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\begin{document}

\subtitle{\itshape
Projektif geometri dersi}
\title{A course in\\
projective 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{September 12--18, 2016\\
Last edited January 9, 2017}
\publishers{%Nesin Mathematics Village\\
{Nesin Matematik K\"oyu}\\
\c Sirince, Sel\c cuk, \.Izmir}
%\begin{abstract}

\maketitle

\addchap{Preface}

This is about a two-week course 
called \emph{Geometriler}
at the Nesin Mathematics Village.
In the first week we read, in my Turkish translation,
some of the lemmas for Euclid's \emph{Porisms}
in Book VII of the \emph{Collection}
of Pappus.
Students presented the lemmas at the board,
and I lectured on modern developments
due to Desargues and others.
In the following week,
we read Lobachevski's
``Geometrical Researches on the Theory of Parallels''
in Halstead's English translation.
Again students presented propositions at the board;
but curved space turned out to be a more difficult concept
than points at infinity.
That week is covered only briefly here,
in the Introduction.

I had given a similar \emph{Geometriler} course
in my department at Mimar Sinan.
That course met two hours a week
for the 13 weeks of the fall semester of 2015--6.
I kept a record like the present one.
Class at the Math Village met two hours a day the first week,
an hour and a half the second,
Monday through Sunday,
except Thursday.
In the second week,
I ended on Friday,
so that I could participate in the
Thales Bulu\c smas\i\
in Milet the next day.

I thank the staff of the Village, 
notably Asl\i\ Can Korkmaz,
for having Pappus and Lobachevski printed and coil-bound
for distribution to students.
I thank all of the workers of the Village,
from Ali Nesin on down,
for making it an excellent place to continue and develop the thoughts
begun more than 2500 years ago, elsewhere in Ionia, by Thales.

%\end{abstract}

\tableofcontents
{\relscale{0.9}\listoffigures}
%\listoftables

\setcounter{chapter}{-1}
\addchap{Introduction}

\addsec{Overview}

Of Pappus's lemmas for Euclid's \emph{Porisms,}
students presented six:
VIII, IV, III, X, XI, and XII, in that order.
Lemmas VIII, XII, and XIII 
are cases of what is now known as Pappus's Theorem,
while Lemmas III, X, and XI are needed to prove XII and XIII.
We skipped Lemma XIII in class,
its proof being similar to that of XII.
Lemma IV is effectively what I shall call the Quadrangle Theorem,
although Coxeter gives it no name \cite[\textbf{14.41,} p.\ 240]{Coxeter-Geometry}.
There is a related theorem called Desargues's Involution Theorem
by Field and Gray \cite[p. 54]{MR869422};
Coxeter describes this as
``the theorem of the quadrangular set''
\cite[\textbf{14.56,} p.\ 246]{Coxeter-Geometry}.

%Coxeter was my main source for the modern treatment of these theorems.

Two students volunteered to present the first two (VIII and IV)
of Pappus's lemmas above.
For the next two lemmas (III and X),
volunteers were not forthcoming,
and so I picked two more students.
When they had fulfilled their assignments,
only two more students were still in class;
I asked them to prepare the last two lemmas (XI and XII)
for the next day.
Class met at 8 \textsc{a.m.},
a difficult time for many.
Nonetheless, 
some absent students did return the next day.

Every presenter of a lemma came more or less prepared for the job,
though sometimes needing help from the audience.
Class was mostly in Turkish,
except on the last day or two,
when only I was speaking: 
with the remaining students' permission,
I switched mostly to English.

In the following week, class was in the afternoon,
as one returning student had begged for it to be.
Most students in the second week were new.
With a couple of notable exceptions,
they did not prepare their presentations well.
Some of them left the Village early, earlier than I did,
without telling me,
and having accepted assignments for the day (Friday)
when they would be gone.
I have doubts about how well even the remaining students 
understood Lobachevski's non-Euclidean conception of parallelism:
they seemed to persist in their Euclidean notions.

On the first day of that second week, 
I reviewed the Euclidean geometry
not requiring the Fifth Postulate
that Loba\-chevski summarizes
in his Theorems 1--15.
This is the geometry of Propositions 1--28 of Book I of the \emph{Elements.}
There is also some solid geometry from Book XI,
though I did not go into this.
I do not know how much the review of Euclidean geometry meant to students who,
unlike those at Mimar Sinan, had not read Euclid in the first place.
I gave away the plot by describing the Poincar\'e half-plane model
for Lobachevskian geometry;
but given the quality of later student presentations,
I have doubts that the model made much sense.
I shall not say more about that second week,
except that, if that part of the course is repeated,
it should probably be coupled with a reading of Euclid;
and then it would need a full week, if not two.

\subsection*{Details}

One may refer to Pappus's Theorem more precisely as Pappus's Hexagon Theorem.
It is the theorem that,
if the vertices of a hexagon
lie alternately on two straight lines,
then the intersection points of the three pairs of opposite sides of the hexagon
lie on a straight line.
See for example 
%\begin{comment}
Figure \ref{fig:Pap-intro},
\begin{figure}
\centering
\psset{unit=1.2cm}
\begin{pspicture}(0,-0.4)(6,3.4)
\pspolygon(0,0)(3,3)(6,0)(0,3)(2,0)(6,3)
\uput[d](0,0){$A$}
\uput[u](3,3){$B$}
\uput[d](6,0){$C$}
\uput[u](0,3){$D$}
\uput[d](2,0){$E$}
\uput[u](6,3){$F$}
\uput[l](1.2,1.2){$G$}
\uput[r](! 2 16 7 div add 12 7 div){$H$}
\uput[u](3,1.5){$K$}
\psline[linestyle=dotted](1.2,1.2)(! 2 16 7 div add 12 7 div)
\psset{linestyle=dashed}
\psline(0,0)(6,0)
\psline(0,3)(6,3)
\end{pspicture}
\caption{Pappus's Hexagon Theorem}\label{fig:Pap-intro}
\end{figure}
which is the same as 
%\end{comment}
Figure \ref{fig:Pap} on page \pageref{fig:Pap}.
Pappus's theorem remains true if some of the opposite sides of the hexagon are parallel,
provided one allows that parallel lines meet on the ``line at infinity'';
See Figure \ref{fig:VIII-2} on page \pageref{fig:VIII-2}
and Figure \ref{fig:Pap-proj} on page \pageref{fig:Pap-proj}.
In any case,
the two straight lines holding the six vertices between them 
can be considered as a degenerate conic section.
Pascal generalized to an arbitrary conic section,
though without proof \cite{Pascal-Conics,Pascal-Coniques}.

The Quadrangle Theorem
is that if a straight line intersects six straight lines,
each of which passes through two of four given points, 
no three of the four being collinear,
then five of the intersection points determine the sixth,
regardless of the choice of the four given points.
Thus in 
%\begin{comment}
Figure \ref{fig:4c-intro},
\begin{figure}
\centering
\psset{unit=5mm}
%\subfloat[Straight line $HK$ is drawn]{\label{fig:4c}
\begin{pspicture}(0,-1)(16,11)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](10,2.5){$D$}
\uput[d](8,2){$C$}
\uput[d](12,3){$E$}
\psdots(0,0)(5.714,1.429)(10,2.5)(8,2)(12,3)
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
\uput[l](7,4){$H$}
\uput[ur](10,4){$K$}
\uput[d](16,4){$L$}
\psline(0,0)(10,10)(5.714,1.429)
\psline(10,10)(10,2.5)
\psline(12,3)(6,6)(8,2)
\psline(7,4)(16,4)%(17,4)
\psline(0,0)(16,4)%(17,4.25)
\psdots[dotstyle=o](10,10)(6,6)(7,4)(10,4)(16,4)
\end{pspicture}
\caption{The Quadrangle Theorem}\label{fig:4c-intro}
\end{figure}
which is the same as 
%\end{comment}
Figure \ref{fig:4c} on page \pageref{fig:4c},
the points $A$, $B$, $C$, $D$, and $E$ on the same straight line
determine the point $L$ on that line,
provided it is understood that $L$ is found by first picking a point $F$ not on the line $AB$,
then picking a point $G$ on the straight line $AF$,
letting $GC$ and $GE$ intersect $FB$ and $FD$ respectively at $H$ and $K$,
and letting $HK$ intersect the original line $AB$.

