle B$}   %B
+\put( 11.5 ,  8.75){$\scriptstyle D$}   %D
+\put( 15.4 ,  0.6 ){$\scriptstyle E$}   %E
+\put(  9.6 ,  4.5 ){\emph{a}}           %a
+\put(  4.8 ,  3.8 ){\emph{b}}           %b
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/130aa311.eepic b/33063-t/images/sources/130aa311.eepic
new file mode 100644
index 0000000..b820d4b
--- /dev/null
+++ b/33063-t/images/sources/130aa311.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}(13,7)
+
+\drawline(1,2)(10.5,2) % AC
+
+\dashline[80]{0.4}(6.75,5.43)(3.05,2) % DA
+\dashline[80]{0.4}(5.75,4.5)(8.45,2)  % BC, len = 3.68
+
+% Ellipse:  u = 5.75  v = 4.5  a = 3.68  b = 3.68  phi = 0.0 Grad
+
+\qbezier[15](2.3501, 3.0917)(2.6089, 2.4669)(3.1478, 1.8978)
+\qbezier[15](3.1478, 1.8978)(3.626, 1.4196)(4.3417, 1.1001)
+\qbezier[15](4.3417, 1.1001)(4.9665, 0.8413)(5.75, 0.82)
+\qbezier[15](5.75, 0.82)(6.4263, 0.82)(7.1583, 1.1001)
+\qbezier[15](7.1583, 1.1001)(7.7831, 1.3589)(8.3522, 1.8978)
+\qbezier[15](8.3522, 1.8978)(8.8304, 2.376)(9.1499, 3.0917)
+
+\put( 2.5, 1.2){$\scriptstyle A$}
+\put( 5.2, 4.7){$\scriptstyle B$}
+\put( 8.4, 1.2){$\scriptstyle C$}
+\put( 6.6, 5.6){$\scriptstyle D$}
+\put(10.6, 1.7){$\scriptstyle E$}
+
+\put( 7.2, 3.3){$\scriptstyle a$}
+\put( 4.0, 3.3){$\scriptstyle b$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/130bb311.eepic b/33063-t/images/sources/130bb311.eepic
new file mode 100644
index 0000000..de63534
--- /dev/null
+++ b/33063-t/images/sources/130bb311.eepic
@@ -0,0 +1,31 @@
+\PGset[0.8em]
+\begin{picture}(12,9)
+
+\drawline(1,3)(11,3) % ED
+\dashline[80]{0.2}(2.3,3)(7,6.4) % C'B m = 0.811
+\dashline[80]{0.4}(6,5.6)(9.7,3) % BC len = 4.522
+\dashline[80]{0.4}(6,5.6)(3.3,3) % BA
+
+% Ellipse:  u = 6.0  v = 5.6  a = 4.522  b = 4.522  phi = 0.0 Grad
+\qbezier[10](1.8222, 3.8695)(2.1402, 3.1018)(2.8025, 2.4025)
+\qbezier[10](2.8025, 2.4025)(3.3901, 1.8149)(4.2695, 1.4222)
+\qbezier[10](4.2695, 1.4222)(5.0373, 1.1042)(6.0, 1.078)
+\qbezier[10](6.0, 1.078)(6.831, 1.078)(7.7305, 1.4222)
+\qbezier[10](7.7305, 1.4222)(8.4982, 1.7402)(9.1975, 2.4025)
+\qbezier[10](9.1975, 2.4025)(9.7851, 2.9901)(10.1778, 3.8695)
+
+
+\put( 3.1, 2.2){$\scriptstyle A$}
+\put( 5.4, 5.8){$\scriptstyle B$}
+\put( 9.7, 2.2){$\scriptstyle C$}
+\put( 1.4, 2.2){$\scriptstyle C'$}
+\put(11.1, 2.7){$\scriptstyle D$}
+\put( 0.2, 2.7){$\scriptstyle E$}
+
+\put( 3.4, 4.3){$\scriptstyle a$}
+\put( 5.0, 3.7){$\scriptstyle b$}
+\put( 8.0, 4.3){$\scriptstyle a$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/130cc311.eepic b/33063-t/images/sources/130cc311.eepic
new file mode 100644
index 0000000..c7d2e2e
--- /dev/null
+++ b/33063-t/images/sources/130cc311.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(9,9)
+
+\drawline(1,3)(8.5,3) % ED
+
+\dashline[80]{0.4}(2.43,3)(4.75,6.68)(7.07,3) % C'BC len = 4.35
+\dashline[80]{0.4}(4.75,6.68)(4.75,3) % BA
+
+% Ellipse:  u = 4.75  v = 5.68  a = 4.35  b = 4.35  phi = 0.0 Grad
+\qbezier[15](1.6741, 3.6041)(2.2393, 3.0388)(3.0853, 2.6611)
+\qbezier[15](3.0853, 2.6611)(3.8239, 2.3552)(4.75, 2.33)
+\qbezier[15](4.75, 2.33)(5.5494, 2.33)(6.4147, 2.6611)
+\qbezier[15](6.4147, 2.6611)(7.1532, 2.967)(7.8259, 3.6041)
+
+\put( 4.4, 2.2){$\scriptstyle A$}
+\put( 4.4, 6.8){$\scriptstyle B$}
+\put( 6.8, 2.2){$\scriptstyle C$}
+\put( 1.7, 2.2){$\scriptstyle C'$}
+\put( 8.6, 2.7){$\scriptstyle D$}
+\put( 0.3, 2.7){$\scriptstyle E$}
+
+\put( 3.1, 4.9){$\scriptstyle a$}
+\put( 5.0, 4.3){$\scriptstyle b$}
+\put( 6.0, 4.9){$\scriptstyle a$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/130dd311.eepic b/33063-t/images/sources/130dd311.eepic
new file mode 100644
index 0000000..4e8fea0
--- /dev/null
+++ b/33063-t/images/sources/130dd311.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(12,9)
+
+\drawline(1,3)(11,3) % ED
+
+\dashline[80]{0.4}(3.1,3)(6,5.4)(8.9,3) % C'BC len = 3.64
+\dashline[80]{0.4}(7.8,3)(5,6.733) % AB m = 1.333
+
+% Ellipse:  u = 6.0  v = 4.4  a = 3.7  b = 3.7  phi = 0.0 Grad
+\qbezier[10](2.5816, 3.9841)(2.8419, 3.3559)(3.3837, 2.7837)
+\qbezier[10](3.3837, 2.7837)(3.8645, 2.3029)(4.5841, 1.9816)
+\qbezier[10](4.5841, 1.9816)(5.2123, 1.7214)(6.0, 1.7)
+\qbezier[10](6.0, 1.7)(6.6799, 1.7)(7.4159, 1.9816)
+\qbezier[10](7.4159, 1.9816)(8.0441, 2.2419)(8.6163, 2.7837)
+\qbezier[10](8.6163, 2.7837)(9.0971, 3.2645)(9.4184, 3.9841)
+
+\put( 7.3, 2.3){$\scriptstyle A$}
+\put( 6.1, 5.4){$\scriptstyle B$}
+\put( 8.7, 2.2){$\scriptstyle C$}
+\put( 2.4, 2.2){$\scriptstyle C'$}
+\put(11.1, 2.7){$\scriptstyle D$}
+\put( 0.3, 2.7){$\scriptstyle E$}
+
+\put( 4.0, 4.3){$\scriptstyle a$}
+\put( 6.6, 3.6){$\scriptstyle b$}
+\put( 7.4, 4.3){$\scriptstyle a$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/131aa312.eepic b/33063-t/images/sources/131aa312.eepic
new file mode 100644
index 0000000..6696b7e
--- /dev/null
+++ b/33063-t/images/sources/131aa312.eepic
@@ -0,0 +1,34 @@
+
+\PGset[0.8em]
+\begin{picture}(30,11)
+
+% a = 7.6, b = 11.4, c = 10.7
+
+\dashline[80]{0.4}(1,1)(11.7,1)(9.724,8.339)(1,1) % ABCA
+
+% Ellipse:  u = 1.0  v = 1.0  a = 11.4  b = 11.4  phi = 5.0 Grad
+\qbezier[20](11.1119, 6.2639)(10.1446, 8.1222)(8.3278, 9.7329)
+% Ellipse:  u = 11.7  v = 1.0  a = 7.6  b = 7.6  phi = 5.0 Grad
+\qbezier[20](11.0376, 8.5711)(9.6463, 8.4494)(8.1907, 7.7413)
+\qbezier[20](8.1907, 7.7413)(6.9519, 7.0964)(5.8781, 5.8852)
+
+
+\drawline(18,5)(28.7,5) % c
+\drawline(19.4,3.5)(27,3.5) % a
+\drawline(17.5,2)(28.9,2) % b
+
+\put( 0.3, 0.7){$\scriptstyle A$}
+\put(11.7, 0.7){$\scriptstyle B$}
+\put( 9.6, 8.5){$\scriptstyle C$}
+
+\put(11.0, 4.4){$\scriptstyle a$}
+\put( 4.5, 4.4){$\scriptstyle b$}
+\put( 6.1, 0.3){$\scriptstyle c$}
+
+\put(23.1, 5.1){$\scriptstyle c$}
+\put(23.1, 3.6){$\scriptstyle a$}
+\put(23.1, 2.1){$\scriptstyle b$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/131bb313.eepic b/33063-t/images/sources/131bb313.eepic
new file mode 100644
index 0000000..de909f4
--- /dev/null
+++ b/33063-t/images/sources/131bb313.eepic
@@ -0,0 +1,46 @@
+\PGset[0.8em]
+\begin{picture}(38,10)
+
+\drawline(1,1)(14,1) % o = 13
+\drawline(3.6,2.5)(12.35,2.5) % m = 7.75
+
+\drawline(8.1,7.6)(5.6,4.4)(10.6,4.4) % C m = 1.28
+
+\dashline[80]{0.4}(19,1)(32,1)(36.771,7.107)(23.771,7.107)(19,1) % ABEHA
+\dashline[80]{0.2}(23.771,7.107)(24.552,8.107) % HD
+
+% Ellipse:  u = 5.6  v = 4.4  a = 3.0  b = 3.0  phi = 0.0 Grad
+\qbezier[10](8.6, 4.4)(8.6, 4.9513)(8.3716, 5.5481)
+\qbezier[10](8.3716, 5.5481)(8.1607, 6.0574)(7.7213, 6.5213)
+\qbezier[10](7.7213, 6.5213)(7.3315, 6.9112)(6.7481, 7.1716)
+\qbezier[10](8.3716, 3.2519)(8.5826, 3.7613)(8.6, 4.4)
+
+% Ellipse:  u = 19.0  v = 1.0  a = 2.678  b = 2.678  phi = 0.0 Grad
+\qbezier[10](21.678, 1.0)(21.678, 1.4921)(21.4741, 2.0248)
+\qbezier[10](21.4741, 2.0248)(21.2858, 2.4795)(20.8936, 2.8936)
+\qbezier[10](20.8936, 2.8936)(20.5456, 3.2416)(20.0248, 3.4741)
+\qbezier[10](21.4741, -0.0248)(21.6625, 0.4298)(21.678, 1.0)
+
+% Ellipse:  u = 23.771  v = 7.107  a = 13.0  b = 13.0  phi = -10.0 Grad
+\qbezier[15](36.5735, 4.8496)(36.9883, 7.2023)(36.4628, 9.9207)
+
+% Ellipse:  u = 32.0  v = 1.0  a = 7.75  b = 7.75  phi = -5.0 Grad
+\qbezier[15](37.9368, 5.9816)(37.0214, 7.0726)(35.5786, 7.8743)
+
+
+\put( 4.8, 4.0){$\scriptstyle C$}
+\put( 7.0, 2.6){$\scriptstyle m$}
+\put( 7.1, 1.1){$\scriptstyle o$}
+      
+\put(18.2, 0.7){$\scriptstyle A$}
+\put(32.2, 0.7){$\scriptstyle B$}
+\put(36.9, 7.0){$\scriptstyle E$}
+\put(22.8, 6.9){$\scriptstyle H$}
+\put(24.6, 8.0){$\scriptstyle D$}
+      
+\put(20.7, 4.3){$\scriptstyle m$}
+\put(26.1, 0.3){$\scriptstyle o$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/132aa314.eepic b/33063-t/images/sources/132aa314.eepic
new file mode 100644
index 0000000..b390cd9
--- /dev/null
+++ b/33063-t/images/sources/132aa314.eepic
@@ -0,0 +1,32 @@
+\PGset[0.8em]
+\begin{picture}(14,14)
+
+% r = 6
+\drawline(1.804,4)(12.196,4)(9.143,12.604)(1.804,4) % BCAB
+
+% Ellipse:  u = 7.0  v = 7.0  a = 6.0  b = 6.0  phi = 0.0 Grad
+\qbezier[20](13.0, 7.0)(13.0, 9.4853)(11.2426, 11.2426)
+\qbezier[20](11.2426, 11.2426)(9.4853, 13.0)(7.0, 13.0)
+\qbezier[20](7.0, 13.0)(4.5147, 13.0)(2.7574, 11.2426)
+\qbezier[20](2.7574, 11.2426)(1.0, 9.4853)(1.0, 7.0)
+\qbezier[20](1.0, 7.0)(1.0, 4.5147)(2.7574, 2.7574)
+\qbezier[20](2.7574, 2.7574)(4.5147, 1.0)(7.0, 1.0)
+\qbezier[20](7.0, 1.0)(9.4853, 1.0)(11.2426, 2.7574)
+\qbezier[20](11.2426, 2.7574)(13.0, 4.5147)(13.0, 7.0)
+
+\dashline[80]{0.2}(1.804,4)(7,7) % BO
+\dashline[80]{0.2}(5.474,8.302)(9,5.294) % EOG, m = -.853
+\dashline[80]{0.2}(7,4)(7,9) % DOF
+
+\put( 9.1,12.7){$\scriptstyle A$}
+\put( 1.0, 3.6){$\scriptstyle B$}
+\put(12.3, 3.6){$\scriptstyle C$}
+\put( 6.5, 3.2){$\scriptstyle D$}
+\put( 4.9, 8.4){$\scriptstyle E$}
+\put( 7.1, 8.7){$\scriptstyle F$}
+\put( 9.1, 4.8){$\scriptstyle G$}
+\put( 7.1, 7.0){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/133aa315.eepic b/33063-t/images/sources/133aa315.eepic
new file mode 100644
index 0000000..d1d0c1a
--- /dev/null
+++ b/33063-t/images/sources/133aa315.eepic
@@ -0,0 +1,35 @@
+
+\PGset[0.8em]
+\begin{picture}(30,9)
+
+\drawline(1,1)(20.3,8.4)(29.4,1)(1,1) % ABCA, mAB=0.383, mBC=-0.813
+
+% mAE = 0.185, mEC = -0.355
+% E = 19.670, 4.454
+\dashline[80]{0.2}(1,1)(19.670,4.454)(29.4,1) % AEC
+\dashline[80]{0.2}(19.670,4.454)(19.670,1) % EH
+
+% Ellipse:  u = 19.67  v = 4.454  a = 3.454  b = 3.454  phi = 0.0 Grad
+\qbezier[20](23.124, 4.454)(23.124, 5.8847)(22.1123, 6.8963)
+\qbezier[20](22.1123, 6.8963)(21.1007, 7.908)(19.67, 7.908)
+\qbezier[20](19.67, 7.908)(18.2393, 7.908)(17.2277, 6.8963)
+\qbezier[20](17.2277, 6.8963)(16.216, 5.8847)(16.216, 4.454)
+\qbezier[20](16.216, 4.454)(16.216, 3.0233)(17.2277, 2.0117)
+\qbezier[20](17.2277, 2.0117)(18.2393, 1.0)(19.67, 1.0)
+\qbezier[20](19.67, 1.0)(21.1007, 1.0)(22.1123, 2.0117)
+\qbezier[20](22.1123, 2.0117)(23.124, 3.0233)(23.124, 4.454)
+
+% mEM = 1.23, len = 3.454, M = 21.849, 7.134
+\dashline[80]{0.2}(18.543,7.719)(19.67,4.454)(21.849,7.134) % KEM
+
+\put( 0.3, 0.7){$\scriptstyle A$}
+\put(20.0, 8.5){$\scriptstyle B$}
+\put(29.5, 0.7){$\scriptstyle C$}
+\put(18.0, 7.8){$\scriptstyle K$}
+\put(19.5, 5.0){$\scriptstyle E$}
+\put(21.9, 7.2){$\scriptstyle M$}
+\put(19.3, 0.2){$\scriptstyle H$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/133bb316.eepic b/33063-t/images/sources/133bb316.eepic
new file mode 100644
index 0000000..bcd1a60
--- /dev/null
+++ b/33063-t/images/sources/133bb316.eepic
@@ -0,0 +1,129 @@
+\PGset[0.8em]
+\begin{picture}(18,17)
+
+% O = 6.4,   9.4
+% P = 9.7,  11.3
+% Q = 9.7,   6.65
+
+\drawline(1,6.295)(12.8,13.086)   % OP produced m=0.576
+\drawline(9.7,16.2)(9.7,1)        % PQ produced m=0
+\drawline(15.5,1.817)(2.4,12.733) % QO produced m=-.833
+
+% bisect through Q
+% mEQO = pi/2 + atan(-.833) = 2.265 rad
+% bisect = 1.133 rad, tan(1.133) = x/y, m = 0.468
+
+% bisect through P
+% mQPB = pi/2 + atan(0.576) = 2.093 rad
+% bisect = 1.047 rad, tan(1.047) = x/y, m = -0.578
+
+% bisect through O
+% mCOP = atan(1/0.833) + atan(1/0.576) = 1.924
+% bisect = 0.962 rad, or vert+0.086, tan(0.086) = x/y, m = 11.6
+
+% interbisection OP =  6.713, 13.027
+% interbisection PQ = 14.146,  8.731
+% interbisection QO =  6.014,  4.925
+
+\dashline[80]{0.2}(5.177,13.915)(16.246,7.517) % OP-PQ m = -0.578
+\dashline[80]{0.2}(15.967,9.583)(2.621,3.337)  % PQ-QO m =  0.468
+\dashline[80]{0.2}(5.789,2.318)(6.938,15.634)  % QO-OP m = 11.591
+
+\dashline[80]{0.2}(6.713,13.027)(4.801,10.732) % r1 = 2.987
+\dashline[80]{0.2}(14.146,8.731)(11.927,12.583) % r2 = 4.445
+\dashline[80]{0.2}(6.014,4.925)(9.7,4.925) % r3 = 3.686
+
+% Ellipse:  u = 6.713  v = 13.027  a = 2.987  b = 2.987  phi = 0.0 Grad
+\qbezier(9.7, 13.027)(9.7, 14.2643)(8.8251, 15.1391)
+\qbezier(3.726, 13.027)(3.726, 11.7897)(4.6009, 10.9149)
+\qbezier(4.6009, 10.9149)(5.4757, 10.04)(6.713, 10.04)
+\qbezier(6.713, 10.04)(7.9503, 10.04)(8.8251, 10.9149)
+\qbezier(8.8251, 10.9149)(9.7, 11.7897)(9.7, 13.027)
+
+% Ellipse:  u = 14.146  v = 8.731  a = 4.445  b = 4.445  phi = 0.0 Grad
+\qbezier(14.146, 13.176)(12.3048, 13.176)(11.0029, 11.8741)
+\qbezier(11.0029, 11.8741)(9.701, 10.5722)(9.701, 8.731)
+\qbezier(9.701, 8.731)(9.701, 6.8898)(11.0029, 5.5879)
+\qbezier(11.0029, 5.5879)(12.3048, 4.286)(14.146, 4.286)
+\qbezier(14.146, 4.286)(15.9872, 4.286)(17.2891, 5.5879)
+
+% Ellipse:  u = 6.014  v = 4.925  a = 3.686  b = 3.686  phi = 0.0 Grad
+\qbezier(9.7, 4.925)(9.7, 6.4518)(8.6204, 7.5314)
+\qbezier(8.6204, 7.5314)(7.5408, 8.611)(6.014, 8.611)
+\qbezier(6.014, 8.611)(4.4872, 8.611)(3.4076, 7.5314)
+\qbezier(3.4076, 7.5314)(2.328, 6.4518)(2.328, 4.925)
+\qbezier(6.014, 1.239)(7.5408, 1.239)(8.6204, 2.3186)
+\qbezier(8.6204, 2.3186)(9.7, 3.3982)(9.7, 4.925)
+
+% mOPQ = 1.048 rad, bisector = 0.524, m = 1.73
+% mPQO = 0.876 rad, bisector = 0.438, m = -2.135
+% intersection = 8.497, 9.219, radius = 1.203
+
+\dashline[80]{0.2}(8.208,8.719)(8.786,9.719) % segment of P bisector
+\dashline[80]{0.2}(8.263,9.719)(8.731,8.719) % segment of Q bisector
+
+% Ellipse:  u = 8.497  v = 9.219  a = 1.203  b = 1.203  phi = 0.0 Grad
+\qbezier(9.7, 9.219)(9.7, 9.7173)(9.3476, 10.0696)
+\qbezier(9.3476, 10.0696)(8.9953, 10.422)(8.497, 10.422)
+\qbezier(8.497, 10.422)(7.9987, 10.422)(7.6464, 10.0696)
+\qbezier(7.6464, 10.0696)(7.294, 9.7173)(7.294, 9.219)
+\qbezier(7.294, 9.219)(7.294, 8.7207)(7.6464, 8.3684)
+\qbezier(7.6464, 8.3684)(7.9987, 8.016)(8.497, 8.016)
+\qbezier(8.497, 8.016)(8.9953, 8.016)(9.3476, 8.3684)
+\qbezier(9.3476, 8.3684)(9.7, 8.7207)(9.7, 9.219)
+
+% bisecting P, r = 2.14, r2 = 3.015
+% F = 9.7, 13.44    F' = 7.846, 10.232
+% Ellipse:  u = 9.7  v = 11.3  a = 2.14  b = 2.14  phi = 0.0 Grad
+\qbezier[10](9.7, 13.44)(9.3067, 13.44)(8.8811, 13.2771)
+\qbezier[10](8.8811, 13.2771)(8.5177, 13.1266)(8.1868, 12.8132)
+\qbezier[10](8.1868, 12.8132)(7.9087, 12.5351)(7.7229, 12.1189)
+\qbezier[10](7.7229, 12.1189)(7.5724, 11.7556)(7.56, 11.3)
+\qbezier[10](7.56, 11.3)(7.56, 10.9067)(7.7229, 10.4811)
+% Ellipse:  u = 9.7  v = 13.44  a = 3.015  b = 3.015  phi = -5.0 Grad
+\qbezier[10](6.6965, 13.7028)(6.6482, 13.1508)(6.8245, 12.5334)
+% Ellipse:  u = 7.846  v = 10.232  a = 3.015  b = 3.015  phi = -10.0 Grad
+\qbezier[10](7.1934, 13.1755)(6.6525, 13.0556)(6.1167, 12.7017)
+
+% bisecting Q, r = 2.14, r2 = 3.28
+% D = 11.344, 5.28  D' = 9.7, 8.79
+% Ellipse:  u = 9.7  v = 6.65  a = 2.14  b = 2.14  phi = 0.0 Grad
+\qbezier[10](11.84, 6.65)(11.84, 7.0433)(11.6771, 7.4689)
+\qbezier[10](11.6771, 7.4689)(11.5266, 7.8323)(11.2132, 8.1632)
+\qbezier[10](11.2132, 8.1632)(10.9351, 8.4413)(10.5189, 8.6271)
+\qbezier[10](10.5189, 8.6271)(10.1556, 8.7776)(9.7, 8.79)
+\qbezier[10](11.2132, 5.1368)(11.4913, 5.4149)(11.6771, 5.8311)
+\qbezier[10](11.6771, 5.8311)(11.8276, 6.1944)(11.84, 6.65)
+% Ellipse:  u = 11.344  v = 5.28  a = 3.28  b = 3.28  phi = 8.0 Grad
+\qbezier[10](13.318, 7.8995)(12.8366, 8.2623)(12.1652, 8.4555)
+% Ellipse:  u = 9.7  v = 8.79  a = 3.28  b = 3.28  phi = 0.0 Grad
+\qbezier[10](12.7303, 7.5348)(12.961, 8.0917)(12.98, 8.79)
+
+
+% bisecting O, r = 2.14, r2 = 3.24
+% A = 4.546, 8.332  A' = 8.044, 8.03
+% Ellipse:  u = 6.4  v = 9.4  a = 2.14  b = 2.14  phi = 5.0 Grad
+\qbezier[10](4.5018, 8.4119)(4.6834, 8.063)(5.0244, 7.7607)
+\qbezier[10](5.0244, 7.7607)(5.3257, 7.5079)(5.7565, 7.359)
+\qbezier[10](5.7565, 7.359)(6.1316, 7.2408)(6.5865, 7.2681)
+\qbezier[10](6.5865, 7.2681)(6.9783, 7.3024)(7.3881, 7.5018)
+\qbezier[10](7.3881, 7.5018)(7.737, 7.6834)(8.0393, 8.0244)
+% Ellipse:  u = 4.546  v = 8.332  a = 3.24  b = 3.24  phi = 15.0 Grad
+\qbezier[10](5.3846, 5.2024)(5.9597, 5.3565)(6.5184, 5.7615)
+% Ellipse:  u = 8.044  v = 8.03  a = 3.24  b = 3.24  phi = -5.0 Grad
+\qbezier[10](5.562, 5.9474)(5.9447, 5.4913)(6.5479, 5.1561)
+
+
+\put( 3.8, 8.3){$\scriptstyle A$}
+\put(11.4,12.7){$\scriptstyle B$}
+\put( 4.0,10.0){$\scriptstyle C$}
+\put(10.6, 4.5){$\scriptstyle D$}
+\put( 9.7, 4.5){$\scriptstyle E$}
+\put( 9.7,13.2){$\scriptstyle F$}
+\put( 5.4, 9.1){$\scriptstyle O$}
+\put( 9.7,11.7){$\scriptstyle P$}
+\put( 9.7, 5.5){$\scriptstyle Q$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/134aa317.eepic b/33063-t/images/sources/134aa317.eepic
new file mode 100644
index 0000000..1fbab1c
--- /dev/null
+++ b/33063-t/images/sources/134aa317.eepic
@@ -0,0 +1,69 @@
+\PGset[0.8em]
+\begin{picture}(28,10)
+
+% r = 4, r2 = 3.4
+% Ellipse:  u = 7.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(11.0, 5.0)(11.0, 6.6569)(9.8284, 7.8284)
+\qbezier(9.8284, 7.8284)(8.6569, 9.0)(7.0, 9.0)
+\qbezier(7.0, 9.0)(5.3431, 9.0)(4.1716, 7.8284)
+\qbezier(4.1716, 7.8284)(3.0, 6.6569)(3.0, 5.0)
+\qbezier(3.0, 5.0)(3.0, 3.3431)(4.1716, 2.1716)
+\qbezier(4.1716, 2.1716)(5.3431, 1.0)(7.0, 1.0)
+\qbezier(7.0, 1.0)(8.6569, 1.0)(9.8284, 2.1716)
+\qbezier(9.8284, 2.1716)(11.0, 3.3431)(11.0, 5.0)
+
+% Ellipse:  u = 18.0  v = 5.0  a = 3.4  b = 3.4  phi = 0.0 Grad
+\qbezier(21.4, 5.0)(21.4, 6.4083)(20.4042, 7.4042)
+\qbezier(20.4042, 7.4042)(19.4083, 8.4)(18.0, 8.4)
+\qbezier(18.0, 8.4)(16.5917, 8.4)(15.5958, 7.4042)
+\qbezier(15.5958, 7.4042)(14.6, 6.4083)(14.6, 5.0)
+\qbezier(14.6, 5.0)(14.6, 3.5917)(15.5958, 2.5958)
+\qbezier(15.5958, 2.5958)(16.5917, 1.6)(18.0, 1.6)
+\qbezier(18.0, 1.6)(19.4083, 1.6)(20.4042, 2.5958)
+\qbezier(20.4042, 2.5958)(21.4, 3.5917)(21.4, 5.0)
+
+\drawline(7,5)(7,9) % OC
+
+\dashline[80]{0.4}(1,9)(11,9) % AM
+
+\dashline[80]{0.2}(7,5)(3,9) % mid=5,7, len=5.657, r=2.829
+
+% Ellipse:  u = 5.0  v = 7.0  a = 2.829  b = 2.829  phi = 20.0 Grad
+\qbezier[15](7.6584, 7.9676)(7.2576, 9.0687)(6.1956, 9.5639)
+\qbezier[15](6.1956, 9.5639)(5.1336, 10.0592)(4.0324, 9.6584)
+\qbezier[15](4.0324, 9.6584)(2.9313, 9.2576)(2.4361, 8.1956)
+\qbezier[15](5.9676, 4.3416)(7.0687, 4.7424)(7.5639, 5.8044)
+\qbezier[15](7.5639, 5.8044)(8.0592, 6.8664)(7.6584, 7.9676)
+
+
+\dashline[80]{0.2}(18,5)(26.5,5) % OE, mid=22.25,5, r=4.25
+
+% Ellipse:  u = 22.25  v = 5.0  a = 4.25  b = 4.25  phi = 0.0 Grad
+\qbezier[20](26.5, 5.0)(26.5, 6.7604)(25.2552, 8.0052)
+\qbezier[20](25.2552, 8.0052)(24.0104, 9.25)(22.25, 9.25)
+\qbezier[20](22.25, 9.25)(20.4896, 9.25)(19.2448, 8.0052)
+\qbezier[20](19.2448, 8.0052)(18.0, 6.7604)(18.0, 5.0)
+\qbezier[20](18.0, 5.0)(18.0, 3.2396)(19.2448, 1.9948)
+\qbezier[20](19.2448, 1.9948)(20.4896, 0.75)(22.25, 0.75)
+\qbezier[20](22.25, 0.75)(24.0104, 0.75)(25.2552, 1.9948)
+\qbezier[20](25.2552, 1.9948)(26.5, 3.2396)(26.5, 5.0)
+
+\dashline[80]{0.4}(26.5,5)(16,9.581) % EM
+\dashline[80]{0.4}(26.5,5)(16,0.419) % EH
+
+\dashline[80]{0.2}(18,5)(19.36,8.116) % OM
+
+
+\put( 0.2, 8.7){$\scriptstyle A$}
+\put( 6.8, 9.2){$\scriptstyle C$}
+\put(11.1, 8.7){$\scriptstyle M$}
+\put( 6.7, 4.2){$\scriptstyle O$}
+
+\put(17.2, 4.7){$\scriptstyle O$}
+\put(18.8, 8.4){$\scriptstyle M$}
+\put(18.8, 1.0){$\scriptstyle H$}
+\put(26.6, 4.7){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/135aa318.eepic b/33063-t/images/sources/135aa318.eepic
new file mode 100644
index 0000000..1549490
--- /dev/null
+++ b/33063-t/images/sources/135aa318.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(16.5,11)
+
+\drawline(1,4.5)(4.643,4.5)(1.893,1.15) % M, m=1.218
+
+\drawline(6,4.5)(12,4.5) % AB
+\dashline[80]{0.2}(9,0.846)(15,8.154) % EB
+
+% F = 9,4.5
+% O = 9, 6.963
+\dashline[80]{0.2}(9,4.5)(9,6.963)(12,4.5) % FOB, OB = 3.882
+
+% Ellipse:  u = 9.0  v = 6.963  a = 3.882  b = 3.882  phi = 0.0 Grad
+\qbezier[20](12.882, 6.963)(12.882, 8.571)(11.745, 9.708)
+\qbezier[20](11.745, 9.708)(10.608, 10.845)(9.0, 10.845)
+\qbezier[20](9.0, 10.845)(7.392, 10.845)(6.255, 9.708)
+\qbezier[20](6.255, 9.708)(5.118, 8.571)(5.118, 6.963)
+\qbezier[20](5.118, 6.963)(5.118, 5.355)(6.255, 4.218)
+\qbezier[20](6.255, 4.218)(7.392, 3.081)(9.0, 3.081)
+\qbezier[20](9.0, 3.081)(10.608, 3.081)(11.745, 4.218)
+\qbezier[20](11.745, 4.218)(12.882, 5.355)(12.882, 6.963)
+
+% given AB and circle O, there is only one point K on O s.t. AKB = M = 50.6deg
+% how do I find that point?
