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The Project Gutenberg EBook Non-Euclidean Geometry, by Henry Manning

This eBook is for the use of anyone anywhere at no cost and with
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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 ***




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and the Project Gutenberg On-line Distributed Proofreading Team.






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\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 FORMULĆ
%%
%% 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 formulć 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 formulć 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 formulć are proved true for any angles. Other
formulć 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 formulć 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 Formulć}

\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 formulć 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 formulć 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 formulć 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 formulć 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 formulć 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 formulć 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
formulć of oblique triangles are obtained from those of right
triangles in the same way as on the sphere, and thus all the formulć
of Plane Trigonometry are obtained from those of Spherical
Trigonometry simply by making this change.

As fundamental formulć 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 formulć 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 formulć 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 formulć 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 formulć 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 formulć:]\[\null\] %%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 formulć are the
same with the omission of the factor $i$. But these formulć are also
very easily obtained from the relation of line and pole, and we
shall produce them in this way.

The formulć 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 formulć:]\[\null\] %%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 formulć 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 formulć 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
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