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\refoot{Lobachevski's Geometry}
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\begin{document}

\title{Lobachevski's Geometry}
\author{David Pierce\\
{Matematik B\"ol\"um\"u}\\
{Mimar Sinan G\"uzel Sanatlar \"Universitesi}\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{dpierce@msgsu.edu.tr}}
\date{August 25, 2017}
\publishers{%Nesin Mathematics Village\\
2017 TMD Lisans ve Lisans\"ust\"u Matematik Yazokulu\\
Nesin Matematik K\"oyu\\
\c Sirince, Sel\c cuk, \.Izmir}
%\begin{abstract}

\maketitle

\addchap*{Preface}

This is a rough draft of my own attempt at retelling the
\emph{Geometrical Researches on the Theory of Parallels,}
by Nicholas Lobachevski.
Though I tried to correct it while composing,
I have not read the present through from the beginning.
A significant feature is the diagrams,
in which the appearance of \emph{parallelism} is preserved,
at the expense of straightness.
In Lobachevski's own diagrams,
the emphasis is the reverse.
Often my diagrams are from the Poincar\'e half-plane or half-space model.
In future I hope to add a section on these models.

In the translation by George Bruce Halstead,
Lobachevski's \emph{Researches} are printed as a supplement to
Roberto Bonola,
\emph{Non-Euclidean Geometry}
(New York: Dover, 1955).

One may read the present document independently,
or for the sake of elucidating Lobachevski himself.
The exposition of Lobachevski is sometimes confusing,
in a way perhaps not absolutely required by the mathematics itself.
The order in which Lobachevski presents his propositions 
(or ``theorems'') seems good,
and I preserve this order.
I do skip the first fifteen propositions,
which summarize what can be known, 
apart from any assumption about parallelism.
Those fifteen propositions
can be found in or inferred from Euclid's \emph{Elements.}
Lobachevski does not cite Euclid specifically for the propositions,
but mentions him when opening his treatise:
\begin{quote}
In geometry I find certain imperfections
which I hold to be the reason why the science,
apart from transition into analytics,
can as yet make no advance 
from that state in which it has come to us from Euclid.
\end{quote}

Book I of Euclid's \emph{Elements}
should not be ``elucidated'' for students.
Students entering my own mathematics department
at Mimar Sinan in Istanbul read this book
(in Turkish translation),
and we hope that they will figure out for themselves
both \emph{what} Euclid is saying about mathematics,
and \emph{whether} he says it in a good way.
The same exercise can be repeated with any writer of mathematics,
in the manner of my \emph{alma mater,}
St John's College in Annapolis and Santa Fe;
but with a modern writer like Lobachevki,
perhaps just understanding his mathematics
is enough of a challenge,
without the stumbling block of obscurities in his proofs.

Before the preparation of the present notes,
I had worked through Lobachevski four times:
\begin{compactenum}[1)]
\item
as a student in the senior mathematics tutorial of St John's College,
1986--7;
\item
as the teacher of an elective upper-level course called 
\emph{Geometriler} at Mimar Sinan, fall semester, 2015--6;
\item
as the teacher of a course called Geometries at the Nesin Mathematics Village, 
\c Sirince, September 12--25, 2016;
\item
  as the teacher of Hyperbolic Geometry in the same place,
  August 7--13, 2017.
\end{compactenum}
The course at Mimar Sinan covered projective geometry in its first half.
In the style of St John's,
and of the Euclid course 
that my own students had taken 
in their first semester in my department,
I had these students present, at the board,
some relevant propositions of Pappus,
which I had translated into Turkish from the Greek.
In the second half of the course,
students presented the propositions of Lobachevski in the same way,
but from the English translation mentioned above.
I kept a detailed record of the course and of the mathematics involved;
the record is on my website;
I have used the record as a source for the present notes.

In the following summer, 
I tried to imitate my Mimar Sinan course in \c Sirince.
Since the Mimar Sinan course had met two hours a week,
and a \c Sirince course met two hours a day,
two weeks in \c Sirince should be enough to cover 
what the Mimar Sinan course did.
I kept a detailed record of the first week,
on projective geometry;
but not of the second week,
on hyperbolic geometry.
I do remember that almost none of the students from the first week
stayed on for the second.
Also, students in the second week were generally not able to understand the propositions of Lobachevsky 
that they were supposed to present---%
if they even tried to understand, and sometimes they did not,
or else they just did not show up for class.

Our last meeting during that second week was on Friday,
since I would attend the Thales Meeting in Miletus the next day,
September 24.
Near the beginning of the week,
I had introduced the Poincar\'e half-plane model 
of Lobachevskian geometry.
This may have been a mistake,
since students continued to show throughout the week
that they did not understand that lines appearing curved in the model
were supposed to be understood as straight.

In the following summer in \c Sirince,
I covered the same material,
but in the standard lecture format.
Perhaps this was better
than expecting students to come to the board,
since \c Sirince is too crowded in both space and time
for students to work well on their own.
In any case,
a number of students in 2017 
seemed more engaged than students in the previous summer.
I treated the two weeks officially as different courses,
and again there was little or no continuity among the students.
In the first week, after the first day,
I lectured in English, because a foreign student started attending;
in the second week, I used Turkish.
Over the course of the second week,
some of the actively questioning students
stopped attending the lectures without warning.
Attendence figures were thus
(dates in July and August on the first line,
numbers attending on the second):
\begin{center}
\begin{tabular}{*3r|*3r||*3r|*3r}
31& 1& 2&4&5&6&7& 8& 9&11&12&13\\\hline
? &12&12&9&6&3&?&15&12&10& 8& 4
\end{tabular}
\end{center}
Numbers for the first day of each week are unknown,
since class rosters had not been distributed,
and I did not make my own lists.
On the other days, figures are only approximate,
since I may not have counted students who came late.

\tableofcontents
\listoffigures

\addchap{Introduction}

By a \textbf{line,}
we shall always understand an \emph{unbounded, straight} line.
We allow ourselves to do everything that Euclid does,
\emph{without} using the postulate that,
when two lines are crossed by another,
and the interior angles on the same side
are less than two right angles,
then the two lines intersect when extended on that side.

The juxtaposition, as $AB$, of two named points
can have five different meanings:
\begin{compactenum}[1)]
\item
the unique line passing through $A$ and $B$,
\item
the same line, \emph{directed} from $A$ to $B$,
\item
the ray that begins at $A$ and passes through $B$,
\item
the line segment bounded by $A$ and $B$,
\item
the same segment, directed from $A$ to $B$.
\end{compactenum}

The \textbf{measure} of an angle is its congruence class.
The measure of a right angle is denoted by $\uppi/2$ or
\begin{equation*}
\frac{\uppi}2.
\end{equation*}
There is no need to understand $\uppi$ as a real number.
However, the measure of every acute angle is $t\uppi/2$
for some real number $t$ in the interval $(0,1)$.
Indeed, since we can bisect angles,
for every acute angle with a measure $\alpha$, 
we can define a sequence $(e_k\colon k\in\N)$
such that, for each $k$,
\begin{equation*}
\sum_{i=1}^ke_i\frac{\uppi}{2^{i+1}}\leq\alpha<
\sum_{i=1}^ke_i\frac{\uppi}{2^{i+1}}+\frac{\uppi}{2^{k+1}}.
\end{equation*}
We make the \textbf{Archimedean} assumption that
the sequence $(e_k\colon k\in\N)$
uniquely determines $\alpha$.
We shall assume \textbf{completeness} also,
in the sense that every such binary sequence
determines the measure of some angle.

Similarly,
the \textbf{length} of a line segment
is its congruence class,
and any two lengths have a ratio, which is a positive real number.
We shall also want that, for every length and every positive real number,
the latter is the ratio of some other length to the given length.

\addchap{The propositions of Lobachevski}

\begin{theorem}{16}
For every length $x$,
there is a measure, denoted by
\begin{equation*}
\Pi(x),
\end{equation*}
such that, in a plane,
under the hypothesis that a segment $AB$ has length $x$
as in Figure \ref{fig:16},
\begin{figure}
\centering
\psset{unit=3cm}
\begin{pspicture}(0,-0.1)(1.5,1.1)
%\psgrid
\psset{PointSymbol=none}
\pstGeonode[PosAngle={-90,90,90,-90}](0,0)A(0,1)B(1.5,1)C(! 1.5 1 1 2.718 1.5 exp div sub)D
\psplot0{1.5}{1 2.718 x neg exp sub}
\ncline AB\tlput{$x$}
\ncline BC
\pstRightAngle[RightAngleSize=0.1] ABC
\pstRotation[RotAngle=-45,PointName=none] AB
\pstMarkAngle[MarkAngleRadius=0.28,LabelSep=0.42]{B'}AB{$\Pi(x)$}
\end{pspicture}
\caption{\propword\ 16. $AD\parallel BC$.}\label{fig:16}
\end{figure}
and $ABC$ is a right angle,
and $AD$ is on the same side of $AB$ that $BC$ is,---%
under this hypothesis,
$BC$ and $AD$ intersect when extended
if and only if the measure of $BAD$ is less than $\Pi(x)$.
There is a ray $AD$ 
such that the measure of $BAD$ is precisely $\Pi(x)$.
\end{theorem}

\begin{proof}
By SAS, the supremum of the measures of all angles $BAX$,
where $X$ is on $BC$ extended,
depends only on $x$
and is $\Pi(x)$ as desired.
By completeness, $\Pi(x)$ is the measure of some angle.
\end{proof}

In the theorem,
when the measure of angle $BAD$ is precisely $\Pi(x)$,
we say $BC$ is \textbf{parallel} to $AD$.
At the moment, while $AD$ is only a directed \emph{line,}
$BC$ must be understood as a \emph{ray.}
In particular, the specification of the point $B$ is important.
We may refer to $\Pi(x)$
as the \textbf{angle of parallelism} of $x$.

In Figure \ref{fig:16},
the ray $AD$ must be understood as being straight,
even though it appears to us as curved.
The same proviso will apply to most figures here.

\begin{theorem}{17}
Two rays of the same directed line
are parallel to the same directed lines.
\end{theorem}

\begin{proof}
We suppose $AB$ is parallel to $CD$,
$ACD$ being a right angle.
There are two cases.
\begin{asparaenum}
\item
In Figure \ref{fig:17-1},
\begin{figure}
\centering
\begin{pspicture}(1,-0.5)(7,3.5)
\psline(7,0)(1,0)(1,3)(7,3)
\psline(1,3)(6,0)
\psline(2,0)(2,3)(4,0)
\uput[u](1,3){$A$}
\uput[r](7,3){$B$}
\uput[d](1,0){$C$}
\uput[r](7,0){$D$}
\uput[u](2,3){$E$}
\uput[ur](2.67,2){$F$}
\uput[d](6,0){$G$}
\uput[d](4,0){$H$}
\uput[d](2,0){$K$}
\end{pspicture}
\caption{\propword\ 17, first case}\label{fig:17-1}
\end{figure}
we drop the perpendicular $EK$ to $CD$
and then draw $EF$ arbitrarily within the angle $BEK$.
Then $AF$, extended, must meet $CD$ at some $G$,
and so $EF$, extended, must cut $KG$.
\item
In Figure \ref{fig:17-2},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(8,3.5)
\psset{PointSymbol=none}
\pstGeonode[PosAngle={90,90,-90}](0,3){E'}(2,3)A(0,0){K'}
\pstHomO[HomCoef=4,PosAngle=90]{E'}A[B]
\pstTranslation[PosAngle=-90]{E'}{K'}{A,B}[C,D]
\pstHomO[HomCoef=0.4,PosAngle=45]AC[F']
\pstInterLL[PosAngle=-90]{E'}{F'}{K'}D{G'}
\pstTranslation[PosAngle=45,PointSymbol=*]{E'}{F'}A[F]
\pstInterLL[PosAngle=-90] AF{K'}DG
\ncline B{E'}
\ncline{E'}{K'}
\ncline{K'}D
\ncline AC
\ncline{E'}{G'}
\ncline AG
\pstMarkAngle{F'}{E'}A{}
\pstMarkAngle FAB{}
\end{pspicture}
\caption{\propword\ 17, second case}\label{fig:17-2}
\end{figure}
we drop the perpendicular $E'K'$ to $CD$,
then draw $E'F'$ arbitrarily in the angle $BE'K'$.
Extended, $E'F'$ must cut $K'C$ or $CA$.
In the latter case, draw $AF$
so that $FAB=F'E'B$;
then $AF$ cuts $CD$ at some $G$,
but also $E'F'$, extended, cannot cut $AF$,
so it cuts $CG$ at some $G'$.
\qedhere
\end{asparaenum}
\end{proof}

Parallelism is now a possible relation of one directed line to another.

\begin{theorem}{18}
The relation of parallelism is symmetric.
\end{theorem}

\begin{proof}
In Figure \ref{fig:18},
\begin{figure}
\centering
\psset{unit=3mm}
\begin{pspicture}(0,-1.5)(22,16.5)
\psline(20,15)(0,15)(0,0)(22,0)
\psline(0,15)(20,0)
\psline(0,0)(16,12)
\psline(0,3)(22,3)
\psline(0,15)(7.2,5.4)
\uput[ul](0,15){$A$}
\uput[r](20,15){$B$}
\uput[dl](0,0){$C$}
\uput[r](22,0){$D$}
\uput[dr](16,12){$E$}
\uput[dr](7.2,5.4){$F$}
\uput[l](0,3){$G$}
\uput[r](22,3){$H$}
\uput[d](20,0){$K$}
\uput[ur](16,3){$L$}
\end{pspicture}
\caption{\propword\ 18}\label{fig:18}

\end{figure}
suppose $AB$ is parallel to $CD$,
and $ACD$ is a right angle,
within which $CE$ is drawn.
Drop the perpendicular $AF$ to $AE$,
let $AG=AF$,
erect $GH$ perpendicular to $AC$,
and let $CAK=FAB$.
Then $AK$ must meet $CD$ at some point $K$,
and so $AK$ cuts $GH$ at some $L$.
By SAS, $CE$ will cut $AB$ at a point whose distance from $AB$
is the length of $GL$.
\end{proof}

We shall show in \propword\ 25
that the relation of parallelism is transitive.

\begin{theorem}{19}
The sum of angles in a triangle is never greater than two right angles.
\end{theorem}

\begin{proof}
In Figure \ref{fig:19},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(8,3.5)
\psline(0,0)(5,0)(8,3)(0,0)(3,3)(5,0)
\uput[d](0,0){$A$}
\uput[u](3,3){$B$}
\uput[d](5,0){$C$}
\uput[75](4,1.5){$D$}
\uput[u](8,3){$E$}
\end{pspicture}
\caption{\propword\ 19}\label{fig:19}
\end{figure}
suppose the least angle of triangle $ABC$ is at $A$.
Bisect $BC$ at $D$,
extend $AD$ to $E$ so that $AD=DE$,
and complete triangle $ACE$.
This has the same angle sum as $ABC$,
but two angles of $ACE$, namely those at $A$ and $E$,
are equal in sum to angle $BAC$,
so one of them is no greater than half.
We can continue this construction
until obtaining a triangle
whose angle sum is that of $ABC$,
but with two angles whose sum 
is less than any pre-assigned positive measure $\alpha$.
(Here we use the Archimedean property of angle measure.)
Thus the angle sum of $ABC$ cannot exceed $\uppi$ by $\alpha$.
Since $\alpha$ is arbitrary,
the angle sum of $ABC$ cannot exceed $\uppi$.
\end{proof}