+
+% 50.6 = atan((x-6)/(y-4.5)) + atan((12-x)/(y-4.5))
+\dashline[80]{0.2}(6,4.5)(10.86,10.37)(12,4.5) % AKB
+
+\put( 5.4, 3.8){$\scriptstyle A$}
+\put(11.8, 3.8){$\scriptstyle B$}
+\put( 9.9, 1.4){$\scriptstyle E$}
+\put( 8.6, 3.8){$\scriptstyle F$}
+\put( 8.8, 7.1){$\scriptstyle O$}
+\put( 3.2, 3.8){$\scriptstyle M$}
+\put(10.9,10.4){$\scriptstyle K$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/137aa147.eepic b/33063-t/images/sources/137aa147.eepic
new file mode 100644
index 0000000..378c5bd
--- /dev/null
+++ b/33063-t/images/sources/137aa147.eepic
@@ -0,0 +1,40 @@
+\PGset[0.8em]
+\begin{picture}(13,11)
+
+% 147
+
+\drawline(1,3.8)(12.6,3.8) % BC
+\drawline(1,8.8)(12.6,8.8) % DE
+\dashline[80]{0.2}(1,6.3)(12.6,6.3) % FG
+
+% M = 3, N = 6.5, P = 4.75, 6.3
+\dashline[80]{0.2}(4.75,1)(4.75,10)
+\dashline[80]{0.2}(3,3.8)(4.75,6.3) % MP, len = 3.052
+\drawline(6.5,3.7)(6.5,3.9) % N
+
+% A = 11.36, 7.55, r = 3.052
+% Ellipse:  u = 11.36  v = 7.55  a = 3.052  b = 3.052  phi = 120.0 Grad
+\qbezier[15](9.834, 10.1931)(8.7392, 9.561)(8.412, 8.3399)
+\qbezier[15](8.412, 8.3399)(8.0848, 7.1188)(8.7169, 6.024)
+\qbezier[15](8.7169, 6.024)(9.349, 4.9292)(10.5701, 4.602)
+\qbezier[15](10.5701, 4.602)(11.7912, 4.2748)(12.886, 4.9069)
+
+% O = 8.576, 6.3
+\dashline[80]{0.2}(11.36,7.55)(8.576,6.3) % AO
+
+
+\put(11.3, 7.2){$\scriptstyle A$}
+\put( 0.7, 3.0){$\scriptstyle B$}
+\put(12.0, 3.0){$\scriptstyle C$}
+\put( 0.7, 9.0){$\scriptstyle D$}
+\put(12.0, 9.0){$\scriptstyle E$}
+\put( 0.7, 6.5){$\scriptstyle F$}
+\put(12.0, 5.6){$\scriptstyle G$}
+\put( 2.4, 3.0){$\scriptstyle M$}
+\put( 6.0, 3.0){$\scriptstyle N$}
+\put( 7.9, 5.6){$\scriptstyle O$}
+\put( 4.9, 6.5){$\scriptstyle P$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/137bb148.eepic b/33063-t/images/sources/137bb148.eepic
new file mode 100644
index 0000000..85fe8c4
--- /dev/null
+++ b/33063-t/images/sources/137bb148.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(23,9)
+
+% C = 15.123, 7.44
+\dashline[80]{0.2}(15.123,7.44)(1,1)(9.53,1) % CEA
+\dashline[80]{0.2}(15.123,7.44)(22.5,1)(16,1) % CFB
+
+\dashline[80]{0.2}(8.062,4.22)(9.758,0.5) % GA
+\dashline[80]{0.2}(18.812,4.22)(15.563,0.5) % HB
+
+\dashline[80]{0.2}(1,7.44)(16,7.44)
+
+\drawline(9.53,1)(16,1)(15.123,7.44)(9.53,1) % ABCA
+\drawline(15.123,1)(15.123,7.44) % CD
+
+\put( 9.0, 0.2){$\scriptstyle A$}
+\put(15.8, 0.2){$\scriptstyle B$}
+\put(14.8, 7.6){$\scriptstyle C$}
+\put(14.7, 0.2){$\scriptstyle D$}
+\put( 0.7, 0.2){$\scriptstyle E$}
+\put(22.0, 0.2){$\scriptstyle F$}
+\put( 7.6, 4.3){$\scriptstyle G$}
+\put(18.8, 4.3){$\scriptstyle H$}
+
+\put( 2.2, 1.1){$\scriptstyle m$}
+\put(21.3, 1.1){$\scriptstyle n$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/138aa158.eepic b/33063-t/images/sources/138aa158.eepic
new file mode 100644
index 0000000..3a288fe
--- /dev/null
+++ b/33063-t/images/sources/138aa158.eepic
@@ -0,0 +1,27 @@
+\PGset[0.8em]
+\begin{picture}(6,6)
+
+% 158
+% P = 1    1
+% Q = 4.5, 1
+% R = 4.9, 3
+% r = 2.6
+\dashline[80]{0.2}(1,1)(4.5,1) % PQ
+\dashline[80]{0.2}(2.75,1)(2.75,5.1) % perp bisector
+\dashline[80]{0.2}(4.9,3)(2.3,3) % Rd
+
+% Ellipse:  u = 4.9  v = 3.0  a = 2.6  b = 2.6  phi = 30.0 Grad
+\qbezier[15](3.6, 5.2517)(2.6673, 4.7132)(2.3886, 3.6729)
+\qbezier[15](2.3886, 3.6729)(2.1099, 2.6327)(2.6483, 1.7)
+\qbezier[15](2.6483, 1.7)(3.1868, 0.7673)(4.2271, 0.4886)
+
+\put( 0.7, 0.2){$\scriptstyle P$}
+\put( 4.5, 0.4){$\scriptstyle Q$}
+\put( 5.0, 2.7){$\scriptstyle R$}
+\put( 2.9, 4.1){$\scriptstyle X$}
+\put( 2.9, 1.6){$\scriptstyle X'$}
+\put( 3.8, 3.2){$\scriptstyle d$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/138bb165.eepic b/33063-t/images/sources/138bb165.eepic
new file mode 100644
index 0000000..251787a
--- /dev/null
+++ b/33063-t/images/sources/138bb165.eepic
@@ -0,0 +1,47 @@
+\PGset[0.8em]
+\begin{picture}(8.5,6)
+
+% 165
+% E = 4,    2.3
+% A = 2,    1
+% D = 6.5,  1
+% C = 1,    4.117  mCD = -.567, mAB = 0.65
+% B = 8,    4.9
+
+\drawline(2,1)(8,4.9) % AB
+\drawline(1,4.117)(6.5,1) %CD
+\dashline[80]{0.2}(4,1)(4,5.7) % YX
+\dashline[80]{0.2}(1.5,2.3)(7.3,2.3) % ZT
+
+% P = 4.786, 3.286, r = 1.9
+% Ellipse:  u = 4.786  v = 3.286  a = 1.9  b = 1.9  phi = 0.0 Grad
+\qbezier[15](6.686, 3.286)(6.686, 4.073)(6.1295, 4.6295)
+\qbezier[15](6.1295, 4.6295)(5.573, 5.186)(4.786, 5.186)
+\qbezier[15](4.786, 5.186)(3.999, 5.186)(3.4425, 4.6295)
+\qbezier[15](3.4425, 4.6295)(2.886, 4.073)(2.886, 3.286)
+\qbezier[15](2.886, 3.286)(2.886, 2.499)(3.4425, 1.9425)
+\qbezier[15](3.4425, 1.9425)(3.999, 1.386)(4.786, 1.386)
+\qbezier[15](4.786, 1.386)(5.573, 1.386)(6.1295, 1.9425)
+\qbezier[15](6.1295, 1.9425)(6.686, 2.499)(6.686, 3.286)
+
+\drawline(4.786,3.286)(4.78,3.286)
+
+\drawline(1,5)(2.9,5) % d
+\drawline(1,4.9)(1,5.1)
+\drawline(2.9,4.9)(2.9,5.1)
+
+\put( 2.0, 0.5){$\scriptstyle A$}
+\put( 7.2, 4.8){$\scriptstyle B$}
+\put( 1.0, 4.2){$\scriptstyle C$}
+\put( 5.8, 0.3){$\scriptstyle D$}
+\put( 4.1, 2.4){$\scriptstyle E$}
+\put( 3.3, 5.0){$\scriptstyle X$}
+\put( 3.3, 0.9){$\scriptstyle Y$}
+\put( 2.6, 1.7){$\scriptstyle Z$}
+\put( 4.5, 3.4){$\scriptstyle P$}
+\put( 6.4, 1.6){$\scriptstyle T$}
+\put( 1.7, 5.1){$\scriptstyle d$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139aa167.eepic b/33063-t/images/sources/139aa167.eepic
new file mode 100644
index 0000000..a3dd4f7
--- /dev/null
+++ b/33063-t/images/sources/139aa167.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(8,5)
+
+% r = 2.75
+% Ellipse:  u = 3.75  v = 3.75  a = 2.75  b = 2.75  phi = 0.0 Grad
+\qbezier[15](1.0, 3.75)(1.0, 2.6109)(1.8055, 1.8055)
+\qbezier[15](1.8055, 1.8055)(2.6109, 1.0)(3.75, 1.0)
+\qbezier[15](3.75, 1.0)(4.8891, 1.0)(5.6945, 1.8055)
+\qbezier[15](5.6945, 1.8055)(6.5, 2.6109)(6.5, 3.75)
+
+\drawline(3.75,3.75)(3.75,1) % AB
+% r/4 = 0.688
+
+\dashline[80]{0.2}(0.5,1.688)(7.0,1.688) % ZT
+\drawline         (0.5,2.376)(7.0,2.376) % DE
+\dashline[80]{0.2}(0.5,3.064)(7.0,3.064) % XY
+
+\put( 3.4, 3.8){$\scriptstyle A$}
+\put( 3.4, 0.2){$\scriptstyle B$}
+\put( 3.8, 1.7){$\scriptstyle C$}
+\put(-0.4, 2.0){$\scriptstyle D$}
+\put( 7.0, 2.0){$\scriptstyle E$}
+\put( 0.0, 3.2){$\scriptstyle X$}
+\put( 6.6, 3.2){$\scriptstyle Y$}
+\put( 1.2, 1.0){$\scriptstyle Z$}
+\put( 5.6, 1.0){$\scriptstyle T$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139bb168.eepic b/33063-t/images/sources/139bb168.eepic
new file mode 100644
index 0000000..b8e30fc
--- /dev/null
+++ b/33063-t/images/sources/139bb168.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+\begin{picture}(7,5)
+
+\drawline(4.5,4.2)(1,1)(6.6,1)(4.5,4.2) % ABCA
+
+% X = midBC = 3.8, mXA = 4.571, F is on that line
+% FD = BD (mBA = 0.914, so Dx = 1+(y-1)/0.914, 
+% mCA = -1.524, so Ex = 4.5+(y-4.2)/-1.524
+% F = 4.06,2.188
+\dashline[80]{0.2}(2.3,2.188)(5.82,2.188) % DE
+\dashline[80]{0.2}(1,1)(5.06,2.576) % BF
+\dashline[80]{0.2}(6.6,1)(3.06,2.656) % CF
+
+\put( 4.2, 4.3){$\scriptstyle A$}
+\put( 0.7, 0.2){$\scriptstyle B$}
+\put( 6.0, 0.2){$\scriptstyle C$}
+\put( 1.5, 2.0){$\scriptstyle D$}
+\put( 5.9, 2.0){$\scriptstyle E$}
+\put( 3.9, 2.3){$\scriptstyle F$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139cc169.eepic b/33063-t/images/sources/139cc169.eepic
new file mode 100644
index 0000000..f6def50
--- /dev/null
+++ b/33063-t/images/sources/139cc169.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\vspace{-1ex}
+\begin{picture}(6,4.5)
+
+\drawline(1,1)(6,1)(2,3.5)(1,1) % BCAB
+
+\drawline(3.65,3.5)(6,3.5) % d
+
+% D = 3.35, 1
+% mAC = -0.625, mAB = 2.5
+% e1: (y-3.5)/(Ex-2) = 2.5$
+% e2: (y-3.5)/(Fx-2) = -0.625$
+% e3: Ex-Fx = 2.35$
+
+% [[y = 2.325, Ex = 3.88, Fx = 1.53]]
+\dashline[80]{0.2}(1.53,2.325)(3.88,2.325)(3.35,1) % FED
+
+\put( 1.7, 3.6){$\scriptstyle A$}
+\put( 0.7, 0.2){$\scriptstyle B$}
+\put( 5.4, 0.2){$\scriptstyle C$}
+\put( 3.0, 0.2){$\scriptstyle D$}
+\put( 4.0, 2.3){$\scriptstyle E$}
+\put( 1.0, 2.3){$\scriptstyle F$}
+
+\put( 4.5, 3.6){$\scriptstyle d$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139dd174.eepic b/33063-t/images/sources/139dd174.eepic
new file mode 100644
index 0000000..3247cdb
--- /dev/null
+++ b/33063-t/images/sources/139dd174.eepic
@@ -0,0 +1,50 @@
+\PGset[0.8em]
+\begin{picture}(15,8)
+
+% R = 3
+% r = 1.5
+\drawline(1,4)(11.5,4) % AO
+
+% M = 6.25, 4
+% Ellipse:  u = 6.25  v = 4.0  a = 1.5  b = 1.5  phi = 0.0 Grad
+\qbezier(7.75, 4.0)(7.75, 4.6213)(7.3107, 5.0607)
+\qbezier(7.3107, 5.0607)(6.8713, 5.5)(6.25, 5.5)
+\qbezier(6.25, 5.5)(5.6287, 5.5)(5.1893, 5.0607)
+\qbezier(5.1893, 5.0607)(4.75, 4.6213)(4.75, 4.0)
+\qbezier(4.75, 4.0)(4.75, 3.3787)(5.1893, 2.9393)
+\qbezier(5.1893, 2.9393)(5.6287, 2.5)(6.25, 2.5)
+\qbezier(6.25, 2.5)(6.8713, 2.5)(7.3107, 2.9393)
+\qbezier(7.3107, 2.9393)(7.75, 3.3787)(7.75, 4.0)
+
+
+% Ellipse:  u = 11.5  v = 4.0  a = 3.0  b = 3.0  phi = 0.0 Grad
+\qbezier(14.5, 4.0)(14.5, 5.2426)(13.6213, 6.1213)
+\qbezier(13.6213, 6.1213)(12.7426, 7.0)(11.5, 7.0)
+\qbezier(11.5, 7.0)(10.2574, 7.0)(9.3787, 6.1213)
+\qbezier(9.3787, 6.1213)(8.5, 5.2426)(8.5, 4.0)
+\qbezier(8.5, 4.0)(8.5, 2.7574)(9.3787, 1.8787)
+\qbezier(9.3787, 1.8787)(10.2574, 1.0)(11.5, 1.0)
+\qbezier(11.5, 1.0)(12.7426, 1.0)(13.6213, 1.8787)
+\qbezier(13.6213, 1.8787)(14.5, 2.7574)(14.5, 4.0)
+
+% B = 9.25, 5.984
+% D = 9.25, 2.016
+\drawline(9.25,5.984)(1,4)(9.25,2.016) % BAD mBA = 0.24
+
+\dashline[80]{0.2}(9.25,5.984)(11.5,4)(9.25,2.016) % BOD
+
+% C = 5.122, 4.989
+% E = 5.122, 3.01
+\dashline[80]{0.2}(5.122,4.989)(6.25,4)(5.122,3.01) % CME
+
+\put( 0.3, 3.7){$\scriptstyle A$}
+\put( 8.6, 6.1){$\scriptstyle B$}
+\put( 4.4, 5.0){$\scriptstyle C$}
+\put( 8.6, 1.3){$\scriptstyle D$}
+\put( 4.4, 2.3){$\scriptstyle E$}
+\put(11.6, 3.7){$\scriptstyle O$}
+\put( 6.1, 3.3){$\scriptstyle M$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139ee175.eepic b/33063-t/images/sources/139ee175.eepic
new file mode 100644
index 0000000..a2997da
--- /dev/null
+++ b/33063-t/images/sources/139ee175.eepic
@@ -0,0 +1,41 @@
+\PGset[0.8em]
+\begin{picture}(12,7)
+
+% r = 2
+
+% Ellipse:  u = 3.0  v = 3.0  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier(5.0, 3.0)(5.0, 3.8284)(4.4142, 4.4142)
+\qbezier(4.4142, 4.4142)(3.8284, 5.0)(3.0, 5.0)
+\qbezier(3.0, 5.0)(2.1716, 5.0)(1.5858, 4.4142)
+\qbezier(1.5858, 4.4142)(1.0, 3.8284)(1.0, 3.0)
+\qbezier(1.0, 3.0)(1.0, 2.1716)(1.5858, 1.5858)
+\qbezier(1.5858, 1.5858)(2.1716, 1.0)(3.0, 1.0)
+\qbezier(3.0, 1.0)(3.8284, 1.0)(4.4142, 1.5858)
+\qbezier(4.4142, 1.5858)(5.0, 2.1716)(5.0, 3.0)
+
+% OM = 6.25
+% Ellipse:  u = 9.25  v = 3.0  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier[15](11.25, 3.0)(11.25, 3.8284)(10.6642, 4.4142)
+\qbezier[15](10.6642, 4.4142)(10.0784, 5.0)(9.25, 5.0)
+\qbezier[15](9.25, 5.0)(8.4216, 5.0)(7.8358, 4.4142)
+\qbezier[15](7.8358, 4.4142)(7.25, 3.8284)(7.25, 3.0)
+\qbezier[15](7.25, 3.0)(7.25, 2.1716)(7.8358, 1.5858)
+\qbezier[15](7.8358, 1.5858)(8.4216, 1.0)(9.25, 1.0)
+\qbezier[15](9.25, 1.0)(10.0784, 1.0)(10.6642, 1.5858)
+\qbezier[15](10.6642, 1.5858)(11.25, 2.1716)(11.25, 3.0)
+
+
+\drawline(2,6)(10.4,6) % AB = 8.4
+\dashline[80]{0.2}(4.579,4.227)(3,3)(9.25,3)(10.829,4.227) % COMD
+\drawline(4.579,4.227)(10.829,4.227) % CD
+
+\put( 1.8, 6.1){$\scriptstyle A$}
+\put(10.0, 6.1){$\scriptstyle B$}
+\put( 4.6, 4.3){$\scriptstyle C$}
+\put(10.8, 4.3){$\scriptstyle D$}
+\put( 2.7, 2.2){$\scriptstyle O$}
+\put( 8.8, 2.2){$\scriptstyle M$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139ff177.eepic b/33063-t/images/sources/139ff177.eepic
new file mode 100644
index 0000000..bd88831
--- /dev/null
+++ b/33063-t/images/sources/139ff177.eepic
@@ -0,0 +1,46 @@
+\PGset[0.8em]
+\begin{picture}(10.5,10.5)
+
+% In a given circle let AOB be a diameter, OC any radius, CD the
+% perpendicular from C to AB.  Upon OC take OM equal to CD.  Find the
+% locus of the point M as OC turns about O.
+
+% r = 4.25
+% Ellipse:  u = 5.25  v = 5.25  a = 4.25  b = 4.25  phi = 0.0 Grad
+\qbezier(9.5, 5.25)(9.5, 7.0104)(8.2552, 8.2552)
+\qbezier(8.2552, 8.2552)(7.0104, 9.5)(5.25, 9.5)
+\qbezier(5.25, 9.5)(3.4896, 9.5)(2.2448, 8.2552)
+\qbezier(2.2448, 8.2552)(1.0, 7.0104)(1.0, 5.25)
+\qbezier(1.0, 5.25)(1.0, 3.4896)(2.2448, 2.2448)
+\qbezier(2.2448, 2.2448)(3.4896, 1.0)(5.25, 1.0)
+\qbezier(5.25, 1.0)(7.0104, 1.0)(8.2552, 2.2448)
+\qbezier(8.2552, 2.2448)(9.5, 3.4896)(9.5, 5.25)
+
+\drawline(1,5.25)(9.5,5.25) % BA
+\dashline[80]{0.2}(5.25,1)(5.25,9.5) % EO
+\drawline(5.25,5.25)(8.255,8.255)(8.255,5.25) % OCD
+
+% Ellipse:  u = 5.25  v = 7.375  a = 2.125  b = 2.125  phi = 0.0 Grad
+\qbezier[10](7.375, 7.375)(7.375, 8.2552)(6.7526, 8.8776)
+\qbezier[10](6.7526, 8.8776)(6.1302, 9.5)(5.25, 9.5)
+\qbezier[10](5.25, 9.5)(4.3698, 9.5)(3.7474, 8.8776)
+\qbezier[10](3.7474, 8.8776)(3.125, 8.2552)(3.125, 7.375)
+\qbezier[10](3.125, 7.375)(3.125, 6.4948)(3.7474, 5.8724)
+\qbezier[10](3.7474, 5.8724)(4.3698, 5.25)(5.25, 5.25)
+\qbezier[10](5.25, 5.25)(6.1302, 5.25)(6.7526, 5.8724)
+\qbezier[10](6.7526, 5.8724)(7.375, 6.4948)(7.375, 7.375)
+
+\dashline[80]{0.2}(5.25,9.5)(7.375,7.375) % EM
+
+\put( 9.5, 5.0){$\scriptstyle A$}
+\put( 0.2, 5.0){$\scriptstyle B$}
+\put( 8.3, 8.3){$\scriptstyle C$}
+\put( 7.8, 4.5){$\scriptstyle D$}
+\put( 5.0, 9.6){$\scriptstyle E$}
+\put( 5.3, 4.5){$\scriptstyle O$}
+\put( 7.3, 6.7){$\scriptstyle M$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/139gg176.eepic b/33063-t/images/sources/139gg176.eepic
new file mode 100644
index 0000000..a706522
--- /dev/null
+++ b/33063-t/images/sources/139gg176.eepic
@@ -0,0 +1,31 @@
+\PGset[0.8em]
+\begin{picture}(7.5,6.5)
+
+% r = 2
+
+% Ellipse:  u = 3.0  v = 3.0  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier[15](5.0, 3.0)(5.0, 3.8284)(4.4142, 4.4142)
+\qbezier[15](4.4142, 4.4142)(3.8284, 5.0)(3.0, 5.0)
+\qbezier[15](3.0, 5.0)(2.1716, 5.0)(1.5858, 4.4142)
+\qbezier[15](1.5858, 4.4142)(1.0, 3.8284)(1.0, 3.0)
+\qbezier[15](1.0, 3.0)(1.0, 2.1716)(1.5858, 1.5858)
+\qbezier[15](1.5858, 1.5858)(2.1716, 1.0)(3.0, 1.0)
+\qbezier[15](3.0, 1.0)(3.8284, 1.0)(4.4142, 1.5858)
+\qbezier[15](4.4142, 1.5858)(5.0, 2.1716)(5.0, 3.0)
+
+\drawline(3,0.5)(3,5.5)
+\drawline(0.5,3)(7.19,3)
+
+% M = 4.735, 3.996
+\dashline[80]{0.2}(3,3)(4.735,3.996) % OM
+
+\drawline(3,4.7)(7.19,3) % AB
+
+\put( 2.2, 4.2){$\scriptstyle A$}
+\put( 6.5, 2.2){$\scriptstyle B$}
+\put( 2.2, 2.3){$\scriptstyle O$}
+\put( 4.8, 4.1){$\scriptstyle M$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/140aa178.eepic b/33063-t/images/sources/140aa178.eepic
new file mode 100644
index 0000000..97a68da
--- /dev/null
+++ b/33063-t/images/sources/140aa178.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}(7,7)
+
+% r = 2.5
+% Ellipse:  u = 3.5  v = 3.5  a = 2.5  b = 2.5  phi = 0.0 Grad
+\qbezier[15](6.0, 5.5)(6.0, 6.5355)(5.2678, 7.2678)
+\qbezier[15](5.2678, 7.2678)(4.5355, 8.0)(3.5, 8.0)
+\qbezier[15](3.5, 8.0)(2.4645, 8.0)(1.7322, 7.2678)
+\qbezier[15](1.7322, 7.2678)(1.0, 6.5355)(1.0, 5.5)
+\qbezier[15](1.0, 5.5)(1.0, 4.4645)(1.7322, 3.7322)
+\qbezier[15](1.7322, 3.7322)(2.4645, 3.0)(3.5, 3.0)
+\qbezier[15](3.5, 3.0)(4.5355, 3.0)(5.2678, 3.7322)
+\qbezier[15](5.2678, 3.7322)(6.0, 4.4645)(6.0, 5.5)
+
+\drawline(1,5.5)(6,5.5) % AOB
+
+% C = 4.82, 5.623
+% D = 4.82, 1.377
+\dashline[80]{0.2}(6,5.5)(4.82,7.623)(3.5,5.5)(4.82,3.377)(6,5.5) % BCODB
+\dashline[80]{0.4}(1,5.5)(4.82,7.623)(4.82,3.377)(1,5.5) % ACDA
+
+\put( 0.3, 5.2){$\scriptstyle A$}
+\put( 6.0, 5.2){$\scriptstyle B$}
+\put( 4.6, 7.7){$\scriptstyle C$}
+\put( 4.4, 2.6){$\scriptstyle D$}
+\put( 2.9, 4.8){$\scriptstyle O$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/140bb182.eepic b/33063-t/images/sources/140bb182.eepic
new file mode 100644
index 0000000..7893016
--- /dev/null
+++ b/33063-t/images/sources/140bb182.eepic
@@ -0,0 +1,21 @@
+\PGset[0.8em]
+\begin{picture}(11,6)
+
+\dashline[80]{0.2}(1,1)(11,1)(6,4.8)(1,1) % EFCE
+\dashline[80]{0.2}(6,4.8)(6,1) % CD
+
+% midEC = 3.5, 2.9
+% mEC = 0.76
+\dashline[80]{0.2}(3.5,2.9)(4.944,1) % A
+\dashline[80]{0.4}(4.944,1)(6,4.8)(7.056,1) % ACB
+\dashline[80]{0.2}(7.056,1)(8.5,2.9) % B
+
+\put( 4.5, 0.2){$\scriptstyle A$}
+\put( 6.7, 0.2){$\scriptstyle B$}
+\put( 5.8, 5.0){$\scriptstyle C$}
+\put( 5.6, 0.2){$\scriptstyle D$}
+\put( 0.8, 0.2){$\scriptstyle E$}
+\put(10.4, 0.2){$\scriptstyle F$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/140cc191.eepic b/33063-t/images/sources/140cc191.eepic
new file mode 100644
index 0000000..b72c94a
--- /dev/null
+++ b/33063-t/images/sources/140cc191.eepic
@@ -0,0 +1,47 @@
+\PGset[0.8em]
+\begin{picture}(10,10)
+
+% r1 = 4
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier[20](9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier[20](7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier[20](5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier[20](2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier[20](1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier[20](2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier[20](5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier[20](7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+\dashline[80]{0.2}(5,9)(1.536,3)(8.464,3)(5,9) % ABCA
+
+% Ellipse:  u = 5.0  v = 5.0  a = 1.2  b = 1.2  phi = 0.0 Grad
+\qbezier[10](6.2, 5.0)(6.2, 5.4971)(5.8485, 5.8485)
+\qbezier[10](5.8485, 5.8485)(5.4971, 6.2)(5.0, 6.2)
+\qbezier[10](5.0, 6.2)(4.5029, 6.2)(4.1515, 5.8485)
+\qbezier[10](4.1515, 5.8485)(3.8, 5.4971)(3.8, 5.0)
+\qbezier[10](3.8, 5.0)(3.8, 4.5029)(4.1515, 4.1515)
+\qbezier[10](4.1515, 4.1515)(4.5029, 3.8)(5.0, 3.8)
+\qbezier[10](5.0, 3.8)(5.4971, 3.8)(5.8485, 4.1515)
+\qbezier[10](5.8485, 4.1515)(6.2, 4.5029)(6.2, 5.0)
+
+\dashline[80]{0.4}(5,7.75)(6.914,3.8)(3.086,3.8)(5,7.75) % PRQP
+
+\dashline[80]{0.2}(5,5)(8.6,6.746) %OF m = .485
+\dashline[80]{0.2}(5,5)(1.4,6.746) %OD
+\dashline[80]{0.2}(5,5)(5,1) % OE
+
+
+\put( 4.7, 9.1){$\scriptstyle A$}
+\put( 0.8, 2.4){$\scriptstyle B$}
+\put( 8.5, 2.4){$\scriptstyle C$}
+\put( 2.8, 6.2){$\scriptstyle D$}
+\put( 5.0, 2.2){$\scriptstyle E$}
+\put( 6.6, 6.2){$\scriptstyle F$}
+\put( 4.7, 7.8){$\scriptstyle P$}
+\put( 2.4, 3.5){$\scriptstyle Q$}
+\put( 6.8, 3.5){$\scriptstyle R$}
+\put( 4.7, 5.2){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/141aa206.eepic b/33063-t/images/sources/141aa206.eepic
new file mode 100644
index 0000000..f44d2f2
--- /dev/null
+++ b/33063-t/images/sources/141aa206.eepic
@@ -0,0 +1,17 @@
+\PGset[0.8em]
+\begin{picture}(8,6)
+
+\dashline[80]{0.2}(3.5,6.7)(1,6.7)(5.2,5) % AEB
+\drawline(3.5,6.7)(6.9,6.7) % AC
+
+\dashline[80]{0.4}(3.5,6.7)(5.2,5)(6.9,6.7)(4.9,8.7) % ABCD
+\dashline[80]{0.4}(5.5,8.7)(3.5,6.7) % DA
+
+\put( 2.9, 6.8){$\scriptstyle A$}
+\put( 4.8, 4.2){$\scriptstyle B$}
+\put( 7.0, 6.4){$\scriptstyle C$}
+\put( 4.9, 8.7){$\scriptstyle D$}
+\put( 0.5, 6.8){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/141bb207.eepic b/33063-t/images/sources/141bb207.eepic
new file mode 100644
index 0000000..c2272d5
--- /dev/null
+++ b/33063-t/images/sources/141bb207.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(11,13)
+
+\drawline(1,6.5)(10,6.5) % AB
+\drawline(3,1)(3,12) % DC
+\dashline[80]{0.4}(7.57,6.5)(7.57,1.93)(3,1.93) % PHQ
+\dashline[80]{0.2}(7.57,1.93)(3,6.5)(7,10.5) % HO(G)
+
+\drawline(3,10.75)(7.57,1.93) % FH m=-1.93
+
+% G = 4.451, 7.95
+\dashline[80]{0.2}(3,7.95)(4.451,7.95)(4.451,6.5) % NGM
+
+\put( 0.3, 6.2){$\scriptstyle A$}
+\put(10.0, 6.2){$\scriptstyle B$}
+\put( 2.7,12.1){$\scriptstyle C$}
+\put( 2.6, 0.2){$\scriptstyle D$}
+\put( 5.2, 6.7){$\scriptstyle E$}
+\put( 2.2,10.4){$\scriptstyle F$}
+\put( 4.2, 8.3){$\scriptstyle G$}
+\put( 4.0, 5.8){$\scriptstyle M$}
+\put( 2.2, 7.6){$\scriptstyle N$}
+\put( 2.2, 5.8){$\scriptstyle O$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/142aa223.eepic b/33063-t/images/sources/142aa223.eepic
new file mode 100644
index 0000000..1d0f9b5
--- /dev/null
+++ b/33063-t/images/sources/142aa223.eepic
@@ -0,0 +1,31 @@
+\PGset[0.8em]
+\begin{picture}(7,7)
+
+% r = 2.5
+% Ellipse:  u = 3.5  v = 3.5  a = 2.5  b = 2.5  phi = 0.0 Grad
+\qbezier[15](6.0, 3.5)(6.0, 4.5355)(5.2678, 5.2678)
+\qbezier[15](5.2678, 5.2678)(4.5355, 6.0)(3.5, 6.0)
+\qbezier[15](3.5, 6.0)(2.4645, 6.0)(1.7322, 5.2678)
+\qbezier[15](1.7322, 5.2678)(1.0, 4.5355)(1.0, 3.5)
+\qbezier[15](1.0, 3.5)(1.0, 2.4645)(1.7322, 1.7322)
+\qbezier[15](1.7322, 1.7322)(2.4645, 1.0)(3.5, 1.0)
+\qbezier[15](3.5, 1.0)(4.5355, 1.0)(5.2678, 1.7322)