In class,
after we saw Pappus's treatment of the Quadrangle and Hexagon Theorems,
I sketched a proof of all cases of the latter by passing to a third dimension and projecting.
I gave another proof by introducing projective coordinates.

Desargues's Theorem is that if the straight lines through corresponding vertices of two triangles
intersect at one point,
then the intersection points of the corresponding sides of the triangles lie on one straight line.
For example, if the triangles are $ABC$ and $DEF$ as in Figure \ref{fig:Des-intro},
\begin{figure}
\centering
\psset{xunit=5.0mm,yunit=5.0mm,dotstyle=o,dotsize=5pt}
\begin{pspicture}(-1.00,-1)(13.00,12)
\psline[linestyle=dotted](12,12)(12,0) % Kappa to Lambda
\psline(0,12)(3,0)   % Eta to Delta
\psline(0,12)(6,0)   % Eta to Zeta
\psline(0,12)(7.5,3) % Eta to Epsilon
\psline(12,6)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Theta to Alpha
\psline(12,6)(3,0)                         % Theta to Delta
\psline(12,12)(! 12 7 19 div mul 
              12 24 7 19 div mul sub) % Kappa to Gamma
\psline(12,12)(6,0)                   % Kappa to Zeta 
\psline(12,0)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Lambda to Alpha
\psline(12,0)(3,0)                         % Lambda to Delta
\psdot(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
\uput[dl](! 12 7 7 36 add div mul
         12 48 7 7 36 add div mul sub){$A$}
\psdot(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
\uput[u](! 12 14 mul 14 14.5 add div 
      12 14.5 14 mul 14 14.5 add div sub){$B$} 
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput[dl](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0)(7.5,3)(6,0) % Delta, Epsilon, Zeta
\uput[d](3,0){$D$}
\uput[dr](7.5,3){$E$}
\uput[d](6,0){$F$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,6)(12,12)(12,0) % Theta, Kappa, Lambda
\uput[r](12,6){$H$}
\uput[r](12,12){$K$}
\uput[dr](12,0){$L$}
\end{pspicture}
\caption{Desargues's Theorem}\label{fig:Des-intro}
\end{figure}
which is the same as Figure \ref{fig:Des} on page \pageref{fig:Des},
then $HKL$ is straight.
I gave the proof attributed to Hessenberg from 1905,
using three applications of Pappus's Theorem,
as sketched in Figure \ref{fig:Des-2} on page \pageref{fig:Des-2}.
Strictly, this proof assumes that all points named in the diagram are distinct;
in the other case, Cronheim gave in 1953
\cite{MR0053531}
the argument sketched in Figure \ref{fig:deg},
\begin{figure}
  \centering
\psset{unit=1.7cm}
  \begin{pspicture}(-2,-2.7)(4,6)
%\psgrid
    \pspolygon[linewidth=1.6pt](0,0)(4,0)(0,4)
\pspolygon[linewidth=1.6pt](1,3)(0.4,0)(0,1)
\psline(0,0)(1,3)
\psline(0,1)(4,0)
\psline(0.4,0)(0,4)
\psline(1,3)(-0.5,0)(0,0)
\psline(1,3)(0,-2)(0,0)
\psline(0.4,0)(-2,6)(0,4)
\uput[ur](1,3){$A$}
\uput[dr](0.4,0){$B$}
\uput{2pt}[210](0,1){$C$}
\uput[dl](0,0){$D$}
\uput[ur](0,4){$E$}
\uput[d](4,0){$F$}
\uput{1pt}[35](0.31,0.92){$G$}
\uput[l](0,-2){$H$}
\uput[l](-2,6){$K$}
\uput[l](-0.5,0){$L$}
\psset{linestyle=dashed}
\psline(0,4)(-0.8,-2.4)(0,0)
\uput[l](-0.8,-2.4){$S$}
\psline(0,1)(! 48 49 div -3 mul 8 div 1 12 49 div 3 mul 8 div add)
\uput[l](! 48 49 div -3 mul 8 div 1 12 49 div 3 mul 8 div add){$R$}
\psset{linestyle=dotted}
\psline(4,0)(-0.8,-2.4)
\psline(0,0)(-2,6)
\psline(0,-2)(-2,6)
  \end{pspicture}
  \caption{Degenerate case of Desargues's Theorem}\label{fig:deg}
  
\end{figure}
where,
by applying Pappus's Theorem to hexagons 
$GCELBA$, $GAELBC$, and $SRDCAF$ in turn,
we have that $FHS$, $DKR$ and finally $LKH$ are straight.

In class,
using Desargue's Theorem and its dual, which is its converse,
I proved again the Quadrangle Theorem. 

Pappus's proofs rely heavily on the proposition known in modern times,
in some countries, as Thales's Theorem \cite{Thales-Theorem}:
a straight line cutting two sides of a triangle cuts them proportionally
if and only if it is parallel to the base.
This is ``equation'' \eqref{eqn:::} on page \pageref{eqn:::} with respect to Figure \ref{fig:VI.2}.
Euclid proves this theorem at the beginning of Book VI of the \emph{Elements,}
using the theory of proportion developed in Book V.
This theory relies on the so-called Archimedean Axiom,
that of two line segments, either can be multiplied so as to exceed the other.
In fact one does not need this axiom,
but one can develop a theory of proportion,
sufficient for establishing Thales's Theorem,
on the basis of Book I of the \emph{Elements.}
One can take Thales's Theorem as a \emph{definition} of proportion,
except that one should fix one of the base angles of the triangle.
But then one can prove, as I did in class,
that the base angles do not matter.
A neat way of proving this is Desargues's Theorem,
applied to Figure \ref{fig:prop} on page \pageref{fig:prop}.
In short, Thales proves Pappus, which proves Desargues, which proves Thales.

In modern axiomatic projective plane geometry,
the theorems of Pappus and Desargues are not equivalent.
In class we proved, 
not exactly their equivalence with Thales's Theorem,
but simply their truth in the geometry of Book I of Euclid's \emph{Elements.} 
They are true in the projective plane over a commutative field;
but the ancient proofs use not just the algebra of fields,
but the geometry of \emph{areas.}
In Lemma VIII, for example,
which is the case of the Hexagon Theorem
when two pairs of opposite sides are parallel,
Pappus's proof relies on adding and subtracting triangles
that are equal because they are on the same base
and between the same parallels.

Following first Descartes \cite{Descartes-Geometrie} 
and then Hilbert \cite{MR0116216},
we can obtain a field from Euclid's geometry.
This may not be the best way to think about that geometry.

\addsec{Origins}

The origin of this course 
is my interest in the origins of mathematics.
This interest goes back at least 
to a tenth-grade geometry class in 1980--1,
where we students were taught to write proofs in the two-column,
statement--reason format.
I did not much care for our textbook, which,
for an example of congruence,
used a photo of a machine stamping out foil trays for TV dinners \cite[p.\ 13]{Weeks-Adkins}.
In ``Commensurability and Symmetry''
\cite{Pierce-com-sym},
I mention how the Weeks--Adkins text confuses equality with sameness.
A geometrical equation like $AB=CD$ 
means not that the segments $AB$ and $CD$ are the same,
but that their \emph{lengths} are the same.
Length is an abstraction from a segment,
as ratio is an abstraction from two segments.
This is why Euclid uses ``equal'' to describe two equal segments,
but ``same'' to describe the ratios of segments in a proportion.
One can maintain the distinction symbolically\label{prop-not}
by writing a proportion as $A:B::C:D$,
rather than as $A/B=C/D$.
I noticed the distinction many years after high school;
but even in tenth grade I thought we should read Euclid.
I went on to read him, along with Homer and Aeschylus and Plato and others,
at St John's College \cite{Pierce-SJC}.

