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+\documentclass[oneside]{book}
+\usepackage[latin1]{inputenc}
+\usepackage[reqno]{amsmath}
+\usepackage{amssymb,graphicx,units,yfonts}
+\begin{document}
+
+
+\thispagestyle{empty}
+\small
+\begin{verbatim}
+
+The Project Gutenberg EBook Non-Euclidean Geometry, by Henry Manning
+
+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: Non-Euclidean Geometry
+
+Author: Henry Manning
+
+Release Date: October 10, 2004 [EBook #13702]
+
+Language: English
+
+Character set encoding: TeX
+
+*** START OF THIS PROJECT GUTENBERG EBOOK NON-EUCLIDEAN GEOMETRY ***
+
+
+
+
+Produced by David Starner, Joshua Hutchinson, John Hagerson, 
+and the Project Gutenberg On-line Distributed Proofreading Team.
+
+
+
+
+
+
+\end{verbatim}
+\normalsize
+\newpage
+
+\frontmatter
+\begin{center}
+\Huge NON-EUCLIDEAN \\
+GEOMETRY \\
+
+\bigskip\bigskip\bigskip\bigskip\bigskip
+\normalsize BY
+
+\bigskip\bigskip \large HENRY PARKER MANNING, \textsc{Ph.D.} \\
+\bigskip \normalsize
+\textsc{Assistant Professor of Pure Mathematics \\
+in Brown University} \\
+
+\vfill
+BOSTON, U.S.A. \\
+\bigskip GINN \& COMPANY, PUBLISHERS. \\
+\bigskip \textgoth{The Athenaeum Press} \\
+\bigskip 1901 \\
+\bigskip\bigskip
+\scriptsize \textsc{Copyright, 1901, by} \\
+HENRY PARKER MANNING \\
+\rule{15mm}{1pt} \\
+\textsc{all rights reserved}
+\end{center}
+
+\chapter{PREFACE}
+
+Non-Euclidean Geometry is now recognized as an important branch of
+Mathematics. Those who teach Geometry should have some knowledge of
+this subject, and all who are interested in Mathematics will find
+much to stimulate them and much for them to enjoy in the novel
+results and views that it presents.
+
+This book is an attempt to give a simple and direct account of the
+Non-Euclidean Geometry, and one which presupposes but little
+knowledge of Mathematics. The first three chapters assume a
+knowledge of only Plane and Solid Geometry and Trigonometry, and the
+entire book can be read by one who has taken the mathematical
+courses commonly given in our colleges.
+
+No special claim to originality can be made for what is published
+here. The propositions have long been established, and in various
+ways. Some of the proofs may be new, but others, as already given by
+writers on this subject, could not be improved. These have come to
+me chiefly through the translations of Professor George Bruce
+Halsted of the University of Texas.
+
+I am particularly indebted to my friend, Arnold B.~Chace, Sc.D., of
+Valley Falls, R.~I., with whom I have studied and discussed the
+subject.\medskip
+
+\hfill HENRY P.~MANNING.
+
+{\footnotesize \textsc{Providence}, January, 1901.}
+
+%% CONTENTS
+%% INTRODUCTION
+%% CHAPTER I: PANGEOMETRY
+%%    I. PROPOSITIONS DEPENDING ONLY ON THE PRINCIPLE OF
+%%       SUPERPOSITION
+%%   II. PROPOSITIONS WHICH ARE TRUE FOR RESTRICTED FIGURES
+%%  III. THE THREE HYPOTHESES
+%%
+%% CHAPTER II: THE HYPERBOLIC GEOMETRY
+%%    I. PARALLEL LINES
+%%   II. BOUNDARY-CURVES AND SURFACES, AND EQUIDISTANT-CURVES
+%%       AND SURFACES
+%%  III. TRIGONOMETRICAL FORMULÆ
+%%
+%% CHAPTER III: THE ELLIPTIC GEOMETRY
+%%
+%% CHAPTER IV: ANALYTIC NON-EUCLIDEAN GEOMETRY
+%%    I. HYPERBOLIC ANALYTIC GEOMETRY
+%%   II. ELLIPTIC ANALYTIC GEOMETRY
+%%  III. ELLIPTIC SOLID ANALYTIC GEOMETRY
+%%
+%% HISTORICAL NOTE
+\tableofcontents
+\mainmatter
+\chapter{INTRODUCTION}
+
+The axioms of Geometry were formerly regarded as laws of thought
+which an intelligent mind could neither deny nor investigate. Not
+only were the axioms to which we have been accustomed found to agree
+with our experience, but it was believed that we could not reason on
+the supposition that any of them are not true, it has been shown,
+however, that it is possible to take a set of axioms, wholly or in
+part contradicting those of Euclid, and build up a Geometry as
+consistent as his.
+
+We shall give the two most important Non-Euclidean
+Geometries.\footnote{See Historical Note, p.~\pageref{histnote}.} In
+these the axioms and definitions are taken as in Euclid, with the
+exception of those relating to parallel lines. Omitting the axiom on
+parallels,\footnote{See p.~\pageref{3hypos}.} we are led to three
+hypotheses; one of these establishes the Geometry of Euclid, while
+each of the other two gives us a series of propositions both
+interesting and useful. Indeed, as long as we can examine but a
+limited portion of the universe, it is not possible to prove that
+the system of Euclid is true, rather than one of the two
+Non-Euclidean Geometries which we are about to describe.
+
+We shall adopt an arrangement which enables us to prove first the
+propositions common to the three Geometries, then to produce a
+series of propositions and the trigonometrical formulæ for each of
+the two Geometries which differ from that of Euclid, and by
+analytical methods to derive some of their most striking properties.
+
+We do not propose to investigate directly the foundations of
+Geometry, nor even to point out all of the assumptions which have
+been made, consciously or unconsciously, in this study. Leaving
+undisturbed that which these Geometries have in common, we are free
+to fix our attention upon their differences. By a concrete
+exposition it may be possible to learn more of the nature of
+Geometry than from abstract theory alone.
+
+\newpage
+Thus we shall employ most of the terms of Geometry without
+repeating the definitions given in our text-books, and assume that
+the figures defined by these terms exist. In particular we assume:
+\smallskip
+
+\begin{enumerate}
+\item[I.] \emph{The existence of straight lines determined by any two
+points, and that the shortest path between two points is a straight
+line.}
+
+\item[II.] \emph{The existence of planes determined by any three points
+not in a straight line, and that a straight line joining any two
+points of a plane lies wholly in the plane.}
+
+\item[III.] \emph{That geometrical figures can be moved about without
+changing their shape or size.}
+
+\item[IV.] \emph{That a point moving along a line from one position
+to another passes through every point of the line between, and that
+a geometrical magnitude, for example, an angle, or the length of a
+portion of a line, varying from one value to another, passes through
+all intermediate values.}
+\end{enumerate}
+
+In some of the propositions the proof will be omitted or only the
+method of proof suggested, where the details can be supplied from
+our common text-books.
+
+\chapter{PANGEOMETRY}
+
+\section{Propositions Depending Only on the Principle of
+Superposition}
+
+\begin{description}
+\item[1.~Theorem.] \emph{If one straight line meets another, the
+sum of the adjacent angles formed is equal to two right angles.}
+
+\item[2.~Theorem.] \emph{If two straight lines intersect, the
+vertical angles are equal.}
+
+\item[3.~Theorem.] \emph{Two triangles are equal if they have a
+side and two adjacent angles, or two sides and the included angle,
+of one equal, respectively, to the corresponding parts of the
+other.}
+
+\item[4.~Theorem.] \emph{In an isosceles triangle the angles
+opposite the equal sides are equal.}
+
+Bisect the angle at the vertex and use (3).
+
+\item[5.~Theorem.] \emph{The perpendiculars erected at the middle
+points of the sides of a triangle meet in a point if two of them
+meet, and this point is the centre of a circle that can be drawn
+through the three vertices of the triangle.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig01.png}
+\end{figure}
+
+\textbf{Proof.} Suppose $EO$ and $FO$ meet at $O$. The triangles
+$AFO$ and $BFO$ are equal by (3). Also, $AEO$ and $CEO$ are equal.
+Hence, $CO$ and $BO$ are equal, being each equal to $AO$. The
+triangle $BCO$ is, therefore, isosceles, and $OD$ if drawn bisecting
+the angle $BOC$ will be perpendicular to $BC$ at its middle point.
+
+\item[6.~Theorem.] \emph{In a circle the radius bisecting an angle at
+the centre is perpendicular to the chord which subtends the angle
+and bisects this chord.}
+
+\item[7.~Theorem.] \emph{Angles at the centre of a circle are
+proportional to the intercepted arcs and may be measured by them.}
+
+\item[8.~Theorem.] \emph{From any point without a line a
+perpendicular to the line can be drawn.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=20mm]{fig02.png}
+\end{figure}
+
+\textbf{Proof.} Let $P'$ be the position which $P$ would take if the
+plane were revolved about $AB$ into coincidence with itself. The
+straight line $PP'$ is then perpendicular to $AB$.
+
+\item[9.~Theorem.] \emph{If oblique lines drawn from a point in a
+perpendicular to a line cut off equal distances from the foot of the
+perpendicular, they are equal and make equal angles with the line
+and with the perpendicular.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=50mm]{fig03.png}
+\end{figure}
+
+\item[10.~Theorem.] \emph{If two lines cut a third at the same angle,
+that is, so that corresponding angles are equal, a line can be drawn
+that is perpendicular to both.}
+
+\textbf{Proof.} Let the angles $FMB$ and $MND$ be equal, and through
+$H$, the middle point of $MN$, draw $LK$ perpendicular to $CD$; then
+$LK$ will also be perpendicular to $AB$. For the two triangles $LMH$
+and $KNH$ are equal by (3).
+
+\item[11.~Theorem.] \emph{If two equal lines in a plane are erected
+perpendicular to a given line, the line joining their extremities
+makes equal angles with them and is bisected at right angles by a
+third perpendicular erected midway between them.}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig04.png}
+\end{figure}
+
+\textbf{Let} $AC$ and $BD$ be perpendicular to $AB$, and suppose
+$AC$ and $BD$ equal. The angles at $C$ and $D$ made with a line
+joining these two points are equal, and the perpendicular $HK$
+erected at the middle point of $AB$ is perpendicular to $CD$ at its
+middle point.
+
+Proved by superposition.
+
+\item[12.~Theorem.] \emph{Given as in the last proposition two
+perpendiculars and a third perpendicular erected midway between
+them; any line cutting this third perpendicular at right angles, if
+it cuts the first two at all, will cut off equal lengths on them and
+make equal angles with them.}
+
+Proved by superposition.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig05.png}
+\end{figure}
+
+\textbf{Corollary.} \emph{The last two propositions hold true if the
+angles at $A$ and $B$ are equal acute or equal obtuse angles, $HK$
+being perpendicular to $AB$ at its middle point. If $AC = BD$, the
+angles at $C$ and $D$ are equal, and $HK$ is perpendicular to $CD$
+at its middle point: or, if $CD$ is perpendicular to $HK$ at any
+point, $K$, and intersects $AC$ and $BD$, it it will cut off equal
+distances on these two lines and make equal angles with them.}
+\end{description}
+
+\section{Propositions Which Are True for Restricted Figures}
+
+The following propositions are true at least for figures whose lines
+do not exceed a certain length. That is, if there is any exception,
+it is in a case where we cannot apply the theorem or some step of
+the proof on account of the length of some of the lines. For
+convenience we shall use the word \emph{restricted} in this sense
+and say that a theorem is true for restricted figures or in any
+restricted portion of the plane.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig06.png}
+\end{figure}
+
+\begin{description}
+\item[1.~Theorem.] \emph{The exterior angle of a triangle is
+greater than either opposite interior angle} (Euclid, I, 16).
+
+\textbf{Proof.} Draw $AD$ from $A$ to the middle point of the
+opposite side and produce it to $E$, making $DE = AD$. The two
+triangles $ADC$ and $EBD$ are equal, and the angle $FBD$, being
+greater than the angle $EBD$, is greater than $C$.
+
+\textbf{Corollary.} \emph{At least two angles of a triangle are
+acute.}
+
+\item[2.~Theorem.] \emph{If two angles of a triangle are equal, the
+opposite sides are equal and the triangle is isosceles.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig07.png}
+\end{figure}
+
+\textbf{Proof.} The perpendicular erected at the middle point of the
+base divides the triangle into two figures which may be made
+to coincide and are equal. This perpendicular, therefore,
+passes through the vertex, and the two sides opposite the
+equal angles of the triangle are equal.
+
+\item[3.~Theorem.] \emph{In a triangle with unequal angles the
+side opposite the greater of the angles is greater than the side
+opposite the smaller; and conversely, if the sides of a triangle are
+unequal the opposite angles are unequal, and the greater angle lies
+opposite the greater side.}
+
+\item[4.~Theorem.] \emph{If two triangles have two sides of one
+equal, respectively, to two sides of the other, but the included
+angle of the first greater than the included angle of the second,
+the third side of the first is greater than the third side of the
+second; and conversely, if two triangles have two sides of one
+equal, respectively, to two sides of the other, but the third side
+of the first greater than the third side of the second, the angle
+opposite the third side of the first is greater than the angle
+opposite the third side of the second.}
+
+\item[5.~Theorem.] \emph{The sum of two lines drawn from any point
+to the extremities of a straight line is greater than the sum of two
+lines similarly drawn but included by them.}
+
+\item[6.~Theorem.] \emph{Through any point one perpendicular only
+can be drawn to a straight line.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=30mm]{fig08.png}
+\end{figure}
+
+\textbf{Proof.} Let $P'$ be the position which $P$ would take if the
+plane were revolved about $AB$ into coincidence with itself. If we
+could have two perpendiculars, $PC$ and $PD$, from $P$ to $AB$, then
+$CP'$ and $DP'$ would be continuations of these lines and we should
+have two different straight lines joining $P$ and $P'$, which is
+impossible.
+
+\textbf{Corollary.} \emph{Two right triangles are equal when the
+hypothenuse and an acute angle of one are equal, respectively, to
+the hypothenuse and an acute angle of the other.}
+
+\item[7.~Theorem.] \emph{The perpendicular is the shortest line that
+can be drawn from a point to a straight line.}
+
+\textbf{Corollary.} \emph{In a right triangle the hypothenuse is
+greater than either of the two sides about the right angle.}
+
+\item[8.~Theorem.] \emph{If oblique lines drawn from a point in a
+perpendicular to a line cut off unequal distances from the foot of
+the perpendicular, they are unequal, and the more remote is the
+greater; and conversely, if two oblique lines drawn from a point in
+a perpendicular are unequal, the greater cuts off a greater distance
+from the foot of the perpendicular.}
+
+\item[9.~Theorem.] \emph{If a perpendicular is erected at the middle
+point of a straight line, any point not in the perpendicular is
+nearer that extremity of the line which is on the same side of the
+perpendicular.}
+
+\textbf{Corollary.} \emph{Two points equidistant from the
+extremities of a straight line determine a perpendicular to the line
+at its middle point.}
+
+\item[10.~Theorem.] \emph{Two triangles are equal when they have
+three sides of one equal, respectively, to three sides of the
+other.}
+
+\item[11.~Theorem.] \emph{If two lines in a plane erected
+perpendicular to a third are unequal, the line joining their
+extremities makes unequal angles with them, the greater angle with
+the shorter perpendicular.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=25mm]{fig09.png}
+\end{figure}
+
+\textbf{Proof.} Suppose $AC > BD$. Produce $BD$, making $BE = AC$.
+Then $BEC = ACE$. But $BDC > BEC$, by (1), and $ACD$ is a part of
+$ACE$. Therefore, all the more $BDC > ACD$.
+
+\item[12.~Theorem.] \emph{If the two angles at $C$ and $D$ are equal,
+the perpendiculars are equal, and if the angles are unequal, the
+perpendiculars are unequal, and the longer perpendicular makes the
+smaller angle.}
+
+\newpage
+\item[13.~Theorem.] \emph{If two lines are perpendicular to a third,
+points on either equidistant from the third are equidistant from the
+other.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=30mm]{fig10.png}
+\end{figure}
+
+\textbf{Proof.} Let $AB$ and $CD$ be perpendicular to $HK$, and on
+$CD$ take any two points, $C$ and $D$, equidistant from $K$; then
+$C$ and $D$ will be equidistant from $AB$. For by superposition we
+can make $D$ fall on $C$, and then $DB$ will coincide with $CA$ by
+(6).
+\end{description}
+
+\medskip The following propositions of Solid Geometry depend directly
+on the preceding and hold true at least for any restricted portion
+of space.
+
+\begin{description}
+\item[14.~Theorem.] \emph{If a line is perpendicular to two
+intersecting lines at their intersection, it is perpendicular to all
+lines of their plane passing through this point.}
+
+\item[15.~Theorem.] \emph{If two planes are perpendicular, a line
+drawn in one perpendicular to their intersection is perpendicular to
+the other, and a line drawn through any point of one perpendicular
+to the other lies entirely in the first.}
+
+\item[16.~Theorem.] \emph{If a line is perpendicular to a plane,
+any plane through that line is perpendicular to the plane.}
+
+\item[17.~Theorem.] \emph{If a plane is perpendicular to each of
+two intersecting planes, it is perpendicular to their intersection.}
+\end{description}
+
+\section{The Three Hypotheses}
+
+The angles at the extremities of two equal perpendiculars are either
+right angles, acute angles, or obtuse angles, at least for
+restricted figures. We shall distinguish the three cases by speaking
+of them as the hypothesis of the right angle, the hypothesis of the
+acute angle, and the hypothesis of the obtuse angle, respectively.
+
+\begin{description}
+\item[1.~Theorem.] \emph{The line joining the extremities of two
+equal perpendiculars is, at least for any restricted portion of the
+plane, equal to, greater than, or less than the line joining their
+feet in the three hypotheses, respectively.}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=35mm]{fig11.png}
+\end{figure}
+
+\textbf{Proof.} Let $AC$ and $BD$ be the two equal perpendiculars
+and $HK$ a third perpendicular erected at the middle point of $AB$.
+Then $HA$ and $KC$ are perpendicular to $HK$, and $KC$ is equal to,
+greater than, or less than $HA$, according as the angle at $C$ is
+equal to, less than, or greater than the angle at $A$ (II, 12).
+Hence, $CD$, the double of $KC$, is equal to, greater than, or less
+than $AB$ in the three hypotheses, respectively.