+\qbezier[15](5.2678, 1.7322)(6.0, 2.4645)(6.0, 3.5)
+
+\dashline[80]{0.2}(3.5,3.5)(5.235,5.3) % OC
+\dashline[80]{0.4}(1.209,2.5)(5.791,2.5)(5.235,5.3)(1.765,5.3)(1.209,2.5) % ABCDA
+\dashline[80]{0.2}(3.5,6)(3.5,1) % diameter
+\dashline[80]{0.2}(5.235,5.3)(5.235,2.5) % CG
+
+\put( 0.5, 1.8){$\scriptstyle A$}
+\put( 5.7, 1.8){$\scriptstyle B$}
+\put( 5.3, 5.4){$\scriptstyle C$}
+\put( 1.2, 5.4){$\scriptstyle D$}
+\put( 3.6, 1.8){$\scriptstyle E$}
+\put( 2.8, 4.6){$\scriptstyle F$}
+\put( 4.6, 1.8){$\scriptstyle G$}
+\put( 2.8, 3.2){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/143aa241.eepic b/33063-t/images/sources/143aa241.eepic
new file mode 100644
index 0000000..a888ecd
--- /dev/null
+++ b/33063-t/images/sources/143aa241.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(7,6)
+
+%% 143aa241
+\drawline(1,1.2)(3,4.9) % AB m=1.85
+\drawline(6.6,1)(5,4.4) % CD m=-2.125
+
+\dashline[80]{0.2}(3.75,1.2)(4.88,3.8) % FE m=2.3
+\dashline[80]{0.2}(1.432,2)(6.129,2) % IG
+
+% H = 4.098,2
+
+\dashline[80]{0.4}(3.78,1)(3.78,5) % bisector
+
+
+\put( 1.1, 0.8){$\scriptstyle A$}
+\put( 2.8, 5.0){$\scriptstyle B$}
+\put( 5.6, 0.8){$\scriptstyle C$}
+\put( 4.7, 4.5){$\scriptstyle D$}
+\put( 4.2, 3.7){$\scriptstyle E$}
+\put( 2.9, 0.8){$\scriptstyle F$}
+\put( 6.2, 1.7){$\scriptstyle G$}
+\put( 4.3, 2.1){$\scriptstyle H$}
+\put( 1.8, 2.1){$\scriptstyle I$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/143bb242.eepic b/33063-t/images/sources/143bb242.eepic
new file mode 100644
index 0000000..14e9f7f
--- /dev/null
+++ b/33063-t/images/sources/143bb242.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(10,8)
+
+\drawline(10,1)(1,1)(6,6.5) % BAC mAC = 1.1
+% P = 5.18,  2.46
+% G = 3.853, 1
+% D = 2.327, 2.46
+
+\dashline[80]{0.2}(3.853,1)(5.18,2.46)(2.327,2.46) % GPD
+
+% F = 6.706, 1, mPF = -.957
+% E = 3.654, 3.92
+\dashline[80]{0.2}(6.706,1)(3.654,3.92)
+
+
+\put( 0.6, 0.2){$\scriptstyle A$}
+\put( 9.3, 0.2){$\scriptstyle B$}
+\put( 5.2, 6.3){$\scriptstyle C$}
+\put( 1.5, 2.3){$\scriptstyle D$}
+\put( 2.9, 3.9){$\scriptstyle E$}
+\put( 6.4, 0.2){$\scriptstyle F$}
+\put( 3.4, 0.2){$\scriptstyle G$}
+\put( 5.3, 2.4){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/143cc243.eepic b/33063-t/images/sources/143cc243.eepic
new file mode 100644
index 0000000..c4dfd89
--- /dev/null
+++ b/33063-t/images/sources/143cc243.eepic
@@ -0,0 +1,20 @@
+\PGset[0.8em]
+\begin{picture}(10,8)
+
+\drawline(1,3.5)(10,3.5) % AB
+
+\dashline[80]{0.2}(3.5,6)(3.5,1)(8,7.86) % PDQ mDQ = 1.524
+
+\dashline[80]{0.2}(3.5,6)(5.14,3.5) % PE
+
+
+\put( 0.7, 2.7){$\scriptstyle A$}
+\put( 9.3, 2.7){$\scriptstyle B$}
+\put( 2.6, 2.7){$\scriptstyle C$}
+\put( 3.1, 0.2){$\scriptstyle D$}
+\put( 5.1, 2.7){$\scriptstyle E$}
+\put( 3.2, 6.1){$\scriptstyle P$}
+\put( 8.0, 7.2){$\scriptstyle Q$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/143dd244.eepic b/33063-t/images/sources/143dd244.eepic
new file mode 100644
index 0000000..5d6dd63
--- /dev/null
+++ b/33063-t/images/sources/143dd244.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+\begin{picture}(10,8)
+
+\drawline(1,3)(10,3) % AB
+
+\dashline[80]{0.2}(2.5,5)(2.5,1)(9.7,6) % PDQ mDQ = .694
+
+\dashline[80]{0.2}(2.5,5)(5.38,3) % PE
+\dashline[80]{0.2}(2.5,1)(8,3)(9.7,6) % DFQ
+\dashline[80]{0.2}(2.5,5)(8,3) % PF
+
+
+\put( 0.6, 2.2){$\scriptstyle A$}
+\put( 9.4, 2.2){$\scriptstyle B$}
+\put( 1.7, 2.2){$\scriptstyle C$}
+\put( 2.1, 0.2){$\scriptstyle D$}
+\put( 5.2, 2.2){$\scriptstyle E$}
+\put( 7.7, 2.2){$\scriptstyle F$}
+\put( 2.2, 5.2){$\scriptstyle P$}
+\put( 9.8, 5.8){$\scriptstyle Q$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/143ee245.eepic b/33063-t/images/sources/143ee245.eepic
new file mode 100644
index 0000000..b99ba86
--- /dev/null
+++ b/33063-t/images/sources/143ee245.eepic
@@ -0,0 +1,117 @@
+\PGset[0.8em]
+\begin{picture}(26,12)
+
+% r  = 3
+% r' = 2
+% r+r' = 5
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier[20](11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier[20](9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier[20](6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier[20](2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier[20](1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier[20](2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier[20](6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier[20](9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 3.0  b = 3.0  phi = 0.0 Grad
+\qbezier(9.0, 6.0)(9.0, 7.2426)(8.1213, 8.1213)
+\qbezier(8.1213, 8.1213)(7.2426, 9.0)(6.0, 9.0)
+\qbezier(6.0, 9.0)(4.7574, 9.0)(3.8787, 8.1213)
+\qbezier(3.8787, 8.1213)(3.0, 7.2426)(3.0, 6.0)
+\qbezier(3.0, 6.0)(3.0, 4.7574)(3.8787, 3.8787)
+\qbezier(3.8787, 3.8787)(4.7574, 3.0)(6.0, 3.0)
+\qbezier(6.0, 3.0)(7.2426, 3.0)(8.1213, 3.8787)
+\qbezier(8.1213, 3.8787)(9.0, 4.7574)(9.0, 6.0)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 1.0  b = 1.0  phi = 0.0 Grad
+\qbezier[10](7.0, 6.0)(7.0, 6.4142)(6.7071, 6.7071)
+\qbezier[10](6.7071, 6.7071)(6.4142, 7.0)(6.0, 7.0)
+\qbezier[10](6.0, 7.0)(5.5858, 7.0)(5.2929, 6.7071)
+\qbezier[10](5.2929, 6.7071)(5.0, 6.4142)(5.0, 6.0)
+\qbezier[10](5.0, 6.0)(5.0, 5.5858)(5.2929, 5.2929)
+\qbezier[10](5.2929, 5.2929)(5.5858, 5.0)(6.0, 5.0)
+\qbezier[10](6.0, 5.0)(6.4142, 5.0)(6.7071, 5.2929)
+\qbezier[10](6.7071, 5.2929)(7.0, 5.5858)(7.0, 6.0)
+
+% Ellipse:  u = 14.5  v = 6.0  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier(16.5, 6.0)(16.5, 6.8284)(15.9142, 7.4142)
+\qbezier(15.9142, 7.4142)(15.3284, 8.0)(14.5, 8.0)
+\qbezier(14.5, 8.0)(13.6716, 8.0)(13.0858, 7.4142)
+\qbezier(13.0858, 7.4142)(12.5, 6.8284)(12.5, 6.0)
+\qbezier(12.5, 6.0)(12.5, 5.1716)(13.0858, 4.5858)
+\qbezier(13.0858, 4.5858)(13.6716, 4.0)(14.5, 4.0)
+\qbezier(14.5, 4.0)(15.3284, 4.0)(15.9142, 4.5858)
+\qbezier(15.9142, 4.5858)(16.5, 5.1716)(16.5, 6.0)
+
+% centerline
+\dashline[80]{0.2}(2.0,6)(25,6)
+
+% O =   6.0, 6
+% O' = 14.5, 6
+
+% tangents from O' (14.5,6) to r-r' circle
+% M = 6.118,6.993
+% N = 6.118,5.007
+\dashline[80]{0.2}(6.118,6.993)(14.5,6)(6.118,5.007) % MO'N
+
+% produce OM and ON to meet r circle
+% A = 6.354, 8.979
+% C = 6.354, 3.021
+\dashline[80]{0.2}(6.354,8.979)(6,6)(6.354,3.021) % AOC
+
+% O'B and O'D parallel to OA and OC
+% B = 14.736,7.986
+% D = 14.736,4.014
+\dashline[80]{0.2}(14.736,7.986)(14.5,6)(14.736,4.014) % BO'D
+
+% AB extended to P and r+r' circle m = -0.118
+\dashline[80]{0.2}(2.379,9.448)(25,6.779)
+
+% CD, same
+\dashline[80]{0.2}(2.379,2.552)(25,5.221)
+
+
+% tangents from O' to r+r' circle
+% R = 8.941, 10.043
+% S = 8.941,  1.957
+\dashline[80]{0.2}(8.941,10.043)(14.5,6)(8.941,1.957) % RO'S
+
+\dashline[80]{0.2}(8.941,10.043)(6,6)(8.941,1.957) % ROS  m=1.375
+
+% E =  7.765, 8.426
+% G =  7.765, 3.574
+
+%e1: (y-6)^2+(x-14.5)^2=2^2
+%e2: (y-8.426)/(x-7.765)=-1/((y-6)/(x-14.5))
+%Why won't this solve?  no matter, EO || O'F
+
+% H = 13.324, 7.617
+% F = 13.324, 4.383
+% extended slightly past F and H
+\dashline[80]{0.2}(7.765,8.426)(15.324,2.929) % EF m=-.727
+\dashline[80]{0.2}(7.765,3.574)(15.324,9.017) % GH
+
+\dashline[80]{0.2}(13.324,4.383)(14.5,6) % FO' m=1.375
+\dashline[80]{0.2}(13.324,7.617)(14.5,6) % HO'
+
+% mOE = 1.375
+
+\put( 6.2, 9.2){$\scriptstyle A$}
+\put(14.8, 8.1){$\scriptstyle B$}
+\put( 6.1, 2.3){$\scriptstyle C$}
+\put(14.9, 3.3){$\scriptstyle D$}
+\put( 8.2, 8.2){$\scriptstyle E$}
+\put(12.5, 3.8){$\scriptstyle F$}
+\put( 8.1, 3.4){$\scriptstyle G$}
+\put(12.5, 7.5){$\scriptstyle H$}
+\put( 5.3, 6.1){$\scriptstyle O$}
+\put(14.7, 6.1){$\scriptstyle O'$}
+\put(25.2, 5.7){$\scriptstyle P$}
+\put(10.8, 5.0){$\scriptstyle Q$}
+\put( 8.9,10.1){$\scriptstyle R$}
+\put( 8.9, 1.4){$\scriptstyle S$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/151aa342.eepic b/33063-t/images/sources/151aa342.eepic
new file mode 100644
index 0000000..de576bd
--- /dev/null
+++ b/33063-t/images/sources/151aa342.eepic
@@ -0,0 +1,50 @@
+\PGset[0.8em]
+\begin{picture}(38,11)
+
+% 12x9
+
+\drawline(7,10)(1,1)(13,1)(7,10) % ABCA
+
+%  6/7 = 0.857
+%  9/7 = 1.286
+
+\dashline[80]{0.2}(1.857,2.286)(12.043,2.286)
+\dashline[80]{0.2}(2.714,3.572)(11.186,3.572)
+
+\drawline     (3.571,4.858)(10.429,4.858) % EF
+
+\dashline[80]{0.2}(4.428,6.144)( 9.472,6.144)
+\dashline[80]{0.2}(5.285,7.430)( 8.615,7.430)
+\dashline[80]{0.2}(6.142,8.716)( 7.758,8.716)
+
+% 21x8.357
+% Ax = Bx + 7
+
+\drawline(23,9.357)(16,1)(37,1)(23,9.357) % ABCA
+
+% Ey=Fy=4.858-.643 = 4.215
+% Ex = 19.53, Fx = 
+\drawline(18.693,4.215)(31.614,4.215) % EF
+
+% Ky = 1.643
+\dashline[80]{0.2}(16.538,1.643)(35.923,1.643) % KH
+
+\put(  6.6, 10.1 ){$\scriptstyle A$}
+\put(  0.7,  0.2 ){$\scriptstyle B$}
+\put( 12.4,  0.2 ){$\scriptstyle C$}
+\put(  2.7,  4.5 ){$\scriptstyle E$}
+\put( 10.7,  4.5 ){$\scriptstyle F$}
+\put(  0.7,  2.0 ){$\scriptstyle M$}
+
+\put( 22.6,  9.5 ){$\scriptstyle A$}
+\put( 15.3,  0.7 ){$\scriptstyle B$}
+\put( 37.2,  0.7 ){$\scriptstyle C$}
+\put( 17.9,  4.2 ){$\scriptstyle E$}
+\put( 31.9,  4.2 ){$\scriptstyle F$}
+\put( 15.8,  1.5 ){$\scriptstyle K$}
+\put( 36.2,  1.5 ){$\scriptstyle H$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/152aa344.eepic b/33063-t/images/sources/152aa344.eepic
new file mode 100644
index 0000000..c648f21
--- /dev/null
+++ b/33063-t/images/sources/152aa344.eepic
@@ -0,0 +1,31 @@
+\PGset[0.8em]
+\begin{picture}(11,8)
+
+% 238x185
+% 9x7
+
+\drawline(1,1)(10,1) % BD
+\drawline(1,3.5)(10,3.5) % HK
+\drawline(1,5.5)(10,5.5) % FG
+\drawline(1,7)(10,7) % AC
+
+
+\drawline(3,7)(2,1) % AB
+\drawline(6,7)(8,1) % CD
+\dashline[80]{0.2}(3,7)(5,1) % AN
+
+
+\put(  2.7,  7.1 ){$\scriptstyle A$}
+\put(  1.5,  0.2 ){$\scriptstyle B$}
+\put(  5.7,  7.1 ){$\scriptstyle C$}
+\put(  7.7,  0.2 ){$\scriptstyle D$}
+\put(  2.0,  5.6 ){$\scriptstyle F$}
+\put(  6.6,  5.6 ){$\scriptstyle G$}
+\put(  1.5,  3.6 ){$\scriptstyle H$}
+\put(  7.2,  3.6 ){$\scriptstyle K$}
+\put(  3.6,  5.6 ){$\scriptstyle L$}
+\put(  4.2,  3.6 ){$\scriptstyle M$}
+\put(  4.7,  0.2 ){$\scriptstyle N$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/153aa345.eepic b/33063-t/images/sources/153aa345.eepic
new file mode 100644
index 0000000..20eda1f
--- /dev/null
+++ b/33063-t/images/sources/153aa345.eepic
@@ -0,0 +1,21 @@
+\PGset[0.8em]
+\begin{picture}(21,11)
+
+% 538x257
+% 19x9
+
+\drawline(8,10)(1,1)(20,1)(8,10) % ABCA
+
+\drawline(2.555,3)(17.333,3)        % EF
+\dashline[80]{0.2}(2.555,3)(16.667,3.5) % EH
+
+
+\put(  7.7, 10.1 ){$\scriptstyle A$}
+\put(  0.2,  0.7 ){$\scriptstyle B$}
+\put( 20.1,  0.7 ){$\scriptstyle C$}
+\put(  1.6,  2.8 ){$\scriptstyle E$}
+\put( 17.6,  2.8 ){$\scriptstyle F$}
+\put( 16.8,  3.6 ){$\scriptstyle H$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/154aa346.eepic b/33063-t/images/sources/154aa346.eepic
new file mode 100644
index 0000000..0c5ce06
--- /dev/null
+++ b/33063-t/images/sources/154aa346.eepic
@@ -0,0 +1,17 @@
+\PGset[0.8em]
+\begin{picture}(25,3)
+
+% 23
+
+\dashline[80]{0.2}( 1,1)(12,1) % M'A
+\drawline     (12,1)(24,1) % AB
+\drawline     (12,1)(12,0.6) % A
+\drawline     (16,1)(16,0.6) % M
+
+\put( 11.6, -0.2 ){$\scriptstyle A$}
+\put( 24.1,  0.9 ){$\scriptstyle B$}
+\put( 15.5, -0.2 ){$\scriptstyle M$}
+\put(  0.0,  0.9 ){$\scriptstyle M'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/154bb346.eepic b/33063-t/images/sources/154bb346.eepic
new file mode 100644
index 0000000..d816cf1
--- /dev/null
+++ b/33063-t/images/sources/154bb346.eepic
@@ -0,0 +1,35 @@
+\PGset[0.8em]
+\begin{picture}(24,3)
+
+% AB = 8
+% M' = A - 3*4 = 1
+% A = 13
+% M = 16
+% B = 21
+% y = 23
+
+\drawline(1,1)(23,1) % xy
+
+\drawline(1,1)(1,1.3) % M'
+\drawline(5,1)(5,1.3)
+\drawline(9,1)(9,1.3)
+
+\drawline(13,1)(13,1.3)
+\drawline(14,1)(14,1.3)
+\drawline(15,1)(15,1.3)
+\drawline(16,1)(16,1.6)
+\drawline(17,1)(17,1.3)
+\drawline(18,1)(18,1.3)
+\drawline(19,1)(19,1.3)
+\drawline(20,1)(20,1.3)
+\drawline(21,1)(21,1.3)
+
+\put(  0.4,  0.8 ){$\scriptstyle x$}
+\put(  0.7,  0.2 ){$\scriptstyle M'$}
+\put( 12.6,  0.2 ){$\scriptstyle A$}
+\put( 15.5,  0.2 ){$\scriptstyle M$}
+\put( 20.6,  0.2 ){$\scriptstyle B$}
+\put( 23.1,  0.8 ){$\scriptstyle y$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/155aa348.eepic b/33063-t/images/sources/155aa348.eepic
new file mode 100644
index 0000000..b10aecc
--- /dev/null
+++ b/33063-t/images/sources/155aa348.eepic
@@ -0,0 +1,27 @@
+\PGset[0.8em]
+\begin{picture}(19,9)
+
+% 520x200
+% 19x7
+
+\drawline(4,1)(19,1)(6.5,6)(4,1) % ABCA
+
+% angle C = tan^-1 x1/y + tan^-1 x2/y
+% y=5, x1=2.5, x2=12.5
+% C = 94.764 deg.
+% ACM = 47.382
+% My = 1, Mx = 1.9+6.5 = 5.9
+\drawline(6.5,6)(8.4,1) % CM
+
+%mBC = 0.364 from (6.5,6)
+%mMC = 2.632 from (4,  1)
+\dashline[80]{0.2}(4,1)(1.394,7.859)(6.5,6) % AEC
+
+\put(  3.7,  0.2 ){$\scriptstyle A$}
+\put( 18.5,  0.2 ){$\scriptstyle B$}
+\put(  6.2,  6.2 ){$\scriptstyle C$}
+\put(  1.0,  8.0 ){$\scriptstyle E$}
+\put(  8.0,  0.2 ){$\scriptstyle M$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/156aa349.eepic b/33063-t/images/sources/156aa349.eepic
new file mode 100644
index 0000000..be259a3
--- /dev/null
+++ b/33063-t/images/sources/156aa349.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(18,10)
+
+% 450x238
+% 16x9
+
+% C = 7.5,7
+% A = 11,1
+
+\drawline(11,1)(17,1)(7.5,7)(11,1) % ABCA
+
+% mAF = mM'C = .923, A=11,1
+% mBC = -.6315, B = 17,1
+\dashline[80]{0.2}(1,1)(11,1)(13.437,3.25) % M'AF
+
+% Ex = 4, Ey = 9.21
+\drawline(1,1)(7.5,7)(4,9.21) % M'CE
+
+
+\put( 10.6,  0.2 ){$\scriptstyle A$}
+\put( 17.1,  0.9 ){$\scriptstyle B$}
+\put(  7.6,  7.1 ){$\scriptstyle C$}
+\put(  3.2,  9.0 ){$\scriptstyle E$}
+\put( 13.4,  3.4 ){$\scriptstyle F$}
+\put(  0.0,  0.9 ){$\scriptstyle M'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/157aa351.eepic b/33063-t/images/sources/157aa351.eepic
new file mode 100644
index 0000000..76fe747
--- /dev/null
+++ b/33063-t/images/sources/157aa351.eepic
@@ -0,0 +1,32 @@
+\PGset[0.8em]
+\begin{picture}(11,8)
+
+% 255x177
+% 9x6
+
+\drawline(1,4)(5,7)(10,5)(8,1)(2,1)(1,4) % ABCDEA
+
+\put(  0.2,  3.7 ){$\scriptstyle A$}
+\put(  4.7,  7.1 ){$\scriptstyle B$}
+\put( 10.2,  4.7 ){$\scriptstyle C$}
+\put(  7.2,  0.2 ){$\scriptstyle D$}
+\put(  1.8,  0.2 ){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
+%
+\PGset[0.6em]
+\begin{picture}(11,7)
+
+% 255x177
+% 9x6
+\drawline(1,4)(5,7)(10,5)(8,1)(2,1)(1,4) % ABCDEA
+
+\put(  0.0,  3.7 ){$\scriptstyle A'$}
+\put(  4.7,  7.1 ){$\scriptstyle B'$}
+\put( 10.2,  4.7 ){$\scriptstyle C'$}
+\put(  7.2,  0.0 ){$\scriptstyle D'$}
+\put(  1.8,  0.0 ){$\scriptstyle E'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/157bb353.eepic b/33063-t/images/sources/157bb353.eepic
new file mode 100644
index 0000000..d7f0c44
--- /dev/null
+++ b/33063-t/images/sources/157bb353.eepic
@@ -0,0 +1,16 @@
+\PGset[0.8em]
+\begin{picture}(34,5)
+
+\drawline(1,1)(7.65,1)(7.65,4.25)(1,4.25)(1,1) % Q
+\drawline(9,1)(15.65,1)(16.65,4.25)(10,4.25)(9,1) % Q'
+\drawline(18,1)(24.65,1)(24.65,4.25)(18,4.25)(18,1) % R
+\drawline(26,1)(33.85,1)(33.85,3.4)(26,3.4)(26,1) % R'
+
+\put( 4.0, 2.2 ){$\scriptstyle Q$}
+\put(12.5, 2.2 ){$\scriptstyle Q'$}
+\put(21.0, 2.2 ){$\scriptstyle R$}
+\put(29.5, 1.8 ){$\scriptstyle R'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/158aa354.eepic b/33063-t/images/sources/158aa354.eepic
new file mode 100644
index 0000000..133f8a3
--- /dev/null
+++ b/33063-t/images/sources/158aa354.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(23,10)
+
+% 284x216
+% 10x8
+
+\drawline(7,9)(1,1)(11,1)(7,9) % ABCA
+
+\dashline[80]{0.2}(2.125,2.5)(10.25,2.5) % EH
+
+% 1.5 shorter
+
+\drawline(18.875,7.5)(14,1)(22.125,1)(18.875,7.5) % ABCA
+
+\put(  6.6,  9.1 ){$\scriptstyle A$}
+\put(  0.3,  0.8 ){$\scriptstyle B$}
+\put( 11.1,  0.8 ){$\scriptstyle C$}
+\put(  1.3,  2.3 ){$\scriptstyle E$}
+\put( 10.4,  2.3 ){$\scriptstyle H$}
+
+\put( 18.5,  7.6 ){$\scriptstyle A'$}
+\put( 13.1,  0.8 ){$\scriptstyle B'$}
+\put( 22.2,  0.8 ){$\scriptstyle C'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/159aa357.eepic b/33063-t/images/sources/159aa357.eepic
new file mode 100644
index 0000000..133f8a3
--- /dev/null
+++ b/33063-t/images/sources/159aa357.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(23,10)
+
+% 284x216
+% 10x8
+
+\drawline(7,9)(1,1)(11,1)(7,9) % ABCA
+
+\dashline[80]{0.2}(2.125,2.5)(10.25,2.5) % EH
+
+% 1.5 shorter
+
+\drawline(18.875,7.5)(14,1)(22.125,1)(18.875,7.5) % ABCA
+
+\put(  6.6,  9.1 ){$\scriptstyle A$}
+\put(  0.3,  0.8 ){$\scriptstyle B$}
+\put( 11.1,  0.8 ){$\scriptstyle C$}
+\put(  1.3,  2.3 ){$\scriptstyle E$}
+\put( 10.4,  2.3 ){$\scriptstyle H$}
+
+\put( 18.5,  7.6 ){$\scriptstyle A'$}
+\put( 13.1,  0.8 ){$\scriptstyle B'$}
+\put( 22.2,  0.8 ){$\scriptstyle C'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/160aa358.eepic b/33063-t/images/sources/160aa358.eepic
new file mode 100644
index 0000000..133f8a3
--- /dev/null
+++ b/33063-t/images/sources/160aa358.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(23,10)
+
+% 284x216
+% 10x8
+
+\drawline(7,9)(1,1)(11,1)(7,9) % ABCA
+
+\dashline[80]{0.2}(2.125,2.5)(10.25,2.5) % EH
+
+% 1.5 shorter
+
+\drawline(18.875,7.5)(14,1)(22.125,1)(18.875,7.5) % ABCA
+
+\put(  6.6,  9.1 ){$\scriptstyle A$}
+\put(  0.3,  0.8 ){$\scriptstyle B$}
+\put( 11.1,  0.8 ){$\scriptstyle C$}
+\put(  1.3,  2.3 ){$\scriptstyle E$}
+\put( 10.4,  2.3 ){$\scriptstyle H$}
+
+\put( 18.5,  7.6 ){$\scriptstyle A'$}
+\put( 13.1,  0.8 ){$\scriptstyle B'$}
+\put( 22.2,  0.8 ){$\scriptstyle C'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/161aa359.eepic b/33063-t/images/sources/161aa359.eepic
new file mode 100644
index 0000000..7c8d963
--- /dev/null
+++ b/33063-t/images/sources/161aa359.eepic
@@ -0,0 +1,63 @@
+\PGset[0.8em]
+\begin{picture}(30,11)
+
+% 313x180
+% 11x7(9)
+
+\drawline(1,1)(12,1)(7,8)(1,1) % ABCA
+
+% (y-10)/(x-12) = 1.167
+% (y-10)/(x-2)  = -1.4
+\drawline(12,10)(2,10)(6.546,3.635)(12,10) % A'B'C'A'
+
+\dashline[80]{0.2}(7,8)(5.571,10) % Cx
+\dashline[80]{0.2}(7,8)(8.714,10) % Cy
+
+\put( 0.3,  0.8 ){$\scriptstyle A$}
+\put(12.0,  0.8 ){$\scriptstyle B$}
+\put( 6.0,  7.6 ){$\scriptstyle C$}
+
+\put(12.0,  9.8 ){$\scriptstyle A'$}
+\put( 1.3,  9.8 ){$\scriptstyle B'$}
+\put( 6.1,  2.8 ){$\scriptstyle C'$}
+
+\put( 6.2,  9.4 ){$\scriptstyle x$}
+\put( 7.8,  9.4 ){$\scriptstyle y$}
+
+% 322x241
+% 12x9
+
+% 143x65
+% 5x2
+
+\drawline(22,9.5)(17,3)(29,3)(22,9.5) % D'E'F'D'
+
+% (y-6)/(x-23) = -.769
+% (y-1)/(x-23) = 1.077
+\drawline(25.709,3.917)(23,6)(23,1)(25.709,3.917) % DEFD
+
+% EH
+% (y-6)/(x-23) = -.769
+% (y-9.5)/(x-22) = 1.3
+\dashline[80]{0.2}(23,6)(20.68,7.784)
+
+% DK
+% (y-1)/(x-23) = 1.077
+% (y-9.5)/(x-22) = -.929
+\dashline[80]{0.2}(25.709,3.917)(26.774,5.065)
+
+\put(25.8,  3.5 ){$\scriptstyle D$}
+\put(22.2,  5.3 ){$\scriptstyle E$}
+\put(22.6,  0.2 ){$\scriptstyle F$}
+
+\put(21.5,  9.6 ){$\scriptstyle D'$}
+\put(16.0,  2.6 ){$\scriptstyle E'$}
+\put(29.0,  2.6 ){$\scriptstyle F'$}
+
+\put(20.0,  7.6 ){$\scriptstyle H$}
+\put(26.8,  5.0 ){$\scriptstyle K$}
+\put(22.2,  2.2 ){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/162aa361.eepic b/33063-t/images/sources/162aa361.eepic
new file mode 100644
index 0000000..dd718df
--- /dev/null
+++ b/33063-t/images/sources/162aa361.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(30,9)
+
+% 10(7.5)x4
+
+\drawline(1,1)(11,1)(8.5,5)(1,1) % ABCA
+\drawline(8.5,5)(8.5,1) % CO
+
+% 422x183
+% 15(11.25)x6.5
+
+\drawline(15,1)(30,1)(26.25,7.5)(15,1) % A'B'C'A'
+\drawline(26.25,7.5)(26.25,1) % C'O'
+
+
+\put( 0.8,  0.2 ){$\scriptstyle A$}
+\put(10.4,  0.2 ){$\scriptstyle B$}
+\put( 8.2,  5.2 ){$\scriptstyle C$}
+\put( 8.2,  0.2 ){$\scriptstyle O$}
+
+\put(14.8,  0.2 ){$\scriptstyle A'$}
+\put(29.4,  0.2 ){$\scriptstyle B'$}
+\put(25.9,  7.7 ){$\scriptstyle C'$}
+\put(25.9,  0.2 ){$\scriptstyle O'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/163aa362.eepic b/33063-t/images/sources/163aa362.eepic
new file mode 100644
index 0000000..dc7b1a5
--- /dev/null
+++ b/33063-t/images/sources/163aa362.eepic
@@ -0,0 +1,56 @@
+\PGset[0.8em]
+\begin{picture}(29,11)
+
+% 374x273
+% 13x10
+
+\drawline(1,1)(14,1) % AE
+\drawline(2,4)(13,4) % A'E'
+
+\drawline(8,11)(1,1)
+\drawline(8,11)(4,1)
+\drawline(8,11)(7,1)
+\drawline(8,11)(11,1)
+\drawline(8,11)(14,1)
+
+\put( 0.7,  0.2 ){$\scriptstyle A$}
+\put( 3.7,  0.2 ){$\scriptstyle B$}
+\put( 6.6,  0.2 ){$\scriptstyle C$}