Let us define the \textbf{defect} of a triangle to be what must be added to the angle sum of the triangle to reach $\uppi$.
By \propword\ 19, the defect of a triangle is positive or zero.
A triangle with positive defect is \textbf{defective.}

\begin{lemma}
If one triangle is divided into two by a line through a vertex,
the defect of the original triangle 
is the sum of the defects of the smaller two triangles.
\end{lemma}

A \textbf{rectangle} is a quadrilateral figure,
each of whose four angles is right.
If such a figure does exist,
a diagonal divides it into two congruent defectless right triangles.

\begin{theorem}{20}
If one triangle has angle measure $\uppi$, then all triangles do.
\end{theorem}

\begin{proof}
Suppose some triangle has no defect.
One of its altitudes has its foot on a side,
thus dividing the triangle into two right triangles,
each having no defect.
One of these is half a rectangle.
By the Archimedean property of lengths,
we can multiply the sides of the rectangle,
so as to exceed the legs of a given right triangle,
as, in Figure \ref{fig:20},
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(3,3.5)
\psline(3,2)(1,0)(3,3)(0,0)(3,0)(3,3)
\uput[dl](0,0){$A$}
\uput[dr](3,0){$B$}
\uput[u](3,3){$C$}
\uput[d](1,0){$D$}
\uput[r](3,2){$E$}
\end{pspicture}
\caption{\propword\ 20}\label{fig:20}
\end{figure}
$AB$ and $BC$ exceed, respectively,
the legs $DB$ and $BE$
of the right triangle $DBE$.
By drawing $CD$,
we can conclude that,
as $ABC$ is defectless,
so must $DBE$ be.
\end{proof}

\begin{theorem}{21}
If there is an acute angle of parallelism,
then there is a defective triangle.
\end{theorem}


\begin{proof}
In Figure \ref{fig:21},
\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}(0,-0.7)(8,4.7)
%\psgrid
\psset{PointSymbol=none}
\pstTriangle[PosAngleA=90,PosAngleB=-90,PosAngleC=-90](0,4)A(0,0)B(3,0)C
\pstInterLC[PosAngle=-90,PointNameB={}] CBCAD{D'}
\ncline AB\tlput{$a$}
\ncline BD
\ncline AD
\psset{SegmentSymbol=MarkHash,MarkAngle=90,linewidth=0.5\pslinewidth}
\pstSegmentMark AC
\pstSegmentMark CD
\pstMarkAngle ACB{$\alpha$}
\pstMarkAngle[MarkAngleRadius=0.5] ACB{}
\psset{MarkAngleRadius=1,LabelSep=1.5}
\pstMarkAngle CAD{$\beta$}
\pstMarkAngle ADC{$\beta$}
\end{pspicture}
\caption{\propword\ 21}\label{fig:21}
\end{figure}
suppose $\Pi(a)<\uppi/2$.
If $DE=AD$,
we have $\alpha\geq2\beta$.
Continuing, as in Figure \ref{fig:21-2},
\begin{figure}
\centering
\psset{unit=5mm}
\begin{pspicture}(0,-1)(17,5)
%\psgrid
\psset{PointSymbol=none}
\pstTriangle[PosAngleA=90,PosAngleB=-90,PosAngleC=-90](0,4)A(0,0)B(3,0)C
\pstInterLC[PosAngle=-90,PointNameB={}] CBCAD{D'}
\pstMarkAngle[SegmentSymbol={}] ACB{$\alpha$}
\pstInterLC[PosAngle=-90,PointNameB={}] CBDAE{E'}
\ncline CE
\ncline AD
\ncline AE
\psset{MarkAngle=90,linewidth=0.5\pslinewidth}
\pstSegmentMark AD
\pstSegmentMark DE
\psset{SegmentSymbol=MarkHash}
\pstSegmentMark AC
\pstSegmentMark CD
\end{pspicture}
\caption{\propword\ 21 continued}\label{fig:21-2}
\end{figure}
we obtain a right triangle with one angle measuring less than
$\uppi/2-\Pi(a)$,
and another angle measuring no greater than $\Pi(a)$;
thus the defect of the triangle is positive.
\end{proof}

\begin{theorem}{22}
If there is a right angle of parallelism,
then there is a defectless triangle.
\end{theorem}

\begin{proof}
In Figure \ref{fig:22},
\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}(0,-0.7)(8,4.7)
%\psgrid
\psset{PointSymbol=none}
\pstTriangle[PosAngleA=90,PosAngleB=-90,PosAngleC=-90](0,4)A(0,0)B(3,0)C
\pstHomO[HomCoef=3]BC[D]
\ncline AB\tlput{$a$}
\ncline BD
\ncline AD
\end{pspicture}
\caption{\propword\ 22}\label{fig:22}
\end{figure}
by placing $D$ on $BC$ far enough away from $C$,
we can make the measure angle $BAD$
as close as we like to $\Pi(a)$.
If this measure is $\uppi/2$,
then the defect of $BAD$ is as small as we like.
Since the defect of $ABC$ is no greater,
this can only be $0$.
\end{proof}

The following are now equivalent.
\begin{compactenum}
\item
There is a defective triangle.
\item
All triangles are defective.
\item
There is an acute angle of parallelism.
\item
All angles of parallelism are acute.
\end{compactenum}
We henceforth assume these.

\begin{theorem}{23}
Every acute angle is the angle of parallelism of some length.
\end{theorem}

\begin{proof}
In Figure \ref{fig:23},
\begin{figure}
\centering
\psset{xunit=15mm}
\begin{pspicture}(0,-0.5)(6,6.5)
\psset{PointSymbol=none}
\def\AB{6 x sub 6 div ln neg 3 mul}
\psplot0{5.2}{\AB}
\pstGeonode[PosAngle=-90](0,0)A(1,0)D(2,0)F(4,0)H(5,0)M(6,0)C
\pstGeonode[PosAngle=135](! 1 5 6 div ln neg 3 mul)E(! 2 4 6 div ln neg 3 mul)G(! 5.2 0.8 6 div ln neg 3 mul)B
\pstGeonode[PointName=none](! 4 2 6 div ln neg 3 mul){H'}(! 5 1 6 div ln neg 3 mul){M'}
\pstGeonode[PosAngle=45](6,6)K(5,3)L
\psline(0,0)(6,0)
\psline(1,0)
\def\EF{4 x add 6 div ln neg 3 mul}
\psplot12{\EF}
\def\GH{2 x add 6 div ln neg 3 mul}
\psplot24{\GH}
\ncline DE
\ncline FG
\ncline H{H'}
\ncline M{M'}
\ncline LC
\ncline CK
\end{pspicture}
\caption{\propword\ 23}\label{fig:23}
\end{figure}
acute angle $BAC$ being given,
we erect the perpendicular $DE$.
If this meets $AB$ at $E$,
we make $DF=AD$
and erect the perpendicular $FG$.
If this meets $AB$ at $G$,
we make $FH=AF$
and erect a perpendicular at $H$, and so on.
The defects of $ADE$ and $FDE$ are equal
(the triangles themselves being congruent),
so the defect of $AFG$ is more than twice that of $ADE$.
Since no defect can exceed $\uppi$,
we must eventually find a point on $AC$
where the perpendicular does not meet $AB$.
By completeness,
there is a closest such point to $A$;
let it be $C$, and let the perpendicular be $CK$.
If we draw $CL$ in the angle $ACK$,
then, dropping the perpendicular $LM$ to $AC$,
we can extend $ML$ to meet $AB$.
Then $CL$, extended, must also meet $AB$.
Thus $CK$ is parallel to $AB$.
\end{proof}

The following does not say parallel lines approach one another
arbitrarily closely.

\begin{theorem}{24}
Prolonged, parallel lines approach one another.
\end{theorem}

\begin{proof}
If equal perpendiculars $AC$ and $BD$ are erected on a line $AB$,
and $CD$ is connected,
as in Figure \ref{fig:24},
\begin{figure}
\centering
\psset{xunit=15mm}
\begin{pspicture}(-2,-0.5)(2,2.5)
\psset{PointSymbol=none}
\pstGeonode[PosAngle=-90](-2,0)A(2,0)B
\pstGeonode[PosAngle=90](-2,2)C(2,2)D
\psplot{-2}2{x dup mul 4 sub 0.1 mul 2 add}
\ncline AB
\ncline AC
\ncline BD
\end{pspicture}
\caption{\propword\ 24}\label{fig:24}
\end{figure}
then, in the quadrilateral $ABDC$,
the angles at $C$ and $D$ are equal and acute,
and so $CD$ cannot be parallel to $AB$,
since the exterior angle at $D$ is obtuse.
\end{proof}

\begin{theorem}{25}
Parallelism is transitive,
whether in the plane or in space.
Indeed, in space,
when each of two intersecting planes contains one of two parallel lines,
the intersection of the planes is parallel to those lines.
\end{theorem}

\begin{proof}
There are two cases in the plane.
Suppose first
$AB\parallel EF$ and $CD\parallel EF$,
as in Figure \ref{fig:25-1}.
\begin{figure}
%\mbox{}\hfill
\subfloat[$AB\parallel EF\And CD\parallel EF$]{\label{fig:25-1}
\psset{unit=4cm}
\begin{pspicture}(-0.6,0.75)(0.7,2.15)
%\psgrid[subgriddiv=20]
\psarc(0,0)1{90}{120}
\psarc(0.5,0){! 7 sqrt 2 div}{90}{! 2 3 sqrt atan 90 add}
\psline(! -0.5 3 sqrt 2 div)(-0.5,2.00)
\psline(0,1)(0,2.00)
\psline(! 0.5 7 sqrt 2 div)(0.5,2.00)
\uput[d](! -0.5 3 sqrt 2 div){$A$}
\uput[u](-0.5,2.00){$B$}
\uput[dr](! 0 1.5 sqrt){$C$}
\uput[u](0,2.00){$D$}
\uput[d](0,1){$G$}
\uput[d](! 0.5 7 sqrt 2 div){$E$}
\uput[u](0.5,2.00){$F$}
\psarc(1,0){! 3 sqrt}{! 1 11 sqrt atan 90 add}{! 3 3 sqrt atan 90 add}
\uput[r](! 0.5 11 sqrt 2 div){$H$}
\uput[ul](!0 2 sqrt){$K$}
\end{pspicture}}
\hfill
\subfloat[$AB\parallel CD\And CD\parallel EF$]{\label{fig:25-2}
\psset{unit=4cm}
\begin{pspicture}(-0.6,0.75)(0.7,2.15)
%\psgrid[subgriddiv=20]
\psarc(0,0)1{90}{120}
\psarc(0.5,0){! 7 sqrt 2 div}{90}{! 2 3 sqrt atan 90 add}
\psline(! -0.5 3 sqrt 2 div)(-0.5,2.00)
\psline(0,1)(0,2.00)
\psline(! 0.5 7 sqrt 2 div)(0.5,2.00)
\uput[d](! -0.5 3 sqrt 2 div){$A$}
\uput[u](-0.5,2.00){$B$}
\uput[dr](! 0 1.5 sqrt){$C$}
\uput[u](0,2.00){$D$}
\uput[d](0,1){$G$}
\uput[d](! 0.5 7 sqrt 2 div){$E$}
\uput[u](0.5,2.00){$F$}
\psarc(1,0){! 3 sqrt}{! 1 11 sqrt atan 90 add}{! 3 3 sqrt atan 90 add}
\uput[r](! 0.5 11 sqrt 2 div){$H$}
\uput[ul](!0 2 sqrt){$K$}
\psarc(2.5,0){! 31 sqrt 2 div}{! 4 15 sqrt atan 90 add}{! 5 6 sqrt atan 90 add}
\uput[dr](0.15,1.50){$L$}
\uput[r](! 0.5 15 sqrt 2 div){$M$}
\end{pspicture}}
%\hfill\mbox{}
\caption{\propword\ 25, planar case}\label{fig:25}
\end{figure}
Let the perpendicular $AE$ to $EF$,
cutting $CD$ at $C$, be dropped.
Since $ACD$ is acute,
when perpendicular $AG$ to $CD$ is dropped,
$AC$ lies within the angle $AGD$.
Another line $AH$ drawn within $AGD$
either cuts $GC$ or else cuts $EF$ at some $H$,
and in the latter case $AH$ must cut $CD$ at some $K$.
Thus $AB\parallel CD$.

Suppose next
$AB\parallel CD$ and $CD\parallel EF$.
Again let the perpendicular $AE$ be dropped to $EF$.
A line $AK$ drawn within angle $BAG$
cuts $CD$ at some $K$.
When $AK$ is extended to $L$,
the line $CL$ within angle $DCE$
must cut $EF$ at some $M$,
and then $AL$ must cut $EM$ at some $H$.
Thus $AB\parallel EF$.