+
+\textbf{Conversely,} if $CD$ is given equal to, greater than, or
+less than $AB$, there is established for this figure the first,
+second, or third hypothesis, respectively.
+
+\textbf{Corollary.} \emph{If a quadrilateral has three right angles,
+the sides adjacent to the fourth angle are equal to, greater than,
+or less than the sides opposite them, according as the fourth angle
+is right, acute, or obtuse.}
+
+\item[2.~Theorem.] \emph{If the hypothesis of a right angle is true
+in a single case in any restricted portion of the plane, it holds
+true in every case and throughout the entire plane.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=75mm]{fig12.png}
+\end{figure}
+
+\textbf{Proof.} We have now a rectangle; that is, a quadrilateral
+with four right angles. By the corollary to the last proposition,
+its opposite sides are equal. Equal rectangles can be placed
+together so as to form a rectangle whose sides shall be any given
+multiples of the corresponding sides of the given rectangle.
+
+Now let $A'B'$ be any given line and $A'C'$ and $B'D'$ two equal
+lines perpendicular to $A'B'$ at its extremities. Divide $A'C'$, if
+necessary, into a number of equal parts so that one of these parts
+shall be less than $AC$, and on $AC$ and $BD$ lay off $AM$ and $BN$
+equal to one of these parts, and draw $MN$. $ABNM$ is a rectangle;
+for otherwise $MN$ would be greater than or less than $AB$ and $CD$,
+and the angles at $M$ and $N$ would all be acute angles or all
+obtuse angles, which is impossible, since their sum is exactly four
+right angles. Again, divide $A'B'$ into a sufficient number of equal
+parts, lay off one of these parts on $AB$ and on $MN$, and form the
+rectangle $APQM$. Rectangles equal to this can be placed together so
+as exactly to cover the figure $A'B'D'C'$, which must therefore
+itself be a rectangle.
+
+\item[3.~Theorem.] \emph{If the hypothesis of the acute angle or the
+hypothesis of the obtuse angle holds true in a single case within a
+restricted portion of the plane, the same hypothesis holds true for
+every case within any such portion of the plane.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=35mm]{fig13.png}
+\end{figure}
+
+\textbf{Proof.} Let $CD$ move along $AC$ and $BD$, always cutting
+off equal distances on these two lines; or, again, let $AC$ and $BD$
+move along on the line $AB$ towards $HK$ or away from $HK$, always
+remaining perpendicular to $AB$ and their feet always at equal
+distances from $H$. The angles at $C$ and $D$ vary continuously and
+must therefore remain acute or obtuse, as the case may be, or at
+some point become right angles. There would then be established the
+hypothesis of the right angle, and the hypothesis of the acute angle
+or of the obtuse angle could not exist even in the single case
+supposed.
+
+The angles at $C$ and $D$ could not become zero nor $180^\circ$ in a
+restricted portion of the plane; for then the three lines $AC$,
+$CD$, and $BD$ would be one and the same straight line.
+
+\item[4.~Theorem.] \emph{The sum of the angles of a triangle, at
+least in any restricted portion of the plane, is equal to, less
+than, or greater than two right angles, in the three hypotheses,
+respectively.}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig14.png}
+\end{figure}
+
+\textbf{Proof.} Given any triangle, $ABD$ (Fig.~1), with right angle
+at $B$, draw $AC$ perpendicular to $AB$ and equal to $BD$. In the
+triangles $ADC$ and $DAB$, $AC=BD$ and $AD$ is common, but $DC$ is
+equal to, greater than, or less than $AB$ in the three hypotheses,
+respectively. Therefore, $DAC$ is equal to, greater than, or less
+than $ADB$ in the three hypotheses, respectively (II, 4). Adding
+$BAD$ to both of these angles, we have $ADB + BAD$ equal to, or
+greater than the right angle $BAC$.
+
+Now at least two angles of any restricted triangle are acute. The
+perpendicular, therefore, from the vertex of the third angle upon
+its opposite side will meet this side within the triangle and divide
+the triangle into two right triangles. Therefore, in any restricted
+triangle the sum of the angles is equal to, less than, or greater
+than two right angles in the three hypotheses, respectively.
+
+We will call the amount by which the angle-sum of a triangle exceeds
+two right angles its excess. The excess of a polygon of $n$ sides is
+the amount by which the sum of its angles exceeds $n-2$ times two
+right angles.
+
+It will not change the excess if we count as additional vertices any
+number of points on the sides, adding to the sum of the angles two
+right angles for each of these points.
+
+\item[5.~Theorem.] \emph{The excess of a polygon is equal to the
+sum of the excesses of any system of triangles into which it may be
+divided.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=30mm]{fig15.png}
+\end{figure}
+
+\textbf{Proof.} If we divide a polygon into two polygons by a
+straight or broken line, we may assume that the two points where it
+meets the boundary are vertices. If the dividing line is a broken
+line, broken at $p$ points, an the sides of the two polygons will be
+the sides of the original polygon, together with the $p+1$ parts
+into which the dividing line is separated by the $p$ points, each
+part counted twice.
+
+\textbf{Let} $S$ be the sum of the angles of the original polygon,
+and $n$ the number of its sides. Let $S'$ and $n'$, $S''$ and $n''$
+have the same meanings for the two polygons into which it is
+divided. Then we have, writing $R$ for the right angle,
+\begin{align*}
+S' + S'' &= S + 4pR, \\
+\intertext{and}
+n' + n'' &= n + 2(p+1). \\
+\intertext{Therefore,}
+S' - 2(n'-2)R + S'' - 2(n''-2)R &= S + 4pR - 2(n+2p-2)R \\
+&= S-2(n-2)R.
+\end{align*}
+
+Any system of triangles into which a polygon may be divided is
+produced by a sufficient number of repetitions of the above process.
+Always the excess of the polygon is equal to the sum of the excesses
+of the parts into which it is divided.
+
+We may extend the notion of excess and apply it to any combination
+of different portions of the plane hounded completely by straight
+lines.
+
+Instead of considering the sum of the angles of a polygon, we may
+take the sum of the exterior angles. The amount by which this sum
+falls short of four right angles equals the excess of the polygon.
+We may speak of it as the deficiency of the exterior angles.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=30mm]{fig16.png}
+\end{figure}
+
+The sum of the exterior angles is the amount by which we turn in
+going completely around the figure, turning at each vertex from one
+side to the next. If we are considering a combination of two or more
+polygons, we must traverse the entire boundary and so as always to
+have the area considered on one side, say on the left.
+
+\item[6.~Theorem.] \emph{The excess of polygons is always zero,
+always negative, or always positive.}
+
+\textbf{Proof.} We know that this theorem is true of restricted
+triangles, but any finite polygon may be divided into a finite
+number of such triangles, and by the last theorem the excess of the
+polygon is equal to the sum of the excesses of the triangles.
+
+When the excess is negative, we may call it deficiency, or speak of
+the excess of the exterior angles.
+
+\textbf{Corollary.} \emph{The excess of a polygon is numerically
+greater than the excess of any part which may be cut off from it by
+straight lines, except in the first hypothesis, when it is zero.}
+\end{description}
+
+\medskip The following theorems apply to the second and third
+hypotheses.
+
+\begin{description}
+\item[7.~Theorem.] \emph{By diminishing the sides of a triangle,
+or even one side while the other two remain less than some fixed
+length, we can diminish its area indefinitely, and the sum of its
+angles will approach two right angles as limit.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=50mm]{fig17.png}
+\end{figure}
+
+\textbf{Proof.} Let $ABDC$ be a quadrilateral with three right
+angles, $A$, $B$, and $C$. A perpendicular moving along $AB$ will
+constantly increase or decrease; for if it could increase a part of
+the way and decrease a part of the way there would be different
+positions where the perpendiculars have the same length; a
+perpendicular midway between them would be perpendicular to $CD$
+also, and we should have a rectangle.
+
+Divide $AB$ into $n$ equal parts, and draw perpendiculars through
+the points of division. The quadrilateral is divided into $n$
+smaller quadrilaterals, which can be applied one to another, having
+a side and two adjacent right angles the same in all. Beginning at
+the end where the perpendicular is the shortest, each quadrilateral
+can be placed entirely within the next. Therefore, the first has its
+area less than $\nicefrac{1}{n}$th of the area of the original
+quadrilateral, and its deficiency or excess less than
+$\nicefrac{1}{n}$th of the deficiency or excess of the whole. Now
+any triangle whose sides are all less than $AC$ or $BD$, and one of
+whose sides is less than one of the subdivisions of $AB$, can be
+placed entirely within this smallest quadrilateral. Such a triangle
+has its area and its deficiency or excess less than
+$\nicefrac{1}{n}$th of the area and of the deficiency or excess of
+the original quadrilateral.
+
+Thus, a triangle has its area and deficiency or excess less than any
+assigned area and deficiency or excess, however small, if at least
+one side is taken sufficiently small, the other two sides not being
+indefinitely large.
+
+\newpage
+\item[8.~Theorem.] \emph{Two triangles having the same deficiency or
+excess have the same area.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig18.png}
+\end{figure}
+
+\textbf{Proof.} Let $AOB$ and $A'OB'$ have the same deficiency or
+excess and an angle of one equal to an angle of the other. If we
+place them together so that the equal angles coincide, the triangles
+will coincide and be entirely equal, or there will be a
+quadrilateral common to the two, and, besides this, two smaller
+triangles having an angle the same in both and the same deficiency
+or excess. Putting these together, we find again a quadrilateral
+common to both and a third pair of triangles having an angle the
+same in both and the same deficiency or excess. We may continue this
+process indefinitely, unless we come to a pair of triangles which
+coincide; for at no time can one triangle of a pair be contained
+entirely within the other, since they have the same deficiency or
+excess.
+
+Let $so$ denote the sum of the sides opposite the equal angles of
+the first two triangles, $sa$ the sum of the adjacent sides, and
+$s'a$ that portion of the adjacent sides counted twice, which is
+common to the two triangles when they are placed together. Writing
+$o'$ and $a'$ for the second pair of triangles, $o''$ and $a'$ for
+the third pair, etc., we have
+\begin{gather*}
+\begin{aligned}
+sa   &= s'a + so',                 & so   &= sa',  \\
+sa'  &= s'a' + so'',               & so'  &= sa'', \\
+sa'' &= s'a'' + so'''\text{, etc.} & so'' &= sa''' \text{, etc.}
+\end{aligned}\\
+\begin{aligned}
+\therefore sa  &= s'a  + s'a''  + s'a^{IV} + \ldots, \\
+           sa' &= s'a' + s'a''' + s'a^{V}  + \ldots.
+\end{aligned}
+\end{gather*}
+
+Therefore, the expressions $s'a$, $s'a'$, $s'a''$, $\cdots$ diminish
+indefinitely. Each of these is made up of a side counted twice from
+one and a side counted twice from the other of a pair of triangles.
+Thus, if we carry the process sufficiently far, the remaining
+triangles can be made to have at least one side as small as we
+please, while all the sides diminish and are less, for example, than
+the longest of the sides of the original triangles. Therefore, the
+areas of the remaining triangles diminish indefinitely, and as the
+difference of the areas remains the same for each pair of triangles,
+this difference must be zero. The triangles of each pair and, in
+particular, the first two triangles have the same area.
+
+\begin{figure}[ht]
+\centering
+\includegraphics[width=70mm]{fig19.png}
+\end{figure}
+
+Let $ABC$ and $DEF$ have the same deficiency or excess, and suppose
+$AC<DF$.  Produce $AC$ to $C'$, making $AC'$ = $DF$. Then there is
+some point, $B'$, on $AB$ between $A$ and $B$ such that $AB'C'$ has
+the same deficiency or excess and the same area as $ABC$. Place
+$AB'C'$ upon $DEF$ so that AC' will coincide with $DF$, and let
+$DE'F$ be the position which it takes. If the triangles do not
+coincide, the vertex of each opposite the common side $DF$ lies
+outside of the other. The two triangles have in common a triangle,
+say $DOF$, and besides this there remain of the two triangles two
+smaller triangles which have one angle the same in both and the same
+deficiency or excess. These two triangles, and therefore the
+original triangles, have the same area.
+
+\item[9.~Theorem.] \emph{The areas of any two triangles are
+proportional to their deficiencies or excesses.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=40mm]{fig20.png}
+\end{figure}
+
+\textbf{Proof.} A triangle may be divided into $n$ smaller triangles
+having equal deficiencies or excesses and equal areas by lines drawn
+from one vertex to points of the opposite side. Each of these
+triangles has for its deficiency or excess $\nicefrac{1}{n}$th of
+the deficiency or excess of the original triangle, and for its area
+$\nicefrac{1}{n}$th of the area of the original triangle.
+
+When the deficiencies or excesses of two triangles are
+commensurable, say in the ratio $m: n$, we can divide them into $m$
+and $n$ smaller triangles, respectively, all having the same
+deficiency or excess and the same area. The areas of the given
+triangles will therefore be in the same ratio, $m : n$.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig21.png}
+\end{figure}
+
+When the deficiencies or excesses of two triangles, $A$ and $B$, are
+not commensurable, we may divide one triangle, $A$, as above, into
+any number of equivalent parts, and take parts equivalent to one of
+these as many times as possible from the other, leaving a remainder
+which has a deficiency or excess less than the deficiency or excess
+of one of these parts. The portion taken from the second triangle
+forms a triangle, $B'$. $A$ and $B'$ have their areas proportional
+to their deficiencies or excesses, these being commensurable. Now
+increase indefinitely the number of parts into which $A$ is divided.
+These parts will diminish indefinitely, and the remainder when we
+take $B'$ from $B$ will diminish indefinitely. The deficiency or
+excess and the area of $B'$ will approach those of $B$, and the
+triangles $A$ and $B$ have their areas and their deficiencies or
+excesses proportional.
+
+\textbf{Corollary.} \emph{The areas of two polygons are to each
+other as their deficiencies or excesses.}
+
+\item[10.~Theorem.] \emph{Given a right triangle with a fixed angle;
+if the sides of the triangle diminish indefinitely, the ratio of the
+opposite side to the hypothenuse and the ratio of the adjacent side
+to the hypothenuse approach as limits the sine and cosine of this
+angle.}
+
+\textbf{Proof.} Lay off on the hypothenuse any number of equal
+lengths. Through the points of division $A_1$, $A_2$, $\cdots$ draw
+perpendiculars $A_1C_1$, $A_2C_2$, $\cdots$ to the base, and to
+these lines produced draw perpendiculars $A_2D_1$, $A_3D_2$,
+$\cdots$ each from the next point of division of the hypothenuse.
+
+\P~The triangles $OA_1C_1$ and $A_2A_1D_1$ are equal (II, 6, Cor.).
+\begin{align*}
+  C_2A_2 &\gtrless C_1D_1 &&\text{and} & C_1C_2 &\lessgtr D_1A_2;\\
+  \intertext{therefore,}
+  \frac{C_2A_2}{OA_2} &\gtrless \frac{C_1A_1}{OA_1}
+  &&\text{and} & \frac{OC_2}{OA_2} &\lessgtr \frac{OC_1}{OA_1},
+\end{align*}
+the upper sign being for the second hypothesis and the lower
+sign for the third hypothesis.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig22.png}
+\end{figure}
+
+\P~Assume
+\begin{equation*}
+  \frac{C_{r-1}A_{r-1}}{OA_{r-1}}
+  \gtrless \frac{C_{r-2}A_{r-2}}{OA_{r-2}}
+  \gtrless \dots
+  \gtrless \frac{C_1A_1}{OA_1},
+\end{equation*}
+and
+\begin{equation*}
+  \frac{OC_{r-1}}{OA_{r-1}}
+  \lessgtr \frac{OC_{r-2}}{OA_{r-2}}
+  \lessgtr \dots
+  \lessgtr \frac{OC_1}{OA_1}.
+\end{equation*}
+
+\P~Since
+\begin{align*}
+  OA_{r-1} &= (r-1) OA_1,\\
+\intertext{and also}
+           &= (r-1) A_{r-1}A_r,
+\end{align*}
+the inequalities
+\begin{equation*}
+  \frac{OC_1}{OA_1} \gtrless \frac{OC_{r-1}}{OA_{r-1}} \quad
+  \text{and}\quad
+  \frac{C_1A_1}{OA_1} \lessgtr \frac{C_{r-1}A_{r-1}}{OA_{r-1}}
+\end{equation*}
+applied to the angle at $A_{r-1}$ become
+\begin{equation*}
+  \frac{A_{r-1}D_{r-1}}{A_{r-1}A_r} \gtrless
+  \frac{C_{r-1}A_{r-1}}{OA_{r-1}}\quad
+  \text{and}\quad
+  \frac{D_{r-1}A_r}{A_{r-1}A_r} \lessgtr
+  \frac{OC_{r-1}}{OA_{r-1}}.
+\end{equation*}
+
+\P~The first of these two inequalities may be written
+\begin{equation*}
+  \frac{A_{r-1}D_{r-1}}{C_{r-1}A_{r-1}} \gtrless
+  \frac{A_{r-1}A_r}{OA_{r-1}}.
+\end{equation*}
+
+\P~Add 1 to both members,
+\begin{align*}
+\frac{C_{r-1}D_{r-1}}{C_{r-1}A_{r-1}} &\gtrless
+                                            \frac{OA_r}{OA_{r-1}}, \\
+\intertext{or}
+\frac{C_{r-1}D_{r-1}}{OA_r} &\gtrless
+                                   \frac{C_{r-1}A{r-1}}{OA_{r-1}}, \\
+\intertext{But}
+C_rA_r &\gtrless C_{r-1}D_{r-1}, \\
+\intertext{therefore}
+\frac{C_rA_r}{OA_r} &\gtrless \frac{C_{r-1}A_{r-1}}{OA_{r-1}}, \\
+\end{align*}
+
+\P~Again,
+\begin{align*}
+C_{r-1}C_r &\lessgtr D_{r-1}A_r. \\
+\intertext{Hence, from the second inequality above, we have}
+\frac{C_{r-1}C_r}{A_{r-1}A_r} &\lessgtr \frac{OC_{r-1}}{OA_{r-1}}, \\
+\intertext{or}
+\frac{C_{r-1}C_r}{OC_{r-1}} &\lessgtr \frac{A_{r-1}A_r}{OA_{r-1}}. \\
+\end{align*}
+
+\P~Add 1 to both members,
+\begin{align*}
+\frac{OC_r}{OC_{r-1}} &\lessgtr \frac{OA_r}{OA_{r-1}}, \\
+\intertext{or}
+\frac{OC_r}{OA_r} &\lessgtr \frac{OC_{r-1}}{OA_{r-1}}
+\end{align*}
+
+The ratios $\dfrac{CA}{OA}$ and $\dfrac{OC}{OA}$ being less than 1,
+and always increasing or always decreasing when the hypothenuse
+decreases, approach definite limits. These limits are continuous
+functions of $A$; if we vary the angle of any right triangle
+continuously, keeping the hypothenuse some fixed length, the other
+two sides will vary continuously, and the limits of their ratios to
+the hypothenuse must, therefore, vary continuously.