+\put(10.6,  0.2 ){$\scriptstyle D$}
+\put(13.5,  0.2 ){$\scriptstyle E$}
+
+\put( 3.0,  3.2 ){$\scriptstyle A'$}
+\put( 5.2,  3.2 ){$\scriptstyle B'$}
+\put( 7.4,  3.2 ){$\scriptstyle C'$}
+\put(10.4,  3.2 ){$\scriptstyle D'$}
+\put(12.8,  3.2 ){$\scriptstyle E'$}
+
+% 350x237
+% 12x9
+
+\drawline(18,1)(29,1) % AE
+\drawline(18,10)(29,10) % E'A'
+
+\drawline(18,1)(29,10) % AA'
+\drawline(21,1)(26,10) % BB'
+\drawline(24,1)(23,10) % CC'
+\drawline(27,1)(20,10) % DD'
+\drawline(29,1)(18,10) % EE'
+
+\put(17.7,  0.2 ){$\scriptstyle A$}
+\put(20.7,  0.2 ){$\scriptstyle B$}
+\put(23.6,  0.2 ){$\scriptstyle C$}
+\put(26.6,  0.2 ){$\scriptstyle D$}
+\put(28.5,  0.2 ){$\scriptstyle E$}
+
+\put(28.7, 10.2 ){$\scriptstyle A'$}
+\put(25.7, 10.2 ){$\scriptstyle B'$}
+\put(22.7, 10.2 ){$\scriptstyle C'$}
+\put(19.7, 10.2 ){$\scriptstyle D'$}
+\put(17.7, 10.2 ){$\scriptstyle E'$}
+
+\put(24.0,  5.2 ){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/164aa363.eepic b/33063-t/images/sources/164aa363.eepic
new file mode 100644
index 0000000..8e697b6
--- /dev/null
+++ b/33063-t/images/sources/164aa363.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(11,12)
+
+% 10x10
+\drawline(1,1)(11,1)    % AE
+\drawline(3.2,5)(9.2,5) % BF
+
+\drawline(3.2,5)(1,1)     % BA
+\drawline(5.5,5)(4.833,1) % DC
+\drawline(9.2,5)(11,1)    % FE
+
+\dashline[80]{0.2}(6.5,11)(3.2,5) % OB
+\dashline[80]{0.2}(6.5,11)(5.5,5) % OD
+\dashline[80]{0.2}(6.5,11)(9.2,5) % OF
+
+
+
+\put( 0.7,  0.2 ){$\scriptstyle A$}
+\put( 2.4,  4.7 ){$\scriptstyle B$}
+\put( 4.4,  0.2 ){$\scriptstyle C$}
+\put( 5.6,  5.1 ){$\scriptstyle D$}
+\put(10.3,  0.2 ){$\scriptstyle E$}
+\put( 9.4,  4.7 ){$\scriptstyle F$}
+\put( 6.3, 11.2 ){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/165aa364.eepic b/33063-t/images/sources/165aa364.eepic
new file mode 100644
index 0000000..f7ddbc5
--- /dev/null
+++ b/33063-t/images/sources/165aa364.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.3,  7.7 ){$\scriptstyle A$}
+\put( 2.7,  0.2 ){$\scriptstyle B$}
+\put(11.2,  0.2 ){$\scriptstyle C$}
+\put(15.0,  6.7 ){$\scriptstyle D$}
+\put( 5.7, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
+\quad\quad
+\PGset[0.5em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.0,  7.5 ){$\scriptstyle A$}
+\put( 2.7, -0.2 ){$\scriptstyle B$}
+\put(10.6, -0.2 ){$\scriptstyle C$}
+\put(15.0,  6.5 ){$\scriptstyle D$}
+\put( 5.5, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/166aa365.eepic b/33063-t/images/sources/166aa365.eepic
new file mode 100644
index 0000000..f7ddbc5
--- /dev/null
+++ b/33063-t/images/sources/166aa365.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.3,  7.7 ){$\scriptstyle A$}
+\put( 2.7,  0.2 ){$\scriptstyle B$}
+\put(11.2,  0.2 ){$\scriptstyle C$}
+\put(15.0,  6.7 ){$\scriptstyle D$}
+\put( 5.7, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
+\quad\quad
+\PGset[0.5em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.0,  7.5 ){$\scriptstyle A$}
+\put( 2.7, -0.2 ){$\scriptstyle B$}
+\put(10.6, -0.2 ){$\scriptstyle C$}
+\put(15.0,  6.5 ){$\scriptstyle D$}
+\put( 5.5, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/167aa366.eepic b/33063-t/images/sources/167aa366.eepic
new file mode 100644
index 0000000..f7ddbc5
--- /dev/null
+++ b/33063-t/images/sources/167aa366.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.3,  7.7 ){$\scriptstyle A$}
+\put( 2.7,  0.2 ){$\scriptstyle B$}
+\put(11.2,  0.2 ){$\scriptstyle C$}
+\put(15.0,  6.7 ){$\scriptstyle D$}
+\put( 5.7, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
+\quad\quad
+\PGset[0.5em]
+\begin{picture}(16,14)
+
+% 378x336
+% 14x12
+
+\drawline(1,8)(3,1)(11.5,1)(15,7)(6,13)(1,8) % ABCDEA
+
+\dashline[80]{0.2}(3,1)(6,13)(11.5,1) % BEC
+
+\put( 0.0,  7.5 ){$\scriptstyle A$}
+\put( 2.7, -0.2 ){$\scriptstyle B$}
+\put(10.6, -0.2 ){$\scriptstyle C$}
+\put(15.0,  6.5 ){$\scriptstyle D$}
+\put( 5.5, 13.2 ){$\scriptstyle E$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/168aa263.eepic b/33063-t/images/sources/168aa263.eepic
new file mode 100644
index 0000000..2200ea9
--- /dev/null
+++ b/33063-t/images/sources/168aa263.eepic
@@ -0,0 +1,38 @@
+\PGset[0.8em]
+\begin{picture}(8,8)
+
+% 6x6
+
+
+% Ellipse:  u = 4.0  v = 4.0  a = 3.0  b = 3.0  phi = 0.0 Grad
+\qbezier(7.0, 4.0)(7.0, 5.2426)(6.1213, 6.1213)
+\qbezier(6.1213, 6.1213)(5.2426, 7.0)(4.0, 7.0)
+\qbezier(4.0, 7.0)(2.7574, 7.0)(1.8787, 6.1213)
+\qbezier(1.8787, 6.1213)(1.0, 5.2426)(1.0, 4.0)
+\qbezier(1.0, 4.0)(1.0, 2.7574)(1.8787, 1.8787)
+\qbezier(1.8787, 1.8787)(2.7574, 1.0)(4.0, 1.0)
+\qbezier(4.0, 1.0)(5.2426, 1.0)(6.1213, 1.8787)
+\qbezier(6.1213, 1.8787)(7.0, 2.7574)(7.0, 4.0) 
+
+% 210, 285, 45, 130
+\drawline(1.402,2.500)(4.776,1.102)(6.121,6.121)(2.072,6.298)(1.402,2.500) % ABCDA
+
+\drawline(1.402,2.500)(6.121,6.121) %AC
+\drawline(4.776,1.102)(2.072,6.298) %BD
+
+% angle ADB = tan^-1 .67/3.798 + tan^-1 2.704/5.196 = 10+27.5 = 37.5 deg
+% mAC = .767
+% (y-2.5)/(x-1.402) = .767
+% (y-6.298)/(x-2.072) = -.839
+\dashline[80]{0.2}(2.072,6.298)(4.117,4.582) % DE
+
+
+\put( 0.8, 1.9 ){$\scriptstyle A$}
+\put( 4.5, 0.3 ){$\scriptstyle B$}
+\put( 6.2, 6.0 ){$\scriptstyle C$}
+\put( 1.4, 6.3 ){$\scriptstyle D$}
+
+\put( 4.2, 4.1 ){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/169aa367.eepic b/33063-t/images/sources/169aa367.eepic
new file mode 100644
index 0000000..03a154d
--- /dev/null
+++ b/33063-t/images/sources/169aa367.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+\begin{picture}(16,9)
+
+% 312x206
+% 11x7
+\drawline(1,1)(12,1)(12,8)(1,1) % AFCA
+
+% -1/mAC = -1.571
+\drawline(12,1)(16.456,1)(12,8) % FBC
+
+
+\put( 0.7, 0.2 ){$\scriptstyle A$}
+\put(15.8, 0.2 ){$\scriptstyle B$}
+\put(11.7, 8.1 ){$\scriptstyle C$}
+\put(11.7, 0.2 ){$\scriptstyle F$}
+
+\put(12.2, 6.5 ){$\scriptstyle a$}
+\put(15.5, 1.2 ){$\scriptstyle b$}
+\put( 2.4, 1.2 ){$\scriptstyle a'$}
+\put(11.0, 6.5 ){$\scriptstyle b'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/170aa370.eepic b/33063-t/images/sources/170aa370.eepic
new file mode 100644
index 0000000..d495b09
--- /dev/null
+++ b/33063-t/images/sources/170aa370.eepic
@@ -0,0 +1,25 @@
+\PGset[0.8em]
+\begin{picture}(12,7)
+
+% 312x206
+% 11x7
+% Ellipse:  u = 6.0  v = 1.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 1.0)(11.0, 3.0711)(9.5355, 4.5355)
+\qbezier(9.5355, 4.5355)(8.0711, 6.0)(6.0, 6.0)
+\qbezier(6.0, 6.0)(3.9289, 6.0)(2.4645, 4.5355)
+\qbezier(2.4645, 4.5355)(1.0, 3.0711)(1.0, 1.0)
+
+\drawline(1,1)(11,1) % AB
+
+\drawline(4.14,1)(4.14,5.641) % DC
+
+\dashline[80]{0.2}(4.14,5.641)(1,1) % CA
+\dashline[80]{0.2}(4.14,5.641)(11,1) % CB
+
+\put( 0.6, 0.2 ){$\scriptstyle A$}
+\put(11.1, 0.7 ){$\scriptstyle B$}
+\put( 3.8, 6.0 ){$\scriptstyle C$}
+\put( 3.8, 0.2 ){$\scriptstyle D$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/171aa371.eepic b/33063-t/images/sources/171aa371.eepic
new file mode 100644
index 0000000..c2b2cc0
--- /dev/null
+++ b/33063-t/images/sources/171aa371.eepic
@@ -0,0 +1,19 @@
+\PGset[0.8em]
+\begin{picture}(16.5,9)
+
+% 312x206
+% 11x7
+\drawline(1,1)(16.456,1)(12,8)(1,1) % ABCA
+
+% -1/mAC = -1.571
+\dashline[80]{0.2}(12,1)(12,8) % FC
+
+
+\put( 0.7, 0.2 ){$\scriptstyle A$}
+\put(15.8, 0.2 ){$\scriptstyle B$}
+\put(11.7, 8.1 ){$\scriptstyle C$}
+\put(11.7, 0.2 ){$\scriptstyle F$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/171bb373.eepic b/33063-t/images/sources/171bb373.eepic
new file mode 100644
index 0000000..5eb46c4
--- /dev/null
+++ b/33063-t/images/sources/171bb373.eepic
@@ -0,0 +1,14 @@
+\PGset[0.8em]
+\begin{picture}( 7,8)
+
+\drawline(1,1)(6.6,1)(6.6,6.6)(1,6.6)(1,1)
+\drawline(1,1)(6.6,6.6)
+
+\put( 0.7, 0.2 ){$\scriptstyle A$}
+\put( 6.0, 0.2 ){$\scriptstyle B$}
+\put( 6.0, 6.7 ){$\scriptstyle C$}
+\put( 0.7, 6.7 ){$\scriptstyle D$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/171cc374.eepic b/33063-t/images/sources/171cc374.eepic
new file mode 100644
index 0000000..03d8e04
--- /dev/null
+++ b/33063-t/images/sources/171cc374.eepic
@@ -0,0 +1,18 @@
+\PGset[0.8em]
+\begin{picture}(14,8)
+
+\drawline(1,1)(13,1) % AB
+\drawline(3.14,6.5)(10.2,3.7) % CD
+\dashline[80]{0.2}(3.14,6.5)(3.14,1)
+\dashline[80]{0.2}(10.2,3.7)(10.2,1)
+
+\put( 0.2, 0.7 ){$\scriptstyle A$}
+\put(13.1, 0.7 ){$\scriptstyle B$}
+\put( 2.8, 6.6 ){$\scriptstyle C$}
+\put( 9.8, 3.9 ){$\scriptstyle D$}
+\put( 2.8, 0.2 ){$\scriptstyle P$}
+\put( 9.8, 0.2 ){$\scriptstyle R$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/172aa375.eepic b/33063-t/images/sources/172aa375.eepic
new file mode 100644
index 0000000..591b7e2
--- /dev/null
+++ b/33063-t/images/sources/172aa375.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(30.5,12)
+
+% 300x262
+% 11x9
+\drawline(6.5,10)(1,1)(12,1)(6.5,10) % ABCA
+
+\dashline[80]{0.2}(6.5,10)(6.5,1) % AD
+
+
+% 436x290
+% 15(5.5)x10
+\drawline(15,11)(20.5,1)(30,1)(15,11) % ABCA
+
+\dashline[80]{0.2}(15,11)(15,1)(20.5,1) % ADB
+
+
+\put( 6.2,10.1 ){$\scriptstyle A$}
+\put( 0.3, 0.8 ){$\scriptstyle B$}
+\put(12.1, 0.8 ){$\scriptstyle C$}
+\put( 6.1, 0.2 ){$\scriptstyle D$}
+
+\put(14.7,11.1 ){$\scriptstyle A$}
+\put(20.2, 0.2 ){$\scriptstyle B$}
+\put(29.7, 0.2 ){$\scriptstyle C$}
+\put(14.7, 0.2 ){$\scriptstyle D$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/173aa376.eepic b/33063-t/images/sources/173aa376.eepic
new file mode 100644
index 0000000..431b257
--- /dev/null
+++ b/33063-t/images/sources/173aa376.eepic
@@ -0,0 +1,18 @@
+\PGset[0.8em]
+\begin{picture}(17,12)
+
+% 436x290
+% 15(5.5)x10
+\drawline(1,11)(6.5,1)(16,1)(1,11) % ACBA
+
+\dashline[80]{0.2}(1,11)(1,1)(6.5,1) % ADC
+
+
+\put(0.7,11.1 ){$\scriptstyle A$}
+\put(6.2, 0.2 ){$\scriptstyle C$}
+\put(16.1, 0.8 ){$\scriptstyle B$}
+\put(0.7, 0.2 ){$\scriptstyle D$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/174aa377.eepic b/33063-t/images/sources/174aa377.eepic
new file mode 100644
index 0000000..1cf2a3c
--- /dev/null
+++ b/33063-t/images/sources/174aa377.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+\begin{picture}(22,8)
+
+% 575x176
+% 20x6
+\drawline(14,7)(1,1)(21,1)(14,7) % ABCA
+
+\drawline(14,7)(11,1) % AM
+
+
+\dashline[80]{0.2}(14,7)(14,1) % AD
+
+
+
+\put(13.7, 7.1 ){$\scriptstyle A$}
+\put( 0.2, 0.7 ){$\scriptstyle B$}
+\put(21.1, 0.7 ){$\scriptstyle C$}
+\put(13.7, 0.2 ){$\scriptstyle D$}
+\put(10.5, 0.2 ){$\scriptstyle M$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/175aa378.eepic b/33063-t/images/sources/175aa378.eepic
new file mode 100644
index 0000000..bec00bb
--- /dev/null
+++ b/33063-t/images/sources/175aa378.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\drawline(1.076,5.132)(9.536,9.536) % MN
+\drawline(4.290,10.698)(10.981,5.564) % PQ
+
+\dashline[80]{0.2}(1.076,5.132)(4.290,10.698) % MP
+\dashline[80]{0.2}(9.536,9.536)(10.981,5.564) % NQ
+
+\put( 0.1, 4.8 ){$\scriptstyle M$}
+\put( 9.6, 9.6 ){$\scriptstyle N$}
+\put( 6.8, 7.4 ){$\scriptstyle O$}
+\put( 3.8,10.8 ){$\scriptstyle P$}
+\put(11.2, 5.3 ){$\scriptstyle Q$}
+
+\put(10.1, 6.3 ){$\scriptstyle a$}
+\put( 9.2, 8.7 ){$\scriptstyle c$}
+\put( 1.8, 5.8 ){$\scriptstyle a'$}
+\put( 4.0, 9.5 ){$\scriptstyle c'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/176aa381.eepic b/33063-t/images/sources/176aa381.eepic
new file mode 100644
index 0000000..b92516a
--- /dev/null
+++ b/33063-t/images/sources/176aa381.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(12,14)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% D at 30deg = 10.33, 8.5
+% mAD = -1.732
+\drawline(10.33,8.5)(7.608,13.214)(1.076,5.132) % DAC
+
+% mAC = 1.237
+\dashline[80]{0.2}(1.076,5.132)(10.33,8.5)(5.815,10.997) % CDB
+
+
+
+\put( 7.3,13.3 ){$\scriptstyle A$}
+\put( 5.3,11.1 ){$\scriptstyle B$}
+\put( 0.3, 4.7 ){$\scriptstyle C$}
+\put(10.4, 8.4 ){$\scriptstyle D$}
+
+\put( 9.3, 9.1 ){$\scriptstyle a$}
+\put( 7.4,12.4 ){$\scriptstyle b$}
+\put( 1.7, 5.5 ){$\scriptstyle a'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/177aa383.eepic b/33063-t/images/sources/177aa383.eepic
new file mode 100644
index 0000000..e05a82a
--- /dev/null
+++ b/33063-t/images/sources/177aa383.eepic
@@ -0,0 +1,35 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier[20](11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier[20](9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier[20](6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier[20](2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier[20](1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier[20](2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier[20](6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier[20](9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+
+\drawline(1.101,7)(10.9,7)(9,10)(1.101,7) % MPNM
+
+% angle MNP = 69.203 + 32.347 = 101.551
+% MNQ = 50.775, mNQ = 3.001
+\dashline[80]{0.2}(9,10)(6,1)(10.9,7) % NQP
+
+
+\put( 0.2, 6.7 ){$\scriptstyle M$}
+\put( 8.9,10.2 ){$\scriptstyle N$}
+\put( 7.0, 6.2 ){$\scriptstyle O$}
+\put(11.0, 6.7 ){$\scriptstyle P$}
+\put( 5.5, 0.2 ){$\scriptstyle Q$}
+
+\put( 6.7, 2.4 ){$\scriptstyle a$}
+\put( 8.9, 9.0 ){$\scriptstyle b$}
+\put( 2.9, 7.1 ){$\scriptstyle a'$}
+\put( 8.1, 9.0 ){$\scriptstyle b'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/178aa384.eepic b/33063-t/images/sources/178aa384.eepic
new file mode 100644
index 0000000..844c57d
--- /dev/null
+++ b/33063-t/images/sources/178aa384.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier[20](11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier[20](9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier[20](6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier[20](2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier[20](1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier[20](2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier[20](6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier[20](9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+
+\drawline(1.101,7)(10.9,7)(9,10)(1.101,7) % MPNM
+
+\drawline(9,7)(9,10)(3,2) % ONP
+
+\dashline[80]{0.2}(3,2)(10.9,7) % PQ
+
+
+\put( 0.2, 6.7 ){$\scriptstyle M$}
+\put( 8.9,10.2 ){$\scriptstyle N$}
+\put( 8.6, 6.2 ){$\scriptstyle O$}
+\put(11.0, 6.7 ){$\scriptstyle Q$}
+\put( 2.2, 1.2 ){$\scriptstyle P$}
+
+\put( 2.9, 7.1 ){$\scriptstyle a$}
+\put( 4.6, 3.4 ){$\scriptstyle a'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/179aa270.eepic b/33063-t/images/sources/179aa270.eepic
new file mode 100644
index 0000000..2c7e9b6
--- /dev/null
+++ b/33063-t/images/sources/179aa270.eepic
@@ -0,0 +1,70 @@
+\PGset[0.8em]
+\begin{picture}(8,6.5)
+
+%   3 
+% 1   2
+
+% A = 0.536, 5.75
+% B = 3.071, 1.714
+% C = 1,     1
+% D = 3,     4.143
+% E = 6.786, 4.643
+
+% 1: C, A, D
+% 2: C, E, D
+% 3: A, E, B
+
+% 1: 1,1;         0.536, 5.75;  3,     4.143
+% 2: 1,1;         6.786, 4.643; 3,     4.143
+% 3: 0.536, 5.75; 6.786, 4.643; 3.071, 1.714
+
+% 1: ( .741, 3.372) +2.386
+% 2: (5.421, 0.395) +4.462
+% 3: (3.602, 4.861) +3.192
+
+% Ellipse:  u = 0.741  v = 3.372  a = 2.386  b = 2.386  phi = 0.0 Grad
+\qbezier(3.127, 3.372)(3.127, 4.3603)(2.4282, 5.0592)
+\qbezier(2.4282, 5.0592)(1.7293, 5.758)(0.741, 5.758)
+\qbezier(0.741, 5.758)(-0.2473, 5.758)(-0.9462, 5.0592)
+\qbezier(0.741, 0.986)(1.7293, 0.986)(2.4282, 1.6848)
+\qbezier(2.4282, 1.6848)(3.127, 2.3837)(3.127, 3.372)
+
+% Ellipse:  u = 5.421  v = 0.395  a = 4.462  b = 4.462  phi = 0.0 Grad
+\qbezier(8.5761, 3.5501)(7.2692, 4.857)(5.421, 4.857)
+\qbezier(5.421, 4.857)(3.5728, 4.857)(2.2659, 3.5501)
+\qbezier(2.2659, 3.5501)(0.959, 2.2432)(0.959, 0.395)
+
+% Ellipse:  u = 3.602  v = 4.861  a = 3.192  b = 3.192  phi = 15.0 Grad
+\qbezier(0.8376, 6.457)(0.1766, 5.312)(0.5188, 4.0348)
+\qbezier(0.5188, 4.0348)(0.861, 2.7577)(2.006, 2.0966)
+\qbezier(2.006, 2.0966)(3.151, 1.4356)(4.4282, 1.7778)
+\qbezier(4.4282, 1.7778)(5.7053, 2.12)(6.3664, 3.265)
+\qbezier(6.3664, 3.265)(7.0274, 4.41)(6.6852, 5.6872)
+
+% re-solving for above circles
+
+% A = 1+3 =  .536, 5.749
+% B = 1+3 = 2.570, 1.840
+% C = 1+2 = 1.000, 1.000
+% D = 1+2 = 2.999, 4.143
+% E = 3+2 = 6.787, 4.643
+% P = 3+2 = 1.476, 2.480
+
+\drawline(0.536, 5.749)(2.570, 1.840) % AB
+\drawline(1.000, 1.000)(2.999, 4.143) % CD
+\drawline(6.787, 4.643)(1.476, 2.480) % EP
+
+
+
+\put(-0.2, 5.8 ){$\scriptstyle A$}
+\put( 2.4, 1.0 ){$\scriptstyle B$}
+\put( 0.2, 0.2 ){$\scriptstyle C$}
+\put( 3.0, 4.5 ){$\scriptstyle D$}
+\put( 6.9, 4.7 ){$\scriptstyle E$}
+\put( 0.6, 2.0 ){$\scriptstyle P$}
+
+\put( 2.2, 2.2 ){$\scriptstyle O$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/180aa280.eepic b/33063-t/images/sources/180aa280.eepic
new file mode 100644
index 0000000..35988c6
--- /dev/null
+++ b/33063-t/images/sources/180aa280.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}( 5, 6)
+
+% 133x100
+% 5x4
+
+\drawline(2.5,4.5)(4,4)(5,1)(1,1)(2.5,4.5) % ABCDA
+
+\drawline(2.5,4.5)(5,1) % AC
+\drawline(4,4)(1,1) % BD
+
+% E = midpoint of AC
+\dashline[80]{0.2}(4,4)(3.75,2.75)(1,1) % BED
+
+% F = midpoint of BD
+\drawline(3.75,2.75)(2.5,2.5) % EF
+
+\put(2.2, 4.6){$\scriptstyle A$}
+\put(3.8, 4.2){$\scriptstyle B$}
+\put(4.5, 0.2){$\scriptstyle C$}
+\put(0.5, 0.2){$\scriptstyle D$}
+\put(3.3, 1.8){$\scriptstyle E$}
+\put(1.8, 2.4){$\scriptstyle F$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/180bb281.eepic b/33063-t/images/sources/180bb281.eepic
new file mode 100644
index 0000000..4e36f5a
--- /dev/null
+++ b/33063-t/images/sources/180bb281.eepic
@@ -0,0 +1,42 @@
+\PGset[0.8em]
+\begin{picture}( 7, 6)
+
+% 133x100
+% 5x4
+
+% Ellipse:  u = 3.0  v = 3.0  a = 2.0  b = 2.0  phi = 0.0 Grad
+\qbezier[20](5.0, 3.0)(5.0, 3.8284)(4.4142, 4.4142)
+\qbezier[20](4.4142, 4.4142)(3.8284, 5.0)(3.0, 5.0)
+\qbezier[20](3.0, 5.0)(2.1716, 5.0)(1.5858, 4.4142)
+\qbezier[20](1.5858, 4.4142)(1.0, 3.8284)(1.0, 3.0)
+\qbezier[20](1.0, 3.0)(1.0, 2.1716)(1.5858, 1.5858)
+\qbezier[20](1.5858, 1.5858)(2.1716, 1.0)(3.0, 1.0)
+\qbezier[20](3.0, 1.0)(3.8284, 1.0)(4.4142, 1.5858)
+\qbezier[20](4.4142, 1.5858)(5.0, 2.1716)(5.0, 3.0)
+
+% y = 1.714
+% Ax = 1.468, Bx = 4.532
+
+% y = 3.464
+% Cx = 4.945
+\drawline(1.468,1.714)(4.532,1.714)(4.945,3.464)(1.468,1.714) % ABCA
+
+\drawline(4.945,3.464)(7.045,4.521) % CH
+
+% CD bisects BCH = 13.279 + 90 + 26.718 = 129.997deg
+% BCD = 64.998 deg, x/y = 1.267, y = 1.75, x = 2.217
+\drawline(4.945,3.464)(7.162,1.714)(4.532,1.714) % CDB
+
+% mCD = -.789
+% mAC = 0.500, mFB = -2.142
+\dashline[80]{0.2}(4.945,3.464)(2.998,5)(4.532,1.714) % CFB
+
+\put(0.9, 1.0){$\scriptstyle A$}
+\put(4.3, 1.0){$\scriptstyle B$}
+\put(4.9, 3.7){$\scriptstyle C$}
+\put(6.8, 1.0){$\scriptstyle D$}
+\put(2.7, 5.2){$\scriptstyle F$}
+\put(7.1, 4.0){$\scriptstyle H$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/181aa385.eepic b/33063-t/images/sources/181aa385.eepic
new file mode 100644
index 0000000..eb8ad3c
--- /dev/null
+++ b/33063-t/images/sources/181aa385.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(19,12)
+
+% m = 7.4
+% n = 5.4
+% p = 3.9
+\drawline(1,3.0)(8.4,3.0) % m
+\drawline(1,2.0)(6.4,2.0) % n
+\drawline(1,1.0)(4.9,1.0) % p
+
+\drawline(1,10.65)(10,10.65) % AB
+
+\dashline[80]{0.2}(1,10.65)(17.5,1) % AX
+
+% mAX = .585
+\dashline[80]{0.2}( 7.387, 6.913)( 4.988,10.65) % CH
+\dashline[80]{0.2}(12.048, 4.187)( 7.898,10.65) % EK
+\dashline[80]{0.2}(15.415, 2.217)(10,    10.65) % FB
+
+
+\put( 0.6,10.8){$\scriptstyle A$}
+\put( 9.7,10.8){$\scriptstyle B$}
+\put( 6.8, 6.2){$\scriptstyle C$}
+\put(11.5, 3.3){$\scriptstyle E$}
+\put(14.8, 1.4){$\scriptstyle F$}
+\put( 4.7,10.8){$\scriptstyle H$}
+\put( 7.6,10.8){$\scriptstyle K$}
+\put(17.6, 0.7){$\scriptstyle X$}
+      
+\put( 3.6, 8.1){$\scriptstyle m$}
+\put( 9.1, 5.0){$\scriptstyle n$}
+\put(13.1, 2.6){$\scriptstyle p$}
+      
+\put( 0.2, 2.7){$\scriptstyle m$}
+\put( 0.3, 1.7){$\scriptstyle n$}
+\put( 0.3, 0.7){$\scriptstyle p$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/182aa386.eepic b/33063-t/images/sources/182aa386.eepic
new file mode 100644
index 0000000..26d52c6
--- /dev/null
+++ b/33063-t/images/sources/182aa386.eepic
@@ -0,0 +1,41 @@
+\PGset[0.8em]
+\begin{picture}(18,12)
+
+% 259x240
+% 9x8.5
+
+\drawline(3,10.5)(16,10.5) % Ax
+\drawline(3,10.5)(12,2) % Ay
+
+% m = 7.3
+% n = 5.4
+% p = 4.3
+\drawline(1,3.0)(8.3,3.0) % m
+\drawline(1,2.0)(6.4,2.0) % n
+\drawline(1,1.0)(5.3,1.0) % p
+
+% mAy = -.944
+% mBD = .707
+\dashline[80]{0.2}(10.3,10.5)(6.127,7.548) % BD
+
+\dashline[80]{0.2}(15.7,10.5)(8.438,5.366) % CF
+      