Suppose finally $AB\parallel CD$,
and two planes containing $AB$ and $CD$ respectively
meet along $EF$,
as in Figure \ref{fig:25-3}.
\begin{figure}
\centering
\psset{xunit=5cm,yunit=2.5cm}
\begin{pspicture}(1.9,-2.2)(4.1,0.2)
%\psgrid
\psset{PointSymbol=none}
\pstGeonode[PosAngle=180,PointName={none,default,none}](0,0)O(2,-1)E(0,-1){E'}
\pstHomO[HomCoef=2]{E'}E[F]
\pstTranslation {E'}OF[B]
\pstHomO[HomCoef=2,PointName=none]O{E'}[C']
\pstTranslation O{C'}B[D]
\pstInterLC[PosAngle=180] OBOE{A'}A
\pstArcOAB OEA
\pstInterLC[PosAngle=-90] {C'}D{C'}A{C''}C
\pstArcOAB{C'}CA
\pstInterLC[PosAngle=90] OAOC{G'}G
\pstArcOAB OCG
\pstMediatorAB[PointName=none,linestyle=none] CEIM
\pstInterLL[PointName=none] IMO{E'}N
\pstArcOAB NCE
\ncline AB
\ncline EF
\ncline CD
\pstHomO[HomCoef=0.8,PosAngle=90]AB[H]
\pstMediatorAB[PointName=none,linestyle=none] EHJP
\pstInterLL[PointName=none] JPO{E'}Q
\pstArcOAB QEH
\end{pspicture}
\caption{\propword\ 25, spatial case}\label{fig:25-3}
\end{figure}
Drop perpendiculars $EA$ to $AB$,
$AC$ to $CD$, and $CG$ to $AB$.
If a line $EH$ is drawn in the angle $AEF$,
then the plane $CEH$
cuts the plane $ACD$
along a line that is bound to cut $AB$ at some point $H$.
Thus $EF\parallel AB$.
\end{proof}

We make use now of spherical geometry,
and later of spherical trigonomety.
On a sphere, we shall let the antipodal point of $X$ be $X'$.
We could take the following theorem to be obvious,
by ``symmetry'';
but like Lobachevski,
we can also establish it by the kind of rearrangement of parts
that Euclid employs.

\begin{theorem}{26}
On a sphere,
triangles whose vertices are respectively antipodal
are equal.
\end{theorem}

\begin{proof}
Given the triangle $ABC$ on the sphere,
we let the perpendicular dropped from the center of the sphere
to the plane of $ABC$ cut the sphere at $D$.
Then $D$ is equidistant from the vertices of $ABC$.
Since $ABD$ is isosceles,
it is congruent to $A'B'D'$,
and likewise for $BCD$ and $CAD$.
Thus $ABC=A'B'C'$.
\end{proof}

Lobachevski treats the total solid angle as $2\uppi$;
we convert to $4\uppi$
(although Lobachevski's convention has its own convenience).
With Lobachevski,
we follow the convention
whereby the measures of the surface angles 
of a spherical triangle $ABC$
are respectively $A$, $B$, and $C$.

\begin{theorem}{27}
At the center of a sphere,
the solid angle subtended by the triangle $ABC$
in the surface of the sphere measures
\begin{equation*}
A+B+C-\uppi.
\end{equation*}
\end{theorem}

\begin{proof}
We use that twice the measure of (for example) $A$
is the sum of $ABC$ and $A'BC$.
Thus
\begin{multline*}
2(A+B+C)\\
=(ABC+A'BC)+(ABC+AB'C)+(ABC+ABC')\\
=2ABC+ABC+A'BC+AB'C+A'B'C\\
=2ABC+2\uppi.	\qedhere
\end{multline*}
\end{proof}


\begin{theorem}{28}
If the intersections of three planes taken in pairs are parallel,
the sum of the angles of the planes taken in pairs is $\uppi$.
\end{theorem}

\begin{proof}
Given parallels $AA'$, $BB'$, and $CC'$ as in Figure \ref{fig:28},
\begin{figure}
  \centering
\psset{unit=33mm}
  \begin{pspicture}(-1,-1.2)(2,1.2)
	%(1,1)
%	\psgrid[subgriddiv=10]
\psset{PointSymbol=none}
	\pstTriangle(0,1)A(-1,0)B(0,-1)C
	\pstGeonode[PosAngle=45](1,0)P
	\pstHomO[HomCoef=1.5]BP[B']
	\ncline AP
	\ncline PC
	\pstInterLL[PointName=none] ACB{B'}X
	\pstTranslation XA{B'}[A']
	\pstTranslation XC{B'}[C']
	\ncline A{A'}
	\ncline B{B'}
	\ncline C{C'}
	\def\SphRad{0.3}
	% triangle DEF
	\pstHomO[HomCoef=\SphRad,PosAngle=180]AC[F]
	\pstInterLC[PointNameA={},PosAngle=90] A{A'}AF{D'}D
	\pstArcOAB AFD
	\psplot0{0.38}{0.6 dup mul x dup mul 1.5 mul sub sqrt neg 1 add}
	\psplot{0.38}{0.6}{0.6 dup mul x dup mul sub 1.5 div sqrt neg 1 add}
	\uput[u](0.38,0.62){$E$}
	% triangle GHK
	\pstHomO[HomCoef=\SphRad,PosAngle=135]PB[H]	
	\psplot{0.4}{0.62}{0.6 dup mul x 1 sub dup mul sub 1.5 div sqrt}
	\psplot{0.4}{0.62}{0.6 dup mul x 1 sub dup mul sub 1.5 div sqrt neg}
	\parametricplot{-0.375}{0.375}{1 0.6 dup mul t dup mul sub 1.5 div sqrt sub t}
	\uput[ur](0.62,0.38){$G$}
	\uput[dr](0.62,-0.38){$K$}
	% triangle LMN
	\pstHomO[HomCoef=\SphRad,PosAngle=180]CA[L]
	\pstInterLC[PointNameA={},PosAngle=-90] C{C'}CL{N'}N
	\pstArcOAB CNL
	\psplot0{0.38}{0.6 dup mul x dup mul 1.5 mul sub sqrt 1 sub}
	\psplot{0.38}{0.6}{0.6 dup mul x dup mul sub 1.5 div sqrt 1 sub}
	\uput[d](0.38,-0.62){$M$}
  \end{pspicture}
  \caption{\propword\ 28}\label{fig:28}
  
\end{figure}
picking a point $P$ along $BB'$,
we construct,
centered at $A$, $P$, and $C$ respectively,
the spherical triangles $DEF$, $GHK$, and $LMN$.
The angle sum of the three planes containing the parallels in pairs
is $D+H+N$.  Also
\begin{align*}
	E+G&=\uppi,&
	F&=L,&
	K+M&=\uppi.
\end{align*}
Thus
\begin{multline*}
	DEF+GHK+LMN\\
	=D+E+F+G+H+K+L+M+N-3\uppi\\
	=D+H+N-\uppi+2F,
\end{multline*}
and so
\begin{equation*}
D+H+N-\uppi=GHK-\bigl(2F-(DEF-LMN)\bigr).
\end{equation*}
The difference $2F-(DEF-LMN)$ is positive,
and $F$ can be as small as desired.
So can the angle $GPH$, by the proof of \propword\ 21,
and this means the solid angle $GHK$
can be as small as desired.
Thus we can only conclude
\begin{equation*}
D+H+N=\uppi.\qedhere
\end{equation*}
\end{proof}

\begin{theorem}{29}
If the perpendicular bisectors of two sides of a
(rectilineal) triangle 
meet at a point,
then the perpendicular bisector of the third side
also passes through that point.
\end{theorem}

\begin{proof}
This is entirely as would be in Euclid.
\end{proof}

\begin{theorem}{30}
If the perpendicular bisectors of two sides of a
(rectilineal) triangle are parallel to one another,
then they are parallel
to the perpendicular bisector of the third side.
\end{theorem}

\begin{proof}
Let the perpendicular bisectors be $DE$, $FG$, and $HK$,
with $HK$ between the others, as in Figure \ref{fig:30}.
\begin{figure}
  \centering
\psset{unit=16mm}
  \begin{pspicture}(-1,1.42)(4.47,4.3)
%	\psgrid
%\psline[linestyle=dashed](-1,0)(4.47,0)
\psarc(0,0)2{60}{120}
\psarc(2.73,0){2.45}{45}{135}
\psarc(1.73,0){3.23}{32.3}{147.7}
\psline(-1,1.73)(-1,4)
\psline[linestyle=dotted](0,2)(0,4)
\psline[linestyle=dotted](1,1.73)(1,4)
\psline(1.73,3.23)(1.73,4)
\psline[linestyle=dotted](2.73,2.45)(2.73,4)
\psline(4.46,1.73)(4.46,4)
\uput[d](-1,1.73){$A$}
\uput[u](-1,4){$A'$}
\uput[d](4.46,1.73){$B$}
\uput[u](4.46,4){$B'$}
\uput[d](1,1.73){$C$}
\uput[u](1,4){$C'$}
\uput[d](0,2){$D$}
\uput[u](0,4){$E$}
\uput[d](2.73,2.45){$F$}
\uput[u](2.73,4){$G$}
\uput[d](1.73,3.23){$H$}
\uput[u](1.73,4){$K$}
\uput[d](1.87,2.26){$a$}
\uput[d](3.60,2.26){$a$}
\uput[d](-0.5,1.94){$b$}
\uput[d](0.5,1.94){$b$}
\uput[u](0.37,2.93){$c$}
\uput[u](3.09,2.93){$c$}
  \end{pspicture}
  \caption{\propword\ 30}\label{fig:30}
  
\end{figure}
There are two cases.
If $DE\parallel FG$,
then, since $HK$ meets neither of these by \propword\ 29,
it must be parallel to them,
as in the proof of \propword\ 25.

Now we assume $HK\parallel FG$.
Let the parallels $AA'$, $BB'$, and $CC'$ to $HK$ be drawn.
As before,
if $DE\nparallel HK$, then it cannot cut $HK$,
so it must cut $AA'$.
Letting the angles of $ABC$ at $B$ and $C$ be
$\beta$ and $\gamma$ respectively, 
we have
\begin{align*}
\beta&=\Pi(a)-\Pi(c),&
\gamma&>\Pi(a)+\Pi(b).	
\end{align*}
If we rotate $CA$ about $C$ into a position $CQ$
as in Figure \ref{fig:30-2},
\begin{figure}
\mbox{}\hfill
\subfloat[]{\label{fig:30-L}
\psset{unit=16mm}
\begin{pspicture}(-0.5,-0.5)(3.5,1.72)
\pspolygon(0,1)(3,1)(1,0)
\psline(1,0)(0.29,1.22)(3,1)
\psline(0,1)(0.29,1.22)
\uput[l](0,1){$A$}
\uput[r](3,1){$B$}
\uput[d](1,0){$C$}
\uput[u](0.29,1.22){$Q$}
\end{pspicture}}
\hfill
\subfloat[]{\label{fig:30-N}
\psset{unit=16mm}
\begin{pspicture}(-0.5,-0.5)(1.5,3.3)
%\psgrid
\pspolygon(0,1)(1,3)(1,0)
\psline(1,0)(0.29,1.22)(1,3)
\psline(0,1)(0.29,1.22)
\uput[l](0,1){$A$}
\uput[u](1,3){$B$}
\uput[d](1,0){$C$}
\uput[r](0.29,1.22){$Q$}
\end{pspicture}}
\hfill\mbox{}
\caption{\propword\ 30 continued}\label{fig:30-2}
\end{figure}
so that $\angle QCB=\Pi(a)+\Pi(b)$,
then $\angle QBC>\Pi(a)-\Pi(c)$.
But we now have a triangle $QBC$ to which the earlier argument applies,
namely that the perpendicular bisector of $QB$
is parallel to the other two.
Thus, letting $QB=2c'$, we have
\begin{gather*}
\Pi(a)-\Pi(c')>\Pi(a)-\Pi(c),\\
\Pi(c')<\Pi(c),\\
c'>c.	
\end{gather*}
But $AC=QC$, and so,
since the greater angle is subtended by the greater side, 
as in Euclid's \emph{Elements} \textsc i.24,
we have $AB>QB$, that is, $c>c'$, contradicting the earlier inequality.
Thus $DE\parallel HK$.
\end{proof}

\begin{theorem}{31}
Given a ray $AB$, for every acute angle measure $\xi$,
we can find $x$ such that $\Pi(x)=\xi$, 
and then, on either side of $AB$,
we can find a point $X$ such that
\begin{align*}
	\angle BAX&=\xi,&\size{AX}&=2x,
\end{align*}
as in Figure \ref{fig:31}.
\begin{figure}
\centering
\psset{unit=5cm}
\begin{pspicture}(1,-0.05)(2,1.05)
%\psgrid
\psarc[linestyle=dashed](1,0)10{90}
\uput[r](2,0){$A$}
\uput[l](1,1){$X$}
\psarc(0,-1){! 5 sqrt}{26.6}{63.4} % 0 to 90 (A to X)
\psarc(0,2)2{-90}{-41.8}       % normal to this
\uput[60](! 3 2 div 3 sqrt 2 div){$Y$}
\psarc(! 0 3 sqrt 3 div neg){! 13 3 div sqrt}{16.1}{43.9} % 0 to 60
\psarc(! 0 2 3 sqrt mul){! 2 3 sqrt mul}{-90}{-59.0} % normal to this
\psarc(! 0 2 3 sqrt add neg){! 13 6 3 sqrt mul add sqrt}{71.9}{78.1} 
                                                  % 60 to 90 (Y to X)
\psarc(! 0 3 3 sqrt sub){! 3 3 sqrt sub}{-90}{-14.7}
\pscircle[fillstyle=solid,fillcolor=white,linestyle=none](0.5,0){0.5}
\psline(2,0)(1,0)
\uput[l](1,0){$B$}
\end{pspicture}
\caption{\propword\ 31.  The horocycle}\label{fig:31}
\end{figure}
The locus of the points $X$
is a curve, the perpendicular bisector of whose every chord
is parallel to $AB$.
\end{theorem}

\begin{proof}
By \propword\ 23, $x$ exists for every $\xi$.
By \propword\ 30, the perpendicular bisector of, for example, $XY$,
is parallel to those of $AX$ and $AY$
and thus to $AB$ itself.
\end{proof}

The curve given by \propword\ 31 is the \textbf{horocycle,}
and $AB$, along with the lines parallel to this, is an \textbf{axis}
of the horocycle.
Though Lobachevski does not use the term,
we may refer to the intersection of an axis with the horocycle
as a \textbf{vertex.}

\begin{theorem}{32}
Every point of the horocycle with a given axis
is approached as close as desired by a circle through the vertex
with center on the axis.
\end{theorem}

\begin{proof}
In Figure \ref{fig:32} then,
\begin{figure}
\centering
\psset{unit=8cm,dotstyle=o}
\begin{pspicture}(-1,-0.03)(0,1.03)
\psarc[linestyle=dashed](0,0)1{90}{180} % the horocycle
\psline(-1,0)(0.2,0)    % AC, axis of the horocycle
\psarc(! 1 3 sqrt){! 3 sqrt}{-150}{-120} % BD
\psarc(! 1 3 sqrt 3 div neg){! 13 3 div sqrt}%
{! 3 3 sqrt mul 5 atan 90 add}%
{! 2 3 sqrt mul 1 atan 90 add} % chord AB
\psarc[linestyle=dotted](-0.25,0){0.75}{90}{180} % circle through A and F
\psarc(1,-1){! 5 sqrt}%
{! 2 sqrt 3 sqrt atan 90 add}%
{! 2 1 atan 90 add} %  chord AF itself
\psarc(! 1 3 3 sqrt mul 1 sub 4 div)%
{! 22 3 3 sqrt mul sub 8 div sqrt}%
{! 3 3 sqrt sub 4 div 2 sqrt atan 180 add}%
{! 3 3 sqrt mul 1 sub 4 atan 180 add} % FE
%\psdots(! 1 2 sqrt sub 3 sqrt 1 sub)% F: (-0.414,0.732)
%(-1,0)% A
%(0,0)% E
%(! -0.5 3 sqrt 2 div)% B
\uput[d](-1,0){$A$}
\uput[120](! -0.5 3 sqrt 2 div){$B$}
\uput{8pt}[-95](! -0.5 3 sqrt 2 div){$\alpha$}
\uput[d](0.2,0){$C$}
\uput[u](! 1 3 sqrt 2 div sub 3 sqrt 1.5 sub){$D$}
\uput[d](0,0){$E$}
\uput[75](! 1 2 sqrt sub 3 sqrt 1 sub){$F$}
\uput{10pt}[255](! 1 2 sqrt sub 3 sqrt 1 sub){$\beta$}
\uput{20pt}[-66](! 1 2 sqrt sub 3 sqrt 1 sub){$\gamma$}
\end{pspicture}
\caption{\propword\ 32}\label{fig:32}
\end{figure}
a horocycle has axes $AC$ and $BD$,
and the circle with center $E$ on $AC$,
and passing through $A$, cuts $BD$ at $F$.
The sum of the angles in triangle $ABF$ is
\begin{equation*}
\alpha-\beta+\alpha+(\uppi-\beta-\gamma),
\end{equation*}
which is $2\alpha-2\beta-\gamma+\uppi$,
but is also less than $\uppi$,
and so
\begin{equation*}
\alpha-\beta\leq\frac12\gamma.
\end{equation*}
Since we can make $\gamma$ as small as desired,
we can make $BF$ the same.
\end{proof}

Since any two segments of two horocycles are comparable,
in the sense that one is congruent to a part or the whole of the other,
the segments have a ratio.