+
+Calling the limits for the moment $sA$ and $cA$, we may extend their
+definition, as in Trigonometry, to any angles, and prove that all
+the formulæ of the sine and cosine hold for these functions. Then
+for certain angles, $30^\circ$, $45^\circ$, $60^\circ$, we can prove
+that they have the same values as the sine and cosine, and their
+values for all other angles as determined from their values for
+these angles will be the same as the corresponding values of the
+sine and cosine.
+
+\begin{figure}[ht]
+\centering
+\includegraphics[width=50mm]{fig23.png}
+\end{figure}
+
+Draw a perpendicular, $CF$, from the right angle $C$ to the
+hypothenuse $AB$. The angle $FCB$ is not equal to $A$, but the
+difference, being proportional to the difference of areas of the two
+triangles $ABC$ and $FBC$, diminishes indefinitely when the sides of
+the triangles diminish. From the relation
+\begin{equation*}
+\frac{AF}{AC}\cdot\frac{AC}{AB} +
+    \frac{FB}{BC}\cdot\frac{BC}{AB} = 1,
+\end{equation*}
+we have, by passing to the limit,
+\begin{equation*}
+(cA)^2 + (sA)^2 = 1.
+\end{equation*}
+
+Let $x$ and $y$ be any two acute angles, and draw the figures used
+to prove the formulas for the sine and cosine of the sum of two
+angles.
+
+The angles $x$ and $y$ remaining fixed, we can imagine all of the
+lines to decrease indefinitely, and the functions $sx$, $ex$, $sy$,
+etc., are the limits of certain ratios of these lines.
+\begin{gather*}
+\begin{aligned}
+\frac{CA}{OA} &= \frac{CE}{OB}\cdot\frac{OB}{OA}
+                                + \frac{EA}{BA}\cdot\frac{BA}{OA}, \\
+\pm \frac{CA}{OA} &= \frac{OD}{OB}\cdot\frac{OB}{OA}
+                                - \frac{CD}{BA}\cdot\frac{BA}{OA}, \\
+\end{aligned} \\
+\left(-\frac{OC}{OA} \text{ in the second figure} \right).
+\end{gather*}
+
+The angles at $M$ are equal in the two triangles $EMB$ and $CMO$,
+and we may write
+\begin{equation*}
+\frac{CM}{OM} = \frac{ME + \delta}{MB}
+                              = \frac{ME \pm CM + \delta}{MB \pm OM},
+\end{equation*}
+where $\delta$ has the limit zero.
+\begin{equation*}
+\therefore \lim \frac{CE}{OB} = \lim\frac{CM}{OM} = sx.
+\end{equation*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig24.png}
+\end{figure}
+
+The angle $EAB$, or $x'$, is not the same as $x$, but differs from
+$x$ only by an amount which is proportional to the difference of the
+areas of the triangles $OMC$ and $MAB$, and which, therefore,
+diminishes indefinitely. Thus, the limits of $sx'$ and $cx'$ are
+$sx$ and $cx$.
+
+Finally, as the two triangles $ACN$ and $BDN$ have the angle $N$ in
+common, we may write
+\begin{equation*}
+\frac{DN}{BN} = \frac{CN + \delta '}{AN}
+                                = \frac{CN - DN + \delta '}{AN - BN},
+\end{equation*}
+where the limit of $\delta'$ is zero.
+\begin{equation*}
+\therefore \lim \frac{CD}{AB} = \lim\frac{CN}{AN} = sx.
+\end{equation*}
+
+Now at the limits our identities become
+\begin{align*}
+s(x+y) &=  sx \cdot cy + cx \cdot sy, \\
+c(x+y) &=  cx \cdot cy - sx \cdot sy.
+\end{align*}
+
+By induction, these formulæ are proved true for any angles. Other
+formulæ sufficient for calculating the values of these functions
+from their values for $30^\circ$, $45^\circ$, and $60^\circ$ are
+obtained from these two by algebraic processes.
+
+If the sides of an isosceles right triangle diminish indefinitely,
+the angle does not remain fixed but approaches $45^\circ$, and the
+ratios of the two sides to the hypothenuse approach as limits $s\
+45^\circ$ and $c\ 45^\circ$. Therefore, these latter are equal, and
+since the sum of their squares is 1, the value of each is
+$\nicefrac{1}{\sqrt{2}}$, the same as the value of the sine and
+cosine of $45^\circ$.
+
+Again, bisect an equilateral triangle and form a triangle in which
+the hypothenuse is twice one of the sides. When the sides diminish,
+preserving this relation, the angles approach $30^\circ$ and
+$60^\circ$. Therefore, the functions, $s$ and $c$, of these angles
+have values which are the same as the corresponding values of the
+sine and cosine of the same angles.
+
+\textbf{Corollary.} \emph{When any plane triangle diminishes
+indefinitely, the relations of the sides and angles approach those
+of the sides and angles of plane triangles in the ordinary geometry
+and trigonometry with which we are familiar.}
+
+\item[11.~Theorem.] \emph{Spherical geometry is the same in the
+three hypotheses, and the formula of spherical trigonometry are
+exactly those of the ordinary spherical trigonometry.}
+
+\textbf{Proof.} On a sphere, arcs of great circles are proportional to
+the angles which they subtend at the centre, and angles on a
+sphere are the same as the diedral angles formed by the planes
+of the great circles which are the sides of the angles. Their
+relations are established by drawing certain plane triangles
+which may be made as small as we please, and therefore may
+be assumed to be like the plane triangles in the hypothesis
+of a right angle. These relations are, therefore, those of the
+ordinary Spherical Trigonometry.
+\end{description}
+
+\medskip The three hypotheses give rise to three systems of Geometry,
+which are called the Parabolic, the Hyperbolic, and the Elliptic
+Geometries. They are also called the Geometries of Euclid, of
+Lobachevsky, and of Riemann. The following considerations exhibit
+some of their chief characteristics.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig25.png}
+\end{figure}
+
+Given $PC$ perpendicular to a line, $CF$; on the latter we take
+\begin{align*}
+  CD    &= PC, \\
+  DD'   &= PD, \\
+  D'D'' &= PD',\ \text{etc.}
+\end{align*}
+
+\label{p27xref}Now if $PC$ is sufficiently short (restricted), it is
+shorter than \emph{any} other line from $P$ to the line $CF$; for
+any line as short as $PC$ or shorter would be included in a
+restricted portion of the plane about the point $P$, for which the
+perpendicular is the shortest distance from the point to the line.
+
+Therefore,
+\begin{align*}
+  PD  &> PC,               &\therefore CD' &> 2CD,\\
+  PD' &> PC,\ \text{etc.;} &          CD'' &> 3CD,\ \text{etc.}
+\end{align*}
+
+Again, in the three hypotheses, respectively,
+\begin{align*}
+  CPD    &\stackrel{=}{\lessgtr} \frac{\pi}{4}, &&\text{and}&
+                         CDP &\stackrel{=}{\lessgtr} \frac{\pi}{4},\\
+  DPD'   &\stackrel{=}{\lessgtr} \tfrac{1}{2} CPD, &&&
+                    CD'P  &\stackrel{=}{\lessgtr} \tfrac{1}{2} CDP,\\
+  D'PD'' &\stackrel{=}{\lessgtr} \tfrac{1}{2} DPD',\ \text{etc.,} &&&
+        CD''P &\stackrel{=}{\lessgtr} \tfrac{1}{2} CD'P,\ \text{etc.}
+\end{align*}
+
+At $P$ we have a series of angles. In the first hypothesis there is
+an infinite number of these angles, and the series forms a
+geometrical progression of ratio $\nicefrac{1}{2}$, whose value is
+exactly $\nicefrac{\pi}{2}$. In the second hypothesis there is also
+an infinite number of these angles, and the terms of the series are
+less than the terms of the geometrical progression. The value of the
+series is, therefore, less than $\nicefrac{\pi}{2}$. In the third
+hypothesis we have a series whose terms are greater than those of
+the geometrical progression, and, therefore, whether the series is
+convergent or divergent, we can get more than $\nicefrac{\pi}{2}$ by
+taking a sufficient number of terms. In other words, we can get a
+right angle or more than a right angle at $P$ by repeating this
+process a certain finite number of times.
+
+The angles at $D$, $D'$, $D''$, $\cdots$ are exactly equal to the
+terms of the series of angles at $P$. In the first two hypotheses
+they approach zero as a limit.
+
+The distances $CD$, $CD'$, $CD''$, $\cdots$ increase each time by
+more than a definite quantity, $CD$; therefore, if we repeat the
+process an unlimited number of times, these distances will increase
+beyond all limit. Thus, in the first and second hypotheses we prove
+that a straight line must be of infinite length.
+
+In the hypothesis of the obtuse angle the line perpendicular to $PC$
+at the point $P$ will intersect $CF$ in a point at a certain finite
+distance from $C$, one of the $D$'s, or some point between. On the
+other side of $PC$ this same perpendicular will intersect $FC$
+produced at the same distance. But we have assumed that two
+different straight lines cannot intersect in two points; therefore,
+for us the third hypothesis cannot be true unless the straight line
+is of finite length returning into itself, and these two points are
+one and the same point, its distance from $C$ in either direction
+being one-half the entire length of the line. In this way, however,
+we can build up a consistent Geometry on the third hypothesis, and
+this Geometry it is which is called the Elliptic Geometry.
+
+The constructions would have been the same, and very nearly all the
+statements would have been the same, if we had taken $CD$ any
+arbitrary length on $CF$.
+
+The restriction which we have placed upon some of the propositions
+of this chapter is necessary in the third hypothesis.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig26.png}
+\end{figure}
+
+Thus, in the proof that the exterior angle of a triangle is greater
+than the opposite interior angle, the line $AD$ drawn through the
+vertex $A$ to the middle point $D$ of the opposite side was produced
+so as to make $AE=2AD$. If $AD$ were greater than half the entire
+length of the straight line determined by $A$ and $D$, this would
+bring the point $E$ past the point $A$, and the angle $CBE$, which
+is equal to the angle $C$, instead of being a part of the exterior
+angle $CBF$, becomes greater than this exterior angle.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig27.png}
+\end{figure}
+
+Again, if two angles of a triangle are equal and the side between
+them is just an entire straight line, it does not follow necessarily
+that the opposite sides are equal. It may be said, however, that the
+opposite sides form one continuous line, and, therefore, this figure
+is not strictly a triangle, but a figure somewhat like a lune. The
+points $A$ and $B$ are the same point, and the angles $A$ and $B$
+are vertical angles.
+
+Finally, though we assume that the shortest path between two points
+is a straight line, it is not always true that a straight line drawn
+between two points is the shortest path between them. We can pass
+from one point to another in two ways on a straight line; namely,
+over each of the two parts into which the two points divide the line
+determined by them. One of these parts will usually be shorter than
+the other, and the longer part will be longer than some paths along
+broken lines or curved lines.
+
+When, however, the straight line is of infinite length, that is, in
+the hypothesis of the right angle and in the hypothesis of the acute
+angle, all the propositions of this chapter hold without
+restriction.
+
+\medskip The Euclidean Geometry is familiar to all. We will now make
+a detailed study of the Geometry of Lobachevsky, and then take up in
+the same way the Elliptic Geometry.
+
+\chapter{THE HYPERBOLIC GEOMETRY}
+
+We have now the hypothesis of the acute angle. Two lines in a plane
+perpendicular to a third diverge on either side of their common
+perpendicular. The sum of the angles of a triangle is less than two
+right angles, and the propositions of the last chapter hold without
+restriction.
+
+\section{Parallel Lines}
+
+From any point, $P$, draw a perpendicular, $PC$, to a given line,
+$AB$, and let $PD$ be any other line from $P$ meeting $CB$ in $D$.
+If $D$ move off indefinitely on $CB$, the line $PD$ will approach a
+limiting position $PE$.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig28.png}
+\end{figure}
+
+$PE$ is said to be parallel to $CB$ at $P$. $PE$ makes with $PC$ an
+angle, $CPE$, which is called the angle of parallelism for the
+perpendicular distance $PC$. It is less than a right angle by an
+amount which is the limit of the deficiency of the triangle $PCD$.
+On the other side of $PC$ we can find another line parallel to $CA$
+and making with $PC$ the same angle of parallelism. We say that $PE$
+is parallel to $AB$ towards that part which is on the same side of
+$PC$ with $PE$. Thus, at any point there are two parallels to a
+line, but only one towards one part of the line. Lines through $P$
+which make with $PC$ an angle greater than the angle of parallelism
+and less than its supplement do not meet $AB$ at all. We write
+$\Pi(p)$ to denote the angle of parallelism for a perpendicular
+distance, $p$.
+
+\begin{description}
+\item[1.~Theorem.] \emph{A straight line maintains its parallelism
+at all points.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig29.png}
+\end{figure}
+
+\textbf{Let} $AB$ be parallel to $CD$ at $E$ and let $F$ be any
+other point of $AB$ on either side of $E$, to prove that $AB$ is
+parallel to $CD$ at $F$.
+
+\textbf{Proof.} To $H$, on $CD$, draw $EH$ and $FH$. If $H$ move off
+indefinitely on $CD$, these two lines will approach positions of
+parallelism with $CD$. But the limiting position of $EH$ is the line
+$AB$ passing through $F$, and if the limiting position of $FH$ were
+some other line, $FK$, $F$ would be the limiting position of $H$,
+the intersection of $EH$ and $FH$.
+
+\item[2.~Theorem.] \emph{If one line is parallel to another, the
+second is parallel to the first.}
+
+\textbf{Given} $AB$ parallel to $CD$, to prove that $CD$ is parallel
+to $AB$.
+
+\textbf{Proof.} Draw $AC$ perpendicular to $CD$. The angle $CAB$
+will be acute; therefore, the perpendicular $CE$ from $C$ to $AB$
+must fall on that side of $A$ towards which the line $AB$ is
+parallel to $CD$ (Chap.~I, II, 1). The angle $ECD$ is then acute and
+less than $CEB$, which is a right angle. That is, we have
+\begin{equation*}
+CAB < ACD, \quad \text{and} \quad CEB > ECD.
+\end{equation*}
+
+If the line $CE$ revolve about the point $C$ to the position of
+$CA$, the angle at $E$ will decrease to the angle $A$, and the angle
+at $C$ will increase to a right angle. There will be some position,
+say $CF$, where these two angles become equal; that is,
+\begin{equation*}
+CFB = FCD.
+\end{equation*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig30.png}
+\end{figure}
+
+Draw $MN$ perpendicular to $CF$ at its middle point and revolve the
+figure about $MN$ as an axis. $CD$ will fall upon the original
+position of $AB$, and $AB$ will fall upon the original position of
+$CD$. Therefore, $CD$ is parallel to $AB$.
+
+\textbf{Corollary.} \emph{$FB$ and $CD$ are both parallel to $MN$.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig31.png}
+\end{figure}
+
+\textbf{Proof.} $FB$ and $CD$ are symmetrically situated with
+respect to $MN$, and cannot intersect $MN$ since they do not
+intersect each other. Draw $FH$ to $H$, on $CD$, intersecting $MN$
+in $K$. If $H$ move off indefinitely on $CD$, $FH$ will approach the
+position of $FB$ as a limit. Now $K$ cannot move off indefinitely
+before $H$ does, for $FK < FH$. But again, when $H$ moves off
+indefinitely, $K$ cannot approach some limiting position at a finite
+distance on $MN$; for $FB$, and therefore $CD$, would then intersect
+$MN$ and each other at this point. Therefore, $H$ and $K$ must move
+off together, and the limiting position of $FH$ must be at the same
+time parallel to $CD$ and $MN$.
+
+In the same way we can prove that any line lying in a plane between
+two parallels must intersect one of them or be parallel to both.
+
+\item[3.~Theorem.] \emph{Two lines parallel to a third towards the
+same part of the third are parallel to each other.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig32.png}
+\end{figure}
+
+\textbf{First,} when they are all in the same plane.
+
+\textbf{Let} $AB$ and $EF$ be parallel to $CD$, to prove that they
+are parallel to each other.
+
+\textbf{Proof.} Suppose $AB$ lies between the other two. To $H$, any
+point on $CD$, draw $AH$ and $EH$, and let $K$ be the point where
+$EH$ intersects $AB$. As $H$ moves off indefinitely on $CD$, $AH$
+and $EH$ approach as limiting positions $AB$ and $EF$. Now $K$
+cannot move off indefinitely before $H$ does, for $EK < EH$. But
+again, when H moves off indefinitely, $K$ cannot approach some
+limiting position at a finite distance on $AB$; for this point would
+be the intersection of $AB$ and $EF$, and the limiting position of
+$H$, whereas $H$ moves off indefinitely on $CD$. Therefore, $H$ and
+$K$ must move off together, and the limiting position of $EH$ must
+be at the same time parallel to $CD$ and $AB$.
+
+If $AB$, lying between the other two, is given parallel to $CD$ and
+$EF$, $EF$ must be parallel to $CD$; for a line through $E$ parallel
+to $CD$ would be parallel to $AB$, and only one line can be drawn
+through $E$ parallel to $AB$ towards the same part.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig33.png}
+\end{figure}
+
+\textbf{Second}, when the lines are not all in the same plane.
+
+\textbf{Let} $AB$ and $CD$ be two parallel lines and let $E$ be any
+point not in their plane.
+
+\textbf{Proof.} To $H$ on $CD$ draw, $AH$ and $EH$. As $H$ moves off
+indefinitely, $AH$ approaches the position of $AB$, and the plane
+$EAH$ the position of the plane $EAB$. Therefore, the limiting
+position of $EH$ is the intersection of the planes $ECD$ and $EAB$.
+The intersection of these planes is, then, parallel to $CD$, and in
+the same way we prove that it is parallel to $AB$.