+\put( 2.6,10.7){$\scriptstyle A$}
+\put(10.0,10.7){$\scriptstyle B$}
+\put(15.4,10.7){$\scriptstyle C$}
+\put( 5.5, 6.8){$\scriptstyle D$}
+\put( 7.8, 4.6){$\scriptstyle F$}
+
+\put(16.2,10.2){$\scriptstyle x$}
+\put(12.2, 1.6){$\scriptstyle y$}
+
+\put( 6.3,10.7){$\scriptstyle m$}
+\put(12.7,10.7){$\scriptstyle n$}
+\put( 3.8, 8.7){$\scriptstyle p$}
+
+\put( 0.2, 2.7){$\scriptstyle m$}
+\put( 0.3, 1.7){$\scriptstyle n$}
+\put( 0.3, 0.7){$\scriptstyle p$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/183aa387.eepic b/33063-t/images/sources/183aa387.eepic
new file mode 100644
index 0000000..61cc09a
--- /dev/null
+++ b/33063-t/images/sources/183aa387.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(21,13)
+
+% m = 5.2
+% n = 6.2
+% m+n = 11.4
+\drawline(14,9)(19.2,9) % m
+\drawline(14,7)(20.2,7) % n
+
+
+% mAC = 1.219
+\dashline[80]{0.2}(1,1)(5.233,11.585)(13.906,1)(1,1) % DAED
+
+\dashline[80]{0.2}(2.931,5.828)(9.956,5.828) % BC
+
+
+
+\put( 4.9,11.7){$\scriptstyle A$}
+\put( 2.2, 5.6){$\scriptstyle B$}
+\put(10.2, 5.6){$\scriptstyle C$}
+\put( 0.2, 0.7){$\scriptstyle D$}
+\put(14.0, 0.7){$\scriptstyle E$}
+
+\put(16.6, 9.2){$\scriptstyle m$}
+\put(17.1, 7.2){$\scriptstyle n$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/184aa388.eepic b/33063-t/images/sources/184aa388.eepic
new file mode 100644
index 0000000..3d80c5f
--- /dev/null
+++ b/33063-t/images/sources/184aa388.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(35,11)
+
+% m =  7.5
+% n = 10.3
+% m+n = 17.8
+\drawline(24,4)(31.5,4) % m
+\drawline(24,2)(34.3,2) % n
+
+\dashline[80]{0.2}(1,1)(20.8,1) % A(B)E
+
+% Ellipse:  u = 9.9  v = 1.0  a = 8.9  b = 8.9  phi = 0.0 Grad
+\qbezier[20](18.8, 1.0)(18.8, 4.6865)(16.1933, 7.2933)
+\qbezier[20](16.1933, 7.2933)(13.5865, 9.9)(9.9, 9.9)
+\qbezier[20](9.9, 9.9)(6.2135, 9.9)(3.6067, 7.2933)
+\qbezier[20](3.6067, 7.2933)(1.0, 4.6865)(1.0, 1.0)
+
+\dashline[80]{0.4}(8.5,1)(8.5,9.789) % CH
+
+\put( 0.3, 0.7){$\scriptstyle A$}
+\put(18.5, 0.2){$\scriptstyle B$}
+\put( 8.2, 0.2){$\scriptstyle C$}
+\put( 8.2,10.0){$\scriptstyle H$}
+\put(20.9, 0.7){$\scriptstyle E$}
+
+\put( 4.2, 0.2){$\scriptstyle m$}
+\put(13.3, 0.2){$\scriptstyle n$}
+
+\put(27.7, 4.2){$\scriptstyle m$}
+\put(29.1, 2.2){$\scriptstyle n$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/185aa390.eepic b/33063-t/images/sources/185aa390.eepic
new file mode 100644
index 0000000..2a76697
--- /dev/null
+++ b/33063-t/images/sources/185aa390.eepic
@@ -0,0 +1,36 @@
+\PGset[0.8em]
+\begin{picture}(35,12)
+
+% AB = 11, BE = 5.5
+% mAE = 0.5, EG = 5.5, AG = 12.298+5.5 = 17.798 (AF = 6.798)
+
+\dashline[80]{0.2}(1,1)(18.798,1) % C'A
+\drawline(18.798,1)(29.798,1) % AB
+
+\dashline[80]{0.2}(29.798,1)(29.798,6.5) % BE
+\dashline[80]{0.2}(18.798,1)(34.717,8.96) % AG
+
+% Ellipse:  u = 29.798  v = 6.5  a = 5.5  b = 5.5  phi = 0.0 Grad
+\qbezier[20](35.298, 6.5)(35.298, 8.7782)(33.6871, 10.3891)
+\qbezier[20](33.6871, 10.3891)(32.0762, 12.0)(29.798, 12.0)
+\qbezier[20](29.798, 12.0)(27.5198, 12.0)(25.9089, 10.3891)
+\qbezier[20](25.9089, 10.3891)(24.298, 8.7782)(24.298, 6.5)
+\qbezier[20](24.298, 6.5)(24.298, 4.2218)(25.9089, 2.6109)
+\qbezier[20](25.9089, 2.6109)(27.5198, 1.0)(29.798, 1.0)
+\qbezier[20](29.798, 1.0)(32.0762, 1.0)(33.6871, 2.6109)
+\qbezier[20](33.6871, 2.6109)(35.298, 4.2218)(35.298, 6.5)
+
+% Ellipse:  u = 18.798  v = 1.0  a = 6.798  b = 6.798  phi = 0.0 Grad
+\qbezier[20](25.596, 1.0)(25.596, 3.8158)(23.6049, 5.8069)
+
+
+\put(18.5, 0.2){$\scriptstyle A$}
+\put(29.5, 0.2){$\scriptstyle B$}
+\put(25.3, 0.2){$\scriptstyle C$}
+\put(29.5, 6.7){$\scriptstyle E$}
+\put(24.3, 3.1){$\scriptstyle F$}
+\put(34.8, 8.8){$\scriptstyle G$}
+\put( 0.2, 0.8){$\scriptstyle C'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/186aa391.eepic b/33063-t/images/sources/186aa391.eepic
new file mode 100644
index 0000000..3618e7a
--- /dev/null
+++ b/33063-t/images/sources/186aa391.eepic
@@ -0,0 +1,37 @@
+\PGset[0.8em]
+\begin{picture}(15,13)
+
+% 354x316
+% 13x11
+
+\drawline(1,7)(3,1)(10.75,1)(14,6.5)(6.5,12)(1,7)
+
+\dashline[80]{0.2}(3,1)(6.5,12)(10.75,1) % BEC
+
+\put( 0.3, 6.7){$\scriptstyle A$}
+\put( 2.7, 0.2){$\scriptstyle B$}
+\put(10.4, 0.2){$\scriptstyle C$}
+\put(14.1, 6.2){$\scriptstyle D$}
+\put( 6.2,12.2){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
+\quad
+\PGset[0.6em]
+\begin{picture}(15,13)
+
+% 354x316
+% 13x11
+
+\drawline(1,7)(3,1)(10.75,1)(14,6.5)(6.5,12)(1,7)
+
+\dashline[80]{0.2}(3,1)(6.5,12)(10.75,1) % BEC
+
+\put( 0.1, 6.7){$\scriptstyle A$}
+\put( 2.7, 0.0){$\scriptstyle B$}
+\put(10.4, 0.0){$\scriptstyle C$}
+\put(14.1, 6.2){$\scriptstyle D$}
+\put( 6.1,12.2){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187aa291.eepic b/33063-t/images/sources/187aa291.eepic
new file mode 100644
index 0000000..3d35669
--- /dev/null
+++ b/33063-t/images/sources/187aa291.eepic
@@ -0,0 +1,32 @@
+\PGset[0.8em]
+\begin{picture}(12,9)
+
+\drawline(1,4)(11.571,4)(6.5,7.857)(1,4) % ABCA
+
+% A =  1,     4
+% B = 11.571, 4      mBC =  -.761
+% C =  6.5,   7.857  mAC =   .701
+% G =  2.643, 2.036  mAG = -1.195
+% H =  3.786, 3.286  mGH =  1.094
+% O =  1.962, 4.674  mOG = -3.874
+\dashline[80]{0.2}(1,4)(2.643,2.036)(3.786,3.286)(1.962,4.674)(2.643,2.036)
+
+% D =  7.496, 4
+% E =  5.094, 6.87
+% F =  9.168, 5.829
+\dashline[80]{0.2}(6.5,7.857)(7.496,4) % CD
+\dashline[80]{0.2}(5.094,6.87)(7.496,4)(9.168,5.829) % EDF
+
+
+\put( 0.2, 3.7){$\scriptstyle A$}
+\put(11.7, 3.7){$\scriptstyle B$}
+\put( 6.2, 8.0){$\scriptstyle C$}
+\put( 6.9, 3.2){$\scriptstyle D$}
+\put( 4.5, 7.0){$\scriptstyle E$}
+\put( 9.2, 5.9){$\scriptstyle F$}
+\put( 2.4, 1.3){$\scriptstyle G$}
+\put( 3.9, 3.0){$\scriptstyle H$}
+\put( 1.6, 4.9){$\scriptstyle O$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187bb292.eepic b/33063-t/images/sources/187bb292.eepic
new file mode 100644
index 0000000..c592d2a
--- /dev/null
+++ b/33063-t/images/sources/187bb292.eepic
@@ -0,0 +1,43 @@
+\PGset[0.8em]
+\begin{picture}(6,7)
+
+% r = 2.5
+% Ellipse:  u = 1.0  v = 1.0  a = 2.5  b = 2.5  phi = 0.0 Grad
+% Ellipse:  u = 3.5  v = 3.5  a = 2.5  b = 2.5  phi = 0.0 Grad
+\qbezier[20](6.0, 3.5)(6.0, 4.5355)(5.2678, 5.2678)
+\qbezier[20](5.2678, 5.2678)(4.5355, 6.0)(3.5, 6.0)
+\qbezier[20](3.5, 6.0)(2.4645, 6.0)(1.7322, 5.2678)
+\qbezier[20](1.7322, 5.2678)(1.0, 4.5355)(1.0, 3.5)
+\qbezier[20](1.0, 3.5)(1.0, 2.4645)(1.7322, 1.7322)
+\qbezier[20](1.7322, 1.7322)(2.4645, 1.0)(3.5, 1.0)
+\qbezier[20](3.5, 1.0)(4.5355, 1.0)(5.2678, 1.7322)
+\qbezier[20](5.2678, 1.7322)(6.0, 2.4645)(6.0, 3.5)
+
+% A = 1.18,  4.431
+% B = 5.82,  4.431
+% C = 4.359, 5.848
+% O = 3.5,   3.5
+\drawline(1.18,4.431)(5.82,4.431)(4.359,5.848)(1.18,4.431) % ABCA
+
+% midpoint of CO = 3.93, 4.674, r=1.25
+% Ellipse:  u = 3.93  v = 4.674  a = 1.25  b = 1.25  phi = 25.0 Grad
+\qbezier[10](4.3575, 5.8486)(3.871, 6.0257)(3.4017, 5.8069)
+\qbezier[10](3.4017, 5.8069)(2.9325, 5.5881)(2.7554, 5.1015)
+\qbezier[10](2.7554, 5.1015)(2.5783, 4.615)(2.7971, 4.1457)
+\qbezier[10](2.7971, 4.1457)(3.0159, 3.6765)(3.5025, 3.4994)
+
+
+\dashline[80]{0.2}(4.359,5.848)(3.5,3.5)(2.704,4.431) % COP
+
+% mCP = .856
+\dashline[80]{0.2}(4.359,5.848)(1.048,3.014) % CD
+
+\put( 0.4, 4.1){$\scriptstyle A$}
+\put( 5.9, 4.1){$\scriptstyle B$}
+\put( 4.1, 6.0){$\scriptstyle C$}
+\put( 0.2, 2.7){$\scriptstyle D$}
+\put( 3.2, 2.8){$\scriptstyle O$}
+\put( 2.2, 3.4){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187cc293.eepic b/33063-t/images/sources/187cc293.eepic
new file mode 100644
index 0000000..1f288bb
--- /dev/null
+++ b/33063-t/images/sources/187cc293.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}(7,9)
+
+% r = 3.5
+% Ellipse:  u = 4.5  v = 4.5  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(8.0, 4.5)(8.0, 5.9497)(6.9749, 6.9749)
+\qbezier(6.9749, 6.9749)(5.9497, 8.0)(4.5, 8.0)
+\qbezier(4.5, 8.0)(3.0503, 8.0)(2.0251, 6.9749)
+\qbezier(2.0251, 6.9749)(1.0, 5.9497)(1.0, 4.5)
+\qbezier(1.0, 4.5)(1.0, 3.0503)(2.0251, 2.0251)
+\qbezier(2.0251, 2.0251)(3.0503, 1.0)(4.5, 1.0)
+\qbezier(4.5, 1.0)(5.9497, 1.0)(6.9749, 2.0251)
+\qbezier(6.9749, 2.0251)(8.0, 3.0503)(8.0, 4.5)
+
+% P = 6.036,5.786
+% n = 2, m = 3.235
+% n:m = 3.5:CA (5.661)
+\dashline[80]{0.2}(4.5,4.5)(8.514,7.861)(2.859,7.592)(7.996,4.674)(4.5,4.5) % OCABO
+
+% mAP = -.568
+
+\put( 2.5, 7.8){$\scriptstyle A$}
+\put( 8.1, 4.4){$\scriptstyle B$}
+\put( 8.6, 7.5){$\scriptstyle C$}
+\put( 4.2, 3.8){$\scriptstyle O$}
+\put( 5.8, 6.0){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187dd294.eepic b/33063-t/images/sources/187dd294.eepic
new file mode 100644
index 0000000..991d603
--- /dev/null
+++ b/33063-t/images/sources/187dd294.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(8,9)
+
+% r = 3.5
+% Ellipse:  u = 4.5  v = 4.5  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(8.0, 4.5)(8.0, 5.9497)(6.9749, 6.9749)
+\qbezier(6.9749, 6.9749)(5.9497, 8.0)(4.5, 8.0)
+\qbezier(4.5, 8.0)(3.0503, 8.0)(2.0251, 6.9749)
+\qbezier(2.0251, 6.9749)(1.0, 5.9497)(1.0, 4.5)
+\qbezier(1.0, 4.5)(1.0, 3.0503)(2.0251, 2.0251)
+\qbezier(2.0251, 2.0251)(3.0503, 1.0)(4.5, 1.0)
+\qbezier(4.5, 1.0)(5.9497, 1.0)(6.9749, 2.0251)
+\qbezier(6.9749, 2.0251)(8.0, 3.0503)(8.0, 4.5)
+
+\dashline[80]{0.2}(4.5,4.5)(6.303,1.5) % OP, m = -1.664
+
+\dashline[80]{0.2}(1.594,2.55)(7.094,2.15) % AB, m = -.072
+
+% C = 5.859, 2.239
+% CP = .862, DP = 1.724
+% D = 5.415, 2.978
+\dashline[80]{0.2}(5.415,2.978)(1.219,3.28)(6.303,1.5) % DEP
+
+\put( 0.9, 2.0){$\scriptstyle A$}
+\put( 7.3, 1.8){$\scriptstyle B$}
+\put( 5.8, 2.3){$\scriptstyle C$}
+\put( 5.5, 3.1){$\scriptstyle D$}
+\put( 0.4, 3.0){$\scriptstyle E$}
+\put( 3.5, 1.6){$\scriptstyle F$}
+\put( 4.2, 4.6){$\scriptstyle O$}
+\put( 6.3, 0.8){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187ee295.eepic b/33063-t/images/sources/187ee295.eepic
new file mode 100644
index 0000000..7130b05
--- /dev/null
+++ b/33063-t/images/sources/187ee295.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(9,8)
+
+% r = 3.25
+% Ellipse:  u = 7.75  v = 4.25  a = 3.25  b = 3.25  phi = 0.0 Grad
+\qbezier(11.0, 4.25)(11.0, 5.5962)(10.0481, 6.5481)
+\qbezier(10.0481, 6.5481)(9.0962, 7.5)(7.75, 7.5)
+\qbezier(7.75, 7.5)(6.4038, 7.5)(5.4519, 6.5481)
+\qbezier(5.4519, 6.5481)(4.5, 5.5962)(4.5, 4.25)
+\qbezier(4.5, 4.25)(4.5, 2.9038)(5.4519, 1.9519)
+\qbezier(5.4519, 1.9519)(6.4038, 1.0)(7.75, 1.0)
+\qbezier(7.75, 1.0)(9.0962, 1.0)(10.0481, 1.9519)
+\qbezier(10.0481, 1.9519)(11.0, 2.9038)(11.0, 4.25)
+
+% P = 1,     3
+% C = 6.758, 1.155   m=PD = 4, n=DC = 2.046, PC = 6.046
+% D = 4.809, 1.779
+\dashline[80]{0.2}(1,3)(6.758,1.155)
+
+% 4 : PA = PA : 6.046, PA = 4.918
+% A = 4.96, 5.917
+% PD/DC = PA/AB, AB = 2.516, PB = 7.434
+\dashline[80]{0.2}(1,3)(6.986,7.409)(6.758,1.155) % PBC
+\dashline[80]{0.2}(4.96,5.917)(4.809,1.779) % AD
+
+\put( 4.3, 6.1){$\scriptstyle A$}
+\put( 6.7, 7.5){$\scriptstyle B$}
+\put( 6.4, 0.3){$\scriptstyle C$}
+\put( 4.3, 1.0){$\scriptstyle D$}
+\put( 0.2, 2.7){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187ff296.eepic b/33063-t/images/sources/187ff296.eepic
new file mode 100644
index 0000000..9b61078
--- /dev/null
+++ b/33063-t/images/sources/187ff296.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(9,8)
+
+% r = 3.25
+% Ellipse:  u = 6.25  v = 4.25  a = 3.25  b = 3.25  phi = 0.0 Grad
+\qbezier(9.5, 4.25)(9.5, 5.5962)(8.5481, 6.5481)
+\qbezier(8.5481, 6.5481)(7.5962, 7.5)(6.25, 7.5)
+\qbezier(6.25, 7.5)(4.9038, 7.5)(3.9519, 6.5481)
+\qbezier(3.9519, 6.5481)(3.0, 5.5962)(3.0, 4.25)
+\qbezier(3.0, 4.25)(3.0, 2.9038)(3.9519, 1.9519)
+\qbezier(3.9519, 1.9519)(4.9038, 1.0)(6.25, 1.0)
+\qbezier(6.25, 1.0)(7.5962, 1.0)(8.5481, 1.9519)
+\qbezier(8.5481, 1.9519)(9.5, 2.9038)(9.5, 4.25)
+
+% P = 1.25, 1
+% PC + CD = PD = 5.
+% PC : CD = CD : 5., PC = 1.91, CD = 3.09
+% C = 2.91, 1
+% A = 3.183,3.175, mPA = 1.125
+% B = 6.958,7.422
+
+\dashline[80]{0.2}(6.958,7.422)(1.25,1)(6.25,1)(6.958,7.422) % BPDB
+
+\dashline[80]{0.2}(3.183,3.175)(2.91,1) % AC
+
+\put( 2.3, 3.1){$\scriptstyle A$}
+\put( 6.7, 7.7){$\scriptstyle B$}
+\put( 2.5, 0.2){$\scriptstyle C$}
+\put( 5.9, 0.2){$\scriptstyle D$}
+\put( 0.5, 0.6){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187gg297.eepic b/33063-t/images/sources/187gg297.eepic
new file mode 100644
index 0000000..533f126
--- /dev/null
+++ b/33063-t/images/sources/187gg297.eepic
@@ -0,0 +1,38 @@
+\PGset[0.8em]
+\begin{picture}(7,9)
+
+% P in the arc subtended by chord AB s.t. PA:PB = m:n (= AD:DB?)
+
+% r = 3.5
+
+% Ellipse:  u = 4.5  v = 4.5  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(8.0, 4.5)(8.0, 5.9497)(6.9749, 6.9749)
+\qbezier(6.9749, 6.9749)(5.9497, 8.0)(4.5, 8.0)
+\qbezier(4.5, 8.0)(3.0503, 8.0)(2.0251, 6.9749)
+\qbezier(2.0251, 6.9749)(1.0, 5.9497)(1.0, 4.5)
+\qbezier(1.0, 4.5)(1.0, 3.0503)(2.0251, 2.0251)
+\qbezier(2.0251, 2.0251)(3.0503, 1.0)(4.5, 1.0)
+\qbezier(4.5, 1.0)(5.9497, 1.0)(6.9749, 2.0251)
+\qbezier(6.9749, 2.0251)(8.0, 3.0503)(8.0, 4.5)
+
+% y = 2.3
+% A = 1.778, 2.3
+% B = 7.222, 2.3
+\drawline(1.778,2.3)(7.222,2.3) % AB
+
+% E = 4.5,2.3
+% C = 4.5,8
+% mCP = -4.222
+% P = 6.07, 1.372
+\dashline[80]{0.2}(4.5,2.3)(4.5,8)(6.07,1.372)(1.778,2.3) % ECPA
+
+\put( 1.2, 1.6){$\scriptstyle A$}
+\put( 7.2, 1.6){$\scriptstyle B$}
+\put( 4.2, 8.1){$\scriptstyle C$}
+\put( 5.9, 2.4){$\scriptstyle D$}
+\put( 3.6, 2.4){$\scriptstyle E$}
+\put( 3.8, 4.2){$\scriptstyle O$}
+\put( 6.0, 0.6){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/187hh298.eepic b/33063-t/images/sources/187hh298.eepic
new file mode 100644
index 0000000..1b84bc8
--- /dev/null
+++ b/33063-t/images/sources/187hh298.eepic
@@ -0,0 +1,52 @@
+\PGset[0.8em]
+\begin{picture}(12,8)
+
+% secant through one of the points of intersection so that
+% the two chords are in the ratio m:n
+
+% r1 = 2.7
+% r2 = 3.5
+% Ellipse:  u = 3.7  v = 4.5  a = 2.7  b = 2.7  phi = 0.0 Grad
+\qbezier(6.4, 4.5)(6.4, 5.6184)(5.6092, 6.4092)
+\qbezier(5.6092, 6.4092)(4.8184, 7.2)(3.7, 7.2)
+\qbezier(3.7, 7.2)(2.5816, 7.2)(1.7908, 6.4092)
+\qbezier(1.7908, 6.4092)(1.0, 5.6184)(1.0, 4.5)
+\qbezier(1.0, 4.5)(1.0, 3.3816)(1.7908, 2.5908)
+\qbezier(1.7908, 2.5908)(2.5816, 1.8)(3.7, 1.8)
+\qbezier(3.7, 1.8)(4.8184, 1.8)(5.6092, 2.5908)
+\qbezier(5.6092, 2.5908)(6.4, 3.3816)(6.4, 4.5)
+
+% Ellipse:  u = 8.2  v = 4.5  a = 3.5  b = 3.5  phi = 0.0 Grad
+\qbezier(11.7, 4.5)(11.7, 5.9497)(10.6749, 6.9749)
+\qbezier(10.6749, 6.9749)(9.6497, 8.0)(8.2, 8.0)
+\qbezier(8.2, 8.0)(6.7503, 8.0)(5.7251, 6.9749)
+\qbezier(5.7251, 6.9749)(4.7, 5.9497)(4.7, 4.5)
+\qbezier(4.7, 4.5)(4.7, 3.0503)(5.7251, 2.0251)
+\qbezier(5.7251, 2.0251)(6.7503, 1.0)(8.2, 1.0)
+\qbezier(8.2, 1.0)(9.6497, 1.0)(10.6749, 2.0251)
+\qbezier(10.6749, 2.0251)(11.7, 3.0503)(11.7, 4.5)
+
+% O =  3.7, 4.5
+% O' = 8.2, 4.5
+% P = 5.399, 6.599
+% A = 5.2, 4.5
+\dashline[80]{0.2}(3.7,4.5)(8.2,4.5) % OO'
+\dashline[80]{0.2}(5.399,6.599)(5.2,4.5) % PA, m = -10.548
+
+\dashline[80]{0.2}(2.427,6.881)(11.345,6.035) % GF
+
+\dashline[80]{0.2}(3.7,4.5)(3.912,6.74) % OD
+\dashline[80]{0.2}(8.2,4.5)(8.372,6.317) % O'E
+
+\put( 1.6, 6.9){$\scriptstyle G$}
+\put( 3.6, 7.4){$\scriptstyle D$}
+\put( 5.1, 7.0){$\scriptstyle P$}
+\put( 8.0, 6.5){$\scriptstyle E$}
+\put(11.4, 5.8){$\scriptstyle F$}
+\put( 3.0, 4.2){$\scriptstyle O$}
+\put( 4.9, 3.8){$\scriptstyle A$}
+\put( 8.3, 4.2){$\scriptstyle O'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/188aa303.eepic b/33063-t/images/sources/188aa303.eepic
new file mode 100644
index 0000000..5274134
--- /dev/null
+++ b/33063-t/images/sources/188aa303.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}(7,6)
+
+% 5x4
+\drawline(1,1)(4,1)(2.5,5)(1,1) % ABCA
+
+\dashline[80]{0.2}(4,1)(6.5,1)(6.5,5)(2.5,5)(2.5,1) % BNMCH
+\dashline[80]{0.2}(6.5,5)(1,1) % MA
+
+% mMA = .727, mBC = -2.667
+% (y-1)/(x-1) = .727    y=.727x   + 0.2
+% (y-1)/(x-4) = -2.667  y=-2.667x + 11.668
+% x = 1.376y-.275     = 3.378
+% y = -3.67y + 12.401 = 2.655
+\dashline[80]{0.2}(1.622,1)(3.378,1)(3.378,2.655)(1.622,2.655)(1.622,1) % DEFGD
+
+\put( 0.5, 0.2){$\scriptstyle A$}
+\put( 3.7, 0.2){$\scriptstyle B$}
+\put( 2.2, 5.1){$\scriptstyle C$}
+\put( 1.3, 0.2){$\scriptstyle D$}
+\put( 3.0, 0.2){$\scriptstyle E$}
+\put( 3.2, 3.0){$\scriptstyle F$}
+\put( 0.9, 2.7){$\scriptstyle G$}
+\put( 2.1, 0.2){$\scriptstyle H$}
+\put( 6.0, 5.2){$\scriptstyle M$}
+\put( 6.0, 0.2){$\scriptstyle N$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/189aa312.eepic b/33063-t/images/sources/189aa312.eepic
new file mode 100644
index 0000000..0f82d6b
--- /dev/null
+++ b/33063-t/images/sources/189aa312.eepic
@@ -0,0 +1,19 @@
+\PGset[0.8em]
+\begin{picture}(8,8)
+
+% 6x6
+\drawline(1,1)(7,1)(7,7)(1,1) % BDCB
+\drawline(7,7)(5,1) % CA
+
+\put( 4.5, 0.2){$\scriptstyle A$}
+\put( 0.5, 0.2){$\scriptstyle B$}
+\put( 6.7, 7.1){$\scriptstyle C$}
+\put( 6.7, 0.2){$\scriptstyle D$}
+
+\put( 3.2, 3.7){$\scriptstyle a$}
+\put( 6.0, 3.0){$\scriptstyle b$}
+\put( 2.5, 0.3){$\scriptstyle c$}
+\put( 7.1, 3.7){$\scriptstyle h$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/189bb313.eepic b/33063-t/images/sources/189bb313.eepic
new file mode 100644
index 0000000..5892315
--- /dev/null
+++ b/33063-t/images/sources/189bb313.eepic
@@ -0,0 +1,24 @@
+\PGset[0.8em]
+\begin{picture}(7.5,7)
+
+% 189bb313
+
+\drawline(1,1)(5.8,5.6)(6.8,1)(1,1) % ACB
+
+\drawline(3.8,1)(5.8,5.6)(5.6,1) % FCD
+
+\put( 0.3, 0.7 ){$\scriptstyle A$}
+\put( 6.8, 0.7 ){$\scriptstyle B$}
+\put( 5.5, 5.7 ){$\scriptstyle C$}
+\put( 5.1, 0.2 ){$\scriptstyle D$}
+\put( 3.3, 0.2 ){$\scriptstyle F$}
+
+\put( 6.3, 3.5 ){$\scriptstyle a$}
+\put( 3.1, 3.5 ){$\scriptstyle b$}
+\put( 2.0, 0.3 ){$\scriptstyle c$}
+\put( 3.5, 2.2 ){$\scriptstyle m$}
+\put( 5.1, 2.2 ){$\scriptstyle h$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/190aa314.eepic b/33063-t/images/sources/190aa314.eepic
new file mode 100644
index 0000000..b23072c
--- /dev/null
+++ b/33063-t/images/sources/190aa314.eepic
@@ -0,0 +1,42 @@
+\PGset[0.8em]
+\begin{picture}(11,11)
+
+% r = 4.5
+% Ellipse:  u = 5.5  v = 5.5  a = 4.5  b = 4.5  phi = 0.0 Grad
+\qbezier[20](10.0, 5.5)(10.0, 7.364)(8.682, 8.682)
+\qbezier[20](8.682, 8.682)(7.364, 10.0)(5.5, 10.0)
+\qbezier[20](5.5, 10.0)(3.636, 10.0)(2.318, 8.682)
+\qbezier[20](2.318, 8.682)(1.0, 7.364)(1.0, 5.5)
+\qbezier[20](1.0, 5.5)(1.0, 3.636)(2.318, 2.318)
+\qbezier[20](2.318, 2.318)(3.636, 1.0)(5.5, 1.0)
+\qbezier[20](5.5, 1.0)(7.364, 1.0)(8.682, 2.318)
+\qbezier[20](8.682, 2.318)(10.0, 3.636)(10.0, 5.5)
+
+% A = 170deg  1.068, 6.281
+% B =  10deg  9.932, 6.281
+% C =  60deg  7.75 , 9.397
+\drawline(1.068,6.281)(9.932,6.281)(7.75,9.397)(1.068,6.281) % ABCA
+
+% ACB = tan^-1 2.144 + tan^-1 0.7 = 64.992+34.992 = 99.987
+% ACD = 49.994, mCE = -3.733
+\drawline(7.75,9.397)(6.915,6.281) % CD
+
+% (y-9.397)/(x-7.75) = 3.733
+% (y-5.5)^2 + (x-5.5)^2 = 4.5^2
+\dashline[80]{0.2}(6.915,6.281)(5.5,1) % DE
+
+\dashline[80]{0.2}(5.5,1)(9.932,6.281) % EB
+
+\put( 0.2, 6.0){$\scriptstyle A$}
+\put(10.1, 6.0){$\scriptstyle B$}
+\put( 7.8, 9.4){$\scriptstyle C$}
+\put( 6.0, 5.5){$\scriptstyle D$}
+\put( 5.0, 0.2){$\scriptstyle E$}
+
+\put( 8.3, 7.5){$\scriptstyle a$}
+\put( 3.9, 7.8){$\scriptstyle b$}
+\put( 3.9, 5.7){$\scriptstyle c$}
+\put( 6.7, 7.2){$\scriptstyle t$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/190bb315.eepic b/33063-t/images/sources/190bb315.eepic
new file mode 100644
index 0000000..612b018
--- /dev/null
+++ b/33063-t/images/sources/190bb315.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(11,11)
+
+% r = 4.5
+% Ellipse:  u = 5.5  v = 5.5  a = 4.5  b = 4.5  phi = 0.0 Grad
+\qbezier(10.0, 5.5)(10.0, 7.364)(8.682, 8.682)
+\qbezier(8.682, 8.682)(7.364, 10.0)(5.5, 10.0)
+\qbezier(5.5, 10.0)(3.636, 10.0)(2.318, 8.682)
+\qbezier(2.318, 8.682)(1.0, 7.364)(1.0, 5.5)
+\qbezier(1.0, 5.5)(1.0, 3.636)(2.318, 2.318)
+\qbezier(2.318, 2.318)(3.636, 1.0)(5.5, 1.0)
+\qbezier(5.5, 1.0)(7.364, 1.0)(8.682, 2.318)
+\qbezier(8.682, 2.318)(10.0, 3.636)(10.0, 5.5)
+
+% A = 170deg  1.068, 6.281
+% B =  10deg  9.932, 6.281
+% C =  60deg  7.75 , 9.397
+\drawline(1.068,6.281)(9.932,6.281)(7.75,9.397)(1.068,6.281) % ABCA
+
+% ACB = tan^-1 2.144 + tan^-1 0.7 = 64.992+34.992 = 99.987
+% ACD = 49.994, mCE = -3.733
+\drawline(7.75,9.397)(7.75,6.281) % CD
+
+% E = 3.25, 1.603
+\dashline[80]{0.2}(3.25,1.603)(9.932,6.281) % EB
+\drawline(3.25,1.603)(7.75,9.397) % EA
+
+\put( 0.2, 6.0){$\scriptstyle B$}