When two parallel line segments are joined by segments of two horocycles of which they are axes,
we may say that a \textbf{rectangle} is formed,
whose \textbf{length} is that of either of the two line segments,
and whose two \textbf{widths} 
are the lengths of the two segments of horocycles.  
Lobachevski does not use such terminology,
but just assigns letters to the lengths that he wants to talk about.

\begin{theorem}{33}
For some unit length,
the ratio of the widths of any rectangle,
the larger to the smaller,
is the power of $\me$ by the ratio of the length of the rectangle to the unit length.
\end{theorem}

\begin{proof}
Let $f_w(x)$ be the greater width
of the rectangle of length $x$ and lesser width $w$.
For all counting numbers $m$ and $n$, we have
\begin{align*}
f_{mw}(x)&=mf_w(x),&	
f_{(1/n)w}&=\frac1nf_w(x),
\end{align*}
and therefore
\begin{equation*}
f_{(m/n)w}(x)=\frac mnf_w(x).
\end{equation*}
By ``continuity'' then, for all positive real numbers $t$,
\begin{equation*}
f_{tw}(x)=tf_w(x),
\end{equation*}
and so for all other widths $w'$,
\begin{equation*}
\frac{f_{w'}(x)}{w'}=\frac{f_w(x)}w.
\end{equation*}
We may now denote this common ratio by $g(x)$;
it is the ratio of widths, greater to less,
of any rectangle of length $x$.
At least, in case this is a ratio of counting numbers,
we have
\begin{align*}
	g(mx)&=g(x)^m,&
	g\left(\frac1nx\right)&=g(x)^{1/n},
\end{align*}
and therefore
\begin{equation*}
g\left(\frac mnx\right)=g(x)^{m/n}.
\end{equation*}
Figure \ref{fig:33}
\begin{figure}
\centering
\psset{unit=4cm}
\begin{pspicture}(0,-1.5625)(2,0)
\psline(0,-1)(0,0)(2,0)(2,-1.5625)
\psline(1,0)(1,-1.25)
\psplot02{1.25 x exp neg}
\psplot01{1.25 x exp neg 0.25 mul}
\psplot01{1.25 x exp neg 0.5 mul}
\psplot01{1.25 x exp neg 0.75 mul}
\psplot12{1.25 x 1 sub exp neg 0.25 mul}
\psplot12{1.25 x 1 sub exp neg 0.5 mul}
\psplot12{1.25 x 1 sub exp neg 0.75 mul}
\psplot12{1.25 x 1 sub exp neg}
\end{pspicture}
\caption{\propword\ 33}\label{fig:33}
\end{figure}
shows $g(2x)=g(x)^2$ where $g(x)=5/4$.
By ``continuity'' then, for all $x$
and for all positive real numbers $t$,
\begin{equation*}
g(tx)=g(x)^t.
\end{equation*}
Since by ``continuity'' one more time,
the equation $g(x)=\me$ is soluble by some length $u$,
we have then
\begin{equation*}
g(tu)=\me^t,
\end{equation*}
as claimed.
\end{proof}

We extend the range of $\Pi$ to obtuse angles by defining
\begin{equation*}
\Pi(x)+\Pi(-x)=\uppi.
\end{equation*}

\begin{theorem}{34}
When a horocycle is rotated about an axis,
so as to generate a surface,
if another axis of the horocycle is selected
when this is in either of two positions,
the two new axes
are axes of a horocycle lying within the surface.
\end{theorem}

\begin{proof}
When the horocycle $AB$,
shown as a dashed line in Figure \ref{fig:34},
\begin{figure}
\centering
\psset{unit=5.4mm}
\begin{pspicture}(-10.5,-1.5)(9.5,22)
%\psgrid[subgriddiv=1]
\psset{Alpha=-70,Beta=55}
%\pstThreeDCoor[linecolor=black,IIIDticks,xMin=-8,xMax=8,yMin=-6,yMax=8,zMin=0,zMax=36]
%\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15](0,0,0){136 sqrt}
\pstThreeDTriangle[linestyle=dashed](-8,-6,6)(6,8,6)(8,-6,6)%ABC
\pstThreeDPut[pOrigin=lt](-8,-6,6){\mybox A}
\pstThreeDPut[pOrigin=lt](-8,-6,24){\mybox{A'}}
\pstThreeDPut[pOrigin=r](6,8,6){\mybox B}
\pstThreeDPut[pOrigin=r](6,8,24){\mybox{B'}}
\pstThreeDPut[pOrigin=lb](8,-6,6){\mybox C}
\pstThreeDPut[pOrigin=lb](8,-6,24){\mybox{C'}}
\pstThreeDPut[pOrigin=rt](-1,1,134 sqrt){\mybox D}
\pstThreeDPut[pOrigin=rt](-1,1,24){\mybox{D'}}
\pstThreeDPut[pOrigin=l](0,-6,10){\mybox E}
\pstThreeDPut[pOrigin=l](0,-6,24){\mybox{E'}}
\pstThreeDPut[pOrigin=lb](0,0,136 sqrt){\mybox F}
\pstThreeDPut[pOrigin=lb](0,0,24){\mybox{F'}}
\pstThreeDPut[pOrigin=rb](7,1,86 sqrt){\mybox G}
\pstThreeDPut[pOrigin=rb](7,1,24){\mybox{G'}}
\pstThreeDPut[pOrigin=b](0,-8,3400 9 div -8 22.667 add dup mul sub sqrt){\mybox{K}}
{\psset{linewidth=1.6pt}
\parametricplotThreeD(-8,8){t -6 100 t dup mul sub sqrt}%AC
\parametricplotThreeD(-8,6){t t 2 add 134 t 1 add dup mul 2 mul sub sqrt}%AB
\parametricplotThreeD(-7,7){7 t 7 div sub t 1 add 86 t dup mul 50 mul 49 div sub sqrt}%BC
}
\parametricplotThreeD(0,1){-8 t mul -6 t mul 136 100 t dup mul mul sub sqrt}
\parametricplotThreeD(0,1){8 t mul -6 t mul 136 100 t dup mul mul sub sqrt}
\parametricplotThreeD(0,1){6 t mul 8 t mul 136 100 t dup mul mul sub sqrt}
\parametricplotThreeD(0,1){0 -6 t mul 136 36 t dup mul mul sub sqrt}
\parametricplotThreeD(0,1){7 t mul t 136 50 t dup mul mul sub sqrt}
\parametricplotThreeD(0,1){t neg t 136 2 t dup mul mul sub sqrt}
\parametricplotThreeD(-6,-8){0 t 3400 9 div t 22.667 add dup mul sub sqrt}%EK
\pstThreeDLine{->}(-8,-6,6)(-8,-6,24)%AA'
\pstThreeDLine{->}(6,8,6)(6,8,24)%BB'
\pstThreeDLine{->}(8,-6,6)(8,-6,24)%CC'
\pstThreeDLine{->}(-1,1,134 sqrt)(-1,1,24)%DD'
\pstThreeDLine{->}(0,-6,10)(0,-6,24)%EE'
\pstThreeDLine{->}(0,0,136 sqrt)(0,0,24)%FF'
\pstThreeDLine{->}(7,1,86 sqrt)(7,1,24)%GG'
\end{pspicture}
\caption{\propword\ 34}\label{fig:34}
\end{figure}
is rotated about the axis $AA'$,
let one of its positions be $AC$,
and let the parallels $BB'$ and $CC'$ to $AA'$ be drawn.
The chord $BC$ having midpoint $G$,
we shall show $B'BG=GCC'$.

Within the plane $A'AB$, 
we erect a perpendicular bisector $DD'$ of $AB$;
it is parallel to $AA'$ (and therefore to $BB'$).
The angle between the planes $A'AB$ and $CAB$ 
being $\Pi(a)$ for some $a$, possibly $0$ or negative,
we erect, in the plane $ABC$,
a perpendicular bisector $DF$ of $AB$ having directed length $a$,
measured into the triangle.
Then
\begin{equation*}
AF=FB.
\end{equation*}
When $FF'$ is erected perpendicular to the plane $ABC$,
it is parallel to $DD'$ and thus to $AA'$.

Within the plane $A'AC$, 
we erect a perpendicular bisector $EE'$ of $AC$;
it is parallel to $AA'$ and thus to $FF'$.
We erect also $EK$ perpendicular to the plane $ABC$.
Then $AE$ is perpendicular to both $EE'$ and $EK$,
which determine a plane in which $EF$ lies---%
assuming for the moment that $F$ is not the point $E$.
Therefore $AE$ is perpendicular to $EF$.
Thus $EF$ is a perpendicular bisector of $AC$,
and so we have
\begin{equation*}
CF=FA=BF.
\end{equation*}
We have this as well if $F$ should happen to be the point $E$.

Finally, letting $G$ be the midpoint of $BC$,
we let the intersection of the planes $ABC$ and $B'BC$ be $GG'$.
As $BB'\parallel FF'$,
it follows that $GG'\parallel BB'$
and similarly $GG'\parallel CC'$.
This yields the desired conclusion.
\end{proof}

In Euclidean spherical trigonometry,
one can reason as follows
about the spherical right triangle $ABC$
in Figure \ref{fig:trig},
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(-1,-0.5)(19,17.5)
%\psgrid[subgriddiv=1]
\psset{Alpha=165,Beta=5}
%\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=15,ysubticks=25](0,0)(19.5,32.5)
\pstThreeDPut[pOrigin=r](0,0,0){\mybox O}
\pstThreeDPut[pOrigin=r](0,18.0,7.5){\mybox B}
\pstThreeDPut[pOrigin=l](15.6,11.7,0){\mybox A}
\pstThreeDPut[pOrigin=rb](0,19.5,0){\mybox C}
\pstThreeDPut[pOrigin=rb](0,32.5,0){\mybox{C'}}
\pstThreeDPut[pOrigin=b](0,32.5,13.5416667){\mybox{B'}}
%%%% radii from O
\pstThreeDLine(0,0,0)(0,19.5,0) % OC
\pstThreeDLine(0,0,0)(15.6,11.7,0) % OA
\pstThreeDLine(0,0,0)(0,18.0,7.5) % OB
{\psset{linestyle=dashed}
\pstThreeDLine(0,19.5,0)(0,32.5,0) % CC'
\pstThreeDLine(0,18.0,7.5)(0,32.5,13.5416667) % BB'
}
%%%% planar triangle AB'C'
{\psset{linestyle=dashed}
\pstThreeDLine(15.6,11.7,0)(0,32.5,0)
\pstThreeDLine(0,32.5,0)(0,32.5,13.5416667)
\pstThreeDLine(15.6,11.7,0)(0,32.5,13.5416667)
}
%%%% spherical triangle ABC
\psset{linewidth=1.6pt}
\pstThreeDCircle[beginAngle=0,endAngle={4 3 atan}]
(0,0,0)(0,19.5,0)(1,0,0)
\pstThreeDCircle[beginAngle=0,endAngle={5 12 atan}]
(0,0,0)(0,19.5,0)(0,0,1)
\pstThreeDCircle[beginAngle=0,endAngle={2929 sqrt 36 atan}]
(0,0,0)(15.6,11.7,0)(0,6.0,2.5)
\end{pspicture}
\caption{Euclidean spherical trigonometry}\label{fig:trig}
\end{figure}
where the angle $C$ is right.
(A reference is Todhunter, \emph{Spherical Geometry,}
fifth edition [London: Macmillan, 1886],
available from Project Gutenberg.)
The center of the sphere being $O$,
and the rectilineal triangle $AB'C'$
lying in a plane tangent to the sphere at $A$,
so that the planar angles $AC'B'$, 
$OC'B'$, and $OAC'$ are right,
we have
\begin{equation*}
\frac{B'C'}{OB'}
=\frac{B'C'}{B'N}\cdot\frac{B'N}{OB'}
\end{equation*}
which means
\begin{equation*}
\sin a=\sin A\sin c,
\end{equation*}
and so by symmetry also
\begin{equation*}
\sin b=\sin B\sin c.
\end{equation*}
Moreover,
\begin{equation*}
\frac{OA}{OB'}=\frac{OA}{OC'}\cdot\frac{OC'}{OB'},
\end{equation*}
which means
\begin{equation*}
\cos c=\cos b\cos a.
\end{equation*}
Therefore, since from
\begin{equation*}
\frac{C'A}{B'A}=\frac{C'A}{OA}\cdot\frac{OA}{B'A}
\end{equation*}
we have
\begin{equation*}
\cos A
=\tan b\cot c=\frac{\sin b}{\sin c}\cdot\frac{\cos c}{\cos b},
\end{equation*}
we can conclude
\begin{equation*}
\cos A=\sin B\cos a,
\end{equation*}
and by symmetry
\begin{equation*}
\cos B=\sin A\cos b.
\end{equation*}


\begin{theorem}{35}
Spherical trigonometry is unchanged by the new postulate on parallels.
\end{theorem}