+
+Now, if $EF$ is given as parallel to one of these two lines towards
+the part towards which they are parallel, it must be the
+intersection of the two planes determined by them and the point $E$,
+and therefore parallel to the other line also.
+
+\newpage
+\item[4.~Theorem.] \emph{Parallel lines continually approach each
+other.}
+
+\textbf{Let} $AB$ and $CD$ be parallel, and from $A$ and $B$, any
+points on $AB$, drop perpendiculars $AC$ and $BD$ to $CD$. Supposing
+that $B$ lies beyond $A$ in the direction of parallelism, we are to
+prove that $BD < AC$.
+
+\textbf{Proof.} At $H$, the middle point of $CD$, erect a
+perpendicular meeting $AB$ in $K$. The angle $BKH$ is an acute
+angle, and the angle $AKH$ is an obtuse angle. Therefore, a
+perpendicular to $HK$ at $K$ must meet $CA$ in some point, $E$,
+between $C$ and $A$ and $DB$ produced in some point, $F$, beyond
+$B$. But $DF = CE$ (Chap.~I, I, 12); therefore, $DB < CA$.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig34.png}
+\end{figure}
+
+\textbf{Corollary.} \emph{If $AB$ and $CD$ are parallel and $AC$
+makes equal angles with them (like $FC$ in 2 above), then $EF$,
+cutting off equal distances on these two lines, $AE = CF$, on the
+side towards which, they are parallel, will be shorter than $AC$.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig35.png}
+\end{figure}
+
+\textbf{Proof.} $MN$, perpendicular to $AC$ at its middle point, is
+parallel to $AB$ and bisects $EF$, the figure being symmetrical with
+respect to $MN$. $EH$, the half of $EF$, is less than $AM$, and
+therefore $EF$ is less than $AC$.
+
+\item[5.~Theorem.] \emph{As the perpendicular distance varies,
+starting from zero and increasing indefinitely, the angle of
+parallelism decreases from a right angle to zero.}
+
+\textbf{Proof.} In the first place the angle of parallelism, which
+is acute as long as the perpendicular distance is positive, will be
+made to differ from a right angle by less than any assigned value if
+we take a perpendicular distance sufficiently small.
+
+\begin{figure}[ht]
+\centering
+\includegraphics[width=70mm]{fig36.png}
+\end{figure}
+
+For, $ADE$ being any given angle as near a right angle as we please,
+we can take a point, $L$, on $DE$ and draw $LR$ perpendicular to
+$DA$ at $R$. The angle $RDL$ must increase to become the angle of
+parallelism for the perpendicular distance $RD$.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig37.png}
+\end{figure}
+
+Now let $p$ be the length of a given perpendicular $PM$, and let
+$\alpha$ be the amount by which its angle of parallelism differs
+from $\nicefrac{\pi}{2}$; that is, say
+\begin{equation*}
+\Pi(p) = \frac{\pi}{2} - \alpha.
+\end{equation*}
+$PM$, being perpendicular to $MN$, and $H$ any point on $MN$, the
+angle $MPH$ approaches as a limit the angle of parallelism,
+$\Pi(p)$, when $H$ moves off indefinitely on $MN$. The line $PH$
+meets the line $MN$ as long as $MPH < \Pi(p)$, and by taking $MPH$
+sufficiently near $\Pi(p)$, but less, we can make the angle $MHP$ as
+small as we please (see p.~\pageref{p27xref}).
+
+In figure on page~\pageref{p38fig}, let $AC$ be perpendicular to
+$AB$, $D$ being any point on $AC$ and $DE$ parallel to $AB$. Draw
+$DK$ beyond $DE$, making with $DE$ an angle, $EDK=\Pi(p)$, and make
+$DK = p$. $TF$, perpendicular to $DK$ at $K$, will be parallel to
+$DE$ and $AB$.
+
+By placing $PM$ of the last figure upon $DKT$, we see that $DC$ will
+meet $KT$ in a point, $G$ if
+\begin{gather*}
+KDC < \Pi(p), \\
+\intertext{that is, if}
+ADE > 2\alpha.
+\end{gather*}
+
+\P~Then in the right triangle $DKG$,
+\begin{gather*}
+DGK + KDG < \frac{\pi}{2}. \\
+\intertext{But}
+ADE + KDG = \frac{\pi}{2} + \alpha; \\
+\intertext{therefore,}
+DGK < ADG - \alpha.
+\end{gather*}
+
+\label{p38fig}\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig38.png}
+\end{figure}
+
+Starting from the point $G$, we can repeat this construction, and
+each time we subtract from the angle of parallelism an amount
+greater than $\alpha$. We can continue this process until the angle
+of parallelism becomes equal to or less than $2\alpha$.
+
+If the point $D$ move along $AC$, $DE$ remaining constantly parallel
+to $AB$, the angle at $D$ will constantly diminish, and by letting
+$D$ move sufficiently far on $AC$ we can reach a point where this
+angle becomes equal to or less than $2\alpha$.
+
+Suppose $D$ is at the point where the angle of parallelism is just
+$2\alpha$. Then, if we draw $DK$ and $TF$ as before, $KT$ will be
+parallel to $DC$. All the parallels to $AB$ lying between $AB$ and
+this position of $TF$ meet $AC$, and as the parallel moves towards
+this position of $TF$, the angle of parallelism at $D$ approaches
+zero, and the point $D$ moves off indefinitely.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig39.png}
+\end{figure}
+
+For an obtuse angle we may take $p$ negative, and we have
+\begin{equation*}
+\Pi(-p) = \pi - \Pi(p).
+\end{equation*}
+
+\item[6.~Theorem.] \emph{The perpendiculars erected at the middle
+points of the sides of a triangle are all parallel if two of them
+are parallel.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=90mm]{fig40.png}
+\end{figure}
+
+\textbf{Let} $A$, $B$, and $C$ be the vertices of the triangle, and
+$D$, $E$, and $F$, respectively, the middle points of the opposite
+sides. Suppose the perpendiculars at $D$ and $E$ are given parallel,
+to prove that the perpendicular at $F$ is parallel to them.
+
+\textbf{Proof.} Draw $CM$ through $C$ parallel to the two given
+parallel perpendiculars. $CM$ forms with the two sides at $C$ angles
+of parallelism $\Pi\left(\dfrac{a}{2}\right)$ and
+$\Pi\left(\dfrac{b}{2}\right)$, of which the angle at $C$ is the sum
+or difference according as $C$ lies between the given perpendiculars
+or on the same side of both. By properly diminishing these angles at
+$C$, keeping the lengths of $CA$ and $CB$ unchanged, we can make the
+perpendiculars at their middle points $D$ and $E$ intersect $CM$,
+and therefore each other, at any distance from $C$ greater than
+$\nicefrac{a}{2}$ and greater than $\nicefrac{b}{2}$.
+
+Let $A'B'C'$ be the triangle so formed, $O$ the point where the two
+given perpendiculars meet, and $C'M'$ the line through $O$. In the
+triangle $A'B'C'$, the three perpendiculars meet at the point $O$
+(Chap.~I, I, 5), Now we can let $O$ move off on $C'M'$, the
+construction remaining the same. That is, we let the lines $C'A'$
+and $C'B'$ rotate about $C'$ without changing their lengths, in such
+a manner that the three perpendiculars $D'O$, $E'O$, and $F'O$ shall
+always pass through $O$. As $O$ moves off indefinitely, the angles
+at $C'$ approach $\Pi\left(\dfrac{a}{2}\right)$ and
+$\Pi\left(\dfrac{b}{2}\right)$ as limits, and the three
+perpendiculars approach positions of parallelism with $C'M'$ and
+with each other. But the triangle $A'B'C'$ approaches as a limit a
+triangle which is equal to $ABC$, having two sides and the included
+angle equal, respectively, to the corresponding parts of the latter.
+Therefore, in $ABC$ the three perpendiculars are all parallel.
+
+\item[7.~Theorem.] \emph{Lines which do not intersect and are not
+parallel have one and only one common perpendicular.}
+
+\textbf{Proof.} Let $AB$ and $CD$ be the two lines, and from $A$,
+any point of $AB$, drop $AC$ perpendicular to $CD$. If $AC$ is not
+itself the common perpendicular, one of the angles which it makes
+with $AB$ will be acute. Let this angle be on the side towards $AB$,
+so that $BAC < \nicefrac{\pi}{2}$. Draw $AE$ parallel to $CD$ on
+this same side of $AC$. The angle $EAC$ is less than $BAC$, since
+$AB$ is not parallel to $CD$ and does not intersect it. Let $AH$ be
+any line drawn in the angle $EAC$, intersecting $CD$ at $H$. If $H$,
+starting from the position of $C$, move off indefinitely on the line
+$CD$, the angle $BAH$ will decrease from the magnitude of the angle
+$BAC$ to the angle $BAE$. The angle $AHC$ will decrease
+\emph{indefinitely} from the magnitude of the angle at $C$, which is
+a right angle and greater than $BAC$. There will be some position
+for which $BAH=AHC$. In this position the line $NM$ through the
+middle point of $AH$ perpendicular to one of the two given lines
+will be perpendicular to the other, as proved in Chap.~I, I, 10.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig41.png}
+\end{figure}
+
+If there were two common perpendiculars we should have a rectangle,
+which is impossible in the Hyperbolic Geometry.
+
+\item[8.~Theorem.] \emph{If the perpendiculars erected at the middle
+points of the sides of a triangle do not meet and are not. parallel,
+they are all perpendicular to a certain line.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=90mm]{fig42.png}
+\end{figure}
+
+\textbf{Proof.} We can draw a line, $AB$, that will be perpendicular
+to two of these lines, and the perpendiculars from the three
+vertices of the triangle upon this line will be equal, by Chap.~I,
+II, 13. A perpendicular to $AB$ erected midway between any two of
+these three is perpendicular to the corresponding side of the
+triangle at its middle point (Chap.~I, I, 11). Thus, all three of
+the perpendiculars erected at the middle points of the sides of the
+triangle are perpendicular to $AB$.
+
+A line is parallel to a plane if it is parallel to its projection on
+the plane.
+
+\item[9.~Theorem.] \emph{A line may be drawn perpendicular to a
+plane and parallel to any line not in the plane.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=100mm]{fig43.png}
+\end{figure}
+
+\textbf{Proof.} Let $AB$ be the given line and $MN$ the plane, if
+$AB$ meets the plane $MN$ at a point, $A$, we take on its projection
+a length, $AC$, such that the angle at $A$ equals $\Pi(AC)$. Then
+$CD$, perpendicular to the plane at $C$, will be parallel to $AB$.
+In the same way, on the other side of the plane a perpendicular can
+be drawn parallel to $BA$ produced.
+
+If $AB$ does not meet $MN$, then at least in one direction it
+diverges from $MN$. Through $H$, any point of the projection of $AB$
+on the plane, we can draw a line, $HK$, parallel to $AB$ towards
+that part of $AB$ which diverges from $MN$, and then draw $CD$
+parallel to this line and perpendicular to the plane.
+
+Unless $AB$ is parallel to $MN$ it will meet the plane at some
+point, or the plane and line will have a common perpendicular, and
+the line will diverge from the plane in both directions. In the
+latter case there are two perpendiculars that are parallel to the
+line, one parallel towards each part of the line.
+
+Two perpendiculars cannot be parallel towards the same part of a
+line; for then they would be parallel to each other, and two lines
+cannot be perpendicular to a plane and parallel to each other.
+\end{description}
+
+\newpage
+\section{Boundary-curves and Surfaces, and Equ\-i\-dist\-ant-curves
+and Surfaces}
+
+Having given the line $AB$, at its extremity, $A$, we take any
+arbitrary angle and produce the side $AC$ so that the perpendicular
+erected at its middle point shall be parallel to $AB$. The locus of
+the point $C$ is a curve which is called oricycle, or
+boundary-curve. $AB$ is its axis.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig44.png}
+\end{figure}
+
+From their definition it follows that all boundary-curves are equal,
+and the boundary-curve is symmetrical with respect to its axis; if
+revolved through two right angles about its axis, it will coincide
+with itself.
+
+\begin{description}
+\item[1.~Theorem.] \emph{Any line parallel to the axis of a
+boundary-curve may be taken for axis.}
+
+\textbf{Let} $AB$ be the axis and $CD$ any line parallel to $AB$, to
+prove that $CD$ may be taken as axis.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig45.png}
+\end{figure}
+
+\textbf{Proof.} Draw $AC$; also to $E$, any other point on the
+curve, draw $AE$ and $CE$. The perpendiculars erected at the middle
+points of $AC$ and of $AE$ are parallel to $AB$ and $CD$ and to each
+other. Therefore, the perpendicular erected at the middle point of
+$CE$, the third side of the triangle $ACE$, is parallel to them and
+to $CD$. $CD$ then may be taken as axis.
+
+\textbf{Corollary.} \emph{The boundary-curve may be slid along on
+itself without altering its shape; that is, it has a constant
+curvature.}
+
+\item[2.~Theorem.] \emph{Two boundary-curves having a common set of
+axes cut off the same distance on each of the axes, and the ratio of
+corresponding arcs depends only on this distance.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig46.png}
+\end{figure}
+
+\textbf{Proof.} Take any two axes and a third axis bisecting the arc
+which the first two intercept on one of the two boundary-curves. By
+revolving the figure about this axis we show that the curves cut off
+equal distances on the two axes.
+
+Let $AA'$, $BB'$, and $CC'$ be any three axes of the two
+boundary-curves $AB$ and $A'B'$; let their common length be $x$ and
+let them intercept arcs $s$ and $t$ on $AB$, $s'$ and $t'$ on
+$A'B'$.
+
+When $s = t$, $s' = t'$, and, in general,
+\begin{equation*}
+\frac{s}{t} = \frac{s'}{t'},
+\end{equation*}
+as we prove, first when $s$ and $t$ are commensurable, and then by
+the method of limits when they are incommensurable. The ratio
+$\nicefrac{s}{s'}$ is, therefore, a constant for the given value of
+$x$.
+
+\P~Write $\nicefrac{s}{s'} = f(x).$
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig47.png}
+\end{figure}
+
+\label{p45ref}\P~From three boundary-curves having the same set of
+axes, we find
+\begin{equation*}
+f(x+y) = f(x)f(y)
+\end{equation*}
+This property is characteristic of the exponential function whose
+general form is $f(x)= e^{ax}$.\label{p45fn}\footnote{Putting $y =
+x$, $2x$, $\cdots$ $(n-1)x$ in succession, we find
+\begin{equation*}
+f(nx) = [f(x)]^n
+\end{equation*}
+for positive integer values of $n$, $x$ being any positive quantity.
+\par Now
+\begin{equation*}
+f\left(\frac{r}{s}x\right) = \left[f\left(\frac{x}{s}\right)\right]^r,
+\end{equation*}
+and this is the $r$th power of the $s$th root of the first member of
+the equation
+\begin{equation*}
+\left[f\left(\frac{x}{s}\right)\right]^s = f(x);\quad \therefore
+f\left(\frac{r}{s}x\right) = [f(x)]^\frac{r}{s}.
+\end{equation*}
+Thus, assuming that $f(x)$ is a continuous function of $x$, we have
+proved that for all real positive values of $x$ and $n$
+\begin{gather*}
+f(nx) = [f(x)]^n, \\
+\intertext{and if we put $x$ for $n$ and $1$ for $x$, we have}
+f(x) = [f(1)]^x.
+\end{gather*}
+\par We will write $f(1) = e^a$; then
+\begin{equation*}
+f(x) = e^{ax}.
+\end{equation*}
+} Therefore, $\nicefrac{s}{s'}= e^{ax}$, the value of $a$ depending
+on the unit of measure (see below p.~\pageref{p76ref}).
+
+\item[3. Theorem.] \emph{The area enclosed by two
+boundary-curves having the same axes and by two of their common axes
+is proportional to the difference of the intercepted arcs.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig48.png}
+\end{figure}
+
+\textbf{Proof.} Let $s$ and $s'$ be the lengths of the intercepted
+arcs, and $l$ the distance measured on an axis between them. Let
+$t$, $t'$, and $k$ be the corresponding quantities for a second
+figure constructed in the same way.
+
+If the corresponding lines in the two figures are all equal, the
+areas are equal, for they can be made to coincide. If only $k = l$,
+the areas are to each other as corresponding arcs, say as $s' : t'$,
+proved first when the arcs are commensurable, and then by the method
+of limits when they are incommensurable.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig49.png}
+\end{figure}
+
+When $l$ and $k$ are commensurable, suppose
+\begin{equation*}
+\frac{l}{m} = \frac{k}{n} = a.
+\end{equation*}
+We can draw a series of boundary-curves at distances equal to $a$ on
+the axes and divide the areas into $m$ and $n$ parts, respectively.
+If $r$ is the ratio of arcs corresponding to the distance $a$, these
+parts will be proportional to the quantities
+\begin{gather*}
+  s',\ s'r,\ s'r^2,\dots s'r^{m-1};\\
+  t',\ t'r,\ t'r^2,\dots t'r^{n-1}.
+\end{gather*}
+
+The two areas are then to each other in the ratio
+\begin{equation*}
+  s' \frac{r^m-1}{r-1} \colon t' \frac{r^n-1}{r-1}.
+\end{equation*}
+
+But
+\begin{equation*}
+  s'r^m = s \text{ and } t'r^n = t,
+\end{equation*}
+so that this is the same as the ratio
+\begin{equation*}
+  s-s' \colon t-t'.
+\end{equation*}
+
+When $l$ and $k$ are incommensurable, we proceed as in other similar
+demonstrations.
+
+This theorem is analogous to the one which we have proved about
+polygons: the area is proportional to the amount of rotation in
+excess of four right angles in going around the figure, for the rate
+of rotation in going along a boundary-curve is constant.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig50.png}
+\end{figure}
+
+The locus of points at a given distance from a straight line is a
+curve which may be called an equidistant-curve. The perpendiculars
+from the different points of this curve upon the base line are equal
+and may be called axes of the curve.
+
+An equidistant-curve fits upon itself when revolved through two
+right angles about one of its axes or when slid along upon itself.
+It has a constant curvature.
+
+It can be proved, exactly as in the case of two boundary-curves
+having the same set of axes, that arcs on an equidistant-curve are
+proportional to the segments cut off by the axes at their
+extremities on the base line or on any other equidistant-curve
+having the same set of axes.