In 2008,
the first course I taught at the Nesin Mathematics Village
was an opportunity to review something I had read at St John's.
Called
``Conic Sections \emph{\`a la} Apollonius of Perga,''
my course reviewed the propositions of Book I of the \emph{Conics}
\cite{MR0053831,MR1660991}
that pertained to the parabola.
I shall say more about this later;
for now I shall note that,
while the course was great for me,
I don't think it meant much for the students who just sat and watched me at the board.
One has to engage with the mathematics for oneself,
especially when it is something so unusual for today as Apollonius.
A good way to do this 
is to have to go to the board and present the mathematics,
as at St John's.

In 2010 at METU in Ankara,
I taught the course called History of Mathematical Concepts
in the manner of St John's.
We studied Euclid, Apollonius, and (briefly) Archimedes
in the first semester;
Al-Khw\=arizm\={\i}, Th\=abit ibn Qurra, Omar Khayy\'am, Cardano, Vi\`ete, Descartes, and Newton in the second
\cite{Pierce-log}.

At METU I loved the content of the course called Fundamentals of Mathematics,
which was required of all first-year students.
I even wrote a text for the course,
a rigorous text that might overwhelm students,
but whose contents I thought at least teachers should know.
In the end I didn't think it was right to try to teach equivalence-relations and proofs to beginning students,
independently from a course of real mathematics.
Moving to Mimar Sinan in 2011,
I was able (in collaboration)
to develop a course in which first-year students 
read and presented the proofs 
that taught mathematics to practically all mathematicians 
until about a century ago.
Among other things,
students would learn the non-trivial
(because non-identical) 
equivalence-relation of congruence.
I did not actually recognize this opportunity
until I had seen the way students tended to confuse 
equality of line-segments with sameness.

Our first-semester Euclid course
is followed by an analytic geometry course.
Pondering the transition from the one course to the other
led to some of the ideas about ratio and proportion
that are worked out in the present course.
My study of Pappus's Theorem arose in this context,
and I was disappointed to find that the \emph{Wikipedia} article called
``Pappus's Hexagon Theorem'' did not provide a precise reference to its namesake.
I rectified this condition on May 13, 2013,
when I added to the article a section called ``Origins,''
giving Pappus's proof.

In order to track down that proof,
I had relied on Heath,
who in \emph{A History of Greek Mathematics}
summarizes most of Pappus's lemmas for Euclid's lost \emph{Porisms}
\cite[p.\ 419--24]{MR654680}.
In this summary,
Heath \emph{may} give the serial numbers of the lemmas as such:
these are the numbers given above as Roman numerals.
Heath \emph{always} gives the numbers of the lemmas 
as propositions within Book VII of Pappus's \emph{Collection,}
according to the enumeration of Hultsch \cite{Pappus}.
Apparently this enumeration was made originally
in the 16th century by Commandino 
in his Latin translation
\cite[pp.\ 62--3, 77]{Jones}.%%%%%
\footnote{Searching for Commandino's name in Jones's book
reveals an interesting tidbit on page 4:
Book III of the \emph{Collection}
is addressed to an otherwise-unknown teacher of mathematics
called Pandrosion.
She must be a woman,
since she is given the feminine form of the 
adjective
\gr{kr'atistos, -h, -on} (``most excellent''); but
``in Commandino's Latin translation her name vanishes, 
leaving the absurdity of the polite epithet
\gr{krat'isth} being treated as a name, `Cratiste'; 
while for no good reason Hultsch alters the text to make the name masculine.''}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

According to Heath,
Pappus's Lemmas XII, XIII, XV, and XVII for the \emph{Porisms,}
or
Propositions 138, 139, 141, and 143 of Book VII,
establish the Hexagon Theorem.
The latter two propositions
can be considered as converses of the former two,
which consider the hexagon lying respectively 
between parallel and intersecting straight lines.

In \emph{Mathematical Thought from Ancient to Modern Times,}
Kline cites only Proposition 139 
as giving Pappus's Theorem
\cite[p.\ 128]{MR0472307}.
This proposition, Lemma XIII,
follows from Lemmas III and X,
as XII follows from XI and X.
For Pappus's Theorem in the most general sense,
one should cite also Proposition 134, Lemma VIII,
which, as we have noted, 
is the case
where two pairs of opposite sides of the hexagon are parallel:
the conclusion is then that the third pair are also parallel.
Heath's summary does not seem to mention this lemma at all.
The omission must be a simple oversight.

%Lemma IV is effectively the Quadrangle Theorem.

In the catalogue of my home department at Mimar Sinan,
there is an elective course called Geometries,
meeting two hours a week.
The course had last been taught in the fall of 2010
when I offered it in the fall of 2015.
For use in the course,
I translated the first 19 of Pappus's 38 lemmas for Euclid's \emph{Porisms.}
I had not thought there was an English version;
but at the end of my work,
I found Jones's.
This helped me to parse a few confusing words.
What I found first, on \emph{Library Genesis,}
was the first volume of Jones's work \cite{Jones};
Professor Jones himself supplied me with Volume II, 
the one with the commentary and diagrams \cite{Jones-2}.

Book VII of Pappus's \emph{Collection}
is an account of the so-called
\emph{Treasury of Analysis} (\gr{>analu'omenos t'opos}).
This \emph{Treasury} 
consisted of works by Euclid, Apollonius, Aristaeus, and Eratosthenes,
most of them now lost.
As a reminder of the wealth of knowledge that is no longer ours,
I ultimately wrote out,
on the back of my Pappus translation,
a table of the contents of the \emph{Treasury.}
Pappus's list of the contents is included by Thomas
in his
\emph{Selections Illustrating the History of Greek Mathematics}
in the Loeb series
\cite[pp.\ 598--601]{MR13:419b}.
Thomas's anthology includes more selections from Pappus's \emph{Collection,}
but none involving the Hexagon Theorem.
He does provide Proposition 130 of Book VII,
that is, Lemma IV for the \emph{Porisms} of Euclid,
the lemma that I am calling the Quadrangle Theorem.

\addsec{Additions}

Here are some further developments of topics of the course.

\subsection*{Involutions}

Lemma IV is that if the points $ABCDEL$ 
in Figure \ref{fig:4c-intro} above
(or in Figure \ref{fig:4c}
on page \pageref{fig:4c})
satisfy the proportion
\begin{equation}\label{eqn:Pap-IV}
  AL\cdot BC:AB\cdot CL::AL\cdot DE:AD\cdot EL,
\end{equation}
then $HKL$ is straight.
Chasles
\cite[p.\ 102]{Chasles}
 rewrites the condition in the form
\begin{equation*}
  BC\cdot AD\cdot EL=AB\cdot DE\cdot CL.
\end{equation*}
We can also write
\begin{equation*}
BC\cdot AD:AB\cdot DE::CL:EL;
\end{equation*}
this makes it easier to see that,
when $ABCDE$ are given,
then some unique $L$ exists so as to satisfy \eqref{eqn:Pap-IV}.
As $L$ varies, the ratio $CL:EL$
takes on all possible values but unity.
If $BC\cdot AD=AB\cdot DE$,
that is,
\begin{equation*}
AB:BC::AD:DE,
\end{equation*}
then $HK\parallel AE$ by Lemma I.
Otherwise,
by Lemma IV, $HK$ must pass through the $L$ that satisfies \eqref{eqn:Pap-IV}.
Thus the converse of the lemma holds as well.
We might speculate whether this converse was one of Euclid's original porisms.
Chasles seems to think it was.

