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diff --git a/33063-t/33063-t.tex b/33063-t/33063-t.tex new file mode 100644 index 0000000..6d6fa4a --- /dev/null +++ b/33063-t/33063-t.tex @@ -0,0 +1,15134 @@ +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % +% % +% 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 *** % +% % +% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % + +\def\ebook{33063} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +%% %% +%% Packages and substitutions: %% +%% %% +%% book: Required. %% +%% inputenc: Standard DP encoding. Required. %% +%% %% +%% ifthen: Logical conditionals. Required. %% +%% %% +%% amsmath: AMS mathematics enhancements. Required. %% +%% amssymb: Additional mathematical symbols. Required. %% +%% %% +%% alltt: Fixed-width font environment. 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\begin{list}{\arabic{Lcount}.} + {\usecounter{Lcount}% + \setlength{\itemsep}{0pt}% + \setlength{\leftmargin}{2\parindent}}} + {\end{list}} + +\makeatother + +%%%%%%%%%%%%%%%%%%%%%%%% START OF DOCUMENT %%%%%%%%%%%%%%%%%%%%%%%%%% + +\begin{document} + +\pagestyle{empty} +\pagenumbering{Alph} + +\phantomsection +\pdfbookmark[-1]{FRONT MATTER.}{FRONT MATTER} + +%%%% PG BOILERPLATE %%%% +\phantomsection +\pdfbookmark[0]{PG BOILERPLATE.}{BOILERPLATE} + +\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} +\end{minipage} +\end{center} +\vfill + +\begin{minipage}{0.85\textwidth} +\small +\phantomsection +\pdfbookmark[0]{TRANSCRIBER'S NOTE.}{TRANSCRIBER'S NOTE} +\subsection*{\centering\normalfont\scshape% +\normalsize\MakeLowercase{\TransNote}}% + +\raggedright +\TransNoteText +\end{minipage} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%% 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 Athenĉum 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.}} + +\begin{PGtext} +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-pdf.pdf or 33063-pdf.zip ***** +This and all associated files of various formats will be found in: + http://www.gutenberg.org/3/3/0/6/33063/ + +Produced by Jeremy Weatherford, Stan Goodman, Kevin Handy +and the Online Distributed Proofreading Team at +http://www.pgdp.net + + +Updated editions will 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