+
+\item[4.~Theorem.] \emph{The boundary-curve is a limiting curve
+between the circle and the equidistant-curve; it may be regarded as
+a circle with infinitely large radius, or as an equidistant-curve
+whose base line is infinitely distant.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig51.png}
+\end{figure}
+
+\textbf{Proof.} Take a line of given length, $AB = 2a$ say, making
+an angle, $A$, with a fixed line, $AC$. Construct another angle at
+$B$ equal to the angle $A$, and draw a perpendicular to $AB$ at its
+middle point, $D$.
+
+If the angle at $A$ is sufficiently small, we have an isosceles
+triangle with $AB$ for base, and its vertex at a point, $F$, on
+$AC$. With $F$ as centre, we can draw a circle through the points
+$A$ and $B$. Now let the angle at $A$ gradually increase, the rest
+of the figure varying so as to keep the construction. $F$ will move
+off indefinitely, and when $A = \Pi(a)$ the three lines $AF$, $BF$,
+and $DF$ will become parallel, and $B$ will become a point on the
+boundary-curve $AB'$, which has $AC$ for axis.
+
+On the other hand, if the angle at $A$ were taken acute, but greater
+than $\Pi(a)$, we should have three lines, $AE$, $BH$, and $DF$,
+perpendicular to a line, $EH$, the base line of an equidistant-curve
+through the points $A$ and $B$. Now let the angle $A$ gradually
+decrease, the rest of the figure varying so as to preserve the
+construction. The quadrilateral $ADFE$, having three right angles
+and the fourth angle $A$ decreasing, must increase in area. We get
+this same movement if we think of $AD$ and $DF$ remaining fixed in
+the plane while $AE$ revolves about $A$, making the angle $A$
+decrease. Thus the only way in which the area of the quadrilateral
+can increase is for $EH$ to move off along on $AC$ and become more
+and more remote from $A$. When $A$ becomes equal to $\Pi (a)$, $BH$
+and $DF$ become parallel to $AC$, and $B$ falls on the
+boundary-curve $AB'$.
+\end{description}
+
+Calling the radius of a circle axis, we find that circles,
+boundary-curves, and equidistant-curves have many properties in
+common:
+
+The perpendicular erected at the middle point of any chord is an
+axis. In particular, a tangent is perpendicular to the axis drawn
+from its point of contact. These are curves cutting at right angles
+a system of lines through a point, a system of parallel lines, and
+the perpendiculars to a given line, respectively.
+
+Two of these curves having the same set of axes cut off equal
+lengths on all these axes, and the ratio of corresponding arcs on
+two such curves is a constant depending only on the way in which
+they divide the axes.
+
+Three points determine one of these curves; that is, through any
+three points not in a straight line we can draw a curve which shall
+be either a circle, a boundary-curve, or an equidistant-curve, and
+through any three points only one such curve can be drawn. Any
+triangle may be inscribed in one and only one of these curves.
+
+Each of these curves can be moved on itself or revolved about any
+axis through $180^\circ$ into coincidence with itself.
+
+A boundary-surface or orisphere is a surface generated by the
+revolution of a boundary-curve about one of its axes.
+
+\begin{description}
+\item[5.~Theorem.] \emph{Any line parallel to the axis of a
+boundary-surface may be regarded as axis.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig52.png}
+\end{figure}
+
+\textbf{Let} $AA'$ be the axis, meeting the surface at $A$, and
+$BB'$ a line parallel to the axis through any other point, $B$, of
+the surface; to prove that $BB'$ may be regarded as axis.
+
+\textbf{Proof.} Let $C$ be a third point on the surface. Draw $CC'$
+through $C$, and through $D$, $E$, and $F$, the middle points of the
+sides of the plane triangle $ABC$, draw $DD'$, $EE'$, and $FF'$ all
+parallel to $AA'$. Finally, let $OO'$ be parallel to these lines and
+perpendicular to the plane $ABC$. The projecting planes of the other
+parallels all pass through $OO'$ (see I, 9).
+
+Since $AA'$ is axis to the surface, $EE'$ and $FF'$ are
+perpendicular to $AC$ and $AB$, respectively. Draw $FK$
+perpendicular to the plane $ABC$ at $F$. It will lie in the
+projecting plane $OFF'$. $AB$, being perpendicular to $FF'$ and to
+$FK$, is perpendicular to this plane, $OFF'$, and therefore to $OF$.
+In the same way we prove that $AC$ is perpendicular to $OE$.
+Therefore, $BC$ is perpendicular to $OD$ (Chap.~I, I, 5). But $OD$
+is the intersection of the plane $ABC$ with the plane $ODD'$. Hence,
+$BC$ is perpendicular to this plane and to $DD'$ (Chap.~I, II, 15).
+
+$DD'$ being parallel to $BB'$ lies in the plane determined by $BB'$
+and $BC$, and in this plane only one perpendicular can be drawn to
+$BC$ at its middle point. Therefore, if we pass any plane through
+$BB'$ and from $B$ draw a chord to any other point, $C$, of its
+intersection with the surface, the perpendicular in this plane to
+$BC$, erected at the middle point of $BC$, will be parallel to
+$BB'$. This proves that the section is a boundary-curve, having
+$BB'$ for axis, and that the surface can be generated by the
+revolution of such a boundary-curve around $BB'$.
+
+Therefore, $BB'$ may be regarded as axis of the surface.
+\end{description}
+
+A plane passed through an axis of a boundary-surface is called a
+principal plane. Every principal plane cuts the surface in a
+boundary-curve. Any other plane cuts the surface in a circle; for
+the surface may be regarded as a surface of revolution having for
+axis of revolution that axis which is perpendicular to the plane.
+This perpendicular may be called the axis of the circle, and the
+point where it meets the surface, the pole of the circle. The pole
+of a circle on a boundary-surface is at the same distance from all
+the points of the circle, distance being measured along
+boundary-lines on the surface.
+
+Any two boundary-surfaces can be made to coincide, and a
+boundary-surface can be moved upon itself, any point to the position
+of any other point, and any boundary-curve through the first point
+to the position of any boundary-curve through the second point. We
+may say that a boundary-surface has a constant curvature, the same
+for all these surfaces. Figures on a boundary-surface can be moved
+about or put upon any other boundary-surface without altering their
+shape or size.
+
+We can develop a Geometry on the boundary-surface. By \emph{line} we
+mean the boundary-curve in which the surface is cut by a principal
+plane. The angle between two lines is the same as the diedral angle
+between the two principal planes which cut out the lines on the
+surface.
+
+\begin{description}
+\item[6.~Theorem.] \emph{Geometry on the boundary-surface is the
+same as the ordinary Euclidean Plane Geometry.}
+
+\textbf{Proof.} On two boundary-surfaces with the same system of
+parallel lines for axes corresponding triangles are similar; that
+is, corresponding angles are equal, having the same measures as the
+diedral angles which cut them out, and corresponding lines are
+proportional by (2). But we can place these figures on the same
+surface; therefore, on one boundary-surface we can have similar
+triangles. Thus, we can diminish the sides of a triangle without
+altering their ratios or the angles. We can do this indefinitely;
+for the ratio of corresponding lines on the two surfaces, being
+expressed by the function $e^{ax}$ of the distance between them, can
+be made as large as we please by taking $x$ sufficiently large. If
+we assume that figures on the boundary-surface become more and more
+like plane figures when we diminish indefinitely their size, it
+follows that a triangle on this surface approaches more and more the
+form of an infinitesimal plane triangle, for which the sum of the
+angles is two right angles, and the angles and sides have the same
+relations as in the Euclidean Plane Geometry. All the formulæ of
+Plane Trigonometry with which we are familiar hold, then, for
+triangles on the boundary-surface.
+
+On the boundary-surface we have the ``hypothesis of the
+right angle''. Rectangles can be formed, and the area of a
+rectangle is proportional to the product of its base and altitude,
+while the area of a triangle is half of the area of a
+rectangle having the same base and altitude.
+\end{description}
+
+An equidistant-surface is a surface generated by the revolution of
+an equ\-i\-dist\-ant-curve about one of its axes. It is the locus of
+points at a given perpendicular distance from a plane. Any
+perpendicular to the plane may be regarded as an axis, and the
+surface is a surface cutting at right angles a system of lines
+perpendicular to the plane. The surface has a constant curvature,
+fitting upon itself in any position.
+
+\section{Trigonometrical Formulæ}
+
+\begin{description}
+\item[1.~Let] $ABC$ be a plane right triangle. Erect $AA'$
+perpendicular to its plane and draw $BB'$ and $CC'$ parallel to
+$AA'$. Draw a boundary-surface through $A$, having these lines for
+axes and forming the boundary-surface triangle $AB''C''$. Also
+construct the spherical triangle about the point $B$.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig53.png}
+\end{figure}
+
+The angle $A$ is the same in the plane triangle and in the
+boundary-surface triangle. The planes through $AA'$ are
+perpendicular to $ABC$. Hence, the spherical triangle has a right
+angle at the vertex which lies on $c$, and $BC$ being perpendicular
+to $CA$ is perpendicular to the plane of $CC'$ and $AA'$. Therefore,
+the plane $BCC'$ is perpendicular to the plane $ACC'$, and the
+diedral whose edge is $BC$ has for plane angle the angle $ACC'= \Pi
+(b)$. Since the boundary-surface triangle is right-angled at $C''$,
+the angle $B''$, or what is the same thing, the diedral whose edge
+is $BB'$, is the complement of the angle $A$.
+
+In the spherical triangle the side opposite the right angle is $\Pi
+(a)$, the two sides about the right angle are $\Pi (c)$ and $B$, and
+the opposite angles are $\Pi (b)$ and $90^\circ-A$.
+
+Applying to these quantities the trigonometrical formulæ for
+spherical right triangles, we get at once the relations that connect
+the sides and angles of plane right triangles.
+
+Produce to quadrants the two sides about the angle whose value is
+the complement of $A$. We form in this way a spherical right
+triangle in which the side opposite the right angle is the
+complement of $\Pi(c)$, the two sides about the right angle are the
+complements of $\Pi(a)$ and $\Pi (b)$, and their opposite angles are
+the complements of $B$ and $A$. From this triangle we deduce the
+following rule for passing from the formulæ of spherical right
+triangles to those of plane triangles:
+
+\emph{Interchange the two angles (or the two sides) and everywhere
+use the complementary function, tailing the corresponding angle of
+parallelism for the sides.}
+
+The formulæ for spherical right triangles are
+\begin{gather*}
+\begin{aligned}
+\sin A &= \frac{\sin a}{\sin c}. & \sin B &= \frac{\sin b}{\sin c}.\\
+\cos A &= \frac{\tan b}{\tan c}. & \cos B &= \frac{\tan a}{\tan c}.\\
+\tan A &= \frac{\tan a}{\sin b}. & \tan B &= \frac{\tan b}{\sin a}.\\
+\sin A &= \frac{\cos B}{\cos b}. & \sin B &= \frac{\cos A}{\cos a}.
+\end{aligned}\\
+\begin{aligned}
+\cos c &= \cos a \cdot \cos b \\
+\cos c &= \cot A \cdot \cot B \\
+\end{aligned}
+\end{gather*}
+
+\newpage
+From these, by the rule given on the previous page, we derive the
+following formulæ for plane right triangles:
+\begin{gather*}
+  \begin{aligned}
+    \cos B &= \frac{\cos\Pi(a)}{\cos\Pi(c)}. &
+    \cos A &= \frac{\cos\Pi(b)}{\cos\Pi(c)}.\\
+    \sin B &= \frac{\cot\Pi(b)}{\cot\Pi(c)}. &
+    \sin A &= \frac{\cot\Pi(a)}{\cot\Pi(c)}.\\
+    \cot B &= \frac{\cot\Pi(a)}{\cos\Pi(b)}. &
+    \cot A &= \frac{\cot\Pi(b)}{\cos\Pi(a)}.\\
+    \cos B &= \frac{\sin A}{\sin\Pi(b)}. &
+    \cos A &= \frac{\sin B}{\sin\Pi(a)}.
+  \end{aligned}\\
+  \begin{aligned}
+    \sin\Pi(c) &= \sin\Pi(a) \sin\Pi(b).\\
+    \sin\Pi(c) &= \tan A \tan B.\footnotemark
+  \end{aligned}
+\end{gather*}
+\footnotetext{We can arrange the parts of a right triangle so as to
+apply Napier's rules; namely, the arrangement would be
+
+\centering
+\includegraphics[width=30mm]{fig54.png}}
+
+We can obtain the formulæ for oblique plane triangles by dropping a
+perpendicular from one vertex upon the opposite side, thus forming
+two right triangles.
+
+\item[2.]~Take the relation
+\begin{equation*}
+  \sin\Pi(a) = \frac{\sin B}{\cos A}.
+\end{equation*}
+
+\textbf{Let} $p$, $q$, and $r$ be the sides of the triangle
+$AB''C''$ of our last demonstration and $p'$, $q'$, and $r$ the
+corresponding sides of the triangle formed in the same way on a
+boundary-surface tangent to the plane $ABC$ at $B$.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig55.png}
+\end{figure}
+
+\begin{align*}
+  \sin B &= \frac{q'}{r}, \\
+  \cos A &= \frac{q}{r} . \\
+  \therefore \sin \Pi(a) &= \frac{q'}{q} .
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=35mm]{fig56.png}
+\end{figure}
+
+\label{p56ref}Now $q$ and $q'$ are corresponding arcs on two
+boundary-curves which have the same set of parallel lines as axes,
+and their distance apart, $x$, is the distance from a boundary-curve
+of the extremity of a tangent of arbitrary length, $a$. Thus, we
+have for corresponding arcs
+\begin{equation*}
+  \frac{s'}{s} = \sin \Pi(a)
+\end{equation*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig57.png}
+\end{figure}
+
+\item[3.]~To $MN$, a given straight line, erect a perpendicular at a
+point, $O$, and on this perpendicular lay off $OA = y$ below $MN$,
+and $OB$ and $BP$ each equal to $x$ above $MN$, $x$ and $y$ being
+any arbitrary lengths. At $P$ draw $PR$ perpendicular to $OP$ and
+extending towards the left, and through $B$ draw $EF$ making with
+$OP$ an angle $\Pi (x)$, and therefore parallel on one side to $ON$
+and on the other side to $PR$. Finally, draw $AK$ and $AH$, the two
+parallels to $EF$ through $A$.
+
+At the point $A$ we have four angles of parallelism:
+\begin{align*}
+ CAK = CAH &= \Pi (AC),       \\
+       OAK &= \Pi (y) ,       \\
+       PAH &= \Pi (y+2x) .    \\
+\intertext{Therefore, }
+   \Pi (y) &= \Pi (AC) + BAC ,\\
+\intertext{and}
+\Pi (y+2x) &= \Pi (AC) - BAC . \\
+\intertext{Now in the right triangle $ABC$}
+\cos \Pi (y + x) &= \frac{\cos \Pi (AC)}{\cos BAC} \\
+\intertext{or}
+  \frac{1-\cos \Pi (y+x)}{1+\cos \Pi (y+x)}
+&= \frac{\cos BAC - \cos \Pi (AC)}{\cos BAC + \cos \Pi (AC)}, \\
+&= \frac{ \sin \tfrac{1}{2} \left[ \Pi (AC) + BAC \right]
+          \sin \tfrac{1}{2} \left[ \Pi (AC) - BAC \right] }
+        { \cos \tfrac{1}{2} \left[ \Pi (AC) + BAC \right]
+          \cos \tfrac{1}{2} \left[ \Pi (AC) - BAC \right] }; \\
+\intertext{whence,}
+\tan^2 \tfrac{1}{2} \Pi (y + x)
+  &= \tan \tfrac{1}{2} \Pi (y) \tan \tfrac{1}{2} \Pi (y + 2x)
+\end{align*}
+
+\P~$\tan \frac{1}{2} \Pi(x)$ is then a function of $x$, say $f(x)$,
+satisfying the condition
+\begin{align*}
+  \left[ f(y + x) \right]^2 & = f(y) f(y+2x) \\
+\intertext{or}
+  \frac{f(y+x)}{f(y)} & = \frac{f(y+2x)}{f(y+x)}
+\end{align*}
+and putting successively in this equation $y + x$, $y + 2x$, etc.,
+for $y$, we may add
+\begin{equation*}
+= \frac{f(y+3x)}{f(y+2x)} = \cdots
+= \frac{f(y+nx)}{f \left[ y+(n-1)x \right]}
+\end{equation*}
+
+\P~$\Pi (0) = \nicefrac{\pi}{2}$ and $\tan \tfrac{1}{2} \Pi (0) =
+1$; therefore, putting $y = 0$ in the first and last of all these
+fractions, we have
+\begin{align*}
+  f(x) & = \frac{f(nx)}{f\left[ (n-1) x\right]} , \\
+\intertext{or}
+  f(nx) & = f\left[(n-1)x \right] f(x). \\
+\therefore f(nx) & = \left[ f(x)\right]^n .
+\end{align*}
+
+This equation is characteristic of the exponential
+function.\footnote{See footnote, p.~\pageref{p45fn}.} $\Pi (x)$
+being an acute angle, $tan \tfrac{1}{2} \Pi (x) < 1$; therefore, we
+may write $f(1) = e^{-a'}$, so that $f(x) = e^{-a'x}$. $a'$ depends
+on the unit of measure; we will take the unit so that $a' = 1$.
+Finally, since $\Pi(-x) = \pi - \Pi (x)$,
+\begin{equation*}
+\tan \tfrac{1}{2} \Pi (-x)
+  = \cot \tfrac{1}{2} \Pi (x)
+  = \left[ \tan \tfrac{1}{2} \Pi (x) \right]^{-1} .
+\end{equation*}
+\newpage
+That is, for all real values of $x$
+\begin{align*}
+    \tan \tfrac{1}{2} \Pi (x) &= e^{-x} ,
+\intertext{or}
+\frac{1-\cos\Pi(x)}{\sin\Pi(x)}
+  &= \cos ix + i\sin ix.\footnotemark
+\end{align*}\label{p59ref}
+\footnotetext{$i$ stands for $\sqrt{-1}$. The best way to get the
+relations between the exponential and trigonometrical functions is
+by their developments in series:
+\begin{align*}
+  e^x &= 1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + \dotsb,\\
+  \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots +
+           (-1)^n\frac{x^{2n}}{2n!} + \dotsb,\\
+  \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots +
+           (-1)^n\frac{x^{2n+1}}{(2n+1)!} + \dotsb.
+\end{align*}
+\par These series are convergent for all values of $x$.