Thomas observes \cite[pp.\ 612--3]{MR13:419b} that 
\begin{quote}
[the converse of Lemma IV]
is one of the ways of expressing the proposition enunciated by Desargues:
\emph{The three pairs of opposite sides of a complete quadrilateral
are cut by any transversal in three pairs of conjugate points of an involution.}
\end{quote}
Following Coxeter, I would call the ``complete quadrilateral'' here a 
\emph{complete quadrangle}
(see Figure \ref{fig:complete}, page \pageref{fig:complete}).
Desargues proves the
\textbf{Involution Theorem}
in his
\emph{Rough Draft of an Essay on the results of taking plane sections of a cone}
\cite[p. 54]{MR869422}.
One way to interpret the theorem is to observe that, 
in Figure \ref{fig:4c-intro} or \ref{fig:4c},
if the points $BCDE$ are conceived of as fixed,
then $A$ determines $L$.
Moreover, this operation \emph{transposes} $A$ and $L$,
and so it is an \emph{involution} of the straight line $BE$.

Desargues proceeds towards the Involution Theorem
by first observing that if seven points $ABCDFGH$ are arranged,
as in Figure \ref{fig:Des-invo},
\begin{figure}
\centering
\psset{unit=5cm}
  \begin{pspicture}(0,-0.1)(2,0)
    \psline(0,0)(2,0)
\psdots(0.5,0)(0.63,0)(0.80,0)(1.26,0)(1.59,0)(2,0)
\psdots[dotstyle=o](0,0)(1,0)
\uput[d](0,0){$A$}
\uput[d](0.5,0){$B$}
\uput[d](0.63,0){$C$}
\uput[d](0.80,0){$D$}
%\uput[d](1,0){$E$}
\uput[d](1.26,0){$F$}
\uput[d](1.59,0){$G$}
\uput[d](2,0){$H$}
  \end{pspicture}
  \caption{Points in involution}\label{fig:Des-invo}
  
\end{figure}
on a straight line so that
\begin{equation}\label{eqn:3}
AB\cdot AH=AC\cdot AG
%AC\cdot AG
=AD\cdot AF,
\end{equation}
then, without reference to $A$, we have
\cite[p.\ 48]{MR869422}
\begin{equation}\label{eqn:Des-invo}
DG\cdot FG:CD\cdot CF
::BG\cdot GH:BC\cdot CH.
\end{equation}
Indeed,
by second equation in \eqref{eqn:3},
\begin{equation}\label{eqn:AGAFADAC}
AG:AF::AD:AC,
\end{equation}
so by addition or subtraction
of the terms of the first ratio to or from the terms of the second,
\begin{equation}\label{eqn:AGAF}
AG:AF::DG:CF.
\end{equation}
Similarly,
after alternating \eqref{eqn:AGAFADAC},
we obtain
\begin{gather*}
  AF:AC::AG:AD,\\
AF:AC::FG:CD.
\end{gather*}
Composing the latter with \eqref{eqn:AGAF} yields
\begin{equation}\label{eqn:AGAC}
AG:AC::DG\cdot FG:CD\cdot CF.
\end{equation}
Replacing $D$ with $B$ and $F$ with $H$
in the second equation of \eqref{eqn:3}
yields the first equation.
Hence we can do the same in \eqref{eqn:AGAC},
%we can interchange $BH$ and $DF$,
obtaining
\begin{equation*}
AG:AC::BG\cdot GH:BC\cdot CH.
\end{equation*}
Eliminating the common ratio from the last two proportions yields
\eqref{eqn:Des-invo}.
For meeting this condition,
the three pairs $BH$, $CG$ and $DF$ of points
are said to be in \textbf{involution,}
by Desargues's definition.


Now suppose these pairs consist of those points 
where a transversal cuts the pairs of opposite sides 
of a complete quadrangle,
as in Thomas's description.
For example, the points $BCDFGH$ could be,
respectively, the points $ABCDEL$ in Figure \ref{fig:4c-intro} or \ref{fig:4c},
or \gr{ABGDEZ} in Figure \ref{fig:IV} on page \pageref{fig:IV}.
Then the converse of Pappus's Lemma IV,
expressed in \eqref{eqn:Pap-IV}, gives us now
\begin{equation}\label{eqn:IV-Des}
  BH\cdot CD:BC\cdot DH::BH\cdot FG:BF\cdot GH.
\end{equation}
This proportion is equivalent to \eqref{eqn:Des-invo}.
However, the best way to show this may not be obvious.
One approach is to introduce the \textbf{cross-ratio,}
though apparently Desargues does not do this \cite[p.\ 52]{MR869422}.
Despite the misgivings expressed above (page \pageref{prop-not}),
we turn to modern notation for ratios.
If $ABCD$ are points on a straight line,
we let
\begin{equation}\label{eqn:cross}
  \frac{AB\cdot CD}{AD\cdot CB}=(ABCD)
\end{equation}
by definition.
Note the pattern of repeated letters on the left.
We could use a different pattern;
we just have to be consistent.
We take line segments to be directed,
so that $CB=-BC$.
Then \eqref{eqn:IV-Des} is equivalent to
\begin{equation}\label{eqn:IV-cross}
  (BHDC)=(BHGF),
\end{equation}
while, since \eqref{eqn:Des-invo} is, in modern notation,
\begin{equation*}
  \frac{DG\cdot FG}{CD\cdot CF}=\frac{BG\cdot GH}{BC\cdot CH},
\end{equation*}
we obtain from this
\begin{equation*}
  \frac{DG\cdot BC}{CD\cdot BG}=\frac{CF\cdot GH}{FG\cdot CH},
\end{equation*}
that is,
\begin{equation}\label{eqn:DGBC}
  (DGBC)=(CFGH).
\end{equation}
But \eqref{eqn:IV-cross}, obtained from Pappus, 
must still hold 
if we permute the pairs $BH$, $CG$, and $DF$.
Sending each pair to the next (and the last to the first),
we obtain from
\eqref{eqn:IV-cross} the equivalent equation
\begin{equation}\label{eqn:CGDB}
  (CGBD)=(CGFH).
\end{equation}
We shall have that this is equivalent to \eqref{eqn:DGBC},
once we understand how cross-ratios are affected
by permutations of their entries.

The cross-ratios that can be formed from $ABCD$
are permuted transitively by a group of order $24$.
Then there are at most $6$ different cross-ratios,
since
\begin{gather*}
  (CDAB)=(ABCD),\\
(BADC)=(ABCD).
\end{gather*}
Moreover, 
from \eqref{eqn:cross} we can read off
\begin{equation*}
  (ADCB)=\frac1{(ABCD)},
\end{equation*}
while
\begin{align*}
  (ACBD)
=\frac{AC\cdot BD}{AD\cdot BC}
&=\frac{(AB+BC)\cdot(BC+CD)}{AD\cdot BC}\\
&=\frac{AC\cdot BC+BC\cdot CD}{AD\cdot BC}-(ABCD)\\
&=1-(ABCD).
\end{align*}
The involutions $x\mapsto 1/x$ and $x\mapsto 1-x$ 
of the set of ratios generate a group of order $6$.
(Here we either exclude the ratios $0$ and $1$,
or allow them along with $\infty$.)
Now we have accounted for all permutations of points.
We have
\begin{align*}
  (CGBD)=(CGFH)
&\iff(CBGD)=(CFGH)\\
&\iff(BCDG)=(CFGH)\\
&\iff(DGBC)=(CFGH),
\end{align*}
that is, \eqref{eqn:CGDB} and \eqref{eqn:DGBC} are equivalent,
as desired.