\newcommand{\myA}{8}%
\newcommand{\myB}{9}%
\newcommand{\myC}{13}%
\newcommand{\myH}{\myC\space dup mul \myA\space dup mul sub
2 \myB\space mul div}%
\newcommand{\myBpp}{-4 \myA\space dup mul mul \myB\space mul % -4a^2b
\myA\space dup mul \myC\space dup mul add % a^2+c^2
mul
5 \myA\space dup mul dup mul mul % 5a^4
4 \myA\space dup mul mul \myB\space dup mul mul add % +4a^2b^2
2 \myA\space dup mul mul \myC\space dup mul mul add % +2a^2c^2
\myC\space dup mul dup mul add % +c^4
div}%
\newcommand{\myCpp}{2 \myA\space mul
\myA\space dup mul \myC\space dup mul add dup mul mul
\myA\space dup mul \myC\space dup mul add dup mul
\myA\space dup mul dup mul 4 mul add div}
\newcommand{\myPhi}{
\myA\space neg % -a
\myA\space dup mul \myC\space dup mul add % a^2+c^2
mul
\myA\space dup mul % a^2
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup mul mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup mul
4 \myA\space dup mul dup mul mul % 4a^4
add
t dup mul mul % times y^2
sub sqrt sub
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup mul
4 \myA\space dup mul dup mul mul % 4a^4
add
div}%
\newcommand{\myD}{\myB\space
\myA\space dup mul \myC\space dup mul add
mul
\myA\space dup mul \myB\space dup mul add \myC\space dup mul add
div neg}%
\newcommand{\mycutoff}{0.6}%
%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%
\begin{proof}
Let $ABC$ be a triangle with right angle at $C$.
As usual, the side opposite vertex $X$ has length $x$.
For some additional lengths $\alpha$ and $\beta$, we have
\begin{align*}
	\angle BAC&=\Pi(\alpha),&\angle ABC&=\Pi(\beta).
\end{align*}
Now we erect $AA'$ perpendicular to the plane of $ABC$,
and we draw $BB'$ and $CC'$ parallel to $AA'$,
as in Figure \ref{fig:35}
\begin{figure}
\centering
\psset{unit=10mm}
\begin{pspicture}(-7,1)(4,8)
%\psgrid[subgriddiv=1]
\psset{Alpha=-60,Beta=15}
%\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-1,xMax=14,yMin=-14,yMax=1,zMin=-8,zMax=8]
%\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=t](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=rt](\myC,0,0){\mybox C}
%%%%%%%%%%%%% The planar triangle ABC
%\psset{linewidth=1.6pt}
\pstThreeDLine[linewidth=1.6pt](\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD[linewidth=1.6pt](\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%%%% BA
\parametricplotThreeD[linewidth=1.6pt](0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
%%%%%%%%% The parallels
%%%% AI
\parametricplotThreeD(0,\mycutoff){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
%%%% CI
\parametricplotThreeD(0,\mycutoff){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
%%%% BI
\parametricplotThreeD(0,\mycutoff){1 t sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  t
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt
  t 1 sub \myB\space mul
  t \myA\space mul
}
%%%%%%%%%%%%%%%% labels
\pstThreeDPut[pOrigin=b](
\myA\space 1 \mycutoff\space dup mul sub sqrt mul,
0,
\mycutoff\space \myA\space mul){\mybox{A'}}
\pstThreeDPut[pOrigin=b](\myC\space dup mul 
\mycutoff\space \myC\space dup mul \myA\space dup mul sub mul sub 
\mycutoff\space dup mul \myA\space dup mul mul sub sqrt,
0,
\mycutoff\space \myA\space mul
){\mybox{C'}}
\pstThreeDPut[pOrigin=b](1 \mycutoff\space sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  \mycutoff\space
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt,
  \mycutoff\space 1 sub \myB\space mul,
  \mycutoff\space \myA\space mul
){\mybox{B'}}
%%%%%%%%%%%%%%%%%%%% AC''
\parametricplotThreeD[linestyle=dashed](\myA,
\myCpp)
{t
0
\myA\space
dup 2 mul t mul t dup mul sub sqrt sub}
%%% label C''
\pstThreeDPut[pOrigin=rb](\myCpp,0,
\myA\space
dup 2 mul \myCpp\space mul \myCpp\space dup mul sub sqrt sub)
{\mybox{C''}}
%%%%%%%%%%%%%%%%%%%%% AB''
\parametricplotThreeD[linestyle=dashed](\myBpp,0){
%%%%%%%%%%%% x
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div
%%%%%%%%%%%% y
t
%%%%%%%%%%%% z
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
}
%%%%%%%%%%%%% label for B''
\pstThreeDPut[pOrigin=lb](
%%%%%%%%%% x
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div,
%%%%%%%%%%%% y
\myBpp\space,
%%%%%%%%%%%% z
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
){\mybox{B''}}
%%%%%%%%%%%%%%%%%%%% B''C''
\parametricplotThreeD[linestyle=dashed](\myBpp,0){
\myA\space dup mul \myC\space dup mul add neg \myPhi\space mul
t
2 \myA\space dup mul mul \myPhi\space mul
\myA\space add
}
%%%%%%%%%%%%%%%% CD
\parametricplotThreeD[linestyle=dotted](\myD,0){
\myC\space
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
t dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
t
\myA\space neg
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
t dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
\myA\space add}
%%%%%%%%%%%% label D
\pstThreeDPut[pOrigin=lb](
\myC\space
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
\myD\space dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul,
\myD\space,
\myA\space neg
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
\myD\space dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
\myA\space add){\mybox D}
\end{pspicture}
\caption{\propword\ 35}\label{fig:35}
\end{figure}
(see the appendix on the actual drawing of the figure).
We shall use the notation
\begin{equation*}
\Pi(x')+\Pi(x)=\frac{\uppi}2.
\end{equation*}
Thus the angles of the planes that meet, two by two, 
in the three parallels are
$\uppi/2$ at $CC'$,
$\Pi(\alpha)$ at $AA'$,
and therefore $\Pi(\alpha')$ at $BB'$,
by \propword\ 28.

Suppose now a sphere centered at $B$
cuts $BB'$, $BA$, and $BC$ respectively
at   $E$,  $F$, and $G$.
In the spherical triangle $EFG$,
\begin{align*}
	g&=\Pi(c),&e&=\Pi(\beta),&f&=\Pi(a),\\
	G&=\Pi(b),&E&=\Pi(\alpha'),&F&=\frac{\uppi}2.
\end{align*}
Conversely, given a spherical triangle with these parameters,
we can recover the planar triangle.
In other words, 
for any ordered quintuple $(a,b,c,\alpha,\beta)$ of lengths,
a right triangle $ABC$ exists as above,
with sides and angles
\begin{equation*}
\bigl(a,b,c,\Pi(\alpha),\Pi(\beta)\bigr),
\end{equation*}
if and only if a spherical right triangle $EFG$ exists as above,
with sides and angles 
\begin{equation*}
\bigl(\Pi(c),\Pi(\beta),\Pi(a),\Pi(b),\Pi(\alpha')\bigr).
\end{equation*}
We can interchange the angles that are not right angles.
Thus the existence of such a spherical right triangle
as was just mentioned is equivalent to the existence of one
with sides and angles 
\begin{equation*}
\bigl(\Pi(\beta),\Pi(c),\Pi(a),\Pi(\alpha'),\Pi(b)\bigr).
\end{equation*}
By what we already saw,
the existence of this last spherical right triangle
is equivalent to the existence of a right triangle 
with sides and angles
\begin{equation*}
\bigl(a,\alpha',\beta,\Pi(b'),\Pi(c)\bigr),
\end{equation*}
simply because the correspondence
\begin{equation*}
	(a,b,c,\alpha,\beta)\leftrightarrow(c,\beta,a,b,\alpha')
\end{equation*}
can also be written as
\begin{equation*}
(a,\alpha',\beta,b',c)\leftrightarrow(\beta,c,a,\alpha',b).
\end{equation*}

We now let $AA'$ be an axis of a horocycle through $A$,
cutting $BB'$ and $B''$ and $CC'$ at $C''$.
If we define
\begin{align*}
	\size{B''C''}&=p,&\size{C''A}&=q,&\size{AB''}&=r,
\end{align*}
then
\begin{align*}
	p&=r\sin\Pi(\alpha),&q&=r\cos\Pi(\alpha),
\end{align*}
by \propword\ 34.
If parallels such as $AA'$ and $CC'$ are given,
and at a point $A$ on one of them a perpendicular is erected,
cutting the other parallel cuts this at $C$,
while the horocycle passing through $A$
(and having the parallels as axes) cuts the other parallel at $C''$,
then the length of $CC''$ and the arc length of $AC''$
are determined by the length of $AC$ alone.
Thus we have functions $f$ and $g$ given by
\begin{align*}
f(b)&=\size{CC''},& g(b)&=q.
\end{align*}
Then also
\begin{align*}
	f(c)&=\size{BB''},&g(c)&=r.
\end{align*}
Now we let a new horosphere with axis $CC''$,
passing through $C$, cut $BB''$ at $D$, so that
\begin{equation*}
BB''=BD+DB''=BD+CC'';
\end{equation*}
and we define
\begin{equation*}
\size{CD}=t.
\end{equation*}
Then
\begin{align*}
	f(a)&=\size{BD},&g(a)&=t.
\end{align*}
Consequently
\begin{equation*}
f(c)=f(a)+f(b).
\end{equation*}
Also, considering the rectangle $CDB''C''$, we have
\begin{equation*}
t=p\me^{f(b)}
=r\sin\Pi(\alpha)\me^{f(b)},
\end{equation*}
and thus
\begin{equation*}
g(a)=g(c)\sin\Pi(\alpha)\me^{f(b)}.
\end{equation*}
By symmetry,
\begin{equation*}
g(b)=g(c)\sin\Pi(\beta)\me^{f(a)},
\end{equation*}
and this gives us
\begin{equation*}
\cos\Pi(\alpha)
=\sin\Pi(\beta)\me^{f(a)}.
\end{equation*}
By the transformation discussed earlier,
\begin{gather*}
\sin\Pi(b)=\cos\Pi(b')
=\sin\Pi(c)\me^{f(a)},\\
\sin\Pi(b)\me^{f(b)}
=\sin\Pi(c)\me^{f(a)}\me^{f(b)}
=\sin\Pi(c)\me^{f(c)},
\end{gather*}
and therefore also, by symmetry again,
\begin{equation*}
\sin\Pi(a)\me^{f(a)}
=\sin\Pi(b)\me^{f(b)}.
\end{equation*}
Since $a$ and $b$ are independent of one another,
and $\sin\Pi(a)\me^{f(a)}$ approaches the limit of $1$ at $0$,
we can conclude
\begin{equation*}
\me^{-f(a)}=\sin\Pi(a).
\end{equation*}
Thus from equations found earlier we obtain
\begin{gather*}
	\sin\Pi(\beta)=\sin\Pi(a)\cos\Pi(\alpha),\\
	\sin\Pi(c)=\sin\Pi(a)\sin\Pi(b).
\end{gather*}
By symmetry, the former yields
\begin{equation*}
	\sin\Pi(\alpha)=\sin\Pi(b)\cos\Pi(\beta).
\end{equation*}
From this, by the more elaborate permutation,
\begin{equation}\label{eqn:cosPb=cosPalphacosPc}
\cos\Pi(b)=\cos\Pi(\alpha)\cos\Pi(c).
\end{equation}
Finally, by interchanging $a$ with $b$ and $\alpha$ with $\beta$ again,
\begin{equation*}
\cos\Pi(a)=\cos\Pi(\beta)\cos\Pi(c).
\end{equation*}
If we relabel the earlier spherical triangle $GEF$ as $ABC$,
then the five equations that we have found become
\begin{gather*}
	\sin b=\sin c\sin B,\\
	\sin a=\sin c\sin A,\\
	\cos B=\sin A\cos b,\\
	\cos A=\sin B\cos a,\\
	\cos c=\cos b\cos a.\qedhere
\end{gather*}
\end{proof}


%\clearpage

\begin{theorem}{36}
For some unit length,
the power of $\me$ by the ratio to that length of any length
is the reciprocal of the tangent 
of half the angle of parallelism of that length:
\begin{equation*}
\tan\frac{\Pi(x)}2=\me^{-x}.
\end{equation*}
\end{theorem}

\begin{proof}
We are given triangle $ABC$ as in Figure \ref{fig:36};
\begin{figure}
\centering
\psset{unit=8cm}
\begin{pspicture}(-0.5,0.82)(0.5,1.55)
\psarc(0,0)1{90}{120}
\psarc(0.5,0){! 7 sqrt 2 div}{90}{! 2 3 sqrt atan 90 add}
\psline(! -0.5 3 sqrt 2 div)(-0.5,1.5)
\psline(0,1)(0,1.5)
\psline(! 0.5 7 sqrt 2 div)(0.5,1.5)
\uput[d](! -0.5 3 sqrt 2 div){$A$}
\uput[u](-0.5,1.5){$A'$}
\uput[dr](! 0 1.5 sqrt){$B$}
\uput[u](0,1.5){$B'$}
\uput[d](0,1){$C$}
\uput[d](! 0.5 7 sqrt 2 div){$D$}
\uput[u](0.5,1.5){$D'$}
\uput[u](! 0.25 0.75 3 sqrt mul){$\beta$}
\uput[r](! 0 2 6 sqrt add 4 div){$a$}
\uput[d](! -0.25 15 sqrt 4 div){$b$}
\uput[u](! -0.25 19 sqrt 4 div){$c$}
\uput{6mm}[70]{45}(! -0.5 3 sqrt 2 div){\smaller$\Pi(c+\beta)$}
\uput{12mm}[36]{30}(! -0.5 3 sqrt 2 div){\smaller$\Pi(\alpha)$}
\uput{3mm}[-120]{25}(! 0 1.5 sqrt){\smaller$\Pi(\beta)$}
\end{pspicture}
\caption{Theorem 36, first case}\label{fig:36}
\end{figure}
as in \propword\ 35,
the angle at $C$ is right,
and for some \emph{distances} $\alpha$ and $\beta$,
the angles at $A$ and $B$ are $\Pi(\alpha)$ and $\Pi(\beta)$ respectively.
The sides opposite $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively.
We first extend $AB$
by the distance $\beta$ to $D$.
Then the perpendicular $DD'$ to $AD$ is parallel to $CB$
(which is in turn extended to $B'$).
If the parallel $AA'$ to $BB'$ is also drawn,
then, considering that $AA'$ is parallel to two different straight lines
to which perpendiculars are dropped from $A$,
we have
\begin{equation}\label{eqn:36}
\Pi(b)=\Pi(\alpha)+\Pi(c+\beta).
\end{equation}
We derive a related equation
by measuring $\beta$ along $BA$ in the other direction.
There are three possibilities here.
If $\beta<c$, we have Figure \ref{fig:36-2},
\begin{figure}
\centering
\psset{unit=11cm}
\begin{pspicture}(-0.5,0.76)(0,1.26)
\psarc(0,0)1{90}{120}   % CA
\psarc(0.5,0){! 7 sqrt 2 div}%
{! 1 6 sqrt atan 90 add}%
{! 2 3 sqrt atan 90 add} % BA
\psarc(-1,0)1{! 0.8 0.6 atan}{60} % A'A
\psline(0,0.8)(! 0 6 sqrt 2 div)
\psarc(-1.5,0){1.5}{! 0.8 2.25 .64 sub sqrt atan}{! 7 sqrt 3 atan}
\uput[l](! -0.5 3 sqrt 2 div){$A$}
\uput[d](-0.4,0.8){$A'$}
\uput[u](! 0 1.5 sqrt){$B$}
\uput[d](0,0.8){$C'$}
\uput[r](0,1){$C$}
\uput[ul](! 7 sqrt 2 div 1.5 atan dup cos 1.5 mul 1.5 sub exch sin 1.5 mul){$D$}
\uput[d](! -1.5 1.61 sqrt add 0.8){$D'$}
\uput[r](! 0 2 6 sqrt add 4 div){$a$}
\uput[d](! -0.25 15 sqrt 4 div){$b$}
\uput[u](! -0.25 19 sqrt 4 div){$c$}
\uput{6mm}[r](! -0.5 3 sqrt 2 div){\smaller$\Pi(b)$}
\uput{8mm}[36]{30}(! -0.5 3 sqrt 2 div){\smaller$\Pi(\alpha)$}
\uput{3mm}[-120]{25}(! 0 1.5 sqrt){\smaller$\Pi(\beta)$}
\end{pspicture}
\caption{Theorem 36, second case, when $\beta<c$}\label{fig:36-2}
\end{figure}
from which we can infer
\begin{equation}\label{eqn:36-2}
  \Pi(c-\beta)=\Pi(\alpha)+\Pi(b).
\end{equation}
In case $\beta=c$, the diagram is as in Figure \ref{fig:36-3},
\begin{figure}
\psset{unit=9cm}
\subfloat[case $\beta=c$]{\label{fig:36-3}
%\psset{unit=10cm}
  \begin{pspicture}(-0.55,0.75)(0.05,1.45)
%	\psgrid[subgriddiv=10]
    \psarc(0,0)1{90}{120}
\psarc(1,0){! 3 sqrt}{! 1 2 sqrt atan 90 add}{150}
\psline(0,0.8)(! 0 2 sqrt)
\psarc(-1,0)1{! 0.8 0.6 atan}{60} % A'A
\uput[l](! -0.5 3 sqrt 2 div){$A$}
\uput[d](-0.4,0.8){$A'$}
\uput[u](! 0 2 sqrt){$B$}
\uput[d](0,0.8){$C'$}
\uput[r](0,1){$C$}
\uput[r](! 0 1 2 sqrt add 2 div){$a$}
\uput[d](! -0.25 15 sqrt 4 div){$b$}
\uput[u](! -0.25 23 sqrt 4 div){$c$}
\uput{6mm}[r](! -0.5 3 sqrt 2 div){\smaller$\Pi(b)$}
\uput{8mm}[40]{30}(! -0.5 3 sqrt 2 div){\smaller$\Pi(\alpha)$}
\uput{3mm}[-120]{35}(! 0 2 sqrt){\smaller$\Pi(\beta)$}
  \end{pspicture}}
\hfill
\subfloat[case $\beta>c$]{\label{fig:36-4}
%\psset{unit=10cm}
  \begin{pspicture}(-0.55,0.75)(0.05,1.45)
	%\psgrid[subgriddiv=10]
%    \psarc(0,0)1{90}{120}
\psarc(0,0){! 1.5 sqrt}{90}{! 1 23 sqrt atan 90 add}
\psarc(1,0){! 3 sqrt}{! 1 2 sqrt atan 90 add}{150}
\psline(0,0.8)(! 0 2 sqrt)
\psarc(-3,0)3{! 4 209 sqrt atan}{! 23 sqrt 11 atan}
\psarc(-1,0)1{! 0.8 0.6 atan}{60} % D'D
\uput[l](! -0.5 3 sqrt 2 div){$D$}
\uput[d](-0.4,0.8){$D'$}
\uput[u](! 0 2 sqrt){$B$}
\uput[d](0,0.8){$C'$}
\uput[r](! 0 6 sqrt 2 div){$C$}
\uput[r](! 0 2 sqrt 6 sqrt 2 div add 2 div){$a$}
\uput[d](! -0.125 95 sqrt 8 div){$b$}
\uput[u](! -0.125 111 sqrt 8 div){$c$}
\uput[ul](! -0.25 23 sqrt 4 div){$A$}
\uput[d](! -3 9 0.64 sub sqrt add 0.8){$A'$}
\uput{4mm}[-30]{-30}(! -0.25 23 sqrt 4 div){\smaller$\Pi(b)$}
\uput{4mm}[25]{25}(! -0.25 23 sqrt 4 div){\smaller$\Pi(\alpha)$}
\uput{4mm}[-100]{80}(! -0.25 23 sqrt 4 div){\smaller$\Pi(\beta-c)$}
\uput{3mm}[-120]{35}(! 0 2 sqrt){\smaller$\Pi(\beta)$}
\uput[ul](! -0.375 71 sqrt 8 div){$\beta-c$}
  \end{pspicture}}
  \caption{Theorem 36, second case, when $\beta\geq c$}
\end{figure}
and then
\begin{equation*}
  \Pi(\alpha)+\Pi(b)=\frac12\uppi;
\end{equation*}
but now $\Pi(c-\beta)=\Pi(0)=\uppi/2$ by definition;
so again \eqref{eqn:36-2} holds.
Finally, if $\beta>c$,
then as in Figure \ref{fig:36-4},
\begin{equation*}
  \Pi(\beta-c)+\Pi(b)+\Pi(\alpha)=\uppi,
\end{equation*}
so \eqref{eqn:36-2} still holds since 
$\uppi-\Pi(\beta-c)=\Pi(c-\beta)$ by definition.