+\put(10.1, 6.0){$\scriptstyle C$}
+\put( 7.8, 9.5){$\scriptstyle A$}
+\put( 7.3, 5.5){$\scriptstyle D$}
+\put( 2.5, 0.9){$\scriptstyle E$}
+
+\put( 8.3, 7.5){$\scriptstyle b$}
+\put( 3.9, 7.8){$\scriptstyle c$}
+\put( 3.9, 5.7){$\scriptstyle a$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/193aa395.eepic b/33063-t/images/sources/193aa395.eepic
new file mode 100644
index 0000000..b06f0ad
--- /dev/null
+++ b/33063-t/images/sources/193aa395.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(38,8)
+
+% AB = 21, AO = 3, AD = 6
+\drawline(1,1)(22,1)(22,7)(1,7)(1,1) % ABCD
+
+% AE = 12, AO = 3, AD = 6
+\drawline(25,1)(37,1)(37,7)(25,7)(25,1) % AEFD
+
+\dashline[80]{0.2}(4,1)(4,7)
+\dashline[80]{0.2}(7,1)(7,7)
+\dashline[80]{0.2}(10,1)(10,7)
+\dashline[80]{0.2}(13,1)(13,7)
+\dashline[80]{0.2}(16,1)(16,7)
+\dashline[80]{0.2}(19,1)(19,7)
+
+\dashline[80]{0.2}(28,1)(28,7)
+\dashline[80]{0.2}(31,1)(31,7)
+\dashline[80]{0.2}(34,1)(34,7)
+
+\put( 0.3, 0.5){$\scriptstyle A$}
+\put(22.1, 0.5){$\scriptstyle B$}
+\put(22.1, 6.7){$\scriptstyle C$}
+\put( 0.2, 6.7){$\scriptstyle D$}
+\put( 3.7, 0.2){$\scriptstyle O$}
+
+\put(24.3, 0.5){$\scriptstyle A$}
+\put(37.1, 0.5){$\scriptstyle E$}
+\put(37.1, 6.7){$\scriptstyle F$}
+\put(24.2, 6.7){$\scriptstyle D$}
+\put(27.7, 0.2){$\scriptstyle O$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/194aa395.eepic b/33063-t/images/sources/194aa395.eepic
new file mode 100644
index 0000000..90929c9
--- /dev/null
+++ b/33063-t/images/sources/194aa395.eepic
@@ -0,0 +1,36 @@
+\PGset[0.8em]
+\begin{picture}(38,8)
+
+% AB = 12
+% AK = 12.5
+\drawline(1,1)(13,1)(13,6.25)(1,6.25)(1,1) % ABCD
+
+\drawline(17,1)(29.8,1)(29.8,6.25)(17,6.25)(17,1) % AEFD
+
+\dashline[80]{0.2}( 3,1)( 3,6.25)
+\dashline[80]{0.2}( 5,1)( 5,6.25)
+\dashline[80]{0.2}( 7,1)( 7,6.25)
+\dashline[80]{0.2}( 9,1)( 9,6.25)
+\dashline[80]{0.2}(11,1)(11,6.25)
+
+\dashline[80]{0.2}(19,1)(19,6.25)
+\dashline[80]{0.2}(21,1)(21,6.25)
+\dashline[80]{0.2}(23,1)(23,6.25)
+\dashline[80]{0.2}(25,1)(25,6.25)
+\dashline[80]{0.2}(27,1)(27,6.25)
+\dashline[80]{0.2}(29,1)(29,6.25)
+
+\put( 0.3, 0.7){$\scriptstyle A$}
+\put(13.1, 0.7){$\scriptstyle B$}
+\put(13.1, 6.0){$\scriptstyle C$}
+\put( 0.2, 6.0){$\scriptstyle D$}
+
+\put(16.3, 0.7){$\scriptstyle A$}
+\put(28.6, 0.2){$\scriptstyle K$}
+\put(29.9, 0.7){$\scriptstyle E$}
+\put(29.9, 6.0){$\scriptstyle F$}
+\put(28.7, 6.4){$\scriptstyle H$}
+\put(16.2, 6.0){$\scriptstyle D$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/195aa397.eepic b/33063-t/images/sources/195aa397.eepic
new file mode 100644
index 0000000..4a395c8
--- /dev/null
+++ b/33063-t/images/sources/195aa397.eepic
@@ -0,0 +1,22 @@
+\PGset[0.8em]
+\begin{picture}(38,8)
+
+\drawline(1,1)(11,1)(11,8)(1,8)(1,1) % R
+\drawline(15,1)(22,1)(22,6)(15,6)(15,1) % R'
+\drawline(26,1)(36,1)(36,6)(26,6)(26,1) % S
+
+\put( 5.5, 4.2){$\scriptstyle R$}
+\put(18.2, 3.2){$\scriptstyle R'$}
+\put(30.5, 3.2){$\scriptstyle S$}
+
+\put( 0.2, 4.2){$\scriptstyle a$}
+\put( 5.7, 0.3){$\scriptstyle b$}
+
+\put(14.2, 3.2){$\scriptstyle a'$}
+\put(18.4, 0.3){$\scriptstyle b'$}
+
+\put(25.2, 3.2){$\scriptstyle a'$}
+\put(30.7, 0.3){$\scriptstyle b$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/196aa398.eepic b/33063-t/images/sources/196aa398.eepic
new file mode 100644
index 0000000..4c5e826
--- /dev/null
+++ b/33063-t/images/sources/196aa398.eepic
@@ -0,0 +1,16 @@
+\PGset[0.8em]
+\begin{picture}(20,10)
+
+\drawline(1,1)(12.3,1)(12.3,9.4)(1,9.4)(1,1) % R
+\drawline(16,1)(19,1)(19,4)(16,4)(16,1) % U
+
+\put( 6.2, 5.0){$\scriptstyle R$}
+\put( 0.3, 5.1){$\scriptstyle a$}
+\put( 6.4, 0.3){$\scriptstyle b$}
+
+\put(17.2, 2.2){$\scriptstyle U$}
+\put(15.3, 2.2){$\scriptstyle 1$}
+\put(17.3, 0.3){$\scriptstyle 1$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/196bb399.eepic b/33063-t/images/sources/196bb399.eepic
new file mode 100644
index 0000000..779daaf
--- /dev/null
+++ b/33063-t/images/sources/196bb399.eepic
@@ -0,0 +1,18 @@
+\PGset[0.8em]
+\begin{picture}(16,10)
+
+\drawline(1,1)(15,1)(15,9)(1,9)(1,1)
+
+\drawline( 3,1)( 3,9)
+\drawline( 5,1)( 5,9)
+\drawline( 7,1)( 7,9)
+\drawline( 9,1)( 9,9)
+\drawline(11,1)(11,9)
+\drawline(13,1)(13,9)
+
+\drawline(1,3)(15,3)
+\drawline(1,5)(15,5)
+\drawline(1,7)(15,7) 
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/197aa400.eepic b/33063-t/images/sources/197aa400.eepic
new file mode 100644
index 0000000..3bc4152
--- /dev/null
+++ b/33063-t/images/sources/197aa400.eepic
@@ -0,0 +1,35 @@
+\PGset[0.8em]
+\begin{picture}(40,8)
+
+\drawline(1,1)(3.25,7)(18.5,7)(16.25,1)(1,1) % AEFDA
+
+\dashline[80]{0.2}(1,1)(1,7)(3.25,7) % ABE
+\dashline[80]{0.2}(16.25,1)(16.25,7) % DC
+
+\drawline(23,1)(31.5,6.5)(38.6,6.5)(30.1,1)(23,1) % AEFDA
+
+\dashline[80]{0.2}(23,1)(23,6.5)(31.5,6.5) % ABE
+\dashline[80]{0.2}(30.1,1)(30.1,6.5) % DC
+
+\put( 0.6, 0.2){$\scriptstyle A$}
+\put( 0.6, 7.1){$\scriptstyle B$}
+\put(16.0, 7.1){$\scriptstyle C$}
+\put(15.8, 0.2){$\scriptstyle D$}
+\put( 3.0, 7.1){$\scriptstyle E$}
+\put(18.2, 7.1){$\scriptstyle F$}
+
+\put(15.6, 3.8){$\scriptstyle a$}
+\put( 7.9, 0.2){$\scriptstyle b$}
+
+\put(22.7, 0.2){$\scriptstyle A$}
+\put(22.7, 6.6){$\scriptstyle B$}
+\put(29.8, 6.6){$\scriptstyle C$}
+\put(29.7, 0.2){$\scriptstyle D$}
+\put(31.2, 6.6){$\scriptstyle E$}
+\put(38.3, 6.6){$\scriptstyle F$}
+
+\put(29.5, 3.2){$\scriptstyle a$}
+\put(26.3, 0.2){$\scriptstyle b$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/198aa403.eepic b/33063-t/images/sources/198aa403.eepic
new file mode 100644
index 0000000..dabfec1
--- /dev/null
+++ b/33063-t/images/sources/198aa403.eepic
@@ -0,0 +1,19 @@
+\PGset[0.8em]
+\begin{picture}(19,9)
+
+\drawline(1,1)(15.3,1)(17.75,7.8)(1,1) % ABCA
+
+\dashline[80]{0.2}(17.75,7.8)(17.75,1)(15.3,1) % CDB
+\dashline[80]{0.2}(1,1)(3.45,7.8)(17.75,7.8) % AHC
+
+\put( 0.5, 0.2){$\scriptstyle A$}
+\put(14.8, 0.2){$\scriptstyle B$}
+\put(17.5, 7.9){$\scriptstyle C$}
+\put(17.3, 0.2){$\scriptstyle D$}
+\put( 3.1, 7.9){$\scriptstyle H$}
+
+\put(17.2, 3.9){$\scriptstyle a$}
+\put( 8.3, 0.2){$\scriptstyle b$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/199aa407.eepic b/33063-t/images/sources/199aa407.eepic
new file mode 100644
index 0000000..a30f73c
--- /dev/null
+++ b/33063-t/images/sources/199aa407.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(19,11.5)
+
+% b=17.75, h=9.4, b'=10, 1/2h = 4.6
+
+\drawline(1,1)(18.75,1)(14.875,10.4)(4.875,10.4)(1,1) % ABCHA
+
+% y = 5.8
+\drawline(2.98,5.8)(16.77,5.8) % OP
+\drawline(9.875,1)(9.875,10.4) % FE
+\drawline(1,1)(14.875,10.4) % AC
+
+
+\put( 0.5, 0.2){$\scriptstyle A$}
+\put(18.3, 0.2){$\scriptstyle B$}
+\put(14.6,10.5){$\scriptstyle C$}
+\put( 9.6,10.5){$\scriptstyle E$}
+\put( 9.5, 0.2){$\scriptstyle F$}
+\put( 4.6,10.5){$\scriptstyle H$}
+\put( 2.2, 5.5){$\scriptstyle O$}
+\put(16.9, 5.5){$\scriptstyle P$}
+
+\put(10.1, 3.0){$\scriptstyle a$}
+\put(14.2, 0.2){$\scriptstyle b$}
+\put(12.2,10.5){$\scriptstyle b'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/199bb409.eepic b/33063-t/images/sources/199bb409.eepic
new file mode 100644
index 0000000..54df8b2
--- /dev/null
+++ b/33063-t/images/sources/199bb409.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+\begin{picture}(19,11)
+
+\drawline(1,4.3)(4.75,8.5)(8.3,10.3)(12.75,8.0)(17.4,4.3)(12.5,1)(5.8,1)(1,4.3) % ABCDEFGA
+
+\dashline[80]{0.2}(1,4.3)(17.4,4.3)
+\dashline[80]{0.2}(4.75,4.3)(4.75,8.5)
+\dashline[80]{0.2}(8.3,4.3)(8.3,10.3)
+\dashline[80]{0.2}(12.75,4.3)(12.75,8)
+\dashline[80]{0.2}(12.5,1)(12.5,4.3)
+\dashline[80]{0.2}(5.8,1)(5.8,4.3)
+
+
+\put( 0.3, 4.0){$\scriptstyle A$}
+\put( 4.3, 8.6){$\scriptstyle B$}
+\put( 8.0,10.4){$\scriptstyle C$}
+\put(12.5, 8.1){$\scriptstyle D$}
+\put(17.5, 4.0){$\scriptstyle E$}
+\put(12.2, 0.2){$\scriptstyle F$}
+\put( 5.5, 0.2){$\scriptstyle G$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/200aa410.eepic b/33063-t/images/sources/200aa410.eepic
new file mode 100644
index 0000000..cf04dc0
--- /dev/null
+++ b/33063-t/images/sources/200aa410.eepic
@@ -0,0 +1,12 @@
+\PGset[0.8em]
+\begin{picture}   (14,14 )
+\drawline     ( 13,13 )( 1, 1 )(13, 1 )( 13,13 )    %A(D)BC(E)A
+\drawline     (  7, 7 )( 13, 4 )              %DE
+\dashline[80]{0.2} ( 13, 4 )(  1, 1 )              %EB
+\put( 12.5  , 13.2 ){$\scriptstyle A$}   %A
+\put( 0.7   ,  0.2 ){$\scriptstyle B$}   %B
+\put(13.3   ,  0.7 ){$\scriptstyle C$}   %C
+\put( 6.3   ,  7   ){$\scriptstyle D$}   %D
+\put(13.3   ,  3.7 ){$\scriptstyle E$}   %E
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/201ab411.eepic b/33063-t/images/sources/201ab411.eepic
new file mode 100644
index 0000000..7263002
--- /dev/null
+++ b/33063-t/images/sources/201ab411.eepic
@@ -0,0 +1,25 @@
+\PGset[0.8em]
+
+\begin{picture}   (12, 6 )
+\drawline     (  1, 1 )( 11, 1) ( 8.333, 5 )( 1, 1 ) % A(O)BCA
+\dashline[80]{0.2}( 8.333, 4.8 )( 8.333, 1) % CO
+
+\put( 0.7  , 0.2 ){$\scriptstyle A$}
+\put( 8.0  , 0.2 ){$\scriptstyle O$}
+\put( 8.0  , 5.2 ){$\scriptstyle C$}
+\put(10.3  , 0.2 ){$\scriptstyle B$}
+\end{picture}
+
+\quad
+
+\begin{picture}   (17, 8 )
+\drawline     (  1, 1 )( 16, 1 )( 12, 7 )( 1, 1 )     % A'(O')B'C'A'
+\dashline[80]{0.2}( 12, 6.8 )( 12, 1 ) % C'O'
+
+\put( 0.7  , 0.2 ){$\scriptstyle A'$}
+\put(11.7  , 0.2 ){$\scriptstyle O'$}
+\put(11.7  , 7.2 ){$\scriptstyle C'$}
+\put(15.3  , 0.2 ){$\scriptstyle B'$}
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/202ab412.eepic b/33063-t/images/sources/202ab412.eepic
new file mode 100644
index 0000000..1d9a4ea
--- /dev/null
+++ b/33063-t/images/sources/202ab412.eepic
@@ -0,0 +1,35 @@
+\PGset[0.8em]
+
+\begin{picture}   (15, 13.6 )
+             %     A          E              D         C         B      A
+\drawline     (  1, 6.8 )( 6.33, 12.6 ) ( 14, 6.8 )( 11, 1 )( 4, 1 )( 1, 6.8 )
+
+\dashline[80]{0.2}( 6.33, 12.6 )(  4, 1 )
+\dashline[80]{0.2}( 6.33, 12.6 )( 11, 1 )
+
+\put( 0.3  , 6.6 ){$\scriptstyle A$}
+\put( 3.7  , 0.2 ){$\scriptstyle B$}
+\put(10.3  , 0.2 ){$\scriptstyle C$}
+\put(14    , 6.6 ){$\scriptstyle D$}
+\put( 6    ,12.8 ){$\scriptstyle E$}
+\end{picture}
+
+\quad
+
+\PGset[0.6em]
+
+\begin{picture}   (15, 13.6 )
+             %     A          E              D         C         B      A
+\drawline     (  1, 6.8 )( 6.33, 12.6 ) ( 14, 6.8 )( 11, 1 )( 4, 1 )( 1, 6.8 )
+
+\dashline[80]{0.2}( 6.33, 12.6 )(  4, 1 )
+\dashline[80]{0.2}( 6.33, 12.6 )( 11, 1 )
+
+\put( 0    , 6.6 ){$\scriptstyle A'$}
+\put( 3.7  , 0.1 ){$\scriptstyle B'$}
+\put(10.3  , 0.1 ){$\scriptstyle C'$}
+\put(14    , 6.6 ){$\scriptstyle D'$}
+\put( 6    ,12.8 ){$\scriptstyle E'$}
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/203aa415.eepic b/33063-t/images/sources/203aa415.eepic
new file mode 100644
index 0000000..04aedc0
--- /dev/null
+++ b/33063-t/images/sources/203aa415.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+
+\begin{picture}   (14, 18 )
+
+\drawline     (  9, 11 )( 3, 8 )( 10, 8 )( 9, 11 ) % ABCA
+\drawline     (  9, 11 )( 6.88, 17.37)( 1.12, 14.37)( 3, 8 ) % AGFB
+\drawline     (  9, 11 )(11.83, 12.41)(12.83, 9.41)(10,8) % AHKC
+\drawline     (  3, 8 )(3, 1)(10,1)(10,8) % BDEC
+
+\dashline[80]{0.2}(  9, 11)(3,1) % AD
+\dashline[80]{0.2}(  9, 11)(9,1) % AL
+\dashline[80]{0.2}( 10,  8)(1.12, 14.37) % CF
+
+\put(  9  , 11.5 ) {$\scriptstyle A$}
+\put(  2  ,  7.7) {$\scriptstyle B$}
+\put( 10.2  ,  7.5 ) {$\scriptstyle C$}
+\put(  2.7  ,  .2 ) {$\scriptstyle D$}
+\put( 9.5  ,  .2 ) {$\scriptstyle E$}
+\put(0.62,14.37) {$\scriptstyle F$}
+\put(6.88,17.37) {$\scriptstyle G$}
+\put(11.83,12.41){$\scriptstyle H$}
+\put(12.83,9.41) {$\scriptstyle K$}
+\put(  8.5  ,  .2 ) {$\scriptstyle L$}
+
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/204aa356.eepic b/33063-t/images/sources/204aa356.eepic
new file mode 100644
index 0000000..5c5f324
--- /dev/null
+++ b/33063-t/images/sources/204aa356.eepic
@@ -0,0 +1,23 @@
+\PGset[0.8em]
+
+\begin{picture}   (9, 9.5)
+
+\drawline     ( 1, 7.5 )( 7.5, 7.5 ) % AC
+
+\dashline[80]{0.2}( 1, 7.5 )(   1,   1 )( 7.5, 1)(7.5, 7.5) % AKGC
+\dashline[80]{0.2}( 5, 7.5 )(5,1)   % BH
+\dashline[80]{0.2}( 1, 3.5 )(7.5,3.5) % DF
+
+\put(  .7  , 7.7 ) {$\scriptstyle A$}
+\put(  4.5  , 7.7 ) {$\scriptstyle B$}
+\put(  7  , 7.7 ) {$\scriptstyle C$}
+\put(  .2  , 3.2 ) {$\scriptstyle D$}
+\put(  5.2  , 2.7 ) {$\scriptstyle E$}
+\put(  7.7  , 3.2 ) {$\scriptstyle F$}
+\put(  .7  , .3 ) {$\scriptstyle K$}
+\put(  4.5  , .3 ) {$\scriptstyle H$}
+\put(  7  , .3 ) {$\scriptstyle G$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/204bb357.eepic b/33063-t/images/sources/204bb357.eepic
new file mode 100644
index 0000000..0946c2b
--- /dev/null
+++ b/33063-t/images/sources/204bb357.eepic
@@ -0,0 +1,22 @@
+\PGset[0.8em]
+\begin{picture}   (9, 12)
+
+\drawline     ( 1, 7.5 )( 7.5, 7.5 ) % AC
+
+\dashline[80]{0.2}( 1, 7.5 )(   1,   1 )( 7.5, 1)(7.5, 7.5) % AGFB
+\dashline[80]{0.2}( 1, 7.5 )(1,10.5)(4,10.5)(4,3.5)   % AHK(C)D
+\dashline[80]{0.2}( 1, 3.5 )(7.5,3.5) % LE
+
+\put(  .2  , 7.2 ) {$\scriptstyle A$}
+\put(  4.2  , 7.7 ) {$\scriptstyle C$}
+\put(  7.5  , 7.2 ) {$\scriptstyle B$}
+\put(  .2  , 3.2 ) {$\scriptstyle L$}
+\put(  7.7  , 3.2 ) {$\scriptstyle E$}
+\put(  .7  , .3 ) {$\scriptstyle G$}
+\put(  7  , .3 ) {$\scriptstyle F$}
+\put(  .7  , 10.7 ) {$\scriptstyle H$}
+\put(  3.7  , 10.7 ) {$\scriptstyle K$}
+\put(  3.7  , 2.7 ) {$\scriptstyle D$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/204cc358.eepic b/33063-t/images/sources/204cc358.eepic
new file mode 100644
index 0000000..157dada
--- /dev/null
+++ b/33063-t/images/sources/204cc358.eepic
@@ -0,0 +1,19 @@
+
+\PGset[0.8em]
+\begin{picture}   (9, 11)
+
+\drawline     ( 1, 7.5 )(   1,   1 )( 7.5, 1)(7.5, 7.5)(1, 7.5) % EABD
+\drawline     ( 7.5, 1 )( 7.5, 3.5 )(   5, 3.5 )(  5, 1 )( 7.5, 1 ) % BFGC
+\dashline[80]{0.2}(1, 7.5)(1, 10)(5, 10)( 5, 7.5 )(5,1)   % EIK(H)G
+
+\put(  .2  , 7.3 ) {$\scriptstyle E$}
+\put(  5.2  , 7.7 ) {$\scriptstyle H$}
+\put(  7.5  , 7.3 ) {$\scriptstyle D$}
+\put(  4.2  , 3.2 ) {$\scriptstyle G$}
+\put(  7.6  , 3.2 ) {$\scriptstyle F$}
+\put(  .7  , .3 ) {$\scriptstyle A$}
+\put(  4.5  , .3 ) {$\scriptstyle C$}
+\put(  7  , .3 ) {$\scriptstyle B$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/206aa417.eepic b/33063-t/images/sources/206aa417.eepic
new file mode 100644
index 0000000..4a2f895
--- /dev/null
+++ b/33063-t/images/sources/206aa417.eepic
@@ -0,0 +1,22 @@
+\PGset[0.7em]
+\begin{picture}   (40, 10)
+
+\drawline     ( 1, 1)( 7, 1)( 7, 7)( 1, 7)( 1, 1) % R
+\drawline     (11, 1)(18, 1)(18, 8)(11, 8)(11, 1) % R'
+
+\dashline[80]{0.4}(31, 1)(39, 1)(39, 9)(31, 9)(31, 1) % S
+
+\dashline[80]{0.2}(21, 8)(21, 1)(29, 1) % BAC
+\dashline[80]{0.2}(21, 6)(27, 1)
+
+\put(  3.5  , 3.5 ) {$\scriptstyle R$}
+\put(  14  ,  4 ) {$\scriptstyle R'$}
+
+\put( 20.1, 5.5 ) {$\scriptstyle B$}
+\put( 20.1, .7 ) {$\scriptstyle A$}
+\put( 26.5, .2 ) {$\scriptstyle C$}
+
+\put( 34.5, 4.5 ) {$\scriptstyle S$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/207aa418.eepic b/33063-t/images/sources/207aa418.eepic
new file mode 100644
index 0000000..41d505b
--- /dev/null
+++ b/33063-t/images/sources/207aa418.eepic
@@ -0,0 +1,29 @@
+
+\PGset[0.7em]
+\begin{picture}   (36, 10)
+
+\drawline     ( 1, 1)( 7, 1)( 7, 7)( 1, 7)( 1, 1) % R
+\drawline     (11, 1)(18, 1)(18, 8)(11, 8)(11, 1) % R'
+
+\dashline[80]{0.4}(31, 1)(35, 1)(35, 5)(31, 5)(31, 1) % S
+
+\dashline[80]{0.2}(21, 8)(21, 1)(28, 1) % (B)A(C)X
+\dashline[80]{0.2}(21, 6)(27, 1) % BC
+
+%\drawline (21,6)(28,3.5)
+%\drawline (21,6)(25,0)
+% arc P1 = (28.0/3.5)  P2 = (21.0/6.0)(B)  P3 = (25.0/0.0)  r = 7.81 (BC)
+\qbezier(28.355, 3.3732)(27.4849, 0.9368)(25.3322, -0.4983)
+
+\put(  3.5  , 3.5 ) {$\scriptstyle R$}
+\put(  14  ,  4 ) {$\scriptstyle R'$}
+
+\put( 20.1, 5.5 ) {$\scriptstyle B$}
+\put( 20.1, .7 ) {$\scriptstyle A$}
+\put( 26.8, .2 ) {$\scriptstyle C$}
+\put( 28, .7 ) {$\scriptstyle X$}
+
+\put( 32.5, 2.5 ) {$\scriptstyle S$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/208aa419.eepic b/33063-t/images/sources/208aa419.eepic
new file mode 100644
index 0000000..04eed7e
--- /dev/null
+++ b/33063-t/images/sources/208aa419.eepic
@@ -0,0 +1,42 @@
+\PGset[0.5em]
+\begin{picture}   (12, 13.6 )
+
+\drawline     (  1, 6.8 )( 6.33, 12.6 ) ( 14, 6.8 )( 11, 1 )( 4, 1 )( 1, 6.8 )
+
+\put( 7.0  , 6.3 ){$\scriptstyle R$}
+\put( 3.7  , -0.3 ){$\scriptstyle A$}
+\put(10.3  , -0.3 ){$\scriptstyle B$}
+\end{picture}
+\quad
+\PGset[0.6em]
+\begin{picture}   (12, 13.6 )
+
+\drawline     (  1, 6.8 )( 6.33, 12.6 ) ( 14, 6.8 )( 11, 1 )( 4, 1 )( 1, 6.8 )
+
+\put( 7.0  , 6.3 ){$\scriptstyle R'$}
+\put( 3.7  , 0.1 ){$\scriptstyle A'$}
+\put(10.3  , 0.1 ){$\scriptstyle B'$}
+\end{picture}
+\quad
+\PGset[0.7em]
+\begin{picture}   (12, 13.6 )
+
+\dashline[80]{0.4}(  1, 6.8 )( 6.33, 12.6 ) ( 14, 6.8 )( 11, 1 )( 4, 1 )( 1, 6.8 )
+
+\put( 7.0  , 6.3 ){$\scriptstyle R''$}
+\put( 3.7  , 0.1 ){$\scriptstyle A''$}
+\put(10.3  , 0.1 ){$\scriptstyle B''$}
+\end{picture}
+\begin{picture}   (10, 6 )
+
+% AB   = 7*0.5em = PO = 5*0.7em
+% A'B' = 7*0.6em = PH = 6*0.7em
+\dashline[80]{0.2}( 4, 1 )(4, 6 )( 10, 1 )( 4, 1 ) % POHP
+
+\put( 3.0  , 5.5 ){$\scriptstyle O$}
+\put( 3.7  , 0.1 ){$\scriptstyle P$}
+\put(9.3  , 0.1 ){$\scriptstyle H$}
+\end{picture}
+
+
+\PGrestore
diff --git a/33063-t/images/sources/209aa420.eepic b/33063-t/images/sources/209aa420.eepic
new file mode 100644
index 0000000..0ad94cb
--- /dev/null
+++ b/33063-t/images/sources/209aa420.eepic
@@ -0,0 +1,27 @@
+\PGset[0.8em]
+\begin{picture}   (20, 12)
+
+%               A    B    C      D      H     E    A
+\drawline     (4,1)(1,6)(4,11)(10,11)(13,6)(10,1)(4,1)
+
+\dashline{0.2}(4,1)(4,11)        % AC
+\dashline{0.2}(1,6)(1,1)         % BI
+\dashline{0.2}(10,11)(10,1)      % DE
+\dashline{0.2}(13,6)(13,1)(4,11) % HFC
+\dashline{0.4}(4,1)(1,1)(4,11)(19,1)(10,1)  % AICKE
+
+\dashline{0.2}(19,1)(10,11)(13,1)% KDF
+
+\put( 3.9  , 0.2 ){$\scriptstyle A$}
+\put( 0.2  , 5.7 ){$\scriptstyle B$}
+\put( 3.9  ,11.2 ){$\scriptstyle C$}
+\put( 9.3  ,11.2 ){$\scriptstyle D$}
+\put( 9.3  , 0.2 ){$\scriptstyle E$}
+\put(12.6  , 0.2 ){$\scriptstyle F$}
+\put(13.1  , 5.7 ){$\scriptstyle H$}
+\put( 0.7  , 0.2 ){$\scriptstyle I$}
+\put(19.0  , 0.7 ){$\scriptstyle K$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/210aa421.eepic b/33063-t/images/sources/210aa421.eepic
new file mode 100644
index 0000000..a6356ab
--- /dev/null
+++ b/33063-t/images/sources/210aa421.eepic
@@ -0,0 +1,47 @@
+\PGset[0.8em]
+\begin{picture}   (13, 8)
+
+\drawline     (1,1)(3,6)(12,6)(10,1)(1,1) % ABCDA
+
+\drawline     (7,6)(7,1) % a
+
+\put( 0.7  , 0.2 ){$\scriptstyle A$}
+\put( 2.9  , 6.2 ){$\scriptstyle B$}
+\put(11.3  , 6.2 ){$\scriptstyle C$}
+\put( 9.2  , 0.2 ){$\scriptstyle D$}
+
+\put( 6.3  , 3.7 ){$\scriptstyle a$} % a = 5
+\put( 4.0  , 0.2 ){$\scriptstyle b$} % b = 9
+
+\end{picture}
+\quad
+\begin{picture}   (8, 8)
+
+% what size is R? 6.7
+\dashline[80]{0.4}(1,1)(7.7,1)(7.7,7.7)(1,7.7)(1,1) % R
+
+\put( 4.1  , 4.1 ){$\scriptstyle R$}
+
+\end{picture}
+\quad
+\begin{picture}   (17, 9)
+
+\dashline[80]{0.2}(1,1)(16,1) % M(NO)X = 15
+
+\dashline[80]{0.2}(6,1)(6,7.7) % NP = 6.7
+
+% Ellipse:  u = 8.0  v = 1.0  a = 7.0  b = 7.0  phi = 0.0 Grad
+\qbezier[20](15.0, 1.0)(15.0, 3.8995)(12.9497, 5.9497)
+\qbezier[20](12.9497, 5.9497)(10.8995, 8.0)(8.0, 8.0)
+\qbezier[20](8.0, 8.0)(5.1005, 8.0)(3.0503, 5.9497)
+\qbezier[20](3.0503, 5.9497)(1.0, 3.8995)(1.0, 1.0)
+
+\put( 0.0  , 1.0){$\scriptstyle M$}
+\put( 5.5  , 0.2){$\scriptstyle N$}
+\put(14.5  , 0.2){$\scriptstyle O$}
+\put( 5.5  , 8.0){$\scriptstyle P$}
+\put(16.2  , 0.7){$\scriptstyle X$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/211aa424.eepic b/33063-t/images/sources/211aa424.eepic
new file mode 100644
index 0000000..999dadc
--- /dev/null
+++ b/33063-t/images/sources/211aa424.eepic
@@ -0,0 +1,35 @@
+\PGset[0.8em]
+\begin{picture}   (7, 7)
+
+\drawline     (1,1)(6,1)(6,6)(1,6)(1,1) % R
+
+\put( 3.2  , 3.2){$\scriptstyle R$}
+
+\end{picture}
+\quad
+% 363x133
+% 13x4.75
+\begin{picture}   (15, 8)
+
+\drawline     (1,1)(14,1) % M(C)N
+
+% Ellipse:  u = 7.5  v = 1.0  a = 6.5  b = 6.5  phi = 0.0 Grad
+\qbezier(14.0, 1.0)(14.0, 3.6924)(12.0962, 5.5962)
+\qbezier(12.0962, 5.5962)(10.1924, 7.5)(7.5, 7.5)
+\qbezier(7.5, 7.5)(4.8076, 7.5)(2.9038, 5.5962)
+\qbezier(2.9038, 5.5962)(1.0, 3.6924)(1.0, 1.0)
+
+\dashline[80]{0.2}(1,1)(1,6)(11.5,6) % MPQ
+\dashline[80]{0.2}(3.3,6)(3.3,1) % SC