+\par Putting $ix$ for $x$, we have
+\begin{gather*}
+  \begin{aligned}
+    e^{ix} &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \dotsb\\
+    &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots +
+    i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dotsb \right);
+  \end{aligned} \\
+  \begin{aligned}
+\emph{i.e.,}\qquad e^{ix}  &= \cos x + i\sin x.\\
+\text{Also}\qquad e^{-ix} &= \cos x - i\sin x.\\
+    \therefore \cos x &= \tfrac{1}{2} \left(e^{ix} + e^{-ix}\right),\\
+    \sin x &= \tfrac{1}{2i} \left(e^{ix} - e^{-ix}\right).
+  \end{aligned}
+\end{gather*}
+\par Again, putting $ix$ for $x$, we have
+\begin{align*}
+  e^x &= \cos ix - i\sin ix,\\
+  e^{-x} &= \cos ix + i\sin ix;\\
+  \intertext{and}
+  \cos ix &= \tfrac{1}{2} \left(e^x + e^{-x}\right),\\
+  \sin ix &= -\tfrac{1}{2i} \left(e^2 - e^{-x}\right).\\
+  \cos ix &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \dotsb,\\
+  \sin ix &= ix \left(1 + \frac{x^2}{3!} + \frac{x^4}{5!}
+                                                     + \dotsb\right).
+\end{align*}
+\par For real values of $x$, $\cos ix$ and $\dfrac{\sin ix}{ix}$ are
+real and positive, and vary from $1$ to $\infty$ as $x$ varies from
+$0$ to $\infty$.
+\par In the equation $\cos^2ix + \sin^2ix = 1$, the first term is real
+and positive for real values of $x$, the second term is real and
+negative; therefore, $\sin ix$ is in absolute value less than $\cos
+ix$, and $\tan ix$ is in absolute value less than $1$. $\tan ix$
+varies in absolute value from $0$ to $1$ as $x$ varies from $0$ to
+$\infty$.}
+
+\P~Changing the sign of $x$, we have
+\begin{align*}
+\frac{1+\cos\Pi(x)}{\sin\Pi(x)} &= \cos ix - i \sin ix, \\
+\intertext{and, adding and subtracting,}
+\frac{1}{\sin\Pi(x)} &= \cos ix, \\
+\cot\Pi(x) &= -i\sin ix.
+\end{align*}
+
+\P~The nature of the angle of parallelism is, therefore, expressed
+by the equations
+\begin{align*}
+\sin\Pi(x) &= \frac{1}{\cos ix}, \\
+\tan\Pi(x) &= \frac{i}{\sin ix}, \\
+\cos\Pi(x) &= \frac{\tan ix}{i}.
+\end{align*}
+
+\item[4.]~Substituting in the formulæ of plane right triangles, we
+find that they reduce to those of spherical right triangles with
+$ia$, $ib$, and $ic$ for $a$, $b$, and $c$, respectively. The
+formulæ of oblique triangles are obtained from those of right
+triangles in the same way as on the sphere, and thus all the formulæ
+of Plane Trigonometry are obtained from those of Spherical
+Trigonometry simply by making this change.
+
+As fundamental formulæ for oblique triangles we write
+\begin{gather*}
+\frac{\sin A}{\sin ia} = \frac{\sin B}{\sin ib}
+                       = \frac{\sin C}{\sin ic}, \\
+\cos ia = \cos ib\, \cos ic + \sin ib\, \sin ic\, \cos A, \\
+\cos A  = -\cos B\, \cos C + \sin B\, \sin C\, \cos ia.
+\end{gather*}
+
+In the notation of the $\Pi$-function, these are
+\begin{gather*}
+\sin A\, \tan\Pi(a) = \sin B\, \tan\Pi(b) = \sin C\, \tan\Pi(c), \\
+\frac{\sin\Pi(b)\,\sin\Pi(c)}{\sin\Pi(a)} =
+                             1-\cos\Pi(b)\,\cos\Pi(c)\,\cos A, \\
+\cos A = -\cos B\, \cos C + \frac{\sin B\,\sin C}{\sin\Pi(a)}.
+\end{gather*}
+
+\item[5.]~Since for very small values of $x$ we have approximately
+\begin{gather*}
+\sin ix = ix, \\
+\cos ix = 1 + \frac{x^2}{2}, \\
+\tan ix = ix, \\
+\intertext{our formulæ for infinitesimal triangles reduce to}
+\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}, \\
+a^2 = b^2 + c^2 - 2bc\,\cos A, \\
+\cos A = -\cos(B+C).
+\end{gather*}
+
+\item[6.]~Triangles on an equidistant-surface are similar to their
+projections on the base plane; that is, they have the same angles
+and their sides are proportional. Thus the formulæ of Plane
+Trigonometry hold for any equidistant-surface if with the letters
+representing the sides we put, besides $i$, a constant factor
+depending on the distance of the surface from the plane.
+\end{description}
+
+\chapter{THE ELLIPTIC GEOMETRY}
+
+In the hypothesis of the obtuse angle a straight line is of finite
+length and returns into itself. This length is the same for all
+lines, since any two lines can be made to coincide. Two straight
+lines always intersect, and two lines perpendicular to a third
+intersect at a point whose distance from the third on either line is
+half the entire length of a straight line.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=70mm]{fig58.png}
+\end{figure}
+
+\begin{description}
+\item{1.}~A straight line does not divide the plane. Starting from
+the point of intersection of two lines and passing along one of them
+a certain finite distance, we come to the intersection point again
+without having crossed the other line. Thus, we can pass from one
+side of the line to the other without having crossed it.
+
+There is one point through which pass all the perpendiculars to a
+given line. It is called the pole of that line, and the line is its
+polar. Its distance from the line is half the entire length of a
+straight line, and the line is the locus of points at this distance
+from its pole. Therefore, if the pole of one line lies on another,
+the pole of the second lies on the first, and the intersection of
+two lines is the pole of the line joining their poles.
+
+The locus of points at a given distance from a given line is a
+circle having its centre at the pole of the line. The straight line
+is a limiting form of a circle when the radius becomes equal to half
+the entire length of a line.
+
+We can draw three lines, each perpendicular to the other two,
+forming a trirectangular triangle. It is also a self-polar triangle;
+each vertex is the pole of the opposite side.
+
+\item[2.]~All the perpendiculars to a plane in space meet at a point
+which is the pole of the plane. It is the centre of a system of
+spheres of which the plane is a limiting form when the radius
+becomes equal to half the entire length of a straight line.
+
+Figures on a plane can be projected from similar figures on any
+sphere which has the pole of the plane for centre. That is, they
+have equal angles and corresponding sides in a constant ratio that
+depends only on the radius of the sphere. Two corresponding angles
+are equal, because they are the same as the diedral angles formed by
+the two planes through the centre of the sphere which cut the sphere
+and the plane in the sides of the angles. Corresponding lines are
+proportional; for if two arcs on the sphere are equal, their
+projections on the plane are equal; and that, in general, two arcs
+have the same ratio as their projections on the plane is proved,
+first when they are commensurable, and by the method of limits when
+they are incommensurable.
+
+Geometry on a plane is, therefore, like Spherical Geometry, but the
+plane corresponds to only half a sphere, just as the diameters of a
+sphere correspond to the points of half the surface. Indeed, the
+points and straight lines of a plane correspond exactly to the lines
+and planes through a point, but we can realize the correspondence
+better that compares the plane with the surface of a sphere. If we
+can imagine that the points on the boundary of a hemisphere at
+opposite extremities of diameters are coincident, the hemisphere
+will correspond to the elliptic plane. There is no particular line
+of the plane that plays the part of boundary. All lines of the plane
+are alike; the plane is unbounded, but not infinite in extent.
+
+The entire straight line corresponds to a semicircle. We will take
+such a unit for measuring length that the entire length of a line
+shall be $\pi$; the formulæ of Spherical Trigonometry will then
+apply without change to our plane. Distances on a line will then
+have the same measure as the angles which they subtend at the pole
+of the line, and the angle between two lines will be equal to the
+distance between their poles. The distance from any point to its
+polar, half the entire length of a straight line, may then be called
+a quadrant.
+
+We can form a self-polar tetraëdron by taking three mutually
+perpendicular planes and the plane which has their intersection for
+pole. The vertices of this tetraëdron are the poles of the opposite
+faces. At each vertex is a trirectangular triedral, and each face is
+a trirectangular triangle.
+
+\item[3.~Theorem.] \emph{All the planes perpendicular to a fixed line
+intersect in another fixed line, called its polar or conjugate. The
+relation is reciprocal, and all the points of either line are at a
+quadrant's distance, from all the points of the other.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig59.png}
+\end{figure}
+
+\textbf{Proof.} Let the two planes perpendicular to the line $AB$ at
+$H$ and $K$ intersect in $CD$. Pass a plane through $AB$ and $R$,
+any point of $CD$. This plane will intersect the two given planes in
+$HR$ and $KR$. $HR$ and $KR$ are perpendicular to $AB$; therefore,
+$R$ is at a quadrant's distance from $H$ and $K$. $R$ is then the
+pole of $AB$ in the plane determined by $AR$ and $R$, and is at a
+quadrant's distance from every point of $AB$. But $R$ is any point
+of $CD$; therefore, any point of either line is at a quadrant's
+distance from each point of the other line, and a point which is at
+a quadrant's distance from one line lies in the other line. Again,
+any point, $H$, of $AB$, being at a quadrant's distance from all the
+points of $CD$, is the pole of $CD$ in the plane determined by it
+and $CD$. Thus, $HR$ and $KR$ are both perpendicular to $CD$, and
+the plane determined by $AB$ and $R$ is perpendicular to $CD$.
+
+\medskip The opposite edges of a self-polar tetraëdron are polar
+lines.
+
+All the lines which intersect a given line at right angles intersect
+its polar at right angles. Therefore, the distances of any point
+from two polar lines are measured on the same straight line and are
+together equal to a quadrant. Two points which are equidistant from
+one line are equidistant from its polar.
+
+The locus of points which are at a given distance from a fixed line
+is a surface of revolution having both this line and its polar as
+axes. We may call it a surface of double revolution. The parallel
+circles about one axis are meridian curves for the other axis. If a
+solid body, or, we may say, all space, move along a straight line
+without rotating about it, it will rotate about the conjugate line
+as an axis without sliding along it. A motion along a straight line
+combined with a rotation about it is called a screw motion. A screw
+motion may then be described as a rotation about each of two
+conjugate lines or as a sliding along each of two conjugate lines.
+
+\item[4.~Theorem.] \emph{In the elliptic geometry there are lines
+not in the same plane which have an infinite number of common
+perpendiculars and are everywhere equidistant.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig60.png}
+\end{figure}
+
+\textbf{Given} any two lines in the same plane and their common
+perpendicular. If we go out on these lines in either direction from
+the perpendicular, they approach each other. Now revolve one of them
+about this perpendicular so that they are no longer in the same
+plane. After a certain amount of rotation the lines will have an
+infinite number of common perpendiculars and be equidistant
+throughout their entire length.
+
+\textbf{Proof.} Let $p$ be the length of the common perpendicular
+$AC$, and take points $B$ and $D$ on the two lines on the same
+side of this perpendicular at a distance, $a$.
+
+$BD<p$, but if $CD$ revolve about $AC$, $BD$ will become longer than
+$p$ by the time $CD$ is revolved through a right angle; for $BCD$
+will then be a right triangle, with $BD$ for hypothenuse and $BC$,
+the hypothenuse of the triangle $ABC$, for one of its sides, so that
+we shall have $BD >BC$ and $BC > AC$.
+
+Suppose, when $CD$ has revolved through an angle, $\theta$, $BD$
+becomes equal to $p$ and takes the position $BD'$. The triangles
+$ABC$ and $D'BC$ are equal, having corresponding sides equal.
+Therefore, $BD'$ is perpendicular to $CD'$. $BD'$ is also
+perpendicular to $BA$; for if we take the diedral $A-BC-D'$ and
+place it upon itself so that the positions of $B$ and $C$ shall be
+interchanged, $A$ will fall on the position of $D'$, and $D'$ on the
+position of $A$, and the angle $D'BA$ must equal the angle $ACD'$.
+Therefore, $BD'$ as well as $CA$ is a common perpendicular to the
+lines $AB$ and $CD'$.
+
+Now at the point $C$ we have a triedral whose three edges are $CB$,
+$CD$, and $CD'$. Moreover, the diedral along the edge $CD$ is a
+right diedral; therefore, the three face angles of the triedral
+satisfy the same relations as do the three sides of a spherical
+right triangle; namely,
+\begin{gather*}
+\cos BCD' = \cos BCD\, \cos DCD' \\
+\intertext{But}
+BCD = \frac{\pi}{2} - ACB \quad\text{and}\quad BCD' = ABC. \\
+\intertext{Hence, this relation may be written}
+\cos ABC = \sin ACB\, \cos \theta.
+\end{gather*}
+
+\P~Again, in the right triangle $ABC$
+\begin{align*}
+\sin ACB &= \frac{\cos ABC}{\cos p} \\
+\therefore \cos \theta &= \cos p, \\
+\intertext{or, since $\theta$ and $p$ are less than
+$\nicefrac{\pi}{2}$,}
+\theta &= p.
+\end{align*}
+
+The angle $\theta$, therefore, does not depend upon $a$. If we take
+any two lines in a plane and turn one about their common
+perpendicular through an angle equal in measure to the length of
+that perpendicular, the two lines will then be everywhere
+equidistant.
+\end{description}
+
+\label{p68ref}As we have no parallel lines in the ordinary sense in
+this Geometry, the name \emph{parallel} has been applied to lines of
+this kind. They have many properties of the parallel lines of
+Euclidean Geometry.
+
+Through any point two lines can be drawn parallel to a given line.
+These are of two kinds, sometimes distinguished as right-wound and
+left-wound. They lie entirely on a surface of double revolution,
+having the given line as axis. The surface is, therefore, a ruled
+surface and has on it two sets of rectilinear generators like the
+hyperboloid of one sheet.
+
+\chapter{ANALYTIC NON-EUCLIDEAN GEOMETRY}
+
+We shall use the ordinary polar coördinates, $\rho$ and $\theta$,
+and for the rectangular coordinates, $x$ and $y$, of a point, we
+shall use the intercepts on the axes made by perpendiculars through
+the point to the axes. The formulæ depend upon the trigonometrical
+relations, and in our two Geometries differ only in the use of the
+imaginary factor $i$ with lengths of lines.
+
+\section{Hyperbolic Analytic Geometry}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig61.png}
+\end{figure}
+
+\begin{description}
+\item[1.~The relations between polar and rectangular coördinates:]
+\[\null\] %%ugly hack
+
+The angles at the origin which the radius vector makes with the axes
+are complementary. From the two right triangles we have
+
+\begin{align*}
+\tan ix &= \cos\theta \tan i\rho, \\
+\tan iy &= \sin\theta \tan i\rho. \\
+\intertext{Therefore,}
+\tan^2 i\rho &= \tan^2 ix + \tan^2 iy, \\
+\tan\theta &= \frac{\tan iy}{\tan ix}.
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig62.png}
+\end{figure}
+
+\item[2. The distance, \boldmath$\delta$, between two points:]
+
+\begin{equation*}
+\cos i\delta = \cos i\rho\, \cos i\rho'
+                + \sin i\rho\, \sin i\rho'\, \cos(\theta'-\theta).
+\end{equation*}
+$\delta$ and one of the points being fixed, this may be regarded as
+the polar equation of a circle.
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig63.png}
+\end{figure}
+
+\item[3. The equation of a line:]\[\null\] %%ugly hack
+
+\textbf{Let} $p$ be the length of the perpendicular from the origin
+upon the line, and $\alpha$ the angle which the perpendicular makes
+with the axis of $x$. From the right triangle formed with this
+perpendicular and $\rho$ we have
+\begin{equation*}
+\tan i\rho\, \cos(\theta-\alpha) = \tan ip.
+\end{equation*}
+This is the polar equation of the line. We get the equation in $x$
+and $y$ by expanding and substituting; namely,
+\begin{equation*}
+\cos\alpha\,\tan ix + \sin\alpha\,\tan iy = \tan ip.
+\end{equation*}
+
+\P~The equation $a\tan ix + b\tan iy = i$ represents a line for
+which
+\begin{equation*}
+a^2 + b^2 = \frac{-1}{\tan^2 ip}.
+\end{equation*}
+Now, for real values of $p$, $-\tan^2ip < 1$ (see footnote,
+p.~\pageref{p59ref}). The line is therefore real if $a$ and $b$ are
+real, and if
+\begin{equation*}
+a^2 + b^2 > 1.
+\end{equation*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig64.png}
+\end{figure}
+
+\item[4. The distance, \boldmath$\delta$, of a point from a line:]
+\[\null\] %%ugly hack
+
+\textbf{Let} the radius vector to the point intersect the line at
+$A$, and let $\rho_1$ be the radius vector to $A$. We have two right
+triangles with equal angles at $A$, and from the expressions for the
+sines of these angles we get the equation
+\begin{equation*}
+\frac{\sin i\delta}{\sin i(\rho-\rho_1)} =
+                                        \frac{\sin ip}{\sin i\rho_1}.
+\end{equation*}
+This equation holds for all points, $x\,y$, of the plane, $\delta$
+being negative when the point is on the same side of the line as the
+origin, and zero when the point is on the line.
+\begin{equation*}
+\sin i\delta
+   =\frac{\sin ip}{\tan i\rho_1}\sin i\rho-\sin ip\,\cos i\rho.
+\end{equation*}
+
+Now,
+\begin{align*}
+\tan i\rho_1 &= \frac{\tan ip}{\cos(\theta-\alpha)}. \\
+\therefore \frac{\sin ip}{\tan i\rho_1}\sin i\rho
+           &=\sin i\rho\,\cos ip\,\cos(\theta-\alpha), \\
+\intertext{and}
+ \sin i\delta &= \cos i\rho\, \cos ip\,
+   \left[\tan i\rho\, \cos(\theta-\alpha) - \tan ip\right].
+\end{align*}
+$\delta$ being fixed, this may be regarded as the polar equation of
+an equidistant-curve.
+
+\begin{figure}[ht]
+\centering
+\includegraphics[width=60mm]{fig65.png}
+\end{figure}
+
+\item[5. The angle between two lines:]\[\null\] %%ugly hack
+
+$\phi$ being the angle which a line makes with the radius vector at
+any point, we have
+\begin{align*}
+ \cos \phi &= \cos ip\, \sin (\theta-\alpha),\\
+ \sin \phi &= \frac{\sin ip}{\sin i\rho}.