The involution of the straight line $BE$
in Figure \ref{fig:4c-intro} or \ref{fig:4c}
that transposes $A$ and $L$
will transpose $B$ and $E$,
and also $C$ and $D$.
Following Coxeter \cite[\textbf{14.56}, p.\ 246]{Coxeter-Geometry},
we obtain this transposition as in Figure \ref{fig:invo},
\begin{figure}
\centering
\psset{unit=2.5mm}
\begin{pspicture}(-1,-1)(36,11)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
%\uput[d](10,2.5){$D$}
%\uput[d](8,2){$C$}
\uput[d](12,3){$E$}
\psdots(0,0)(5.714,1.429)(12,3)%(10,2.5)(8,2)
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
%\uput[l](7,4){$H$}
%\uput[ur](10,4){$K$}
\uput[d](16,4){$L$}
\psline(0,0)(10,10)(5.714,1.429)
%\psline(10,10)(10,2.5)
\psline(12,3)(6,6)%(8,2)
%\psline(7,4)(16,4)
\psline(0,0)(16,4)
\psdots(10,10)(6,6)(16,4)%(7,4)(10,4)
\end{pspicture}

\begin{pspicture}(-1,-1)(36,11.7)
%\psgrid
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](12,3){$E$}
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
\uput[d](16,4){$L$}
\uput[105](7.82,5.64){$M$}
\uput[-85](7.6,5.2){$N$}
\psline(10,10)(5.714,1.429)
\psline(0,0)(16,4)
\psset{linestyle=dotted}
\psline(0,0)(10,10)
\psline(12,3)(6,6)(16,4)
\psline(6,6)(5.714,1.429)
\psdots(0,0)(5.714,1.429)(12,3)(10,10)(6,6)(16,4)(7.82,5.64)(7.6,5.2)
\uput[r](22,4){$ABEL\overset G{\underset{\wedge}=}FBNM$}
\end{pspicture}

\begin{pspicture}(-1,-2)(36,11.7)
\uput[d](5.714,1.429){$B$}
\uput[d](12,3){$E$}
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
\uput[d](16,4){$L$}
\uput[105](7.82,5.64){$M$}
\uput[-85](7.6,5.2){$N$}
\uput[d](22,-2){$P$}
\psline(5.714,1.429)(10,10)
\psline(6,6)(22,-2)
\psset{linestyle=dotted}
\psline(10,10)(22,-2)
\psline(5.714,1.429)(16,4)(6,6)
\psline(16,4)(7.6,5.2)
\psdots(5.714,1.429)(12,3)(10,10)(6,6)(16,4)(7.82,5.64)(7.6,5.2)(22,-2)
\uput[r](22,4){$FBNM\underset{\wedge}{\overset L=}PENG$}
\end{pspicture}

\begin{pspicture}(-1,-3.7)(36,11.7)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](12,3){$E$}
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
\uput[d](16,4){$L$}
\uput[-85](7.6,5.2){$N$}
\uput[d](22,-2){$P$}
\psline(0,0)(16,4)
\psline(22,-2)(6,6)
\psset{linestyle=dotted}
\psline(10,10)(22,-2)
\psline(10,10)(0,0)
\psline(10,10)(5.714,1.429)
\psline(10,10)(12,3)
\psdots(0,0)(5.714,1.429)(12,3)(10,10)(6,6)(16,4)(7.6,5.2)(22,-2)
\uput[r](22,4){$PENG\underset{\wedge}{\overset F=}LEBA$}
\end{pspicture}

\caption{An involution}\label{fig:invo}
\end{figure}
as a composition of three projections of one straight line onto another.
In Coxeter's notation,
$ABEL\overset G{\underset{\wedge}=}FBNM$
means $AGF$, 
%$BGF$, 
$EGN$, and $LGM$ are all straight
(and implicitly $ABEL$ and $FBNM$ are straight).
It follows from Lemma IV that this transformation is an involution
and is uniquely determined by the pairs $AL$ and $BE$.

\subsection*{Locus problems}

Thomas's anthology \cite[600--3]{MR13:419b}
includes the account by Pappus 
of five- and six-line locus problems
that Descartes quotes in the \emph{Geometry}
\cite[pp.\ 18--21]{Descartes-Geometry}.
Pappus suggests no solution to such problems;
but later in the \emph{Geometry,} 
Descartes solves a special case of the five-line problem,
where four of the straight lines---%
say $\ell_0$, $\ell_1$, $\ell_2$, and $\ell_3$---%
are parallel to one another,
each a distance $a$ from the previous,
while the fifth line---$\ell_4$---is perpendicular to them.
What is the locus of points such that the product of their distances
to $\ell_0$, $\ell_1$, and $\ell_3$
is equal to the product of $a$
with the distances to $\ell_2$ and $\ell_4$?
One can write down an \emph{equation} for the locus,
and Descartes does.
Effectively letting the $x$- and $y$-axes be $\ell_2$ and $\ell_4$,
the positive direction of the latter being from $\ell_3$ towards $\ell_0$,
Descartes obtains
\begin{equation*}
y^3-2ay^2-a^2y+2a^3=axy.
\end{equation*}
This may allow us to plot points on the desired locus,
as in Figure \ref{fig:5-line};
\begin{figure}
\centering
\begin{pspicture}*(-4,-2.7)(1.5,3)
\psline(-2,-2)(-2,3)
\psline(-1,-2)(-1,3)
\psline(0,-2)(0,3)
\psline(1,-2)(1,3)
\psline(-3,0)(1.5,0)
\uput[d](-2,-2){$\ell_0$}
\uput[d](-1,-2){$\ell_1$}
\uput[d](-0,-2){$\ell_2$}
\uput[d]( 1,-2){$\ell_3$}
\uput[l](-3, 0){$\ell_4$}
\psplot{-3}{-0.25}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\psplot{0.25}{1.5}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\end{pspicture}
\caption{A five-line locus}\label{fig:5-line}
\end{figure}
but we could already do that.
The equation is thus not a solution to the locus problem,
since it does not tell us what the locus \emph{is.}
But Descartes shows that the locus is traced
by the intersection of a moving parabola
with a straight line passing through one fixed point
and one point that moves with the parabola.
See Figure \ref{fig:5-line-sol}.
\begin{figure}
\centering
\begin{pspicture}*(-4,-2.7)(2,3)
\psline(-2,-2)(-2,3)
\psline(-1,-2)(-1,3)
\psline(0,-2)(0,3)
\psline(1,-2)(1,3)
\psline(-3,0)(2,0)
\uput[d](-2,-2){$\ell_0$}
\uput[d](-1,-2){$\ell_1$}
\uput[d](-0,-2){$\ell_2$}
\uput[d]( 1,-2){$\ell_3$}
\uput[l](-3, 0){$\ell_4$}
\psplot{-3}{2}{x dup mul 0.5 add}
\psline(-3,-0.75)(2,3)
\psdots[dotstyle=o](-2,0)(0,1.5)
\psset{linestyle=dotted}
\psplot{-3}{-0.25}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\psplot{0.25}{2}{x dup dup dup mul exch 2 mul add 1 sub exch 2 exch div sub}
\end{pspicture}
\caption{Solution of the five-line locus problem}\label{fig:5-line-sol}
\end{figure}
The parabola
has axis sliding along $\ell_2$,
and $a$ is its \emph{latus rectum.}%%%%
\footnote{See my article ``Abscissas and Ordinates'' \cite{MR3312989}
for more than you ever imagined wanting to know about the term \emph{latus rectum.}}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
The straight line passes through the intersection of $\ell_0$ and $\ell_4$
and through the point on the axis of the parabola whose distance from the vertex is $a$.


Descartes's solution is apparently one that Pappus would recognize as such.
Thus Descartes's algebraic methods would seem to represent an advance,
and not just a different way of doing mathematics.