From the system of \eqref{eqn:36} and \eqref{eqn:36-2},
we obtain
\begin{gather*}
  2\Pi(b)=\Pi(c-\beta)+\Pi(c+\beta),\\
  2\Pi(\alpha)=\Pi(c-\beta)-\Pi(c+\beta),
\end{gather*}
which yield immediately
\begin{equation*}
  \frac{\cos\Pi(b)}
       {\cos\Pi(\alpha)}
=\frac{\cos\left(\frac12\Pi(c-\beta)+\frac12\Pi(c+\beta)\right)}
      {\cos\left(\frac12\Pi(c-\beta)-\frac12\Pi(c+\beta)\right)}.
\end{equation*}
Now, from the proof of Theorem 35, 
we use \eqref{eqn:cosPb=cosPalphacosPc}
%, namely $\cos\Pi(\alpha)\cos\Pi(c)=\cos\Pi(b)$,
to obtain
\begin{equation}\label{eqn:36-stop}
\cos\Pi(c)=\frac{\cos\left(\frac12\Pi(c-\beta)+\frac12\Pi(c+\beta)\right)}
{\cos\left(\frac12\Pi(c-\beta)-\frac12\Pi(c+\beta)\right)}.
\end{equation}
From this, we shall obtain
\begin{equation}\label{eqn:L}
\left(\tan\frac{\Pi(c)}2\right)^2=\tan\frac{\Pi(c-\beta)}2\cdot\tan\frac{\Pi(c+\beta)}2.
\end{equation}
Lobachevski does not give a derivation,
but if we write \eqref{eqn:36-stop} as
\begin{equation}\label{eqn:cos}
\cos\theta=\frac{\cos(\phi+\psi)}{\cos(\phi-\psi)},
\end{equation}
then, since
\begin{equation*}
\tan\frac{\theta}2=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta},
\end{equation*}
so that
\begin{equation*}
\left(\tan\frac{\theta}2\right)^2=\frac{1+\cos\theta}{1-\cos\theta},
\end{equation*}
we obtain from \eqref{eqn:cos}
\begin{multline*}
\left(\tan\frac{\theta}2\right)^2
=\frac{\cos(\phi-\psi)-\cos(\phi+\psi)}{\cos(\phi-\psi)+\cos(\phi+\psi)}
=\frac{\sin\phi\cdot\sin\psi}{\cos\phi\cdot\cos\psi}\\
=\tan\phi\cdot\tan\psi.
\end{multline*}
Now Lobachevski proposes replacing $\beta$ with $c$, $2c$, $3c$, and so forth.
One can do this; that is, one can use induction to obtain
\begin{equation*}
  \left(\tan\frac{\Pi(c)}2\right)^n=\tan\frac{\Pi(nc)}2.
\end{equation*}
But it seems neater to me to rewrite \eqref{eqn:L} as
\begin{equation*}
  \frac{\tan(\Pi(c)/2)}{\tan(\Pi(c-\beta)/2)}
= \frac{\tan(\Pi(c+\beta)/2)}{\tan(\Pi(c)/2)};
\end{equation*}
for then
\begin{equation*}
  \left(\tan\frac{\Pi(c)}2\right)^n=\prod_{k=1}^n
\frac{\tan\bigl(\Pi(kc)/2\bigr)}{\tan\Bigl(\Pi\bigl((k-1)c\bigr)/2\Bigr)}
=\tan\frac{\Pi(nc)}2
\end{equation*}
since $\tan\bigl(\Pi(0)/2\bigr)=1$.
By continuity,
\begin{equation*}
  \left(\tan\frac{\Pi(c)}2\right)^t=\tan\frac{\Pi(tc)}2.
\end{equation*}
for all positive real numbers $t$.
For some unit length $u$ we have
\begin{equation*}
\tan\frac{\Pi(u)}2=\me^{-1},
\end{equation*}
and then
\begin{equation*}
\tan\frac{\Pi(tu)}2=\me^{-t}.\qedhere
\end{equation*}
\end{proof}

\begin{theorem}{37}{}
[Equations for solving arbitrary planar triangles.]
\end{theorem}

\appendix

\addchap{Drawing a figure}

I record here the process of reproducing,
with precision,
the diagrams of Lobachevski,
and specifically his Figure 28,
used for [Theorem] 35.
Drawing such a figure tests the computational limits,
both of myself when working by hand,
and of the PostScript program. 
The figure shows a rectilineal triangle $ABC$
in which the angle at $C$ is right.
The line $AA'$ is perpendicular to the plane of $ABC$,
and $BB'$ and $CC'$ parallel to $AA'$.
The three parallels are axes of a horosphere passing through $A$,
and the horosphere cuts $BB'$ and $CC'$ at $B''$ and $C''$ respectively.
The solid angle at $B$ bounded by $BA$, $BC$, and $BB'$ is also considered,
as is the horocycle with axis $CC'$ passing through $C$.


\addsec{The triangle}

I propose to depict $ABC$ as being flat to us,
the leg $AC$ being straight to us.
This leg then should be perpendicular the bounding plane
of a Poincar\'e half-plane.
I take that plane to be the $yz$ plane,
in the coordinate system of Figure \ref{fig:triangle},
\renewcommand{\myA}{9}%
\renewcommand{\myB}{12}%
\renewcommand{\myC}{13}%
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(-19,-6)(3,5)
%\psgrid[subgriddiv=1]
%\psset{Alpha=-70,Beta=55}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=15,yMin=-15,yMax=2,zMin=-4,zMax=4]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=b](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=lb](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=b](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=lt](\myC,0,0){\mybox c}
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDPut[pOrigin=lb](0,\myH\space neg,0){\mybox{-h}}
\pstThreeDDot(0,\myH\space neg,0)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%\parametricplotThreeD(\myB\space neg,0){\myA\space dup mul \myH\space dup mul add t \myH\space add dup mul sub sqrt t 0}%BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
\end{pspicture}
\caption{The triangle}\label{fig:triangle}
\end{figure}
and $ABC$ can be as shown.
Thus $AC$ is along the $x$ axis,
and $BC$ is an arc centered at the origin.
Thus there are positive parameters $a$, $b$, and $c$ such that
\begin{align*}
  A&=(a,0,0),&C&=(c,0,0),&B&=(\sqrt{c^2-b^2},-b,0).
\end{align*}
Also $AB$ is an arc centered at a point $H$ on the $y$ axis.
We may let
\begin{equation*}
H=(0,-h,0),
\end{equation*}
and then
\begin{gather*}
	a^2+h^2=c^2-b^2+(b-h)^2,\\
	h=\frac{c^2-a^2}{2b}.
\end{gather*}
The circle of which $AB$ is an arc is given by
\begin{gather*}
x^2+(y+h)^2=a^2+h^2,\\
x^2=a^2-2hy-y^2,
\end{gather*}
so that, as an arc, 
\begin{equation*}
AB=\left\{
\left(\sqrt{a^2+t(c^2-a^2)-t^2b^2},-bt,0\right)\colon 0\leq t\leq1\right\}.
\end{equation*}

\addsec{The horosphere}

The horosphere tangent at $A$ to $ABC$
is the real sphere of radius $a$ having center at $N$,
where
\begin{equation*}
N=(a,0,a),
\end{equation*}
as in Figure \ref{fig:horosphere}.
\begin{figure}
\centering
%\psset{unit=4.6mm}
\psset{unit=3mm}
\begin{pspicture}(-17.5,-9)(6.5,15)
%\psgrid[subgriddiv=1]
\psset{spotY=150}%
\psset{Alpha=-25,Beta=-30}
\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15](\myA,0,\myA){\myA} %horosphere
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-5,xMax=18,yMin=-15,yMax=32,zMin=0,zMax=14]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=rb](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=rb](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=t](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=lb](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=t](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=rb](\myC,0,0){\mybox c}
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%\parametricplotThreeD(\myB\space neg,0){\myA\space dup mul \myH\space dup mul add t \myH\space add dup mul sub sqrt t 0}%BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
\pstThreeDPut[pOrigin=r](0,0,\myA){\mybox I}
\pstThreeDPut[pOrigin=lb](0,0,\myA){\mybox a}
\pstThreeDDot(0,0,\myA)
\pstThreeDPut[pOrigin=r](\myA,0,\myA){\mybox N}
\pstThreeDDot(\myA,0,\myA)
\end{pspicture}
\caption{The horosphere}\label{fig:horosphere}
\end{figure}
This sphere is tangent to the $yz$ plane at a point $I$, and so 
\begin{equation*}
I=(0,0,a).
\end{equation*}

\addsec{The easy parallel}

The name of the point $I$ can be understood to stand for
``infinity.''
The parallels $AA'$, $BB'$, and $CC'$
at the vertices of $ABC$
will for us be arcs of circles
passing through $I$,
with centers on the $yz$ plane,
in planes parallel to the $x$ axis.
We shall concentrate on the arcs $AI$, $BI$, and $CI$;
if they are all parametrized similarly,
then they can be cut off uniformly at the points $A'$, $B'$, and $C'$.

The arcs $AI$ and $CI$
have centers on the $z$ axis.
The center of $AI$ is just the origin,
so the arc is as in Figure \ref{fig:easy}.
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(-19,-6)(3,10)
%\psgrid[subgriddiv=1]
%\psset{Alpha=-70,Beta=55}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=15,yMin=-15,yMax=2,zMin=-2,zMax=11]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rb](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
%\pstThreeDPut[pOrigin=t](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=lb](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=b](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=lt](\myC,0,0){\mybox c}
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%\parametricplotThreeD(\myB\space neg,0){\myA\space dup mul \myH\space dup mul add t \myH\space add dup mul sub sqrt t 0}%BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
\pstThreeDPut[pOrigin=rb](0,0,\myA){\mybox I}
\pstThreeDPut[pOrigin=lt](0,0,\myA){\mybox a}
%\pstThreeDDot(0,0,\myA)
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
\end{pspicture}
\caption{The easy parallel}\label{fig:easy}
\end{figure}

\addsec{The middle parallel}

A direction vector of the line $CI$ is $(c,0,-a)$,
and the midpoint of $CI$ is $(c/2,0,a/2)$, 
and so the perpendicular bisector of $CI$ is
\begin{gather*}
c\left(x-\frac c2\right)-a\left(z-\frac a2\right)=0,\\
	2cx-2az=c^2-a^2,
\end{gather*}
which cuts the $z$ axis at $-(c^2-a^2)/2a$.
Thus the center of the arc $CI$ is $M$, given by
\begin{equation*}
M=\left(0,0,-\frac{c^2-a^2}{2a}\right),
\end{equation*}
as in Figure \ref{fig:middle}.
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(-19,-6)(3,10.5)
%\psgrid[subgriddiv=1]
%\psset{Alpha=-70,Beta=55}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=15,yMin=-15,yMax=2,zMin=-7,zMax=11]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rb](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
%\pstThreeDPut[pOrigin=t](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=lb](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=b](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=lt](\myC,0,0){\mybox c}
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%%%% AB
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%AB
%%%% AI
\pstThreeDPut[pOrigin=rb](0,0,\myA){\mybox I}
\pstThreeDPut[pOrigin=lt](0,0,\myA){\mybox a}
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
%%%%% CI
\pstThreeDPut[pOrigin=r](0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg){\mybox M}
\pstThreeDDot(0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg)
\parametricplotThreeD(0,0.999999){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
\end{pspicture}
\caption{The middle parallel}\label{fig:middle}
\end{figure}
The radius $MI$ is then
$a+(c^2-a^2)/2a$,
%, which is $(a^2+c^2)/2a$.
so the points of the arc $CI$ are thus
\begin{equation*}
\left(\sqrt{\left(a+\frac{c^2-a^2}{2a}\right)^2
-\left(ta+\frac{c^2-a^2}{2a}\right)^2},0,ta\right),
\end{equation*}
where $0\leq t\leq1$.
The radicand simplifies to
\begin{gather*}
a^2+c^2-a^2-t^2a^2-t(c^2-a^2),\\
	c^2-t(c^2-a^2)-t^2a^2,
\end{gather*}
so that
\begin{equation*}
CI=\left\{\left(\sqrt{c^2-t(c^2-a^2)-t^2a^2},0,ta\right)\colon
0\leq t\leq1\right\}.
\end{equation*}