+\end{align*}
+
+For two lines intersecting at this point,
+\begin{align*}
+ \sin \phi_1\, \sin \phi_2 &= \frac{\sin ip_1\, \sin ip_2}{\sin^2 i\rho}\\
+ &= \sin ip_1\, \sin ip_2 + \frac{\sin ip_1\, \sin ip_2}{\tan^2 i\rho}.
+\end{align*}
+
+Now, from the equation of the line
+\begin{align*}
+\frac{\sin ip_1}{\tan i\rho} &= \cos ip_1\, \cos(\theta-\alpha_1),\\
+\frac{\sin ip_2}{\tan i\rho} &= \cos ip_2\, \cos(\theta-\alpha_2);\\
+\intertext{so that}
+\sin \phi_1 \sin \phi_2 &= \sin ip_1\, \sin ip_2
+   + \cos ip_1\, \cos ip_2\, \cos(\theta-\alpha_1)\,\cos(\theta-\alpha_2).\\
+\intertext{Again,}
+\cos \phi_1\, \cos \phi_2 &= \cos ip_1\, \cos ip_2\, \sin
+  (\theta-\alpha_1)\,\sin (\theta-\alpha_2). \\
+\intertext{Adding these equations, we have}
+\cos (\phi_2-\phi_1) &= \sin ip_1\, \sin ip_2 +
+        \cos ip_1\, \cos ip_2\, \cos (\alpha_2-\alpha_1).
+\end{align*}
+
+\P~Two lines are perpendicular if
+\begin{align*}
+\cos (\alpha_2-\alpha_1) + \tan ip_1\, \tan ip_2 &= 0. \\
+\intertext{The lines}
+ a \tan ix + b \tan iy &= i,\\
+ a' \tan ix + b' \tan iy &= i, \\
+\intertext{are perpendicular if}
+aa' + bb' &=1.
+\end{align*}
+
+\item[6. The equation of a circle in $x$ and $y$:]
+
+\begin{align*}
+ \sin i\rho\, \cos \theta &= \cos i\rho\, \tan ix,\\
+ \sin i\rho\, \cos \theta &= \cos i\rho\, \tan iy;
+\intertext{also,}
+ \cos i\rho &= \frac{1}{\sqrt{1+\tan^2 i\rho}} =
+\frac{1}{\sqrt{1+\tan^2 ix+\tan^2 iy}}.
+\end{align*}
+
+The equation of a circle may, therefore, be written
+\begin{multline*}
+  (1 + \tan^2ix + \tan^2iy)
+  (1 + \tan^2ix' + \tan^2iy') \cos^2i\delta\\
+  = (1 + \tan ix \tan ix' + \tan iy \tan iy')^2.
+\end{multline*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig66.png}
+\end{figure}
+
+\item[7. The equation of a boundary-curve:]\[\null\] %%ugly hack
+
+\textbf{Let} the axis of the boundary-curve which passes through the
+origin make an angle, $\alpha$, with the axis of $x$, and let the
+point where the boundary-curve cuts this axis be at a distance, $k$,
+from the origin, positive if the origin is on the convex side of the
+curve, negative if the origin is on the concave side of the curve.
+The boundary-curve is the limiting position of a circle whose
+centre, on this axis, moves off indefinitely.
+
+$\rho'$ being the radius vector to the centre, the radius of the
+circle is $\rho'-k$, and its equation may be written
+\begin{equation*}
+  \cos i(\rho'-k) = \cos i\rho\, \cos i\rho'
+  + \sin i\rho\, \sin i\rho'\, \cos(\theta-\alpha),
+\end{equation*}
+or, expanding and dividing by $\cos i\rho'$,
+\begin{equation*}
+  \cos ik + \tan i\rho'\, \sin ik
+  = \cos i\rho + \sin i\rho\, \tan i\rho'\, \cos(\theta-\alpha).
+\end{equation*}
+
+\P~Now, let $\rho'$ increase indefinitely. $\tan i\rho'$ tends to
+the limit $i$, so that the limit of the first member of the equation
+is
+\begin{equation*}
+  \cos ik + i\sin ik, \text{ or } e^{-k},
+\end{equation*}
+and the polar equation of the curve is
+\begin{equation*}
+  e^{-k} = \cos i\rho \left[1 + i\tan i\rho\,\cos(\theta-\alpha) \right];
+\end{equation*}
+or, in $x\,y$ coördinates,
+\begin{equation*}
+  (1 + \tan^2ix + \tan^2iy) e^{-2k}
+     = (1 + i\cos\alpha\,\tan ix + i\sin\alpha\,\tan iy)^2.
+\end{equation*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig67.png}
+\end{figure}
+
+\textbf{Let} $k$ be negative and equal, say, to $-b$, and let
+$\alpha=0$; also, let $a$ be the ordinate of the point $A$ where the
+curve cuts the axis of $y$.
+
+Substituting in the equation, we find
+\begin{equation*}
+  e^b = \cos ia.
+\end{equation*}
+
+Through $A$ draw a line parallel to the axis of $x$, and, therefore,
+making an angle, $\Pi(a)$, with the axis of $y$. If we draw a
+boundary-curve through the origin having the same set of parallel
+lines for axes, so that the two boundary-curves cut off a distance,
+$b$, on these axes, we know that the ratio of corresponding arcs is
+\begin{gather*}
+  \frac{s'}{s} = \sin\Pi(a) = \frac{1}{\cos ia};
+\tag{See p.~\pageref{p56ref}.} \\
+\intertext{therefore,}
+  \frac{s'}{s} = e^{-b}.\tag{See p.~\pageref{p45ref}.}
+\end{gather*}
+
+\item[8.~The equation of an equidistant-curve:]\label{p76ref}
+\[\null\] %%ugly hack
+
+The polar equation of~(4) reduced to an equation in $x$ and $y$
+takes the form
+\begin{multline*}
+  (1 + \tan^2ix + \tan^2iy) \sin^2i\delta \\
+  = \cos^2ip (\cos\alpha\,\tan ix + \sin\alpha\, \tan iy -\tan ip)^2.
+\end{multline*}
+
+\newpage
+\item[9.~Comparison of the three equations:]\[\null\] %%ugly hack
+
+The equation
+\begin{equation*}
+  (1 + \tan^2ix + \tan^2iy)c^2
+     = -(i-a\tan ix -b\tan iy)^2
+\end{equation*}
+represents a circle, a boundary-curve, or an equidistant-curve,
+according as $a^2+b^2 < 1$, $=1$, $>1$, respectively.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig68.png}
+\end{figure}
+
+\item[10.~Differential formulæ:]\[\null\] %%ugly hack
+
+Suppose we have an isosceles triangle in which the angle $A$ at the
+vertex diminishes indefinitely. In the formula
+\begin{equation*}
+  \frac{\sin A}{\sin ia} = \frac{\sin C}{\sin ic}
+\end{equation*}
+we may put for
+\begin{center}
+\begin{tabular}{ccc}
+  $\sin A$, & $\sin ia$, & $\sin C$;\\
+  $A$, & $ia$, & $1$,\\
+\end{tabular}\\
+\end{center}
+respectively. Therefore,
+\begin{equation*}
+  ia = \sin ic \cdot A.\tag{I.}
+\end{equation*}
+
+\textbf{Corollary.} \emph{In a circle of radius $\mathbf{r}$, the
+ratio of any arc to the angle subtended at the centre is $\sin
+\mathbf{ir}$.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig69.png}
+\end{figure}
+
+Again, in the right triangle $ABC$, let the hypothenuse $c$ revolve
+about the vertex $A$. Differentiating the equation
+\begin{align*}
+\sin A &= \frac{\cos B}{\cos ib},  \\
+\intertext{where $b$ is constant, we have }
+\cos Ad\,A &= -\,\frac{\sin B\,dB}{\cos ib}.  \\
+\intertext{But }
+\sin B &= \frac{\cos A}{\cos ia};  \\
+\therefore dB &= -\cos ia\, \cos ib\, dA,  \\
+\intertext{or (II.) }
+dB &= -\cos ic\, dA.
+\end{align*}
+
+\P~Now, using polar coördinates, we have an infinitesimal right
+triangle whose hypothenuse, $ds$, makes an angle, say $\phi$, with
+the radius vector (see figure on page~\pageref{p78fig}). The two
+sides about the right angle are $d\rho$ and $\frac{\sin i\rho}{i}
+d\theta$; therefore,
+\begin{align*}
+ds^2 &= d\rho^2 - \sin^2i\rho\, d\theta^2,  \\
+\tan\phi &= \frac{\sin i\rho}{i}\, \frac{d\theta}{d\rho}.
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig70.png}
+\end{figure}
+
+\label{p78fig}For two arcs cutting at right angles, let $d'$ denote
+differentiation along the second arc:
+\begin{align*}
+  \frac{\sin i\rho}{i}\, \frac{d\theta}{\rho}
+  &= -\, \frac{i}{\sin i\rho}\, \frac{d'\rho}{d'\theta},\\
+\intertext{or}
+  \frac{d\rho}{d\theta}\, \frac{d'\rho}{d'\theta}
+  &= \sin^2i\rho.
+\end{align*}
+
+\item[11.~Area:]\[\null\] %%ugly hack
+
+It equals
+\begin{equation*}
+  \iint \frac{\sin i\rho}{i}\, d\rho\, d\theta.\footnotemark
+\end{equation*}
+\footnotetext{The unit of area being so chosen that the area of an
+infinitesimal rectangle may be expressed as the product of its base
+and altitude.}
+
+We will consider only the case where the origin is within the area
+to be computed and where each radius vector meets the bounding curve
+once, and only once.
+
+Integrating with respect to $\rho$, from $\rho=0$, we have
+\begin{align*}
+  \int_0^{2\pi} (\cos i\rho - 1) d\theta,\\
+  \intertext{or}
+  \int_0^{2\pi} \cos i\rho\, d\theta - 2\pi.
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig71.png}
+\end{figure}
+
+Suppose $P$ and $P'$ are two ``consecutive'' points on the curve,
+$PM$ and $P'M'$ the tangents at these points, and $\phi$ the angle
+which the tangent makes with the radius vector. The angle $MP'M'$
+indicates the amount of turning or rotation at these points as we go
+around the curve.
+
+\P~Now, by~(II.),
+\begin{equation*}
+  MP'M' = d\phi + \cos i\rho\, d\theta.
+\end{equation*}
+
+\label{p79ref}\P~In going around the curve, $\phi$ may vary but
+finally returns to its original value. That is, for our curve
+\begin{gather*}
+  \int d\phi = 0, \\
+\intertext{and the amount of rotation is}
+  \int_0^{2\pi} \cos i\rho\, d\theta.
+\end{gather*}
+
+Hence, the area is equal to the excess over four right angles in the
+amount of rotation as we go around the curve. This theorem can be
+extended to any finite area.
+
+\item[12. A modified system of coördinates:]\[\null\] %%ugly hack
+
+Our equations take simple forms if we write $iu$ for $\tan ix$, $iv$
+for $\tan iy$, $ir$ for $\tan i\rho$, and so on for all lengths of
+lines.
+
+Thus, we have
+\begin{align*}
+u^2 + v^2 &= r^2.\footnotemark \\
+\intertext{The equation of a line is}
+au + bv &= 1, \\
+\intertext{and the equation}
+(1 - u^2 - v^2)c^2 &= (1 - au - bv)^2
+\end{align*}
+represents a circle, a boundary-curve, or an equidistant-curve,
+according as $a^2+b^2<1$, $=1$, $>1$, respectively.
+\footnotetext{If we draw a quadrilateral with three right angles and
+the diagonal to the acute angle, and use $a$, $b$, and $c$ in the
+same way that $u$, $v$, and $r$ are used above, the five parts
+lettered in the figure have the relations of a right triangle in the
+Euclidean Geometry; \emph{e.g.},
+\begin{equation*}
+a^2 + b^2 = c^2 , \sin A = \frac{a}{c}, \text{etc}.
+\end{equation*}\includegraphics[width=25mm]{fig73.png}}\label{p80fn}
+\end{description}
+
+\section{Elliptic Analytic Geometry}
+
+The Elliptic Analytic Geometry may be developed just as we have
+developed the Hyperbolic Analytic Geometry, and the formulæ are the
+same with the omission of the factor $i$. But these formulæ are also
+very easily obtained from the relation of line and pole, and we
+shall produce them in this way.
+
+The formulæ of Elliptic Plane Analytic Geometry may be applied to a
+sphere in any of our three Geometries.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig72.png}
+\end{figure}
+
+\begin{description}
+\item[1. The relations between polar and rectangular coördinates:]
+
+\begin{gather*}
+\tan x = \cos\theta\, \tan\rho, \quad \tan y = \sin\theta\, \tan\rho; \\
+\intertext{therefore,}
+\begin{aligned}
+  \tan^2\rho &= \tan^2x + \tan^2y,\\
+  \tan\theta &= \frac{\tan y}{\tan x}.\footnotemark
+\end{aligned}
+\end{gather*}
+\footnotetext{The line which has the origin for pole forms with the
+coördinate axes a trirectangular triangle, and $x$, $y$, and
+$\theta$ may be regarded as representing the directions of the given
+point from its three vertices.
+\par On a sphere, if we take as origin the pole of the equator, $\rho$
+and $\theta$ are colatitude and longitude. $x$ and $y$, one with its
+sign changed, are the ``bearings'' of the point from two points
+$90^\circ$ apart on the equator.}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig74.png}
+\end{figure}
+
+\item[2. The distance, \boldmath$\delta$, between two points:]
+
+\begin{equation*}
+\cos\delta = \cos\rho\,\cos\rho' +
+                             \sin\rho\,\sin\rho'\cos(\theta'-\theta).
+\end{equation*}
+
+This may be regarded as the polar equation of a circle of radius
+$\delta$, $\rho'$ and $\theta'$ being the polar coördinates of the
+centre.
+
+Now,
+\begin{align*}
+  \sin\rho\,\cos\theta &= \cos\rho\,\tan x,\\
+  \sin\rho\,\sin\theta &= \cos\rho\,\tan y;
+\intertext{also,}
+  \cos\rho = \frac{1}{\sqrt{1+\tan^2\rho}}
+  &= \frac{1}{\sqrt{1+\tan^2x+\tan^2y}}
+\end{align*}
+The equation of a circle in rectangular coördinates may, therefore,
+be written
+\begin{multline*}
+  (1 + \tan^2x + \tan^2y) (1 + \tan^2x' + \tan^2y') \cos^2\delta\\
+  = (1 + \tan x\tan x' + \tan y\tan y')^2.
+\end{multline*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig75.png}
+\end{figure}
+
+\item[3. The equation of a line:]\[\null\] %%ugly hack
+
+When $\delta = \nicefrac{\pi}{2}$, the circle becomes a straight
+line. For this we have, therefore, the equation
+\begin{equation*}
+  \tan x \tan x' + \tan y \tan y' + 1 = 0.
+\end{equation*}
+$x'y'$ is the pole of the line.
+
+From the equation
+\begin{align*}
+\tan\rho\, \cos(\theta-\alpha) &= \tan p, \\
+\intertext{or}
+\cos\alpha\, \tan x + \sin\alpha\,\tan y &= \tan p, \\
+\intertext{we find}
+\tan x' &= -\frac{\cos\alpha}{\tan p},\\
+\tan y' &= -\frac{\sin\alpha}{\tan p},
+\end{align*}
+as can be shown geometrically, the polar coördinates of this point
+being
+\begin{equation*}
+p+\frac{\pi}{2},\ \alpha.
+\end{equation*}
+
+The equation
+\begin{equation*}
+  a\tan x + b\tan y + 1 = 0
+\end{equation*}
+represents a real line for any real values of $a$ and $b$.
+
+\newpage
+\item[4. The distance, \boldmath$\delta$, of a point from a
+straight line:]\[\null\] %%ugly hack
+
+This is the complement of the distance between the point and the
+pole of the line; it is expressed by the equation
+\begin{align*}
+  \sin\delta &= -\cos\rho\,\sin p
+                            + \sin\rho\,\cos p\,\cos(\theta-\alpha)\\
+  &= \cos\rho\,\cos p
+                 \left[\tan\rho\,\cos(\theta-\alpha) - \tan p\right].
+\end{align*}
+
+\item[5. The angle, \boldmath$\phi$, between two lines:]
+\[\null\] %%ugly hack
+
+This is equal to the distance between their poles; therefore,
+\begin{equation*}
+  \cos\phi = \sin p\,\sin p' + \cos p\,\cos p'\,\cos(\alpha'-\alpha).
+\end{equation*}
+
+The two lines
+\begin{align*}
+  a\phantom{'}\tan x + b\phantom{'}\tan y + 1 &= 0,\\
+  a'\tan x + b'\tan y + 1 &= 0\\
+  \intertext{are perpendicular if}
+  aa' + bb' + 1 &= 0.
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig76.png}
+\end{figure}
+
+\item[6. Differential formulæ:]\[\null\] %%ugly hack
+
+The formula
+\begin{equation*}
+  \frac{\sin A}{\sin a} = \frac{\sin C}{\sin c}
+\end{equation*}
+becomes, when $A$ diminishes indefinitely,
+\begin{equation*}
+  a = \sin c \cdot A.\tag{I.}
+\end{equation*}
+
+\newpage
+\label{p83ref}\textbf{Corollary.} \emph{In a circle of radius
+$\mathbf{r}$, the ratio of any arc, to the angle subtended at the
+centre is $\sin \mathbf{r}$.}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig77.png}
+\end{figure}
+
+From the right triangle $ABC$, if $b$ remain fixed, we get, by
+differentiating the equation
+\begin{align*}
+  \sin A &= \frac{\cos B}{\cos b}, \\
+  dB &= -\cos c \,dA.\tag{II.}
+\end{align*}
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig78.png}
+\end{figure}
+
+\newpage
+Thus, we have for differential formulæ in polar coördinates
+\begin{align*}
+  ds^2 &= d\rho^2 + \sin^2\rho\,d\theta^2,\\
+  \tan\phi &= \sin\rho\, \frac{d\theta}{d\rho};\footnotemark
+\intertext{and for two arcs cutting at right angles}
+  \frac{d\rho}{d\theta}\, \frac{d'\rho}{d'\theta} &= -\sin^2\rho.