\subsection*{Ancients and moderns}

%A theme of the course is the difference between ancient and modern mathematics.
%I have taken up that theme in mentioning Descartes.
%Ancient mathematics might be called diagrammatic, while 
Modern mathematics is \emph{literal}
in the sense of relying on the letters that might appear in a diagram,
rather than the diagram itself.
The diagram is an integral part of ancient proofs.
Pappus may put points of a diagram in a list,
without giving all of their relations.
For example, the enunciation of Lemma I of our text is,
\begin{quote}
\gr{>'Estw katagraf`h}\hfill Let the diagram be\\
\gr{<h ABGDEZH,}\hfill \gr{ABGDEZH},\\
\gr{ka`i >'estw}\hfill and let it be that\\
\gr{<ws <h AZ pr`os t`hn ZH,}\hfill as \gr{AZ} is to \gr{ZH},\\
\gr{o<'utws <h AD pr`os t`hn DG,}\hfill so is \gr{AD} to \gr{DG};\\
\gr{ka`i >epeze'uqjw <h JK;}\hfill and let \gr{JK} have been joined;\\{}
\gr{<'oti}\hfill [I say] that\\
\gr{par'allhl'os >estin <h JK t~h| AG.}\hfill \gr{JK} is parallel to \gr{AG}.
\end{quote}
Reconstruct the diagram from that!
One needs to know, first of all, 
that \gr{AZHDG} are on a straight line.
\gr{ABE} are also on a straight line,
and one needs to know that \gr L is to be on this line
when \gr{ZL} is constructed parallel to \gr{BD}.
We can infer the locations of \gr J and \gr K
from Pappus's assertion that,
as \gr{EB} is to \gr{BL},
so are \gr{EK} to \gr{KZ}
and \gr{EJ} to \gr{JH};
but still we must understand that each ratio involves collinear points.
In his diagrams,
Hultsch gives five possible configurations
meeting these conditions;
my translation gives a sixth,
as in Figure \ref{fig:Lemma-I},
\begin{figure}
\centering
\psset{unit=5mm}
\begin{pspicture}(0,-1)(12,9)
\pspolygon[linewidth=2pt](4,2)(8,4)(10,8)(4,8)
\psline[linewidth=2pt](0,0)(12,0)
\psline(0,0)(4,2)
\psline(4,0)(4,2)
\psline(2,0)(10,8)
\psline(6,0)(8,4)
\psline(4,8)(12,0)
\psset{linestyle=dotted}
\psline(6,0)(12,6)
\psline(12,0)(12,6)(8,4)
\uput[d](0,0){\gr A}
\uput[ul](4,2){\gr B}
\uput[d](4,0){\gr G}
\uput[d](2,0){\gr D}
\uput[u](8,4){\gr E}
\uput[d](6,0){\gr Z}
\uput[d](12,0){\gr H}
\uput[ul](4,8){\gr J}
\uput[ur](10,8){\gr K}
\uput[ur](12,6){\gr L}
\end{pspicture}
\caption{A diagram for Lemma I of Pappus}\label{fig:Lemma-I}
\end{figure}
where \gr{BEKJ} is emboldened
to indicate that the lemma is a version of the Quadrangle Theorem 
where one of the six straight lines 
through the vertices of the quadrangle 
is parallel to a straight line cutting the others.


Netz observes that Greek mathematics \emph{never}
simply declares what letters in a diagram stand for
\cite[pp.\ 24--5]{MR1683176}:
\begin{quote}
Nowhere in Greek mathematics do we find a moment of specification \emph{per se,}
a moment whose purpose is to make sure that the attribution of letters in the diagram is fixed.
\end{quote}
It may well be Descartes, he says,
who first fixes such attributions.
I note an early passage in \emph{La G\'eom\'etrie}
\cite[p.\ 2]{Descartes-Geometrie}:
\begin{quote}
Mais souvent on n'a pas besoin de tracer ainsi ces lignes sur le papier,
et il suffit de les d\'esigner par quelques lettres,
chacune par une seule.
Comme pour ajouter la ligne BD \`a GH,
je nomme l'une $a$ et l'autre $b$,
et \'ecris $a+b$; et $a-b$ pour soustraire $b$ de $a$;
et $ab$ pour les multiplier l'une par l'autre\lips
\end{quote}
We have already seen how Descartes's abbreviations
make solutions to ancient unsolved problems possible.

Nonetheless, 
it it possible to take cleverness with notation too far.
The third proof of Pappus's Theorem given in the course
is taken from Coxeter \cite[p.\ 236]{Coxeter-Geometry},
and it may appeal to modern Cartesian sensibilities;
but it gives less sense of \emph{why} the theorem is true
than the second proof, 
which involves passing to a third dimension and 
\emph{projecting,} starting from the diagram of Lemma VIII.
Coxeter's diagram for Pappus's Theorem is labelled as in Figure \ref{fig:Pap-Cox};
\begin{figure}
\centering
\psset{unit=1cm}
\begin{pspicture}(0,-0.4)(6,3.4)
\pspolygon(0,0)(3,3)(6,0)(0,3)(2,0)(6,3)
\uput[d](0,0){$A_2$}
\uput[u](3,3){$B_1$}
\uput[d](6,0){$C_2$}
\uput[u](0,3){$A_1$}
\uput[d](2,0){$B_2$}
\uput[u](6,3){$C_1$}
\uput[l](1.2,1.2){$C_3$}
\uput[r](! 2 16 7 div add 12 7 div){$A_3$}
\uput[u](3,1.5){$B_3$}
\psline[linestyle=dotted](1.2,1.2)(! 2 16 7 div add 12 7 div)
\psset{linestyle=dashed}
\psline(0,0)(6,0)
\psline(0,3)(6,3)
\end{pspicture}
\caption{Pappus's Theorem in modern notation}\label{fig:Pap-Cox}
\end{figure}
it allows us to observe, 
once for all,
that $A_iB_jC_k$ is straight
whenever $\{i,j,k\}=\{1,2,3\}$.
If one is going to be this clever,
it seems to me,
one might as well go all the way 
and let the indices be $0$, $1$, and $2$,
so that one can use for the set of them the notation $3$.

I do not know whether Pappus (or Euclid) 
recognized a single theorem lying behind Lemmas VIII, XII, and XIII.
He may have recognized a similarity between the lemmas,
but not a satisfactory way to prove the lemmas once for all.
He may have thought it important to treat each case individually.

There is a point of view that blurs the distinctions between the three conic sections;
but it not necessarily the best point of view.
When Omar Khayy\=am used conic sections to solve what we should call cubic equations,
he had to consider several cases,
depending on what we should call the signs of the coefficients of the equations
\cite[pp.\ 556--60]{Katz}.
For example, for the case where 
``a cube and sides are equal to squares and numbers,''
we can write the problem as the equation
\begin{equation*}
x^3+b^2x=cx^2+b^2d,
\end{equation*}
which we manipulate into
\begin{equation*}
\frac{x^2}{b^2}=\frac{d-x}{x-c}.
\end{equation*}
We can define $y$ in terms of a solution so that
\begin{equation*}
\frac xb=\frac y{x-c}=\frac{d-x}y.
\end{equation*}
Thus we solve the original equation
by finding the intersection of the two conics given by
\begin{align*}
	x^2-cx&=by,&
  y^2+(x-c)\cdot(x-d)&=0.
\end{align*}
These are normally a parabola and a circle;
however, if we have allowed negative coefficients,
then we may have had to let $b$ be imaginary.
This matters if we want to construct real solutions.

In Rule Four of the posthumously published 
\emph{Rules for the Direction of the Mind}
\cite[373, p.\ 17]{Descartes-Eng},
Descartes writes of a method that is
\begin{quote}
so useful\lips that without it the pursuit of learning would,
I think, be more harmful than profitable.
Hence I can readily believe that the great minds of the past
were to some extent aware of it,
guided to it even by nature alone\lips
This is our experience in the simplest of sciences,
arithmetic and geometry:
we are well aware that the geometers of antiquity
employed a sort of analysis which they went on to apply
to the solution of every problem,
though they begrudged revealing it to posterity.
At the present time a sort of arithmetic called ``algebra''
is flourishing, and this is achieving for numbers
what the ancients did for figures.
\end{quote}
We have already observed that Book VII of Pappus's \emph{Collection}
concerned the \emph{Treasury of Analysis.}
The term ``treasury'' is a modern flourish;
but our word ``analysis'' 
is a transliteration of the Greek of writers like Pappus.
It means freeing up, or dissolving.
As Pappus describes it,
mathematical analysis is assuming what you are trying to find,
so that you can work backwards to see how to get there.
We do this today by giving what we want to find a name $x$.