+
+\put( 0.1  , 0.7){$\scriptstyle M$}
+\put( 3.0  , 0.2){$\scriptstyle C$}
+\put(14.2  , 0.7){$\scriptstyle N$}
+\put( 0.2  , 5.5){$\scriptstyle P$}
+\put( 2.8  , 6.2){$\scriptstyle S$}
+\put(12.0  , 5.8){$\scriptstyle Q$}
+
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/212aa425.eepic b/33063-t/images/sources/212aa425.eepic
new file mode 100644
index 0000000..ac2d066
--- /dev/null
+++ b/33063-t/images/sources/212aa425.eepic
@@ -0,0 +1,55 @@
+\PGset[0.8em]
+\begin{picture}   (9, 9)
+
+\drawline     (1,1)(8,1)(8,8)(1,8)(1,1) % R
+
+\put( 4.2  , 4.2){$\scriptstyle R$}
+
+\end{picture}
+\quad
+% diam = 8
+\begin{picture}   (10, 13)
+
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier(7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier(5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier(2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier(1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier(2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier(5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier(7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+\drawline          (1,5)(9,5) % MN
+\drawline          (1,5)(1,12)(6.985,1.527) % MSB
+% 1,12 extended through 5,5 by length=4
+% slope = 4/7
+% 16 = x^2 + (.5714x)^2
+% x = 3.473
+% y = 1.985
+% y/x = 4/7
+% y = .5714x
+% (ax)^2 = 
+
+\put( 0.1  , 4.7){$\scriptstyle M$}
+\put( 3.0  , 9.0){$\scriptstyle C$}
+\put( 9.0  , 4.7){$\scriptstyle N$}
+\put( 7.2  , 0.7){$\scriptstyle B$}
+\put( 0.8  , 12.1){$\scriptstyle S$}
+
+\end{picture}
+\quad
+\begin{picture}   (17, 9)
+
+% SB = base = 12
+% SC = alti =  4
+\dashline[80]{0.4}(1,1)(13,1)(16,5)(4,5)(1,1) % R
+\dashline[80]{0.2}(8,5)(8,1)
+
+%\dashline{0.2}
+
+\put( 4.9  , 2.9 ){$\scriptstyle R'$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/213aa426.eepic b/33063-t/images/sources/213aa426.eepic
new file mode 100644
index 0000000..2f1f237
--- /dev/null
+++ b/33063-t/images/sources/213aa426.eepic
@@ -0,0 +1,33 @@
+\PGset[0.7em]
+\begin{picture}   (10, 10)
+
+\drawline     (2.5,1)(1,6)(5,9)(9,6)(7.5,1)(2.5,1) % P
+
+\put( 4.2  , 4.2){$\scriptstyle P$}
+\put( 2.2  , 0.2){$\scriptstyle A$}
+\put( 6.7  , 0.2){$\scriptstyle B$}
+
+\end{picture}
+\quad
+\PGset[0.8em]
+\begin{picture}   (10, 9)
+% 260x230
+% 8x8
+
+\drawline     (2.5,1)(1,4.5)(2.5,8)(7.5,8)(9,4.5)(7.5,1)(2.5,1)
+
+\put ( 4.7  , 4.2){$\scriptstyle Q$}
+
+\end{picture}
+\quad
+\begin{picture}   (10, 10)
+
+\dashline[80]{0.4}(2.5,1)(1,6)(5,9)(9,6)(7.5,1)(2.5,1) % P
+
+\put( 4.2  , 4.2){$\scriptstyle P'$}
+\put( 2.2  , 0.2){$\scriptstyle A'$}
+\put( 6.7  , 0.2){$\scriptstyle B'$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/214aa427.eepic b/33063-t/images/sources/214aa427.eepic
new file mode 100644
index 0000000..6e0214e
--- /dev/null
+++ b/33063-t/images/sources/214aa427.eepic
@@ -0,0 +1,71 @@
+
+\PGset[0.8em]
+\begin{picture}   (9, 9)
+
+\drawline     (1,1)(8,1)(8,8)(1,8)(1,1) % R
+
+\put( 4.2  , 4.2){$\scriptstyle R$}
+
+\end{picture}
+\quad
+\PGset[0.8em]
+\begin{picture}   (6, 5)
+
+% m = 6, n = 4
+
+\drawline     (1,4)(7,4) % m
+\drawline     (1,2)(5,2) % n
+
+\put ( 0.2  , 3.7){$\scriptstyle m$}
+\put ( 0.2  , 1.7){$\scriptstyle n$}
+
+\end{picture}
+\quad
+\begin{picture}   (14, 14)
+
+\dashline[80]{0.2}(1,7)(12.668,7) % A(B)C
+
+\dashline[80]{0.2}(1,7)(13,1) % Ay, angle CAy = 45deg., y/x=0.5
+% AE = m = 6 = sqrt(x^2+y^2)
+% 36 = 4y^2 + y^2
+% Ey = 7-2.683, Ex = 1+5.366
+% E = (6.366, 4.317)
+
+% EF = n = 4 = sqrt(x^2+y^2)
+% 16 = 4y^2+y^2
+% Fy = 4.317-1.789, Fx = 6.366+3.578
+% F = (9.944, 2.528)
+
+% y = (13,1)
+
+\dashline[80]{0.2}(6.366, 4.317)(8,7) % EB, m=1.642
+
+% Cy=7, so y=4.472 = 1.642x, x = 2.724+Fx
+\dashline[80]{0.2}(9.944, 2.528)(12.668,7)   %FC, m=mEB
+
+% Ellipse:  u = 6.834  v = 7.0  a = 5.834  b = 5.834  phi = 0.0 Grad
+\qbezier[20](12.668, 7.0)(12.668, 9.4165)(10.9593, 11.1253)
+\qbezier[20](10.9593, 11.1253)(9.2505, 12.834)(6.834, 12.834)
+\qbezier[20](6.834, 12.834)(4.4175, 12.834)(2.7087, 11.1253)
+\qbezier[20](2.7087, 11.1253)(1.0, 9.4165)(1.0, 7.0)
+
+\dashline[80]{0.2}(8,7)(8,12.7) % BD
+
+
+\put( 0.2  , 6.7){$\scriptstyle A$}
+\put( 8.2  , 6.0){$\scriptstyle B$}
+\put(12.9  , 6.7){$\scriptstyle C$}
+\put( 8.2  ,12.9){$\scriptstyle D$}
+\put( 6.0  , 3.5){$\scriptstyle E$}
+\put( 9.5  , 1.7){$\scriptstyle F$}
+
+\put( 4.2  , 7.2){$\scriptstyle a$}
+\put(10.0  , 7.2){$\scriptstyle b$}
+\put( 3.0  , 5.1){$\scriptstyle m$}
+\put( 7.8  , 2.8){$\scriptstyle n$}
+\put( 7.2  , 9.0){$\scriptstyle x$}
+\put(13.3  , 1.0){$\scriptstyle y$}
+
+\end{picture}
+
+\PGrestore
diff --git a/33063-t/images/sources/215aa428.eepic b/33063-t/images/sources/215aa428.eepic
new file mode 100644
index 0000000..04fa9c3
--- /dev/null
+++ b/33063-t/images/sources/215aa428.eepic
@@ -0,0 +1,32 @@
+\PGset[0.8em]
+\begin{picture}   (10, 10)
+
+\drawline     (2,1)(1,6)(5,9)(9,6)(7,1)(2,1) % R
+
+\put( 4.2  , 4.2){$\scriptstyle R$}
+\put( 2.0  , 0.2){$\scriptstyle A$}
+\put( 6.2  , 0.2){$\scriptstyle B$}
+
+\end{picture}
+\begin{picture} (10, 5)
+
+\drawline     (1,4)(9,4) % m
+\drawline     (1,2)(8,2) % n
+
+\put( 0.1, 3.7 ){$\scriptstyle m$}
+\put( 0.3, 1.7 ){$\scriptstyle n$}
+
+\end{picture}
+\PGset[0.7em]
+\begin{picture}   (10, 10)
+
+\dashline[80]{0.4}(2,1)(1,6)(5,9)(9,6)(7,1)(2,1) % R
+
+\put( 4.2  , 4.2){$\scriptstyle S$}
+\put( 2.0  , 0.2){$\scriptstyle A'$}
+\put( 6.2  , 0.2){$\scriptstyle B'$}
+
+\end{picture}
+
+
+\PGrestore
diff --git a/33063-t/images/sources/217aa404.eepic b/33063-t/images/sources/217aa404.eepic
new file mode 100644
index 0000000..f55d065
--- /dev/null
+++ b/33063-t/images/sources/217aa404.eepic
@@ -0,0 +1,16 @@
+\PGset[0.8em]
+\begin{picture}   (9, 9)
+
+\drawline     (1,1)(5,8)(9,1)(1,1) % ABCA
+\drawline     (5,1)(5,8)           % DB
+
+\put( 1.0  , 0.2){$\scriptstyle A$}
+\put( 4.7  , 8.2){$\scriptstyle B$}
+\put( 8.2  , 0.2){$\scriptstyle C$}
+\put( 4.7  , 0.2){$\scriptstyle D$}
+\put( 2.3  , 4.5){$\scriptstyle a$}
+\put( 2.7  , 0.2){$\scriptstyle a/2$}
+\put( 5.2  , 3.7){$\scriptstyle h$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/217bb405.eepic b/33063-t/images/sources/217bb405.eepic
new file mode 100644
index 0000000..299c3bb
--- /dev/null
+++ b/33063-t/images/sources/217bb405.eepic
@@ -0,0 +1,17 @@
+\PGset[0.8em]
+\begin{picture}   (10, 7)
+
+\drawline     (1,1)(7,6)(9,1)(1,1) % ABCA
+\drawline     (7,1)(7,6)           % DB
+
+\put( 0.2  , 0.7){$\scriptstyle A$}
+\put( 6.7  , 6.2){$\scriptstyle B$}
+\put( 9.2  , 0.7){$\scriptstyle C$}
+\put( 6.7  , 0.2){$\scriptstyle D$}
+\put( 8.0  , 4.0){$\scriptstyle a$}
+\put( 3.7  , 0.2){$\scriptstyle b$}
+\put( 4.0  , 4.0){$\scriptstyle c$}
+\put( 6.2  , 2.5){$\scriptstyle h$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/217cc406.eepic b/33063-t/images/sources/217cc406.eepic
new file mode 100644
index 0000000..d7aab4f
--- /dev/null
+++ b/33063-t/images/sources/217cc406.eepic
@@ -0,0 +1,45 @@
+\PGset[0.8em]
+\begin{picture}   (12,12)
+
+% circle is 10x10 at 6,6
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier[20](11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier[20](9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier[20](6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier[20](2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier[20](1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier[20](2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier[20](6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier[20](9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% By = Cy = 7 (center+1)
+% 5 = sqrt(x^2 + 1^2), x = 4.899
+% Bx = 1.101, Cx = 10.899
+% Ax = 9 (center+3), Ay = 10
+
+\drawline(9,10)(1.101,7)(10.899,7)(9,10) % ABCA
+
+\drawline(9,7)(9,10) % DA
+
+% mAC = 3/1.899 = 1.58
+% 1/mAC = 0.6329, 0.6329x = y
+% 25 = x^2 + (0.6329x)^2 = 1.4x^2
+% 4.226, 2.674
+% E = (3.326, 1.774)
+
+\dashline[80]{0.2}(10.899,7)(3.326,1.774) % CE
+\drawline     (3.326,1.774)(9,10)     % EA
+
+\put( 9.5  ,10.0){$\scriptstyle A$}
+\put( 0.2  , 6.7){$\scriptstyle B$}
+\put(11.2  , 6.7){$\scriptstyle C$}
+\put( 8.5  , 6.2){$\scriptstyle D$}
+\put( 2.8  , 1.0){$\scriptstyle E$}
+
+\put( 3.8  , 6.2){$\scriptstyle a$}
+\put( 9.7  , 7.7){$\scriptstyle b$}
+\put( 5.8  , 9.0){$\scriptstyle c$}
+\put( 8.5  , 7.7){$\scriptstyle h$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/220aa430.eepic b/33063-t/images/sources/220aa430.eepic
new file mode 100644
index 0000000..cb6e16e
--- /dev/null
+++ b/33063-t/images/sources/220aa430.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}   (12,12)
+
+% sum of angles = 1260 deg., each angle = 140deg (40deg. at center)
+% so, we need 9 points spaced 40 deg. apart around a circle
+% radius = 5
+
+% x=6-5cos(theta), y=6-5sin(theta) for theta=0-360 by 40deg
+\drawline(1, 6)(2.170, 2.786)(5.132, 1.076)(8.5, 1.670)(10.699, 4.290)
+ (10.699, 7.710)(8.5, 10.330)(5.132, 10.924)(2.170, 9.214)(1,6)
+
+\put( 0.3, 5.5 ){$\scriptstyle A$}
+\put( 1.1, 9.0 ){$\scriptstyle B$}
+\put( 4.8,11.0 ){$\scriptstyle C$}
+\put( 8.5,10.5 ){$\scriptstyle D$}
+\put(10.7, 3.7 ){$\scriptstyle F$}
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/221aa431.eepic b/33063-t/images/sources/221aa431.eepic
new file mode 100644
index 0000000..15f2334
--- /dev/null
+++ b/33063-t/images/sources/221aa431.eepic
@@ -0,0 +1,48 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% x=6+5cos(theta), y=6+5sin(theta) for theta=234+72
+% ABCDEA
+\drawline(3.061,1.955)(8.939, 1.955)(10.755,7.545)(6, 11)(1.245,7.545)(3.061,1.955)
+
+% outer radius=5, inner radius=4.045
+
+\dashline[80]{0.2}(6,6)( 3.061, 1.955) %AO
+\dashline[80]{0.2}(6,6)( 8.939, 1.955) %BO
+\dashline[80]{0.2}(6,6)(10.755, 7.545) %CO
+\dashline[80]{0.2}(6,6)( 6,    11    ) %DO
+\dashline[80]{0.2}(6,6)( 1.245, 7.549) %EO
+
+\dashline[80]{0.2}(6,1.955)(6,10.045) % DF
+
+\put( 2.5, 1.1 ){$\scriptstyle A$}
+\put( 9.0, 1.1 ){$\scriptstyle B$}
+\put(11.0, 7.5 ){$\scriptstyle C$}
+\put( 5.6,11.2 ){$\scriptstyle D$}
+\put( 0.4, 7.5 ){$\scriptstyle E$}
+\put( 5.8, 1.1 ){$\scriptstyle F$}
+\put( 6.1, 6.4 ){$\scriptstyle O$}
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier[20](11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier[20](9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier[20](6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier[20](2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier[20](1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier[20](2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier[20](6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier[20](9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 4.045  b = 4.045  phi = 0.0 Grad
+\qbezier[20](10.045, 6.0)(10.045, 7.6755)(8.8602, 8.8602)
+\qbezier[20](8.8602, 8.8602)(7.6755, 10.045)(6.0, 10.045)
+\qbezier[20](6.0, 10.045)(4.3245, 10.045)(3.1398, 8.8602)
+\qbezier[20](3.1398, 8.8602)(1.955, 7.6755)(1.955, 6.0)
+\qbezier[20](1.955, 6.0)(1.955, 4.3245)(3.1398, 3.1398)
+\qbezier[20](3.1398, 3.1398)(4.3245, 1.955)(6.0, 1.955)
+\qbezier[20](6.0, 1.955)(7.6755, 1.955)(8.8602, 3.1398)
+\qbezier[20](8.8602, 3.1398)(10.045, 4.3245)(10.045, 6.0)
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/223aa439.eepic b/33063-t/images/sources/223aa439.eepic
new file mode 100644
index 0000000..c003497
--- /dev/null
+++ b/33063-t/images/sources/223aa439.eepic
@@ -0,0 +1,40 @@
+\PGset[0.8em]
+\begin{picture}(14,15)
+
+% inner radius = 6, center = 7,8
+% outer radius = 7.416
+
+% x=7+6cos(theta), y=8+6sin(theta) for theta=234+72
+% ABCDEA
+\drawline(3.472,3.146)(10.522,3.152)(12.706,9.854)(7.000,14.000)(1.300,9.848)(3.472,3.146)
+
+% x=7+7.416cos(theta), y=8+7.416sin(theta) for theta=270+72
+% FGHJKF
+\drawline(7.000,.584)(14.045,5.716)(11.360,13.999)(2.647,13.992)(-.052,5.709)(7.000,.584)
+
+\put ( 2.8, 2.5 ){$\scriptstyle A$}
+\put (10.5, 2.5 ){$\scriptstyle B$}
+\put (12.7, 9.8 ){$\scriptstyle C$}
+\put ( 6.6,14.1 ){$\scriptstyle D$}
+\put ( 0.4, 9.5 ){$\scriptstyle E$}
+
+\put ( 6.6, 0.0 ){$\scriptstyle F$}
+\put (14.1, 5.4 ){$\scriptstyle G$}
+\put (11.0,14.1 ){$\scriptstyle H$}
+\put ( 2.6,14.1 ){$\scriptstyle J$}
+\put (-0.7, 5.4 ){$\scriptstyle K$}
+
+% Ellipse:  u = 7.0  v = 8.0  a = 6.0  b = 6.0  phi = 0.0 Grad
+\qbezier(13.0, 8.0)(13.0, 10.4853)(11.2426, 12.2426)
+\qbezier(11.2426, 12.2426)(9.4853, 14.0)(7.0, 14.0)
+\qbezier(7.0, 14.0)(4.5147, 14.0)(2.7574, 12.2426)
+\qbezier(2.7574, 12.2426)(1.0, 10.4853)(1.0, 8.0)
+\qbezier(1.0, 8.0)(1.0, 5.5147)(2.7574, 3.7574)
+\qbezier(2.7574, 3.7574)(4.5147, 2.0)(7.0, 2.0)
+\qbezier(7.0, 2.0)(9.4853, 2.0)(11.2426, 3.7574)
+\qbezier(11.2426, 3.7574)(13.0, 5.5147)(13.0, 8.0)
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/224aa441.eepic b/33063-t/images/sources/224aa441.eepic
new file mode 100644
index 0000000..54e7a09
--- /dev/null
+++ b/33063-t/images/sources/224aa441.eepic
@@ -0,0 +1,55 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% outer radius = 5
+% x=6+5cos(theta), y=6+5sin(theta) for theta=234+72
+% ABCDEA
+\drawline(3.060,1.955)(8.935,1.960)(10.755,7.545)(6.000,11.000)(1.250,7.540)(3.060,1.955)
+
+\drawline(6,6)(3.060,1.955)
+\drawline(6,6)(8.935,1.960)
+\drawline(6,6)(10.755,7.545)
+\drawline(6,6)(6.000,11.000)
+\drawline(6,6)(1.250,7.540)
+
+% inner radius = 4.045
+% x=6+4.045cos(theta), y=6+4.045sin(theta) for theta=234+72
+% FGHJKF
+\drawline(3.622,2.728)(8.374,2.732)(9.846,7.249)(6.000,10.045)(2.158,7.245)(3.622,2.728)
+
+\drawline(6,11)(6,1.955)
+\drawline(1.25,7.54)(9.842,4.755)
+
+\put ( 2.8, 1.2 ){$\scriptstyle A'$}
+\put ( 8.5, 1.2 ){$\scriptstyle B'$}
+\put (11.0, 7.2 ){$\scriptstyle C'$}
+\put ( 5.8,11.1 ){$\scriptstyle D'$}
+\put ( 0.4, 7.2 ){$\scriptstyle E'$}
+
+\put ( 6.2, 6.2 ){$\scriptstyle O$}
+
+\put ( 3.4, 2.0 ){$\scriptscriptstyle A$}
+\put ( 8.0, 2.0 ){$\scriptscriptstyle B$}
+\put ( 9.8, 7.4 ){$\scriptscriptstyle C$}
+\put ( 6.0,10.2 ){$\scriptscriptstyle D$}
+\put ( 1.6, 7.4 ){$\scriptscriptstyle E$}
+
+\put ( 5.6, 1.2 ){$\scriptstyle M$}
+\put ( 9.9, 4.4 ){$\scriptstyle N$}
+\put ( 8.5, 9.3 ){$\scriptstyle P$}
+\put ( 2.8, 9.3 ){$\scriptstyle Q$}
+\put ( 1.2, 4.4 ){$\scriptstyle R$}
+
+% Ellipse:  u = 6.0  v = 6.0  a = 4.045  b = 4.045  phi = 0.0 Grad
+\qbezier(10.045, 6.0)(10.045, 7.6755)(8.8602, 8.8602)
+\qbezier(8.8602, 8.8602)(7.6755, 10.045)(6.0, 10.045)
+\qbezier(6.0, 10.045)(4.3245, 10.045)(3.1398, 8.8602)
+\qbezier(3.1398, 8.8602)(1.955, 7.6755)(1.955, 6.0)
+\qbezier(1.955, 6.0)(1.955, 4.3245)(3.1398, 3.1398)
+\qbezier(3.1398, 3.1398)(4.3245, 1.955)(6.0, 1.955)
+\qbezier(6.0, 1.955)(7.6755, 1.955)(8.8602, 3.1398)
+\qbezier(8.8602, 3.1398)(10.045, 4.3245)(10.045, 6.0)
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/224bb442.eepic b/33063-t/images/sources/224bb442.eepic
new file mode 100644
index 0000000..7be7589
--- /dev/null
+++ b/33063-t/images/sources/224bb442.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(10,10)
+
+% radius = 4
+% x=5+4cos(theta), y=5+4sin(theta) for theta=45+90
+% BADCB
+\drawline(7.828,7.828)(2.176,7.828)(2.172,2.172)(7.824,2.172)(7.828,7.828)
+
+% x=5+4cos(theta), y=5+4sin(theta) for theta=45+45
+% BEAKDHCFB
+\drawline(7.828,7.828)(5.000,9.000)(2.176,7.828)(1.000,5.000)(2.172,2.172)(5.000,1.000)(7.824,2.172)(8.996,5.000)(7.828,7.828)
+
+\put ( 1.6, 7.9 ){$\scriptstyle A$}
+\put ( 7.9, 7.9 ){$\scriptstyle B$}
+\put ( 7.8, 1.6 ){$\scriptstyle C$}
+\put ( 1.4, 1.6 ){$\scriptstyle D$}
+\put ( 4.7, 9.2 ){$\scriptstyle E$}
+\put ( 9.1, 4.7 ){$\scriptstyle F$}
+\put ( 4.5, 0.3 ){$\scriptstyle H$}
+\put ( 0.2, 4.7 ){$\scriptstyle K$}
+
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier(7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier(5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier(2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier(1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier(2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier(5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier(7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/224cc443.eepic b/33063-t/images/sources/224cc443.eepic
new file mode 100644
index 0000000..076c8e7
--- /dev/null
+++ b/33063-t/images/sources/224cc443.eepic
@@ -0,0 +1,44 @@
+\PGset[0.8em]
+\begin{picture}(10,10)
+
+% inside radius = 4, middle=4.329, outside = 5.657
+% x=5+5.657cos(theta), y=5+5.657sin(theta) for theta=225+90
+% ABCDA
+\drawline(1.001,1.001)(8.993,1.001)(8.999,8.999)(1.007,8.999)(1.001,1.001)
+
+% x=5+4.329cos(theta), y=5+4.329sin(theta) for theta=247.5+45
+% EFGHIKLM
+\drawline(3.342,1.005)(6.653,1.005)(8.995,3.347)(8.999,6.653)(6.658,8.995)(3.347,8.995)(1.005,6.653)(1.001,3.347)(3.342,1.005)
+
+\drawline(5,5)( 9,5)
+
+\put ( 0.8, 9.1 ){$\scriptstyle D$}
+\put ( 3.0, 9.1 ){$\scriptstyle K$}
+\put ( 6.5, 9.1 ){$\scriptstyle I$}
+\put ( 8.3, 9.1 ){$\scriptstyle C$}
+
+\put ( 0.1, 6.3 ){$\scriptstyle L$}
+\put ( 9.1, 6.3 ){$\scriptstyle H$}
+\put ( 0.1, 3.1 ){$\scriptstyle M$}
+\put ( 9.1, 3.1 ){$\scriptstyle G$}
+
+\put ( 0.8, 0.3 ){$\scriptstyle A$}
+\put ( 3.0, 0.3 ){$\scriptstyle E$}
+\put ( 6.5, 0.3 ){$\scriptstyle F$}
+\put ( 8.3, 0.3 ){$\scriptstyle B$}
+
+\put ( 4.7, 5.1 ){$\scriptstyle O$}
+
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier(7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier(5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier(2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier(1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier(2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier(5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier(7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/225aa445.eepic b/33063-t/images/sources/225aa445.eepic
new file mode 100644
index 0000000..f660d5a
--- /dev/null
+++ b/33063-t/images/sources/225aa445.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(20.5,11)
+
+% small r = 4, large r = 5
+% x=5+4cos(theta), y=5+4sin(theta) for theta=225+72
+% ABCDEA
+\drawline(2.6484,1.7640)(7.3504,1.7640)(8.8044,6.2360)(5.0004,8.9996)(1.1964,6.2360)(2.6484,1.7640)
+% x=15+5cos(theta), y=5+5sin(theta) for theta=225+72
+% ABCDEA
+\drawline(12.0605,.9550)(17.9380,.9550)(19.7555,6.5450)(15.0005,9.9995)(10.2455,6.5450)(12.0605,.9550)
+
+
+\put(2.2, 1.0){$\scriptstyle A$}
+\put(6.9, 1.0){$\scriptstyle B$}
+\put(8.8, 5.8){$\scriptstyle C$}
+\put(4.7, 9.1){$\scriptstyle D$}
+\put(0.6, 5.9){$\scriptstyle E$}
+\put(4.7, 5.2){$\scriptstyle Q$}
+
+\put(12.0,  .2){$\scriptstyle A'$}
+\put(17.1,  .2){$\scriptstyle B'$}
+\put(19.9, 6.6){$\scriptstyle C'$}
+\put(14.8,10.2){$\scriptstyle D'$}
+\put( 9.4, 6.6){$\scriptstyle E'$}
+\put(14.8, 5.2){$\scriptstyle Q'$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/226aa447.eepic b/33063-t/images/sources/226aa447.eepic
new file mode 100644
index 0000000..712016c
--- /dev/null
+++ b/33063-t/images/sources/226aa447.eepic
@@ -0,0 +1,41 @@
+
+\PGset[0.8em]
+\begin{picture}(20.5,15.5)
+
+% small r = 4, large r = 5
+% x=5+4cos(theta), y=5+4sin(theta) for theta=225+72
+% ABCDEA
+\drawline(2.6484,1.7640)(7.3504,1.7640)(8.8044,6.2360)(5.0004,8.9996)(1.1964,6.2360)(2.6484,1.7640)
+% x=15+5cos(theta), y=5+5sin(theta) for theta=225+72
+% ABCDEA
+\drawline(12.0605,.9550)(17.9380,.9550)(19.7555,6.5450)(15.0005,9.9995)(10.2455,6.5450)(12.0605,.9550)
+
+\dashline[80]{0.2}(5,5)(2.6484,1.7640) % OA
+\dashline[80]{0.2}(5,5)(5.0,   1.7640) % OM
+\dashline[80]{0.2}(5,5)(7.3504,1.7640) % OB
+
+\dashline[80]{0.2}(15,5)(12.0605,0.9550) % OA
+\dashline[80]{0.2}(15,5)(15.0,   0.9550) % OM
+\dashline[80]{0.2}(15,5)(17.9380,0.9550) % OB
+
+
+\put(2.2, 1.0){$\scriptstyle A$}
+\put(6.9, 1.0){$\scriptstyle B$}
+\put(8.8, 5.8){$\scriptstyle C$}
+\put(4.7, 9.1){$\scriptstyle D$}
+\put(0.6, 5.9){$\scriptstyle E$}
+\put(4.7, 5.2){$\scriptstyle O$}
+\put(4.4, 1.0){$\scriptstyle M$}
+
+\put(12.0,  .2){$\scriptstyle A'$}
+\put(17.1,  .2){$\scriptstyle B'$}
+\put(19.9, 6.6){$\scriptstyle C'$}
+\put(14.8,10.2){$\scriptstyle D'$}
+\put( 9.4, 6.6){$\scriptstyle E'$}
+\put(14.8, 5.2){$\scriptstyle O'$}
+\put(14.5, 0.2){$\scriptstyle M'$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/227aa449.eepic b/33063-t/images/sources/227aa449.eepic
new file mode 100644
index 0000000..955e7db
--- /dev/null
+++ b/33063-t/images/sources/227aa449.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(10,10)
+
+% radius = 4, hex radius = 3.4644
+% x=5+3.4644cos(theta), y=5+3.4644sin(theta) for theta=30+60
+% AB...