+\end{align*}
+\footnotetext{If $\phi$ is constant, as in the logarithmic spiral of
+Euclidean Geometry, we can integrate this equation; namely,
+  \begin{align*}
+    \tan\phi\, \frac{d\rho}{\sin\rho} &= d\theta.\\
+    \therefore \tan\phi\, \log\tan\frac{\rho}{2} &= \theta+c,\\
+    \intertext{or}
+    \tan\frac{\rho}{2} &= e^\frac{\theta+c}{\tan\phi}.\\
+    \intertext{Writing $c'$ for $e^\frac{c}{\tan\phi}$, this is}
+    \tan\frac{\rho}{2} &= c' e^\frac{\theta}{\tan\phi}.
+  \end{align*}
+  On the sphere this is the curve called the loxodrome.}
+
+The formula for area is\footnote{The unit of area being properly chosen.}
+\begin{equation*}
+  \iint \sin\rho\, d\rho\, d\theta.
+\end{equation*}
+We integrate first with respect to $\rho$, and if the area contains
+the origin and each radius vector meets the curve once, and only
+once, our expression becomes
+\begin{equation*}
+  2\pi - \int_0^{2\pi} \cos\rho\, d\theta.
+\end{equation*}
+
+The entire rotation in going around the curve is found as on
+page~\pageref{p79ref}, and is
+\begin{equation*}
+  \int_0^{2\pi} \cos\rho\, d\theta.
+\end{equation*}
+Thus the area is equal to the amount by which this rotation is less
+than four right angles.
+
+For example, the area of a circle of radius $\rho$ is
+$2\pi(1-\cos\rho)$, and the amount of turning in going around it is
+$2\pi\cos\rho$. The area of the entire plane is $2\pi$.
+
+\item[7. A modified system of coördinates:]\[\null\] %%ugly hack
+
+Writing $u$ for $\tan x$, $v$ for $\tan y$, $r$ for $\tan\rho$,
+etc., we have
+\begin{align*}
+  u^2 + v^2 &= r^2.\footnotemark \\
+\intertext{The equation of a line then becomes}
+  au + bv + 1 &= 0, \\
+\intertext{and the equation of a circle}
+  (1 + u^2 + v^2)c^2 &= (1 + au + bv)^2.
+\end{align*}
+\footnotetext{The footnote on page~\pageref{p80fn} applies here
+also.}
+\end{description}
+
+\section{Elliptic Solid Analytic Geometry}
+
+We will develop far enough to get the equation of the surface of
+double revolution.
+
+\begin{figure}[h]
+\centering
+\includegraphics[width=60mm]{fig79.png}
+\end{figure}
+
+\begin{description}
+\item[1. Coördinates, lines, and planes:]\[\null\] %%ugly hack
+
+Draw three planes through the point perpendicular to the axes. For
+co\-ör\-din\-ates $x$, $y$, $z$, we take the intercepts which these
+planes make on the axes.
+
+The lines of intersection of these three planes are perpendicular to
+the coördinate planes (Chap.~I, II, 16 and 17); in fact, all the
+face angles in the figure are right angles except those at $P$ and
+the three angles $BA'C$, $CB'A$, and $AC'B$, which are obtuse
+angles.
+
+Let $\rho$ be the radius vector to the point $P$, and $\alpha$,
+$\beta$, and $\gamma$ the three angles which it makes with the three
+axes. Then
+\begin{align*}
+  \cos^2\alpha + \cos^2\beta + \cos^2\gamma &= 1,\\
+  \cos\alpha &= \frac{\tan x}{\tan\rho},\ \text{etc.;}\\
+  \tan^2x + \tan^2y + \tan^2z &= \tan^2\rho.
+\end{align*}
+
+For the angle between two lines intersecting at the origin
+\begin{equation*}
+  \cos\theta = \cos\alpha\,\cos\alpha' + \cos\beta\,\cos\beta'
+  + \cos\gamma\,\cos\gamma'.
+\end{equation*}
+
+The angle subtended at the origin by the two points $xyz$ and
+$x'y'z'$ is given by the equation
+\begin{equation*}
+  \cos\theta = \frac{\tan x\, \tan x' + \tan y\, \tan y'
+    + \tan z\, \tan z'}{\tan\rho\, \tan\rho'}.
+\end{equation*}
+
+For the distance between two points
+\begin{equation*}
+  \cos\delta = \cos\rho\,\cos\rho' + \sin\rho\,\sin\rho'\,\cos\theta.
+\end{equation*}
+
+This gives us the equation of a sphere, and for $\delta =
+\nicefrac{\pi}{2}$ the equation of a plane. The latter in
+rectangular coördinates is
+\begin{equation*}
+  \tan x\, \tan x' + \tan y\, \tan y' + \tan z\, \tan z' + 1 = 0.
+\end{equation*}
+
+Let $p$ be the length of the perpendicular from the origin upon the
+plane, and $\alpha$, $\beta$, $\gamma$ the angles which this
+perpendicular makes with the axes. Then we have for its pole
+\begin{align*}
+  \rho' &= p + \frac{\pi}{2},\\
+  \tan x' = \tan\rho'\,\cos\alpha &= -\,\frac{\cos\alpha}{\tan p},\
+  \text{etc.;}
+\end{align*}
+hence, the equation of the plane may be written
+\begin{equation*}
+\cos\alpha\,\tan x + \cos\beta\,\tan y + \cos\gamma\,\tan z
+   = \tan p.
+\end{equation*}
+
+\newpage
+\item[2. The surface of double revolution:]\[\null\] %%ugly hack
+
+Take one of its axes for the axis of $z$, suppose $k$ the distance
+of the surface from this axis, and let $\theta$ denote the angle
+which the plane through the point $P$ and the axis of $z$ makes with
+the plane of $xz$. We may call $z$ and $\theta$ latitude and
+longitude.
+
+Produce $OA$ and $CB$. They will meet at a distance,
+$\nicefrac{\pi}{2}$, from the axis of $z$ in a point, $O'$, on the
+other axis of the surface, and there form an angle that is equal in
+measure to $z$.
+
+From the right triangle $O'AB$
+\begin{align*}
+\cos z &= \frac{\tan OA}{\tan O'B}.\\
+\intertext{But}
+\tan O'A &= \cot x,\\
+\intertext{and}
+\tan O'B &= \cot CB = \frac{\cot k}{\cos\theta}.\\
+\intertext{Therefore,}
+\cos z &= \frac{\tan k\,\cos\theta}{\tan x},\\
+\intertext{or}
+\tan x &= \frac{\tan k\,\cos\theta}{\cos z}.\\
+\intertext{Similarly,}
+\tan y &= \frac{\tan k\,\sin\theta}{\cos z}.\\
+\intertext{Squaring and adding, we have for the equation of the
+  surface}
+\tan^2x + \tan^2y &= \tan^2k\sec^2z.
+\end{align*}
+
+\newpage
+\begin{figure}[h]
+\centering
+\includegraphics[width=80mm]{fig80.png}
+\end{figure}
+
+For the length of the chord joining two points on the surface, we
+have
+\begin{equation*}
+\cos\delta = \cos\rho\,\cos\rho'
+  (1 + \tan x\,\tan x' + \tan y\,\tan y' + \tan z\,\tan z').
+\end{equation*}
+Now,
+\begin{align*}
+\tan^{2}\rho &= \tan^{2}k\,\sec^{2}z + \tan^{2}z; \\
+\intertext{therefore,}
+\sec^{2}\rho &= \sec^{2}k\,\sec^{2}z,\\
+\intertext{or}
+\cos\rho &= \cos k\,\cos z. \\
+\intertext{That is, in terms of $z$, $z'$, $\theta$, and $\theta'$,
+we have}
+\cos\delta &= \cos^2k\,\cos(z'-z) + \sin^2k\,\cos(\theta'-\theta). \\
+\intertext{From this we can get an expression for $ds$, the
+differential element of length on the surface:}
+\cos ds &= \cos^2k\,\cos dz + \sin^2k\,\cos d\theta, \\
+\intertext{or, since}
+\cos ds &= 1 - \frac{ds^2}{2},\ \text{etc.,} \\
+ds^2 &= \cos^2k\, dz^2 + \sin^2k\, d\theta^2.
+\end{align*}
+
+$z$ and $\theta$ are proportional to the distances measured along
+the two systems of circles. These circles cut at right angles, and
+may be used to give us a system of rectangular coördinates on the
+surface. The actual lengths along these two systems of circles are
+$\theta\sin k$ and $z\cos k$ (see Cor.~p.~\pageref{p83ref}). If,
+therefore, we write
+\begin{equation*}
+\alpha = \theta\sin k,\quad \beta = z\cos k,
+\end{equation*}
+we shall have a rectangular system on the surface where the
+coördinates are the distances measured along these two systems of
+circles which cut at right angles.
+
+The formula now becomes
+\begin{equation*}
+ds^2 = d\alpha^2 + d\beta^2.
+\end{equation*}
+
+An equation of the first degree in $\alpha$ and $\beta$ represents a
+curve which enjoys on this surface all the properties of the
+straight line in the plane of the Euclidean Geometry. Through any
+two points one, and only one, such line can be drawn, because two
+sets of coördinates are just sufficient to determine the
+coefficients of an equation of the first degree. The shortest
+distance between two points on the surface is measured on such a
+line. For, the distance between two points on a path represented by
+an equation in $\alpha$ and $\beta$ is the same as the distance
+between the corresponding points and on the corresponding path in a
+Euclidean plane in which we take $\alpha$ and $\beta$ for
+rectangular coördinates. It must, therefore, be the shortest when
+the path is represented by an equation of the first degree in
+$\alpha$ and $\beta$. Such a line on a surface is called a
+\emph{geodesic line}, or, so far as the surface is concerned, a
+straight line. The distance between any two points measured on one
+of these lines is expressed by the formula
+\begin{equation*}
+d = \sqrt{(\alpha-\alpha')^2 + (\beta-\beta')^2}.
+\end{equation*}
+
+Triangles formed of these lines have all the properties of plane
+triangles in the Euclidean Geometry: the sum of the angles is $\pi$,
+etc. In fact this surface has the same relation to elliptic space
+that the boundary-surface has to hyperbolic space.
+
+The normal form of the equation of a line is
+\begin{equation*}
+\alpha\cos\omega + \beta\sin\omega = p.
+\end{equation*}
+
+The rectilinear generators of the surface make a constant angle,
+$\pm k$, with all the circles drawn around the axis which is polar
+to the axis of $z$. These generators are then ``straight lines'' on
+the surface, and their equation takes the form
+\begin{equation*}
+\alpha\cos k \pm \beta\sin k = p.
+\end{equation*}
+\end{description}
+
+\chapter{HISTORICAL NOTE}
+
+The history of Non-Euclidean Geometry has been so well and so often
+written that we will give only a brief outline.\label{3hypos}
+
+There is one axiom of Euclid that is somewhat complicated in its
+expression and does not seem to be, like the rest, a simple
+elementary fact. It is this:\footnote{See article on the axioms of
+Euclid by Paul Tannery. \emph{Bulletin des Sciences Mathématiques},
+1884.}
+
+\begin{quotation}\emph{If two lines are cut by a third, and the sum of the
+interior angles on the same side of the cutting line is less than
+two right angles, the lines will meet on that side when sufficiently
+produced.}\end{quotation}
+
+Attempts were made by many mathematicians, notably by Legendre, to
+give a proof of this proposition; that is, to show that it is a
+necessary consequence of the simpler axioms preceding it. Legendre
+proved that the sum of the angles of a triangle can never exceed two
+right angles, and that if there is a single triangle in which this
+sum is equal to two right angles, the same is true of all triangles.
+This was, of course, on the supposition that a line is of infinite
+length. He could not, however, prove that there exists a triangle
+the sum of whose angles is two right angles.\footnote{See, for
+example, the twelfth edition of his \emph{Éléments de Géométrie},
+Livre~I, Proposition~XIX, and Note~II. See also a statement by Klein
+in an article on the Non-Euclidean Geometry in the second volume of
+the first series of the \emph{Bulletin des Sciences Mathématiques}.}
+
+At last some mathematicians began to believe that this statement was
+not capable of proof, that an equally consistent Geometry could be
+built up if we suppose it not always true, and, finally, that all of
+the postulates of Euclid were only hypotheses which our experience
+had led us to accept as true, but which could be replaced by
+contrary statements in the development of a logical Geometry.
+
+The beginnings of this theory have sometimes been ascribed to Gauss,
+but it is known now that a paper was written by
+Lambert,\footnote{See \emph{American Mathematical Monthly},
+July--August, 1895.} in 1766, in which he maintains that the
+parallel axiom needs proof, and gives some of the characteristics of
+Geometries in which this axiom does not hold. Even as long ago as
+1733 a book was published by an Italian, Saccheri, in which he gives
+a complete system of Non-Euclidean Geometry, and then saves himself
+and his book by asserting dogmatically that these other hypotheses
+are false. It is his method of treatment that has been taken as the
+basis of the first chapter of this book.\footnote{The translation of
+Saccheri by Halsted has been appearing in the \emph{American
+Mathematical Monthly}.}
+
+Gauss was seeking to prove the axiom of parallels for many years,
+and he may have discovered some of the theorems which are
+consequences of the denial of this axiom, but he never published
+anything on the subject.
+
+Lobachevsky, in Russia, and Johann Bolyai, in Hungary, first
+asserted and proved that the axiom of parallels is not necessarily
+true. They were entirely independent of each other in their work,
+and each is entitled to the full credit of this discovery. Their
+results were published about 1830.
+
+It was a long time before these discoveries attracted much notice.
+Meanwhile, other lines of investigation were carried on which were
+afterwards to throw much light on our subject, not, indeed, as
+explanations, but by their striking analogies.
+
+Thus, within a year or two of each other, in the same journal
+(Crelle) appeared an article by Lobachevsky giving the results of
+his investigations, and a memoir by Minding on surfaces on which he
+found that the formulæ of Spherical Trigonometry hold if we put $ia$
+for $a$, etc. Yet these two papers had been published thirty years
+before their connection was noticed (by Beltrami).\label{histnote}
+
+Again, Cayley, in 1859, in the \emph{Philosophical Transactions},
+published his Sixth Memoir on Quantics, in which he developed a
+projective theory of measurement and showed how metrical properties
+can be treated as projective by considering the anharmonic relations
+of any figures with a certain special figure that he called the
+absolute. In 1872 Klein took up this theory and showed that it gave
+a perfect image of the Non-Euclidean Geometry.
+
+It has also been shown that we can get our Non-Euclidean Geometries
+if we think of a unit of measure varying according to a certain law
+as it moves about in a plane or in space.
+
+The older workers in these fields discovered only the Geometry in
+which the hypothesis of the acute angle is assumed. It did not occur
+to them to investigate the assumption that a line is of finite
+length. The Elliptic Geometry was left to be discovered by Riemann,
+who, in 1854, took up a study of the foundations of Geometry. He
+studied it from a very different point of view, an abstract
+algebraic point of view, considering not our space and geometrical
+figures, except by way of illustration, but a system of variables.
+He investigated the question, What is the nature of a function of
+these variables which can be called element of length or distance?
+and found that in the simplest cases it must be the square root of a
+quadratic function of the differentials of the variables whose
+coefficients may themselves be functions of the variables. By taking
+different forms of the quadratic expressions we get an infinite
+number of these different kinds of Geometry, but in most of them we
+lose the axiom that bodies may be moved about without changing their
+size or shape.
+
+Two more names should be included in this sketch,---Helmholtz and
+Clifford. These did much to make the subject popular by articles in
+scientific journals. To Clifford we owe the theory of parallels in
+elliptic space, as explained on page~\pageref{p68ref}. He showed
+that we can have in this Geometry a finite surface on which the
+Euclidean Geometry holds true.\footnote{Some of the more interesting
+accounts of Non-Euclidean Geometry are: \emph{Encyclopedia
+Britannica}, article ``Measurement'', by Sir Robert Ball.
+\emph{Revue Générale des Sciences}, 1891, ``Les Géométries
+Non-Euclidean,'' by Poincaré. \emph{Bulletin of the American
+Mathematical Society}, May and June, 1900, ``Lobachevsky's
+\emph{Geometry}'', by Frederick S.~Woods. \emph{Mathematische
+Annalen}, Bd.~xlix, p.~149, 1897, and \emph{Bulletin des Sciences
+Mathématiques}, 1897, ``Letters of Gauss and Bolyai''; particularly
+interesting is one letter in which Gauss gives a formula for the
+area of a triangle on the hypothesis that we can draw three mutually
+parallel lines enclosing a finite area always the same. The last two
+articles refer to the publications of Professors Engel and Stäckel,
+which give in German a full history of the theory of parallels and
+the writings and lives of Lobachevsky and Bolyai. See also the
+translations by Prof. George Bruce Halsted of Lobachevsky and Bolyai
+and of an address by Professor Vasiliev.}
+
+The chief lesson of Non-Euclidean Geometry is that the axioms of
+Geometry are only deductions from our experience, like the theories
+of physical science. For the mathematician, they are hypotheses
+whose truth or falsity does not concern him, but only the
+philosopher. He may take them in any form he pleases and on them
+build his Geometry, and the Geometries so obtained have their
+applications in other branches of mathematics.
+
+The ``axiom'', so far as this word is applied to these geometrical
+propositions, is not ``self-evident'', and is not necessarily true.
+If a certain statement can be proved,---that is, if it is a
+necessary consequence of axioms already adopted,---then it should
+not be called an axiom. When two or more mutually contradictory
+statements are equally consistent with all the axioms that have
+already been accepted, then we are at liberty to take either of
+them, and the statement which we choose becomes for our Geometry an
+axiom. Our Geometry is a study of the consequences of this axiom.
+
+The assumptions which distinguish the three kinds of Geometry that
+we have been studying may be expressed in different forms. We may
+say that one or two or no parallels can be drawn through a point;
+or, that the sum of the angles of a triangle is equal to, less than,
+or greater than two right angles; or, that a straight line has two
+real points, one real point, or no real point at infinity; or, that
+in a plane we can have similar figures or we cannot have similar
+figures, and a straight line is of finite or infinite length, etc.
+But any of these forms determines the nature of the Geometry, and
+the others are deducible from it.
+
+\newpage
+\chapter{PROJECT GUTENBERG "SMALL PRINT"}
+\small
+\pagenumbering{gobble}
+\begin{verbatim}
+
+
+
+
+
+End Project Gutenberg's Non-Euclidean Geometry, by Henry Manning
+
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