\addsec{The hard parallel}

The arc $BI$ has center $K$ on the line through $I$
and the foot $L$ of the perpendicular dropped from $B$ 
to the $yz$ plane,
as in Figure \ref{fig:setup}.
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(-18,-7)(3,10)
%\psgrid[subgriddiv=1]
\psset{Alpha=60,Beta=30}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=15,yMin=-15,yMax=2,zMin=-2,zMax=11]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=b](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=t](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=b](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=b](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=lt](\myC,0,0){\mybox c}
%\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
%\pstThreeDDot(0,\myH\space neg,0)
\pstThreeDPut[pOrigin=rb](0,0,\myA){\mybox I}
\pstThreeDLine(0,0,\myA)(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=lt](0,0,\myA){\mybox a}
\pstThreeDDot(0,0,\myA)
%%%% K
\pstThreeDPut[pOrigin=rb](0,
\myA\space dup mul \myC\space dup mul add 
\myB\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div,
\myC\space dup mul \myB\space dup mul 2 mul sub \myA\space dup mul sub
\myA\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div){\mybox K}
\pstThreeDDot(0,
\myA\space dup mul \myC\space dup mul add 
\myB\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div,
\myC\space dup mul \myB\space dup mul 2 mul sub \myA\space dup mul sub
\myA\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
%\parametricplotThreeD(\myB\space neg,0){\myA\space dup mul \myH\space dup mul add t \myH\space add dup mul sub sqrt t 0}%BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
\end{pspicture}
\caption{Setup for the hard parallel}\label{fig:setup}
\end{figure}
Thus
\begin{align*}
	K&=(0,-k_1,k_2),&
L&=(0,-b,0),
\end{align*}
and the line $IL$ is given by
\begin{align*}
x&=0,&ay-bz&=-ab.  
\end{align*}
To find the parameters $k_1$ and $k_2$,
we observe that a direction vector
and the midpoint of $BI$ are respectively
\begin{align*}
&(\sqrt{c^2-b^2},-b,-a),&
&\left(\frac{\sqrt{c^2-b^2}}2,-\frac b2,\frac a2\right),    
\end{align*}
so the perpendicular bisecting plane is
\begin{gather*}
  \sqrt{c^2-b^2}
\left(x-\frac{\sqrt{c^2-b^2}}2\right)
-b
\left(y+\frac b2\right)
-a\left(z-\frac a2\right)=0,\\
  2\sqrt{c^2-b^2}x
  -2by-2az=c^2-a^2.
\end{gather*}
Thus we obtain $K$ by solving simultaneously
\begin{equation*}
  \left\{
  \begin{array}{*5c}
    -ay&+&bz&=&ab\\
    -2by&-&2az&=&c^2-a^2.
  \end{array}\right.
\end{equation*}
We may use Cramer's Rule:
\begin{gather*}
  \begin{vmatrix}
    ab&b\\c^2-a^2&-2a
  \end{vmatrix}
  =-b  \begin{vmatrix}
    a&-1\\c^2-a^2&2a
  \end{vmatrix}
  =-b(a^2+c^2),\\
  \begin{vmatrix}
    -a&ab\\-2b&c^2-a^2
  \end{vmatrix}
  =a  \begin{vmatrix}
    1&b\\-2b&a^2-c^2
  \end{vmatrix}
  =a(a^2+2b^2-c^2),\\
	  \begin{vmatrix}
    -a&b\\-2b&-2a
  \end{vmatrix}
=  2\begin{vmatrix}
    a&-b\\b&a
  \end{vmatrix}
  =2(a^2+b^2),
\end{gather*}
and so 
\begin{align*}
k_1&=\frac{b(a^2+c^2)}{2(a^2+b^2)},&
k_2&=\frac{a(a^2+2b^2-c^2)}{2(a^2+b^2)}.
\end{align*}
The point $(0,-k_1,k_2)$ or $K$ lies on the \emph{ray} $IL$.
The segment $LI$ consists of the points
\begin{equation*}
\bigl(0,(t-1)b,ta\bigr),
\end{equation*}
where $0\leq t\leq 1$.
The arc from $B$ to $I$
with center at $K$
has the radius $r$,
this being the distance from $B$ to $K$
and so satisfying
\begin{align*}
r^2
&=c^2-b^2+(b-k_1)^2+k_2{}^2\\
&=c^2-2bk_1+k_1{}^2+k_2{}^2.
\end{align*}
The points of the arc $IB$ are thus
\begin{equation*}
\left(\sqrt{r^2-\bigl((1-t)b-k_1\bigr)^2-(ta-k_2)^2},(t-1)b,ta\right).
\end{equation*}
The radicand simplifies to
\begin{gather*}
	c^2-2bk_1-(1-t)^2b^2+2(1-t)bk_1-t^2a^2+2tak_2,\\
	c^2-2bk_1-b^2+2tb^2-t^2b^2+2bk_1-2tbk_1-t^2a^2+2tak_2,\\
	c^2-b^2+2t(b^2-bk_1+ak_2)-t^2(a^2+b^2),	
\end{gather*}
and here
\begin{gather*}
	2(bk_1-ak_2)
	=\frac{b^2(a^2+c^2)}{a^2+b^2}	
	-\frac{a^2(a^2+2b^2-c^2)}{a^2+b^2}\\
	=\frac{
	a^2b^2+b^2c^2
	-a^4-2a^2b^2+a^2c^2}{a^2+b^2}\\
	=\frac{a^2c^2
	+b^2c^2
	-a^4-a^2b^2}{a^2+b^2}
	=c^2-a^2,
\end{gather*}
so that
\begin{align*}
  2(b^2-bk_1+ak_2)
  &=a^2+2b^2-c^2\\
&=a^2+b^2-(c^2-b^2),
\end{align*}
and therefore the radicand is
\begin{gather*}
c^2-b^2+t\bigl(a^2+b^2-(c^2-b^2)\bigr)-t^2(a^2+b^2),\\
(1-t)\bigl((c^2-b^2)+t(a^2+b^2)\bigr).
\end{gather*}
Thus
\begin{multline*}
BI=
\left\{\left(\sqrt{(1-t)\bigl((c^2-b^2)+t(a^2+b^2)\bigr)},
(t-1)b,ta\right)\colon\right.\\
\left.\vphantom{\sqrt{\bigl(\bigr)}}
0\leq t\leq 1\right\},	
\end{multline*}
as in Figure \ref{fig:parallels}.
\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}*(-14,-3.5)(1.5,12)
%\psgrid[subgriddiv=1]
\psset{Beta=15}
\pstThreeDPut[pOrigin=rb](0,0,\myA){\mybox I}
\pstThreeDPut[pOrigin=lt](0,0,\myA){\mybox a}
%%%%%%%%%%%%%% The label of the point K
\pstThreeDPut[pOrigin=b](0,
\myA\space dup mul \myC\space dup mul add 
\myB\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div,
\myC\space dup mul \myB\space dup mul 2 mul sub \myA\space dup mul sub
\myA\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div){\mybox K}
%%%%%%%%%%%%%%%  The point K itself
\pstThreeDDot(0,
\myA\space dup mul \myC\space dup mul add 
\myB\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div,
\myC\space dup mul \myB\space dup mul 2 mul sub \myA\space dup mul sub
\myA\space mul neg 
2 div 
\myA\space dup mul \myB\space dup mul add div)%K
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-1,xMax=14,yMin=-14,yMax=1,zMin=-8,zMax=11]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rb](\myA,0,0){\mybox A}
\pstThreeDPut[pOrigin=lt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=l](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=t](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=b](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=rb](\myC,0,0){\mybox C}
\pstThreeDPut[pOrigin=lt](\myC,0,0){\mybox c}
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\pstThreeDLine(0,0,\myA)(0,\myB\space neg,0)
\pstThreeDDot(0,0,\myA)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
\parametricplotThreeD(0,0.999999){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
%%%% BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
%%%%%%%%%%%%%%%%%%% BI
\parametricplotThreeD(0,0.999999){1 t sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  t
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt
  t 1 sub \myB\space mul
  t \myA\space mul
}
\end{pspicture}
\caption{All of the parallels}\label{fig:parallels}
\end{figure}

\addsec{The easy plane}

We now want to know where the horosphere with center $I$
passing through $A$ cuts the three planes $ACI$, $CBI$, and $BAI$.
Again,
the points where the horosphere cuts the three parallels $AA'$, $BB'$, and $CC'$ are $A$, $B''$, and $C''$ respectively.
The plane of $ACI$ is in the $xz$ plane,
and the intersection of the horosphere with this plane is given by
\begin{gather*}
  (x-a)^2+(z-a)^2=a^2,\\
  x^2-2ax=-a^2+2az-z^2,
\end{gather*}
while by our earlier computations the circle of $CI$ is given by
\begin{equation*}
  	x^2=c^2-\frac{c^2-a^2}az-z^2.
\end{equation*}
To find $C''$,
we eliminate the squares, obtaining
\begin{equation*}
  2ax=a^2+c^2-\left(2a+\frac{c^2-a^2}a\right)z
  =\frac{a^2+c^2}a(a-z),
\end{equation*}
which defines (with $y=0$) a straight line through $(0,0,a)$,
as we already expect.
Plugging into the first equation for a circle,
and assuming $x\neq0$, we have
\begin{gather*}
  a^2=(x-a)^2+\frac{4a^4}{(a^2+c^2)^2}x^2\\
  =\left(1+\frac{4a^4}{(a^2+c^2)^2}\right)x^2-2ax+a^2,\\
  2a=\left(1+\frac{4a^4}{(a^2+c^2)^2}\right)x,\\
  x=\frac{2a(a^2+c^2)^2}{(a^2+c^2)^2+4a^4}.
\end{gather*}
Thus
\begin{equation*}
AC''=\left\{\left(t,0,a-\sqrt{2at-t^2}\right)\colon
a\leq t\leq\frac{2a(a^2+c^2)^2}{(a^2+c^2)^2+4a^4}\right\},
\end{equation*}
as in Figure \ref{fig:easy-plane}.

\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}*(-13,-1)(2,10.2)
%\psgrid[subgriddiv=1]
\psset{Alpha=-30,Beta=15,spotX=170,spotY=220}
\pstThreeDPut[pOrigin=l](0,0,\myA){\mybox I}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-1,xMax=14,yMin=-14,yMax=1,zMin=-2,zMax=10]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rt](\myA,0,0){\mybox A}
%\pstThreeDPut[pOrigin=rt](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=b](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=b](0,\myB\space neg,0){\mybox L}
\pstThreeDPut[pOrigin=lt](0,\myB\space neg,0){\mybox{-b}}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=rb](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\pstThreeDPut[pOrigin=rt](\myC,0,0){\mybox C}
%\pstThreeDPut[pOrigin=rt](\myC,0,0){\mybox c}
\pstThreeDDot(0,0,\myA)
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
\parametricplotThreeD(0,0.999999){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
%%%% BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
%%%%%%%%%%%%%%%%%%% BI
\parametricplotThreeD(0,0.999999){1 t sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  t
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt
  t 1 sub \myB\space mul
  t \myA\space mul
}
%%%%%%%%%%%%%%%%%%%% center N of horosphere
\pstThreeDPut[pOrigin=r](\myA,0,\myA){\mybox N}
\pstThreeDDot[drawCoor=true](\myA,0,\myA)
%%%%%%%%%%%%%%%%%%%% AC''
\parametricplotThreeD[linestyle=dashed](\myA,
\myCpp)
{t
0
\myA\space
dup 2 mul t mul t dup mul sub sqrt sub}
\parametricplotThreeD[linestyle=dashed,linewidth=0.8pt](0,\myA)
{t
0
\myA\space
dup 2 mul t mul t dup mul sub sqrt sub}
\pstThreeDPut[pOrigin=rb](\myCpp,0,
\myA\space
dup 2 mul \myCpp\space mul \myCpp\space dup mul sub sqrt sub)
{\mybox{C''}}
\end{pspicture}
\caption{The easy plane}\label{fig:easy-plane}
\end{figure}