+\drawline(9.0000,5.0000)(7.0004,8.4640)(3.0008,8.4640)(1.0004,5.0000)(2.9996,1.5360)(6.9992,1.5360)(9.0000,5.0000)
+
+\drawline(5,5)(3,1.536) %OA
+\drawline(5,5)(5,1.536) %OP
+
+\put(4.6, 5.2){$\scriptstyle O$}
+
+\put(2.4, 0.6){$\scriptstyle A$}
+\put(6.6, 0.6){$\scriptstyle B$}
+\put(5.1, 1.6){$\scriptstyle P$}
+
+
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier(7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier(5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier(2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier(1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier(2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier(5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier(7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/228aa451.eepic b/33063-t/images/sources/228aa451.eepic
new file mode 100644
index 0000000..554ee07
--- /dev/null
+++ b/33063-t/images/sources/228aa451.eepic
@@ -0,0 +1,39 @@
+\PGset[0.8em]
+\begin{picture}(11,7)
+
+% radius = 5
+% Ellipse:  u = 6.0  v = 1.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 1.0)(11.0, 3.0711)(9.5355, 4.5355)
+\qbezier(9.5355, 4.5355)(8.0711, 6.0)(6.0, 6.0)
+\qbezier(6.0, 6.0)(3.9289, 6.0)(2.4645, 4.5355)
+\qbezier(2.4645, 4.5355)(1.0, 3.0711)(1.0, 1.0)
+
+% Ellipse:  u = 6.0  v = -1.0  a = 5.3851  b = 5.3851  phi = 22.0 Grad
+\qbezier(10.993, 1.0173)(10.1574, 3.0855)(8.1041, 3.957)
+\qbezier(8.1041, 3.957)(6.0509, 4.8286)(3.9827, 3.993)
+\qbezier(3.9827, 3.993)(1.9145, 3.1574)(1.043, 1.1041)
+
+\drawline(1,1)(11,1)
+
+% O = 6,-1, r=5.3851
+% EO = 6.2452, EC^2 + r^2 = EO^2, ET = 3.1627
+% mid = 3.835,1.25 r=3.1126
+% (x-6)^2 + (y+1)^2 = 5.3851^2
+% (x-3.835)^2 + (y-1.25)^2 = 3.1126^2
+% C=4.726, 4.232
+% (x-6)^2 + (y-1)^2 = 5^2
+% (x-4.726)=4.1755(y-4.232)
+% F = 8.728,5.190
+\dashline[80]{0.2}(1.6695,3.5)(8.728,5.190) % EF
+
+\put( 1.0, 0.3){$\scriptstyle A$}
+\put(10.1, 0.3){$\scriptstyle B$}
+\put( 5.1, 4.6){$\scriptstyle C$}
+\put( 5.4, 6.3){$\scriptstyle D$}
+\put( 0.8, 3.3){$\scriptstyle E$}
+\put( 8.9, 5.2){$\scriptstyle F$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/229aa454.eepic b/33063-t/images/sources/229aa454.eepic
new file mode 100644
index 0000000..fe0aca5
--- /dev/null
+++ b/33063-t/images/sources/229aa454.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(14,14)
+
+% radius = 6
+% Ellipse:  u = 7.0  v = 7.0  a = 6.0  b = 6.0  phi = 0.0 Grad
+\qbezier(13.0, 7.0)(13.0, 9.4853)(11.2426, 11.2426)
+\qbezier(11.2426, 11.2426)(9.4853, 13.0)(7.0, 13.0)
+\qbezier(7.0, 13.0)(4.5147, 13.0)(2.7574, 11.2426)
+\qbezier(2.7574, 11.2426)(1.0, 9.4853)(1.0, 7.0)
+\qbezier(1.0, 7.0)(1.0, 4.5147)(2.7574, 2.7574)
+\qbezier(2.7574, 2.7574)(4.5147, 1.0)(7.0, 1.0)
+\qbezier(7.0, 1.0)(9.4853, 1.0)(11.2426, 2.7574)
+\qbezier(11.2426, 2.7574)(13.0, 4.5147)(13.0, 7.0)
+
+\drawline(7,7)(4,1)(10,1)(7,7) % OA'B'O
+\drawline(7,7)(7,1) % OC
+
+% (x-7)^2 + (y-7)^2 = 6^2
+% (y-7)=2(x-7)
+\drawline(4.3167, 1.6334)(9.6833,1.6334) % AB
+
+
+\put( 6.7, 7.0){$\scriptstyle O$}
+
+\put( 3.4, 1.2){$\scriptstyle A$}
+\put(10.0, 1.2){$\scriptstyle B$}
+\put( 3.6, 0.2){$\scriptstyle A'$}
+\put( 9.2, 0.2){$\scriptstyle B'$}
+
+\put( 6.6, 0.2){$\scriptstyle C$}
+\put( 7.1, 1.8){$\scriptstyle D$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/231aa455.eepic b/33063-t/images/sources/231aa455.eepic
new file mode 100644
index 0000000..39ab80e
--- /dev/null
+++ b/33063-t/images/sources/231aa455.eepic
@@ -0,0 +1,37 @@
+\PGset[0.8em]
+\begin{picture}(26,14)
+
+% radius = 6
+% Ellipse:  u = 7.0  v = 7.0  a = 6.0  b = 6.0  phi = 0.0 Grad
+\qbezier(13.0, 7.0)(13.0, 9.4853)(11.2426, 11.2426)
+\qbezier(11.2426, 11.2426)(9.4853, 13.0)(7.0, 13.0)
+\qbezier(7.0, 13.0)(4.5147, 13.0)(2.7574, 11.2426)
+\qbezier(2.7574, 11.2426)(1.0, 9.4853)(1.0, 7.0)
+\qbezier(1.0, 7.0)(1.0, 4.5147)(2.7574, 2.7574)
+\qbezier(2.7574, 2.7574)(4.5147, 1.0)(7.0, 1.0)
+\qbezier(7.0, 1.0)(9.4853, 1.0)(11.2426, 2.7574)
+\qbezier(11.2426, 2.7574)(13.0, 4.5147)(13.0, 7.0)
+
+% Ellipse:  u = 20.0  v = 7.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(25.0, 7.0)(25.0, 9.0711)(23.5355, 10.5355)
+\qbezier(23.5355, 10.5355)(22.0711, 12.0)(20.0, 12.0)
+\qbezier(20.0, 12.0)(17.9289, 12.0)(16.4645, 10.5355)
+\qbezier(16.4645, 10.5355)(15.0, 9.0711)(15.0, 7.0)
+\qbezier(15.0, 7.0)(15.0, 4.9289)(16.4645, 3.4645)
+\qbezier(16.4645, 3.4645)(17.9289, 2.0)(20.0, 2.0)
+\qbezier(20.0, 2.0)(22.0711, 2.0)(23.5355, 3.4645)
+\qbezier(23.5355, 3.4645)(25.0, 4.9289)(25.0, 7.0)
+
+\dashline[80]{0.2}(13.0000,7.0000)(11.2432,11.2426)(7.0006,12.9994)(2.7586,11.2426)(1.0006,7.0000)(2.7568,2.7574)(6.9994,1.0000)(11.2414,2.7574)(13.0000,7.0000)
+
+\drawline(7,7)(7,1)
+
+\dashline[80]{0.2}(25.0000,7.0000)(23.5360,10.5355)(20.0005,11.9995)(16.4655,10.5355)(15.0005,7.0000)(16.4640,3.4645)(19.9995,2.0000)(23.5345,3.4645)(25.0000,7.0000)
+
+\drawline(20,7)(20,2)
+
+\put( 6.7,  7.2){$\scriptstyle Q$}
+\put( 19.7, 7.2){$\scriptstyle Q'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/232aa459.eepic b/33063-t/images/sources/232aa459.eepic
new file mode 100644
index 0000000..d9c2b01
--- /dev/null
+++ b/33063-t/images/sources/232aa459.eepic
@@ -0,0 +1,26 @@
+\PGset[0.8em]
+\begin{picture}(14,13)
+
+% radius = 6
+\drawline(13.0000,7.0000)(10.0006,12.1960)(4.0012,12.1960)(1.0006,7.0000)(3.9994,1.8040)(9.9988,1.8040)(13.0000,7.0000)
+
+\dashline[80]{0.2}(7,7)(4, 1.804) % OA
+\dashline[80]{0.2}(7,7)(10,1.804) % OB
+\dashline[80]{0.2}(7,7)(13,7)     % OC
+\dashline[80]{0.2}(7,7)(10, 12.196) % OD
+\dashline[80]{0.2}(7,7)(4, 12.196)  % OE
+\dashline[80]{0.2}(7,7)(1, 7)       % OF
+
+\drawline(7,7)(7,1.804) % OM
+
+\put( 3.9, 0.9){$\scriptstyle A$}
+\put( 9.2, 0.9){$\scriptstyle B$}
+\put(13.1, 6.7){$\scriptstyle C$}
+\put( 9.2,12.3){$\scriptstyle D$}
+\put( 3.9,12.3){$\scriptstyle E$}
+\put( 0.3, 6.7){$\scriptstyle F$}
+\put( 6.7, 7.4){$\scriptstyle O$}
+\put( 6.5, 0.9){$\scriptstyle M$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/233aa461.eepic b/33063-t/images/sources/233aa461.eepic
new file mode 100644
index 0000000..e31f0e0
--- /dev/null
+++ b/33063-t/images/sources/233aa461.eepic
@@ -0,0 +1,30 @@
+\PGset[0.8em]
+\begin{picture}(11,11)
+
+% circle rad=4
+% Ellipse:  u = 5.0  v = 5.0  a = 4.0  b = 4.0  phi = 0.0 Grad
+\qbezier(9.0, 5.0)(9.0, 6.6569)(7.8284, 7.8284)
+\qbezier(7.8284, 7.8284)(6.6569, 9.0)(5.0, 9.0)
+\qbezier(5.0, 9.0)(3.3431, 9.0)(2.1716, 7.8284)
+\qbezier(2.1716, 7.8284)(1.0, 6.6569)(1.0, 5.0)
+\qbezier(1.0, 5.0)(1.0, 3.3431)(2.1716, 2.1716)
+\qbezier(2.1716, 2.1716)(3.3431, 1.0)(5.0, 1.0)
+\qbezier(5.0, 1.0)(6.6569, 1.0)(7.8284, 2.1716)
+\qbezier(7.8284, 2.1716)(9.0, 3.3431)(9.0, 5.0)
+
+% pent rad=4.9437
+\dashline[80]{0.2}(5.0004,9.9432)(.2991,6.5276)(2.0936,1.0006)(7.9049,1.0006)(9.7019,6.5276)(5.0004,9.9437)
+
+\drawline(5,5)(5,1)
+
+\put( 4.6, 5.2){$\scriptstyle O$}
+\put( 4.5, 0.2){$\scriptstyle M$}
+
+\put(-0.5, 6.2){$\scriptstyle A$}
+\put( 2.0, 0.2){$\scriptstyle B$}
+\put( 7.2, 0.2){$\scriptstyle C$}
+\put( 9.8, 6.2){$\scriptstyle D$}
+\put( 4.6,10.1){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/234aa466.eepic b/33063-t/images/sources/234aa466.eepic
new file mode 100644
index 0000000..d82d6b0
--- /dev/null
+++ b/33063-t/images/sources/234aa466.eepic
@@ -0,0 +1,38 @@
+
+\PGset[0.8em]
+\begin{picture}(14.5,8)
+
+% 73x182
+% base = 5, h = 6.5
+
+% A =  1.0, 1.0
+% B =  6.0, 1.0
+% C =  3.5, 7.5
+\drawline(1,1)(6,1)(3.5,7.5)(1,1)
+
+% arc
+% P1 = (1.0/1.0)  P2 = (3.5/7.5)  P3 = (6.0/1.0)  r = 6.9641
+\qbezier(1.0, 1.0001)(3.5, 0.0386)(6.0, 1.0001)
+
+% base2 = 4.7, h2 = 5.8
+% A' =  9.0, 1.0
+% B' = 13.7, 1.0
+% C' = 11.35,6.8
+\drawline(9,1)(13.7,1)(11.35,6.8)(9,1)
+
+% arc
+% P1 = (9.0/1.0)  P2 = (11.35/6.8)  P3 = (13.7/1.0)  r = 6.2579
+\qbezier(9.0, 1.0001)(11.35, 0.0479)(13.7, 1.0001)
+
+\put(  0.2,  0.6  ){$\scriptstyle A$}
+\put(  6.0,  0.6  ){$\scriptstyle B$}
+\put(  3.3,  7.7  ){$\scriptstyle C$}
+\put(  3.1, -0.2  ){$\scriptstyle P$}
+
+\put(  8.0,  0.6  ){$\scriptstyle A'$}
+\put( 13.7,  0.6  ){$\scriptstyle B'$}
+\put( 11.1,  7.0  ){$\scriptstyle C'$}
+\put( 11.0, -0.2  ){$\scriptstyle P'$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/235aa467.eepic b/33063-t/images/sources/235aa467.eepic
new file mode 100644
index 0000000..45df60c
--- /dev/null
+++ b/33063-t/images/sources/235aa467.eepic
@@ -0,0 +1,28 @@
+
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% radius=5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\dashline[80]{0.4}(9.5360,9.5355)(2.4655,9.5355)(2.4640,2.4645)(9.5345,2.4645)(9.5360,9.5355)
+
+\dashline[80]{0.2}(9.5360,9.5355)(2.4640,2.4645)
+\dashline[80]{0.2}(2.4655,9.5355)(9.5345,2.4645)
+
+\put(  5.7,  6.5  ){$\scriptstyle O$}
+\put(  1.8,  1.8  ){$\scriptstyle A$}
+\put(  1.8,  9.6  ){$\scriptstyle B$}
+\put(  9.6,  9.6  ){$\scriptstyle C$}
+\put(  9.6,  1.8  ){$\scriptstyle D$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/236aa469.eepic b/33063-t/images/sources/236aa469.eepic
new file mode 100644
index 0000000..0abf208
--- /dev/null
+++ b/33063-t/images/sources/236aa469.eepic
@@ -0,0 +1,30 @@
+
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% radius=5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\dashline[80]{0.4}(11.0000,6.0000)(8.5005,10.3300)(3.5010,10.3300)(1.0005,6.0000)(3.4995,1.6700)(8.4990,1.6700)(11.0000,6.0000)
+
+\dashline[80]{0.2}(8.5005,10.3300)(6,6)(11,6) % FOC
+
+\put(  5.3,  5.7  ){$\scriptstyle O$}
+
+\put(  2.7,  1.0  ){$\scriptstyle A$}
+\put(  8.8,  1.0  ){$\scriptstyle B$}
+\put( 11.1,  5.7  ){$\scriptstyle C$}
+\put(  2.7, 10.4  ){$\scriptstyle D$}
+\put(  0.1,  5.7  ){$\scriptstyle E$}
+\put(  8.8, 10.4  ){$\scriptstyle F$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/237aa472.eepic b/33063-t/images/sources/237aa472.eepic
new file mode 100644
index 0000000..8686f50
--- /dev/null
+++ b/33063-t/images/sources/237aa472.eepic
@@ -0,0 +1,34 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% radius=5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% hard to see unless the dashes are closer together
+\dashline[80]{0.4}(10.7555,7.5450)(8.9395,10.0450)(6.0005,10.9995)(3.0620,10.0450)(1.2455,7.5450)(1.2445,4.4550)(3.0605,1.9550)(5.9995,1.0000)(8.9380,1.9550)(10.7545,4.4550)(10.7555,7.5450)
+
+\dashline[80]{0.2}(5.9995,1)(6,6)(3.0605,1.955) % BOC
+
+% OS = 3.09, m=1.376, S = O - (1.817,2.5) = (4.183,3.5)
+
+\dashline[80]{0.2}(5.9995,1)(4.183,3.5) % BS
+
+\put(  5.8, 6.2  ){$\scriptstyle O$}
+
+\put(  5.6, 0.2  ){$\scriptstyle B$}
+\put(  2.5, 1.1  ){$\scriptstyle C$}
+\put(  0.4, 4.1  ){$\scriptstyle F$}
+\put(  3.6, 3.4  ){$\scriptstyle S$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/239aa475.eepic b/33063-t/images/sources/239aa475.eepic
new file mode 100644
index 0000000..0f1b7cb
--- /dev/null
+++ b/33063-t/images/sources/239aa475.eepic
@@ -0,0 +1,37 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+% radius = 5, side of decagon = 3.0899, side of pentedecagon = 2.0785
+
+% Ey = Hy, EH = 5, both on the circle r=5, Ex = 6-2.5, Hx = 6+2.5
+% Ey = Hy = 6-4.33 = 1.67
+\dashline[80]{0.2}(3.5,1.67)(8.5,1.67)(6.523,1.029)(3.5,1.67) % EHFE
+
+
+\dashline[80]{0.2}(8.4990,1.6700)(10.0445,3.0615)(10.8905,4.9605)(10.8905,7.0395)(10.0455,8.9385)
+
+\put(  5.7, 5.7  ){$\scriptstyle Q$}
+
+\put( 10.0, 2.5  ){$\scriptstyle A$}
+\put( 11.0, 4.8  ){$\scriptstyle B$}
+\put( 11.0, 6.8  ){$\scriptstyle C$}
+\put( 10.3, 8.7  ){$\scriptstyle D$}
+
+\put(  2.8, 0.9  ){$\scriptstyle E$}
+\put(  6.2, 0.2  ){$\scriptstyle F$}
+\put(  8.5, 0.9  ){$\scriptstyle H$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/240aa477.eepic b/33063-t/images/sources/240aa477.eepic
new file mode 100644
index 0000000..0d8a11f
--- /dev/null
+++ b/33063-t/images/sources/240aa477.eepic
@@ -0,0 +1,47 @@
+\PGset[0.8em]
+\begin{picture}(24,11)
+
+% radius1 = 5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\dashline[80]{0.4}(11.0000,6.0000)(8.5005,10.3300)(3.5010,10.3300)(1.0005,6.0000)(3.4995,1.6700)(8.4990,1.6700)(11.0000,6.0000)
+
+\dashline[80]{0.2}(3.501,10.33)(6,6)(8.5005,10.33) % C'O'D'
+
+% radius2 = 4.5
+\drawline(22.5000,6.0000)(20.2504,9.8970)(15.7509,9.8970)(13.5005,6.0000)(15.7496,2.1030)(20.2491,2.1030)(22.5000,6.0000)
+
+\dashline[80]{0.2}(15.7509,9.8)(18,6)(20.2504,9.8) % COD
+
+\put(  5.6, 5.2  ){$\scriptstyle O'$}
+
+\put(  2.7, 0.7  ){$\scriptstyle A'$}
+\put(  0.0, 5.7  ){$\scriptstyle B'$}
+\put(  2.8,10.5  ){$\scriptstyle C'$}
+\put(  8.4,10.5  ){$\scriptstyle D'$}
+\put( 11.2, 5.7  ){$\scriptstyle E'$}
+\put(  8.3, 0.7  ){$\scriptstyle F'$}
+
+\put( 17.6, 5.2  ){$\scriptstyle O$}
+
+\put( 15.2, 1.2  ){$\scriptstyle A$}
+\put( 12.7, 5.7  ){$\scriptstyle B$}
+\put( 15.5,10.1  ){$\scriptstyle C$}
+\put( 20.0,10.1  ){$\scriptstyle D$}
+\put( 22.6, 5.7  ){$\scriptstyle E$}
+\put( 20.0, 1.2  ){$\scriptstyle F$}
+
+
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/241aa478.eepic b/33063-t/images/sources/241aa478.eepic
new file mode 100644
index 0000000..8640567
--- /dev/null
+++ b/33063-t/images/sources/241aa478.eepic
@@ -0,0 +1,33 @@
+
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% radius = 5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\dashline[80]{0.2}(6,11)(6,1) % DH
+
+% side of inscribed pentagon = 5sin(36) = 2.9385
+\drawline(6,11)(3.062,10.045)(8.9395,10.045) % DAB
+
+\dashline[80]{0.2}(6,6)(3.062,10.045)(6,1) % OAH
+
+
+\put(  6.2, 5.7  ){$\scriptstyle O$}
+
+\put(  2.4,10.2  ){$\scriptstyle A$}
+\put(  9.2,10.0  ){$\scriptstyle B$}
+\put(  6.2, 9.2  ){$\scriptstyle C$}
+\put(  5.7,11.2  ){$\scriptstyle D$}
+\put(  5.6, 0.2  ){$\scriptstyle H$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/242aa480.eepic b/33063-t/images/sources/242aa480.eepic
new file mode 100644
index 0000000..62bc84a
--- /dev/null
+++ b/33063-t/images/sources/242aa480.eepic
@@ -0,0 +1,18 @@
+\PGset[0.8em]
+\begin{picture}(12,12)
+
+% radius = 5
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\drawline(1,6)(11,6)
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/243aa484.eepic b/33063-t/images/sources/243aa484.eepic
new file mode 100644
index 0000000..6dc8fdb
--- /dev/null
+++ b/33063-t/images/sources/243aa484.eepic
@@ -0,0 +1,20 @@
+\PGset[0.8em]
+\begin{picture}(17,9)
+
+\dashline[80]{0.2}(1,6)(1,1)(5,1) % EDB
+\dashline[80]{0.2}(9,7)(9,1)      % ED
+
+\drawline(5,1)(1,6)(16,1) % BEC
+\drawline(5,1)(5,8)(16,1) % BAC
+\drawline(5,1)(9,7)(16,1)(5,1) % BECB
+
+\put( 4.7, 8.1 ){$\scriptstyle A$}
+\put( 4.5, 0.2 ){$\scriptstyle B$}
+\put(16.2, 0.2 ){$\scriptstyle C$}
+\put( 0.5, 0.2 ){$\scriptstyle D$}
+\put( 8.5, 0.2 ){$\scriptstyle D$}
+\put( 0.7, 6.1 ){$\scriptstyle E$}
+\put( 8.7, 7.1 ){$\scriptstyle E$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/244aa485.eepic b/33063-t/images/sources/244aa485.eepic
new file mode 100644
index 0000000..02716e3
--- /dev/null
+++ b/33063-t/images/sources/244aa485.eepic
@@ -0,0 +1,43 @@
+\PGset[0.8em]
+\begin{picture}(16,18)
+
+% AB = 14, AE=EB=7
+% CE = 4
+% H=+17, P = 1, A=9
+
+\drawline(1,9)(8,13)(15,9)(1,9) % ACB
+
+\dashline[80]{0.2}(8,13)(8,9) % CE
+\dashline[80]{0.2}(8,13)(15,17)(15,13) % CHK
+\dashline[80]{0.2}(8,13)(15,13)(15,1.8063)  % CKP
+
+% perim = 14+2*8.0622 = 30.1244
+% 16.1244 = AD+DB
+% AD = 11.1244, DB = 5
+% A = 1,9, B = 15,9
+
+% (x-1)^2+(y-9)^2 = 11.1244^2
+% (x-15)^2+(y-9)^2 = 5^2
+% x = 11.5269, y = 5.40315
+
+\drawline(1,9)(11.5269,5.40315)(15,9) % ADB
+
+\dashline[80]{0.2}(11.5269,9)(11.5269,5.40315)(15,5.40315) % FDM
+
+\dashline[80]{0.2}(11.5269,5.40315)(15,1.8063)(1,9) % DPA
+
+
+
+\put( 0.2, 8.6 ){$\scriptstyle A$}
+\put(15.1, 8.6 ){$\scriptstyle B$}
+\put( 7.6,13.2 ){$\scriptstyle C$}
+\put(11.0, 4.7 ){$\scriptstyle D$}
+\put( 7.7, 8.1 ){$\scriptstyle E$}
+\put(11.2, 9.2 ){$\scriptstyle F$}
+\put(15.1,16.8 ){$\scriptstyle H$}
+\put(15.1,12.7 ){$\scriptstyle K$}
+\put(15.1, 5.1 ){$\scriptstyle M$}
+\put(15.1, 1.4 ){$\scriptstyle P$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/245aa486.eepic b/33063-t/images/sources/245aa486.eepic
new file mode 100644
index 0000000..1fdee2a
--- /dev/null
+++ b/33063-t/images/sources/245aa486.eepic
@@ -0,0 +1,29 @@
+\PGset[0.8em]
+\begin{picture}(24,11)
+
+% MN = 22
+\drawline(1,1)(23,1) % MN
+
+% Ellipse:  u = 12.0  v = 1.0  a = 9.0  b = 9.0  phi = 0.0 Grad
+\qbezier(21.0, 1.0)(21.0, 4.7279)(18.364, 7.364)
+\qbezier(18.364, 7.364)(15.7279, 10.0)(12.0, 10.0)
+\qbezier(12.0, 10.0)(8.2721, 10.0)(5.636, 7.364)
+\qbezier(5.636, 7.364)(3.0, 4.7279)(3.0, 1.0)
+
+% B = 135, C = 95, D = 30
+\drawline(3,1)(5.6379,7.3639)(11.2179,9.9649)(19.7949,5.5)(21,1) % ABCDE
+
+\dashline[80]{0.2}(3,1)(11.2179,9.9649)(21,1) % ACE
+
+
+\put( 2.5, 0.2 ){$\scriptstyle A$}
+\put( 4.8, 7.2 ){$\scriptstyle B$}
+\put(10.9,10.2 ){$\scriptstyle C$}
+\put(19.9, 5.5 ){$\scriptstyle D$}
+\put(20.5, 0.2 ){$\scriptstyle E$}
+
+\put( 0.0, 0.7 ){$\scriptstyle M$}
+\put(23.1, 0.7 ){$\scriptstyle N$}
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/246aa487.eepic b/33063-t/images/sources/246aa487.eepic
new file mode 100644
index 0000000..38cbc32
--- /dev/null
+++ b/33063-t/images/sources/246aa487.eepic
@@ -0,0 +1,62 @@
+\PGset[0.8em]
+\begin{picture}(24,12)
+
+% Ellipse:  u = 6.0  v = 6.0  a = 5.0  b = 5.0  phi = 0.0 Grad
+\qbezier(11.0, 6.0)(11.0, 8.0711)(9.5355, 9.5355)
+\qbezier(9.5355, 9.5355)(8.0711, 11.0)(6.0, 11.0)
+\qbezier(6.0, 11.0)(3.9289, 11.0)(2.4645, 9.5355)
+\qbezier(2.4645, 9.5355)(1.0, 8.0711)(1.0, 6.0)
+\qbezier(1.0, 6.0)(1.0, 3.9289)(2.4645, 2.4645)
+\qbezier(2.4645, 2.4645)(3.9289, 1.0)(6.0, 1.0)
+\qbezier(6.0, 1.0)(8.0711, 1.0)(9.5355, 2.4645)
+\qbezier(9.5355, 2.4645)(11.0, 3.9289)(11.0, 6.0)
+
+\dashline[80]{0.4}(6,11)(6,1) % HA
+
+\dashline[80]{0.4}(1.171,7.294)(6,11)(8.5005,10.33) % CHD
+
+\drawline(6,1)(1.6695,3.5)(1.171,7.294)(8.5005,10.33)(10.829,4.706)(6,1) % ABCDEA
+
+% irregular polygons are icky...
+% AB = 5, BC = 3.8266, CD = 7.9333, DE = 6.0869, EA = 6.0871
+
+% 190x132 = 4.7/6.78 = .6932, y=.6932x
+% x = 6.52, y=4.5196
+
+% A = 19.52,  0.5196
+% (x-19.52)^2+(y-.5196^2)=5^2
+% (x-14)^2+(y-6)^2=3.8266^2
+% B = 14.721, 2.242
+% C = 14.00,  6
+% D = 20.52, 10.5196
+% (x-20.52)^2+(y-10.5196)^2=6.0869^2
+% (x-19.52)^2+(y-.5196)^2=6.0871^2
+% E = 23.438, 5.178
+
+\drawline(19.52,0.5196)(14.721,2.242)(14,6)(20.52,10.5196)(23.438,5.178)(19.52,0.5196) % ABCDEA
+
+% CH = 6.0871, DH = 2.5887
+% (x-14)^2 + (y-6)^2 = 6.0871^2
+% (x-20.52)^2 + (y-10.5196)^2 = 2.5887^2
+% H = 17.934, 10.645
+\dashline[80]{0.2}(14,6)(17.934,10.645)(20.52,10.5196) % C'H'D'
+
+\dashline[80]{0.2}(17.934, 10.645)(19.52,0.5196) % H'A'
+
+\put(  5.7,  0.2 ){$\scriptstyle A$}
+\put(  0.7,  2.9 ){$\scriptstyle B$}
+\put(  0.3,  7.0 ){$\scriptstyle C$}
+\put(  8.5, 10.4 ){$\scriptstyle D$}
+\put( 11.0,  4.2 ){$\scriptstyle E$}
+\put(  5.7, 11.1 ){$\scriptstyle H$}
+
+\put( 19.8,  0.0 ){$\scriptstyle A'$}
+\put( 13.6,  2.0 ){$\scriptstyle B'$}
+\put( 13.0,  5.5 ){$\scriptstyle C'$}
+\put( 20.6, 10.6 ){$\scriptstyle D'$}
+\put( 23.5,  4.8 ){$\scriptstyle E'$}
+\put( 17.6, 10.8 ){$\scriptstyle H'$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/247aa488.eepic b/33063-t/images/sources/247aa488.eepic
new file mode 100644
index 0000000..e39e89a
--- /dev/null
+++ b/33063-t/images/sources/247aa488.eepic
@@ -0,0 +1,28 @@
+\PGset[0.8em]
+\begin{picture}(13,14)
+
+% radius = 6
+
+\drawline(7.0006,12.9994)(1.8046,9.9994)(1.8034,4.0000)(6.9994,1.0000)(12.1954,4.0006)(12.1966,10.0000)(7.0006,13.0000)
+
+% sidelen = 6sin(30) = 6
+% AB+BC = AK+AC = 12
+% AK = 2AC
+% A = 1.8046, 10
+% C = 12.1966,10
+% (x-1.8046)^2+(y-10)^2 = 8^2
+% (x-12.1966)^2+(y-10)^2 = 4^2
+% K = 9.31,12.769
+
+\dashline[80]{0.2}(1.8046,10)(9.31,12.769)(12.1966,10)(1.8046,10) % AKCA
+
+
+\put(  1.0,  9.8 ){$\scriptstyle A$}
+\put(  6.6, 13.2 ){$\scriptstyle B$}
+\put( 12.3,  9.8 ){$\scriptstyle C$}
+\put( 12.3,  3.8 ){$\scriptstyle D$}
+\put(  9.0, 12.9 ){$\scriptstyle K$}
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/248aa490.eepic b/33063-t/images/sources/248aa490.eepic
new file mode 100644
index 0000000..ec99b1d
--- /dev/null
+++ b/33063-t/images/sources/248aa490.eepic
@@ -0,0 +1,33 @@
+\PGset[0.8em]
+\begin{picture}(21,9)
+
+% 235x190
+% 8x7
+
+\drawline(1,1)(9,1)(5,8)(1,1) % ABC
+
+% C = 5,8
+% D = 6.5,1
+% AD = CE = 5.5
+% CD = DE = 7.1589
+% (x-5)^2+(y-8)^2 = 5.5^2
+% (x-7)^2+(y-1)^2 = 7.159^2
+% E = 0.756, 4.502
+\dashline[80]{0.2}(5,8)(6.5,1)(0.756,4.502)(5,8) % CDEC
+
+
+\drawline(14,1)(20,1)(20,7)(14,7)(14,1) % Q'
+
+\put(  0.5,  0.2 ){$\scriptstyle A$}
+\put(  8.5,  0.2 ){$\scriptstyle B$}
+\put(  4.7,  8.2 ){$\scriptstyle C$}
+\put(  6.0,  0.2 ){$\scriptstyle D$}
+\put(  0.0,  4.2 ){$\scriptstyle E$}
+
+\put(  4.7,  3.8 ){$\scriptstyle Q$}
+\put( 16.7,  3.7 ){$\scriptstyle Q'$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/249aa491.eepic b/33063-t/images/sources/249aa491.eepic
new file mode 100644
index 0000000..b219077
--- /dev/null
+++ b/33063-t/images/sources/249aa491.eepic
@@ -0,0 +1,24 @@
+\PGset[0.8em]
+\begin{picture}(33,11)
+
+% pentagon Q' radius = 5
+% sidelen = 5.877, P = 29.385
+% area = 2.5 x P = 73.4625
+% square Q radius = 8
+% square Q' radius = 7.346
+
+\drawline(11.0000,6.0000)(8.5005,10.3300)(3.5010,10.3300)(1.0005,6.0000)(3.4995,1.6700)(8.4990,1.6700)(11.0000,6.0000) % Q'
+
+\drawline(14,1.67)(22,1.67)(22,9.67)(14,9.67)(14,1.67) % Q
+
+\drawline(25,1.67)(32.35,1.67)(32.35,9.02)(25,9.02)(25,1.67) % Q''
+
+
+\put(  5.7,  5.7 ){$\scriptstyle Q$}
+\put( 17.7,  5.4 ){$\scriptstyle Q'$}
+\put( 28.3,  5.2 ){$\scriptstyle Q''$}
+
+
+
+\end{picture}
+\PGrestore
diff --git a/33063-t/images/sources/config.tex b/33063-t/images/sources/config.tex
new file mode 100644
index 0000000..e479871
--- /dev/null
+++ b/33063-t/images/sources/config.tex
@@ -0,0 +1,10 @@
+% figures
+% Create \savelength to use in saving and restoring \unitlength
+%    just in case some other value is being used between diagrams
+\newlength{\PGsavelength}
+
+% \PGset saves current \unitlength in \PGsavelength, and sets \unitlength to {arg}
+\newcommand{\PGset}[1][\unitlength]{\setlength{\PGsavelength}{\unitlength}\setlength{\unitlength}{#1}}
+
+% \PGrestore sets \unitlength back to the value saved in \PGsavelength
+\newcommand{\PGrestore}{\setlength{\unitlength}{\PGsavelength}}
diff --git a/33063-t/images/sources/woodcutsmall.eps b/33063-t/images/sources/woodcutsmall.eps
new file mode 100644
index 0000000..2b4ea46
--- /dev/null
+++ b/33063-t/images/sources/woodcutsmall.eps
@@ -0,0 +1,4690 @@
+%!PS-Adobe-3.0 EPSF-3.0
+%%Creator: (ImageMagick)
+%%Title: (woodcutsmall.eps)
+%%CreationDate: (Thu Sep 15 19:15:31 2005)
+%%BoundingBox: 0 0 171 144
+%%HiResBoundingBox: 0 0 171.26 144
+%%DocumentData: Clean7Bit
+%%LanguageLevel: 1
+%%Pages: 1
+%%EndComments
+
+%%BeginDefaults
+%%EndDefaults
+
+%%BeginProlog
+%
+% Display a color image.  The image is displayed in color on
+% Postscript viewers or printers that support color, otherwise
+% it is displayed as grayscale.
+%
+/DirectClassPacket
+{
+  %
+  % Get a DirectClass packet.
+  %
+  % Parameters:
+  %   red.
+  %   green.
+  %   blue.
+  %   length: number of pixels minus one of this color (optional).
+  %
+  currentfile color_packet readhexstring pop pop
+  compression 0 eq
+  {
+    /number_pixels 3 def
+  }
+  {
+    currentfile byte readhexstring pop 0 get
+    /number_pixels exch 1 add 3 mul def
+  } ifelse
+  0 3 number_pixels 1 sub
+  {
+    pixels exch color_packet putinterval
+  } for
+  pixels 0 number_pixels getinterval
+} bind def
+
+/DirectClassImage
+{
+  %
+  % Display a DirectClass image.
+  %
+  systemdict /colorimage known
+  {
+    columns rows 8
+    [
+      columns 0 0
+      rows neg 0 rows
+    ]
+    { DirectClassPacket } false 3 colorimage
+  }
+  {
+    %
+    % No colorimage operator;  convert to grayscale.
+    %
+    columns rows 8
+    [
+      columns 0 0
+      rows neg 0 rows
+    ]
+    { GrayDirectClassPacket } image
+  } ifelse
+} bind def
+
+/GrayDirectClassPacket
+{
+  %
+  % Get a DirectClass packet;  convert to grayscale.
+  %
+  % Parameters:
+  %   red
+  %   green
+  %   blue
+  %   length: number of pixels minus one of this color (optional).
+  %
+  currentfile color_packet readhexstring pop pop
+  color_packet 0 get 0.299 mul
+  color_packet 1 get 0.587 mul add
+  color_packet 2 get 0.114 mul add
+  cvi
+  /gray_packet exch def
+  compression 0 eq
+  {
+    /number_pixels 1 def
+  }
+  {
+    currentfile byte readhexstring pop 0 get
+    /number_pixels exch 1 add def
+  } ifelse
+  0 1 number_pixels 1 sub
+  {
+    pixels exch gray_packet put
+  } for
+  pixels 0 number_pixels getinterval
+} bind def
+
+/GrayPseudoClassPacket
+{
+  %
+  % Get a PseudoClass packet;  convert to grayscale.
+  %
+  % Parameters:
+  %   index: index into the colormap.
+  %   length: number of pixels minus one of this color (optional).
+  %
+  currentfile byte readhexstring pop 0 get
+  /offset exch 3 mul def
+  /color_packet colormap offset 3 getinterval def
+  color_packet 0 get 0.299 mul
+  color_packet 1 get 0.587 mul add
+  color_packet 2 get 0.114 mul add
+  cvi
+  /gray_packet exch def
+  compression 0 eq
+  {
+    /number_pixels 1 def
+  }
+  {
+    currentfile byte readhexstring pop 0 get
+    /number_pixels exch 1 add def
+  } ifelse
+  0 1 number_pixels 1 sub
+  {
+    pixels exch gray_packet put
+  } for
+  pixels 0 number_pixels getinterval
+} bind def
+
+/PseudoClassPacket
+{
+  %
+  % Get a PseudoClass packet.
+  %
+  % Parameters:
+  %   index: index into the colormap.
+  %   length: number of pixels minus one of this color (optional).
+  %
+  currentfile byte readhexstring pop 0 get
+  /offset exch 3 mul def
+  /color_packet colormap offset 3 getinterval def
+  compression 0 eq
+  {
+    /number_pixels 3 def
+  }
+  {
+    currentfile byte readhexstring pop 0 get
+    /number_pixels exch 1 add 3 mul def
+  } ifelse
+  0 3 number_pixels 1 sub
+  {
+    pixels exch color_packet putinterval
+  } for
+  pixels 0 number_pixels getinterval
+} bind def
+
+/PseudoClassImage
+{
+  %
+  % Display a PseudoClass image.
+  %
+  % Parameters:
+  %   class: 0-PseudoClass or 1-Grayscale.
+  %
+  currentfile buffer readline pop
+  token pop /class exch def pop
+  class 0 gt
+  {
+    currentfile buffer readline pop
+    token pop /depth exch def pop
+    /grays columns 8 add depth sub depth mul 8 idiv string def
+    columns rows depth
+    [
+      columns 0 0
+      rows neg 0 rows
+    ]
+    { currentfile grays readhexstring pop } image
+  }
+  {
+    %
+    % Parameters:
+    %   colors: number of colors in the colormap.
+    %   colormap: red, green, blue color packets.
+    %
+    currentfile buffer readline pop
+    token pop /colors exch def pop
+    /colors colors 3 mul def
+    /colormap colors string def
+    currentfile colormap readhexstring pop pop
+    systemdict /colorimage known
+    {
+      columns rows 8
+      [
+        columns 0 0
+        rows neg 0 rows
+      ]
+      { PseudoClassPacket } false 3 colorimage
+    }
+    {
+      %
+      % No colorimage operator;  convert to grayscale.
+      %
+      columns rows 8
+      [
+        columns 0 0
+        rows neg 0 rows
+      ]
+      { GrayPseudoClassPacket } image
+    } ifelse
+  } ifelse
+} bind def
+
+/DisplayImage
+{
+  %
+  % Display a DirectClass or PseudoClass image.
+  %
+  % Parameters:
+  %   x & y translation.
+  %   x & y scale.
+  %   label pointsize.
+  %   image label.
+  %   image columns & rows.
+  %   class: 0-DirectClass or 1-PseudoClass.
+  %   compression: 0-none or 1-RunlengthEncoded.
+  %   hex color packets.
+  %
+  gsave
+  /buffer 512 string def
+  /byte 1 string def
+  /color_packet 3 string def
+  /pixels 768 string def
+
+  currentfile buffer readline pop
+  token pop /x exch def
+  token pop /y exch def pop
+  x y translate
+  currentfile buffer readline pop
+  token pop /x exch def
+  token pop /y exch def pop
+  currentfile buffer readline pop
+  token pop /pointsize exch def pop
+  /Times-Roman findfont pointsize scalefont setfont
+  x y scale
+  currentfile buffer readline pop
+  token pop /columns exch def
+  token pop /rows exch def pop
+  currentfile buffer readline pop
+  token pop /class exch def pop
+  currentfile buffer readline pop
+  token pop /compression exch def pop
+  class 0 gt { PseudoClassImage } { DirectClassImage } ifelse
+  grestore
+} bind def
+%%EndProlog
+%%Page:  1 1
+%%PageBoundingBox: 0 0 171 144
+userdict begin
+DisplayImage
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+%%PageTrailer
+%%Trailer
+%%EOF
diff --git a/33063-t/images/woodcutsmall.pdf b/33063-t/images/woodcutsmall.pdf
new file mode 100644
index 0000000..6e04c40
--- /dev/null
+++ b/33063-t/images/woodcutsmall.pdf
diff --git a/LICENSE.txt b/LICENSE.txt
new file mode 100644
index 0000000..6312041
--- /dev/null
+++ b/LICENSE.txt
@@ -0,0 +1,11 @@
+This eBook, including all associated images, markup, improvements,
+metadata, and any other content or labor, has been confirmed to be
+in the PUBLIC DOMAIN IN THE UNITED STATES.
+
+Procedures for determining public domain status are described in
+the "Copyright How-To" at https://www.gutenberg.org.
+
+No investigation has been made concerning possible copyrights in
+jurisdictions other than the United States. Anyone seeking to utilize
+this eBook outside of the United States should confirm copyright
+status under the laws that apply to them.
diff --git a/README.md b/README.md
new file mode 100644
index 0000000..00ba34a
--- /dev/null
+++ b/README.md
@@ -0,0 +1,2 @@
+Project Gutenberg (https://www.gutenberg.org) public repository for
+eBook #33063 (https://www.gutenberg.org/ebooks/33063)