Today we think of conic sections as having \emph{axes}:
one for the parabola, and to each for the ellipse and hyperbola.
The notion comes from Apollonius;
but for him, an axis is just a special case of a \emph{diameter.}
A diameter of a conic section bisects the chords of the section
that are parallel to a certain straight line.
This straight line is the tangent drawn
at a point where the diameter meets the section.
Apollonius shows that \emph{every} straight line through the center of an ellipse or hyperbola is a diameter in this sense;
and every straight line parallel to the axis is a diameter of a parabola.
As far as I can tell,
it is difficult to show this
by the method's of Descartes's analytic geometry.
Like Euclid's and Pappus's proofs,
Apollonius's proofs rely on \emph{areas.}
There are areas of scalene triangles
and non-rectangular parallelograms.
Earlier I discussed a locus problem in terms of
\emph{distances} from several given straight lines.
For Pappus, what is involved
is not distances as such,
but the lengths of segments
drawn to the given lines at given angles,
which are not necessarily right angles.
The apparently greater generality is trivial.
This is why Descartes can solve a five-line problem using algebra.
But if one is going to prove that
a straight line parallel to the axis of a parabola
is a diameter, one cannot just treat all angles as right.
Apollonius may have had a secret weapon in coming up with his propositions about conic sections;
but I don't think it was Cartesian analysis.
Such at least was my impression
when going through Apollonius for my course in \c Sirince, 
in 2008.

\addchap[Monday]{Monday, September 12, 2016}

\addsec[Quadrangle Theorem: statement]{Statement of the Quadrangle Theorem}

Suppose five points, $A$ through $E$, 
fall on a straight line as in Figure \ref{fig:4a},
\begin{figure}[b!]
\centering
\psset{unit=3.2mm}% was 2.9mm
\subfloat[Point $F$ is chosen]{\label{fig:4a}
\begin{pspicture}(0,-1.5)(12,11)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](10,2.5){$D$}
\uput[d](8,2){$C$}
\uput[d](12,3){$E$}
\psdots(0,0)(5.714,1.429)(10,2.5)(8,2)(12,3)
\uput[u](10,10){$F$}
\psline(0,0)(10,10)(5.714,1.429)
\psline(10,10)(10,2.5)
\psline(0,0)(12,3)%(17,4.25)
\psdots[dotstyle=o](10,10)
\end{pspicture}
}
\qquad
\subfloat[Point $G$ is chosen]{\label{fig:4b}
\begin{pspicture}(0,-1.5)(12,11)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](10,2.5){$D$}
\uput[d](8,2){$C$}
\uput[d](12,3){$E$}
\psdots(0,0)(5.714,1.429)(10,2.5)(8,2)(12,3)
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
%\uput[l](7,4){$H$}
%\uput[ur](10,4){$K$}
\psline(0,0)(10,10)(5.714,1.429)
\psline(10,10)(10,2.5)
\psline(12,3)(6,6)(8,2)
\psline(0,0)(12,3)%(17,4.25)
\psdots[dotstyle=o](10,10)(6,6)%(7,4)(10,4)
\end{pspicture}
}
\caption{The Quadrangle Theorem set up}
\end{figure}
and $F$ is a random point not on the straight line.
Join $FA$, $FB$, and $FD$.
Now let $G$ be a random point on $FA$,
as in Figure \ref{fig:4b},
and join $GC$ and $GE$.
Supposing these two straight lines cross $FB$ and $FD$ at $H$ and $K$ respectively,
join $HK$ as in Figure \ref{fig:4c}.
\begin{figure}
\centering
\psset{unit=5mm}
%\subfloat[Straight line $HK$ is drawn]{\label{fig:4c}
\begin{pspicture}(0,-1)(16,11)
\uput[d](0,0){$A$}
\uput[d](5.714,1.429){$B$}
\uput[d](10,2.5){$D$}
\uput[d](8,2){$C$}
\uput[d](12,3){$E$}
\psdots(0,0)(5.714,1.429)(10,2.5)(8,2)(12,3)
\uput[u](10,10){$F$}
\uput[ul](6,6){$G$}
\uput[l](7,4){$H$}
\uput[ur](10,4){$K$}
\uput[d](16,4){$L$}
\psline(0,0)(10,10)(5.714,1.429)
\psline(10,10)(10,2.5)
\psline(12,3)(6,6)(8,2)
\psline(7,4)(16,4)%(17,4)
\psline(0,0)(16,4)%(17,4.25)
\psdots[dotstyle=o](10,10)(6,6)(7,4)(10,4)(16,4)
\end{pspicture}
\caption{The Quadrangle Theorem}\label{fig:4c}
\end{figure}
If this straight line crosses the original straight line $AB$ at $L$,
then $L$ depends only on the original five points,
not on $F$ or $G$.
Let us call this result the \textbf{Quadrangle Theorem.}%
\index{complete!--- quadrangle}%
\index{Theorem!Quadrangle ---}
It is about how the straight line $AB$
crosses the six straight lines
that pass through pairs of the four points $F$, $G$, $H$, and $K$.
Any such collection of four points,
no three of which are collinear,
together with the six straight lines that they determine,
as in Figure \ref{fig:quadrangle},
\begin{figure}
\mbox{}\hfill
\subfloat[A complete quadrangle]{\label{fig:quadrangle}
\psset{yunit=4.5mm,xunit=6mm,dotstyle=o}
\begin{pspicture}(2,4)(10,12)
\pspolygon(9,6)(6,12)(3,6)(6,4)
\psline(3,6)(9,6)
\psline(6,4)(6,12)
\psdots(9,6)(6,12)(3,6)(6,4)
\end{pspicture}}
\hfill
\subfloat[A complete quadrilateral]{\label{fig:quadrilateral}
\psset{unit=3mm,dotstyle=o}
\begin{pspicture}(-2,0)(14,12)
\psline(9,6)(0,0)(6,12)(12,0)(3,6)
\psdots(9,6)(0,0)(6,12)(12,0)(3,6)(6,4)
\end{pspicture}}
\hfill\mbox{}
\caption{``Complete'' figures}\label{fig:complete}
\end{figure}
is called a \textbf{complete quadrangle}\label{quadrangle} 
(\emph{tam d\"ortgen}).%
\index{complete!--- quadrangle}
Similarly,
any collection of four straight lines,
no three passing through the same point,
together with the six points at the intersections of pairs of these six straight lines,
as in Figure \ref{fig:quadrilateral},
is a \textbf{complete quadrilateral}
(\emph{tam d\"ortkenar}).

As stated,
the Quadrangle Theorem
is a consequence of Lemma IV
of our text of Pappus.%
\label{IV}%
\index{Lemmas of Pappus!4@IV}
Lemmas I, II, V, VI, and VII treat other cases,
such as when $HK$ in Figure \ref{fig:4c} is parallel to $AE$.

\addsec[Desargues's Theorem: statement]{Statement of Desargues's Theorem}

There is an alternative proof of the Quadrangle Theorem
based on Desargues's Theorem.
Desargues was a contemporary of Descartes,
and the theorem named for him is that,
if the straight lines through corresponding vertices of two triangles
intersect at a common point,
as $AD$, $BE$, and $CF$ meet at $H$
in Figure \ref{fig:Des},
\begin{figure}
\centering
\psset{xunit=5.0mm,yunit=5.0mm,dotstyle=o,dotsize=5pt}
\begin{pspicture}(-1.00,-1)(13.00,12)
\psline[linestyle=dotted](12,12)(12,0) % Kappa to Lambda
\psline(0,12)(3,0)   % Eta to Delta
\psline(0,12)(6,0)   % Eta to Zeta
\psline(0,12)(7.5,3) % Eta to Epsilon
\psline(12,6)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Theta to Alpha
\psline(12,6)(3,0)                         % Theta to Delta
\psline(12,12)(! 12 7 19 div mul 
              12 24 7 19 div mul sub) % Kappa to Gamma
\psline(12,12)(6,0)                   % Kappa to Zeta 
\psline(12,0)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Lambda to Alpha
\psline(12,0)(3,0)                         % Lambda to Delta
\psdot(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
\uput[dl](! 12 7 7 36 add div mul
         12 48 7 7 36 add div mul sub){$A$}
\psdot(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
\uput[u](! 12 14 mul 14 14.5 add div 
      12 14.5 14 mul 14 14.5 add div sub){$B$} 
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput[dl](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0)(7.5,3)(6,0) % Delta, Epsilon, Zeta
\uput[d](3,0){$D$}
\uput[dr](7.5,3){$E$}
\uput[d](6,0){$F$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,6)(12,12)(12,0) % Theta, Kappa, Lambda
\uput[r](12,6){$H$}
\uput[r](12,12){$K$}
\uput[dr](12,0){$L$}
\end{pspicture}
\caption{Desargues's Theorem}\label{fig:Des}
\end{figure}
then the intersection points of corresponding sides of the triangles
lie in a single straight line.
If two pairs of corresponding sides are parallel,
this means the third pair must be parallel as well.