\addsec{The middle plane}

The arc $AB''$ lies in the intersection of two spheres:
\begin{compactenum}[(1)]
\item
the horosphere,
which is the sphere with center $N$ passing through $A$,
and therefore also $I$,
given by
\begin{equation*}
(x-a)^2+y^2+(z-a)^2=a^2,
\end{equation*}
and
\item
the sphere in which the arcs $AB$ and $AI$ lie,
which is just the sphere with center $H$ passing through $A$.
\end{compactenum}
The intersection of the two spheres
is then a circle lying in the plane through $A$
normal to $HN$ and thus to $(a,h,a)$.
The plane of the circle is thus given by
\begin{equation*}
a(x-a)+hy+az=0.
\end{equation*}
Using this to eliminate $x-a$
from the equation of the horosphere,
we obtain
\begin{gather*}
	(az+hy)^2+a^2y^2+a^2(z-a)^2=a^4,\\
	2a^2z^2+2a(hy-a^2)z+(a^2+h^2)y^2=0,\\
	\begin{aligned}
		2az
		&=a^2-hy\mp\sqrt{(hy-a^2)^2-2(a^2+h^2)y^2}\\
		&=a^2-hy\mp\sqrt{a^4-2a^2hy-(2a^2+h^2)y^2}.
	\end{aligned}
\end{gather*}
Rewriting the plane equation as
\begin{equation*}
2ax=2(a^2-hy)-2az,
\end{equation*}
from the expression for $2az$ we obtain
\begin{equation*}
2ax=a^2-hy\pm\sqrt{a^4-2a^2hy-(2a^2+h^2)y^2}.
\end{equation*}
Now we know the points of the circle
in terms of their projections onto the $y$ axis.
Along the arc $AB''$,
the value of $z$ is the less of the two found.
For the range of $y$,
we need the $y$ coordinate of $B''$.
The point $B''$ itself is at the intersection of the circle
with the arc $BI$.
This arc lies in a plane with $L$,
and so this plane is given by
\begin{gather*}
-ay+bz=ab,\\
bz=ab+ay.
\end{gather*}
We use this in the earlier expression for $2az$,
which we first rewrite as
\begin{equation*}
2abz=a^2b-bhy\mp\sqrt{a^4b^2-2a^2b^2hy-(2a^2+h^2)b^2y^2}.
\end{equation*}
Thus the $y$ coordinate of $B''$ satisfies
\begin{gather*}
2a^2(b+y)=a^2b-bhy\mp\sqrt{a^4b^2-2a^2b^2hy-(2a^2+h^2)b^2y^2},\\
\bigl(a^2b+(2a^2+bh)y\bigr)^2=a^4b^2-2a^2b^2hy-(2a^2+h^2)b^2y^2,
\end{gather*}
Performing all of the multiplications gives
\begin{multline*}
a^4b^2+4a^4by+2a^2b^2hy+4a^4y^2+4a^2bhy^2+b^2h^2y^2\\
=a^4b^2-2a^2b^2hy-2a^2b^2y^2-b^2h^2y^2,
\end{multline*}
and so
\begin{gather*}
(4a^4+2a^2b^2+4a^2bh+2b^2h^2)y^2
+(4a^4b+4a^2b^2h)y=0,\\
y=-2a^2b\frac{a^2+bh}{2a^4+a^2b^2+2a^2bh+b^2h^2}
\end{gather*}
since $y\neq0$.
The expression seems to overwhelm PostScript
or else my own powers of programming it.
Since $h$ is a derived parameter,
or an dependent variable,
and $2bh=c^2-a^2$, we have
\begin{gather*}
y=-2a^2b\frac{4a^2+2(c^2-a^2)}{8a^4+4a^2b^2+4a^2(c^2-a^2)+(c^2-a^2)^2}\\
=-4a^2b\frac{a^2+c^2}{5a^4+4a^2b^2+2a^2c^2+c^4}.
\end{gather*}
This lets us obtain $AB''$ as in Figure \ref{fig:middle-plane}.
\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}*%(-14,-5)(1.5,12.5)
(-1.5,-5)(14,12.5)
%\psgrid[subgriddiv=1]
%\psset{Alpha=60,Beta=-30,spotX=140,spotY=30}
\psset{Alpha=-120,Beta=30,spotX=60,spotY=160}
%\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15,fillstyle=none](0,\myH\space neg,0){\myA\space dup mul \myH\space dup mul add sqrt} % other sphere
%\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15,fillstyle=vlines](\myA,0,\myA){\myA} %horosphere
\pstThreeDPut[pOrigin=r](0,0,\myA){\mybox I}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=20,yMin=-17,yMax=1,zMin=-7,zMax=11]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rb](\myA,0,0){\mybox A}
%\pstThreeDPut[pOrigin=lb](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=t](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=rt](0,\myB\space neg,0){\mybox L}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=rt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\pstThreeDPut[pOrigin=t](\myC,0,0){\mybox C}
%\pstThreeDPut[pOrigin=lb](\myC,0,0){\mybox c}
\pstThreeDDot(0,0,\myA)
\pstThreeDPut[pOrigin=r](\myCpp,0,
\myA\space
dup 2 mul \myCpp\space mul \myCpp\space dup mul sub sqrt sub)
{\mybox{C''}}
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
%%%%%%%%%%%%%%%% CI
\pstThreeDPut[pOrigin=r](0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg){\mybox M}
\pstThreeDDot(0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg)
\parametricplotThreeD(0,0.999999){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
%%%% BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
%%%%%%%%%%%%%%%%%%% BI
\parametricplotThreeD(0,0.999999){1 t sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  t
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt
  t 1 sub \myB\space mul
  t \myA\space mul
}
%%%%%%%%%%%%%%%%%%%% center N of horosphere
\pstThreeDPut[pOrigin=r](\myA,0,\myA){\mybox N}
\pstThreeDDot[drawCoor=true](\myA,0,\myA)
%%%%%%%%%%%%%%%%%%%% AC''
\parametricplotThreeD[linestyle=dashed](\myA,
\myCpp)
{t
0
\myA\space
dup 2 mul t mul t dup mul sub sqrt sub}
%%%%%% arc of horosphere in xz plane from A to I
\parametricplotThreeD[linestyle=dashed,linewidth=0.8pt](0,\myA)
{t 0 \myA\space dup 2 mul t mul t dup mul sub sqrt sub}
%%%%%%%%%%%%%%%%%%%%% AB''
\parametricplotThreeD[linestyle=dashed](\myBpp,0){
%%%%%%%%%%%% x
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div
%%%%%%%%%%%% y
t
%%%%%%%%%%%% z
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
}
%%%%%%%%%%%%% label for B''
\pstThreeDPut[pOrigin=lb](
%%%%%%%%%% x
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div,
%%%%%%%%%%%% y
\myBpp\space,
%%%%%%%%%%%% z
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
){\mybox{B''}}
\end{pspicture}
\caption{The middle plane}\label{fig:middle-plane}
\end{figure}

\addsec{The third plane}

Like $AC''$,
$C''B''$ is the intersection of two circles with centers in the $xz$ plane: again the horosphere, whose center is $N$ and that passes through $I$;
but now also the sphere with center $M$ that passes through $I$.
Since the $z$ coordinate of $M$ is $-(c^2-a^2)/2a$,
the plane of intersection of the spheres is
\begin{equation*}
ax+\left(a+\frac{c^2-a^2}{2a}\right)(z-a)=0
\end{equation*}
or rather
\begin{equation*}
2a^2x+(a^2+c^2)(z-a)=0.
\end{equation*}
Eliminating $x$ from the equation of the horosphere gives
\begin{gather*}
	\bigl((a^2+c^2)(z-a)+2a^3\bigr)^2
	+4a^4\bigl(y^2+(z-a)^2\bigr)=4a^6,\\
	\bigl((a^2+c^2)^2+4a^4\bigr)(z-a)^2
	+4a^3(a^2+c^2)(z-a)
	+4a^4y^2=0,
	\end{gather*}
and so
\begin{multline*}
	\frac{z-a}2\\
	=\frac{-a^3(a^2+c^2)
	\mp\sqrt{a^6(a^2+c^2)^2-a^4\bigl((a^2+c^2)^2+4a^4\bigr)y^2}}
	{(a^2+c^2)^2+4a^4}\\
		=a^2\frac{-a(a^2+c^2)
	\mp\sqrt{a^2(a^2+c^2)^2-\bigl((a^2+c^2)^2+4a^4\bigr)y^2}}
	{(a^2+c^2)^2+4a^4}\\
	=a^2\Phi,
\end{multline*}
say;
from the plane equation then,
\begin{equation*}
x=-(a^2+c^2)\Phi.
\end{equation*}
Since we already know where $B''$ itself is,
we can draw $C''B''$ as in Figure \ref{fig:third-plane}.

\begin{figure}
\centering
\psset{unit=8mm}
\begin{pspicture}*(-8.5,-5)(5,12)
%\psgrid[subgriddiv=1]
\psset{Alpha=-60,Beta=15,spotX=150,spotY=210}
%\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15,fillstyle=none](0,\myH\space neg,0){\myA\space dup mul \myH\space dup mul add sqrt} % other sphere
%\pstThreeDSphere[linewidth=0.4pt,SegmentColor={[cmyk]{0,0,0,0}},increment=15,fillstyle=vlines](\myA,0,\myA){\myA} %horosphere
\pstThreeDPut[pOrigin=lb](0,0,\myA){\mybox I}
\pstThreeDCoor[linewidth=\myaxiswidth,linestyle=\myaxisstyle,linecolor=black,IIIDticks,xMin=-2,xMax=15,yMin=-17,yMax=5,zMin=-7,zMax=12]
\pstThreeDPlaneGrid[planeGrid=xy,linewidth=0.4pt,xsubticks=13,ysubticks=13](0,-13)(13,0)
\pstThreeDPut[pOrigin=rt](\myA,0,0){\mybox A}
%\pstThreeDPut[pOrigin=lb](\myA,0,0){\mybox a}
\pstThreeDPut[pOrigin=t](\myC\space dup mul \myB\space dup mul sub sqrt,\myB\space neg,0){\mybox B}
\pstThreeDPut[pOrigin=rt](0,\myB\space neg,0){\mybox L}
\pstThreeDDot(0,\myB\space neg,0)
\pstThreeDPut[pOrigin=lt](0,\myH\space neg,0){\mybox H}
\pstThreeDDot(0,\myH\space neg,0)
\pstThreeDPut[pOrigin=rt](\myC,0,0){\mybox C}
%\pstThreeDPut[pOrigin=lb](\myC,0,0){\mybox c}
\pstThreeDDot(0,0,\myA)
\pstThreeDPut[pOrigin=rb](\myCpp,0,
\myA\space
dup 2 mul \myCpp\space mul \myCpp\space dup mul sub sqrt sub)
{\mybox{C''}}
\psset{linewidth=1.6pt}
\pstThreeDLine(\myA,0,0)(\myC,0,0)%AC
\parametricplotThreeD(\myB\space neg,0){\myC\space dup mul t dup mul sub sqrt t 0}%BC
\parametricplotThreeD(0,0.999999){\myA\space 1 t dup mul sub sqrt mul 0 t \myA\space mul}%AI
%%%%%%%%%%%%%%%% CI
\pstThreeDPut[pOrigin=r](0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg){\mybox M}
\pstThreeDDot(0,0,\myC\space dup mul \myA\space dup mul sub 2 \myA\space mul div neg)
\parametricplotThreeD(0,0.999999){\myC\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul sub 
t dup mul \myA\space dup mul mul sub sqrt 
0 
t \myA\space mul}%CI
%%%% BA
\parametricplotThreeD(0,1){\myA\space dup mul 
t \myC\space dup mul \myA\space dup mul sub mul add 
t dup mul \myB\space dup mul mul sub sqrt 
t \myB\space mul neg 
0}%BA
%%%%%%%%%%%%%%%%%%% BI
\parametricplotThreeD(0,0.999999){1 t sub % 1-t
  \myC\space dup mul \myB\space dup mul sub % c^2-b^2
  t
  \myA\space dup mul \myB\space dup mul add % a^2+b^2
  mul add mul sqrt
  t 1 sub \myB\space mul
  t \myA\space mul
}
%%%%%%%%%%%%%%%%%%%% center N of horosphere
\pstThreeDPut[pOrigin=r](\myA,0,\myA){\mybox N}
\pstThreeDDot[drawCoor=true](\myA,0,\myA)
%%%%%%%%%%%%%%%%%%%% AC''
\parametricplotThreeD[linestyle=dashed](\myA,
\myCpp)
{t
0
\myA\space
dup 2 mul t mul t dup mul sub sqrt sub}
%%%%%% arc of horosphere in xz plane from A to I
%\parametricplotThreeD[linestyle=dashed,linewidth=0.8pt](0,\myA)
%{t 0 \myA\space dup 2 mul t mul t dup mul sub sqrt sub}
%%%%%%%%%%%%%%%%%%%%% AB''
\parametricplotThreeD[linestyle=dashed](\myBpp,0){
%%%%%%%%%%%% x
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div
%%%%%%%%%%%% y
t
%%%%%%%%%%%% z
\myA\space dup mul \myH\space t mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul t mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
t dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
}
%%%%%%%%%%%%% label for B''
\pstThreeDPut[pOrigin=lb](
%%%%%%%%%% x
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
add
2 \myA\space mul div,
%%%%%%%%%%%% y
\myBpp\space,
%%%%%%%%%%%% z
\myA\space dup mul \myH\space \myBpp\space mul sub % a^2-hy
\myA\space dup mul dup mul % a^4
2 \myA\space dup mul mul \myH\space mul \myBpp\space mul % 2a^2hy
sub
2 \myA\space dup mul mul \myH\space dup mul add % 2a^2+h^2
\myBpp\space dup mul % y^2
mul sub
sqrt
sub
2 \myA\space mul div
){\mybox{B''}}
%%%%%%%%%%%%%%%%%%%% B''C''
\parametricplotThreeD[linestyle=dashed](\myBpp,0){
\myA\space dup mul \myC\space dup mul add neg \myPhi\space mul
t
2 \myA\space dup mul mul \myPhi\space mul
\myA\space add
}
%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% new horocycle %%%%
%%%%%%%%%%%%%%%%%%%%%%%%%
\pstThreeDDot(\myA\space dup mul \myC\space dup mul add
2 \myA\space mul div,0,\myA)
\pstThreeDPut[pOrigin=r](\myA\space dup mul \myC\space dup mul add
2 \myA\space mul div,0,\myA){\mybox P}
%%%%%%%%%%%%%%%% CD
\parametricplotThreeD[linestyle=dotted](\myD,0){
\myC\space
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
t dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
t
\myA\space neg
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
t dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
\myA\space add}
%%%%%%%%%%%% label D
\pstThreeDPut[pOrigin=t](
\myC\space
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
\myD\space dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul,
\myD\space,
\myA\space neg
\myA\space dup mul \myC\space dup mul add % a^2+c^2
dup dup mul
\myA\space dup mul \myC\space dup mul add % a^2+c^2
4 mul
\myD\space dup mul mul sub sqrt
add
\myA\space dup mul \myC\space dup mul add % a^2+c^2
2 mul div mul
\myA\space add){\mybox D}
\end{pspicture}
\caption{The third plane}\label{fig:third-plane}
\end{figure}
Finally, we consider a new horocycle,
still tangent to the $yz$ plane at $I$,
but now passing through $C$.
The $x$ coordinate of its center $P$ is thus given by
\begin{gather*}
x^2=(c-x)^2+a^2,\\
	2cx=a^2+c^2,
\end{gather*}
so
\begin{equation*}
P=\left(\frac{a^2+c^2}{2c},0,a\right),
\end{equation*}
and the new horosphere itself is given by
\begin{equation*}
\left(x-\frac{a^2+c^2}{2c}\right)^2+y^2+(z-a)^2
=\left(\frac{a^2+c^2}{2c}\right)^2.
\end{equation*}
This cuts $BI$ at a point $D$,
and we want to draw the arc $CD$
in the intersection of the new horosphere 
with the sphere having center $M$ 
and also passing through $I$ (and $C$).
Thus $CD$ is an arc of a circle in the plane given by
\begin{gather*}
	\frac{a^2+c^2}{2c}x+\frac{a^2+c^2}{2a}(z-a)=0,\\
	ax+cz=ac.
\end{gather*}
Eliminating now $z-a$ from the horosphere equation gives
\begin{gather*}
	\bigl(2cx-(a^2+c^2)\bigr)^2+4c^2y^2+4a^2x^2=(a^2+c^2)^2,\\
	4(a^2+c^2)x^2-4c(a^2+c^2)x+4c^2y^2=0,\\
	x=c\frac{a^2+c^2\pm\sqrt{(a^2+c^2)^2-4(a^2+c^2)y^2}}{2(a^2+c^2)}
	=c\Psi,\\
	z-a=-a\Psi.
\end{gather*}
The point $D$ lies as before in the plane $BIL$,
given by
\begin{equation*}
b(z-a)=ay,
\end{equation*}
and so the $y$ coordinate of $D$ satisfies
\begin{gather*}
	y=-b\Psi,\\
	2y=-b\mp b\frac{\sqrt{(a^2+c^2)^2-4(a^2+c^2)y^2}}{a^2+c^2},\\
	(a^2+c^2)^2(2y+b)^2=b^2\bigl((a^2+c^2)^2-4(a^2+c^2)y^2\bigr),\\
	(a^2+c^2)(a^2+b^2+c^2)y^2
	+(a^2+c^2)^2by=0,\\
	y=-b\frac{a^2+c^2}{a^2+b^2+c^2}.
\end{gather*}




\end{document}