\addsec[Pappus's Theorem: statement]{Statement of Pappus's Theorem}

We shall prove Desargues's Theorem
by means of \textbf{Pappus's Hexagon Theorem.}%
\index{Theorem!Pappus's Hexagon ---}
This is the theorem that if the vertices of a hexagon
lie alternately on two straight lines,
as do the vertices of $ABCDEF$ in Figure \ref{fig:Pap},
\begin{figure}
\centering
\psset{unit=1.2cm}
\begin{pspicture}(0,-0.4)(6,3.4)
\pspolygon(0,0)(3,3)(6,0)(0,3)(2,0)(6,3)
\uput[d](0,0){$A$}
\uput[u](3,3){$B$}
\uput[d](6,0){$C$}
\uput[u](0,3){$D$}
\uput[d](2,0){$E$}
\uput[u](6,3){$F$}
\uput[l](1.2,1.2){$G$}
\uput[r](! 2 16 7 div add 12 7 div){$H$}
\uput[u](3,1.5){$K$}
\psline[linestyle=dotted](1.2,1.2)(! 2 16 7 div add 12 7 div)
\psset{linestyle=dashed}
\psline(0,0)(6,0)
\psline(0,3)(6,3)
\end{pspicture}
\caption{Pappus's Hexagon Theorem}\label{fig:Pap}
\end{figure}
then the points of intersection,
as $G$, $H$, and $K$,
of the three pairs of opposite sides of the hexagon
also lie on a straight line.
Again there are other cases,
as when two of the pairs of opposite sides are parallel;
here the third pair must be parallel as well.

Pappus's Theorem is Lemmas VIII, XII and XIII in our text.
In the proofs of XII and XIII,
Lemmas III, X, and XI are used.
Lemma VIII needs only the theorem of Euclid
(namely I.37 and 39 of the \emph{Elements} \cite{MR17:814b})
that when two triangles have the same base,
as in Figure \ref{fig:tri},
\begin{figure}
  \centering
  \begin{pspicture}(0,-0.5)(5,2.5)
    \pspolygon(0,0)(3,0)(0,2)
\psline(0,0)(5,2)(3,0)
\psline[linestyle=dotted](0,2)(5,2)
\uput[d](0,0){$A$}
\uput[d](3,0){$B$}
\uput[u](0,2){$C$}
\uput[u](5,2){$D$}
  \end{pspicture}
  \caption{Triangles on the same base}\label{fig:tri}
  
\end{figure}
then the triangles are equal to one another
just in case
the straight line joining the apices of the triangles is parallel to their common base.

\addsec{Equality and sameness}

Equality of triangles means not congruence of the triangles,
but sameness of their areas.
In fact Euclid's proof that \emph{parallelograms}
on the same base and in the same parallels are equal
is by cutting the parallelograms into congruent pieces,
as in Figure \ref{fig:parallelograms},
\begin{figure}
\centering
\psset{unit=8mm}
\begin{pspicture}(-1,-0.625)(8,5.625)
\pspolygon(0,0)(4,0)(8,5)(-1,5)
\psline(-0.8,4)(0,5)
\psline(0,0)(4,5)
\psline(4,0)(3,5)
\psline(7.2,4)(7,5)
\psline[linestyle=dotted](-1,5)(4,0)(4,5)
\uput[d](0,0){$A$}
\uput[d](4,0){$B$}
\uput[u](-1,5){$C$}
\uput[u](3,5){$D$}
\uput[u](4,5){$E$}
\uput[u](8,5){$F$}
\uput[l](3.2,4){$G$}
\uput[l](-0.8,4){$H$}
\uput[u](0,5){$K$}
\uput[dr](7.2,4){$L$}
\uput[u](7,5){$M$}
\end{pspicture}
\caption{Equal parallelograms}\label{fig:parallelograms}
\end{figure}
where parallelograms $ABDC$ and $ABFE$ are equal,
because, when $CH=DG$ and $FL=EG$,
while $CK$ and $FM$ are both equal to $DE$,
then triangles $CHK$ and $MLF$ are congruent to one another
(and to $DGE$),
and the pentagons $AGDKH$ and $BLMEG$ are congruent to one another.
Being half of equal parallelograms,
triangles $ABC$ and $ABE$
are now equal to one another.

\addsec[Thales's Theorem: statement]{Statement of Thales's Theorem}

After Lemma VIII,
Pappus's proofs of other cases of the Hexa\-gon Theorem
use what is known as \textbf{Thales's Theorem.}
This is that
if the straight line $DE$ cuts the sides of triangle $ABC$ as in Figure \ref{fig:VI.2},
\begin{figure}
\centering\psset{unit=1.5cm}
\begin{pspicture}(0,-0.33)(3,2.33)
\pspolygon(0,0)(3,0)(1,2)
\psline(0.6,1.2)(1.8,1.2)
\uput[u](1,2){$A$}
\uput[d](0,0){$B$}
\uput[d](3,0){$C$}
\uput[ul](0.6,1.2){$D$}
\uput[ur](1.8,1.2){$E$}
\psline[linestyle=dotted](0.6,1.2)(1.8,0)
\uput[d](1.8,0){$F$}
\end{pspicture}
\caption{Thales's Theorem}\label{fig:VI.2}
\end{figure}
then
\begin{equation}\label{eqn:::}
DE\parallel BC\iff AD:DB::AE:EC.
\end{equation}
We obtain the alternative formulation
\begin{equation*}
DE\parallel BC\iff AD:AB::DE:BC
\end{equation*}
by drawing $DF$ parallel to $AC$
and using a rule like
\begin{equation*}
a:b::c:d\iff a+b:b::c+d:d
\end{equation*}

The proportion $AD:DB::AE:EC$
is an expression not of the \emph{equality} of the ratios $AD:DB$
and $AE:EC$,
but of their \emph{sameness}.
An equation of bounded straight lines,
like $CH=DG$ earlier,
says that the \emph{lengths} of the straight lines are the same.
But the length of a line cannot be drawn in a diagram;
it is an abstraction from the line itself.
Equality of bounded straight lines is an equivalence relation,
and for the sake of having a formal definition, 
we can understand the length of $CH$ 
as the corresponding equivalence class,
consisting of all of the straight lines
that are equal to $CH$.
Similarly, the ratio of two straight lines is abstract
and cannot be drawn in a diagram.
We shall discuss proportions later today
and more formally on Wednesday.

\addsec{Thales himself}

There is little evidence
that Thales knew, in full generality,
the theorem named for him.
I have learned this while preparing for the Thales Meeting
to be held on Saturday, September 24,
in Thales's home town of Miletus.
Thales supposedly measured the heights of the Pyramids
by considering their shadows;
but he may have done this just when his own shadow
was as long as a person is tall,
since in this case the height of the pyramid
would be the same as the length of its own shadow
(as measured from the center of the base of course).

Thales may have recognized that two triangles are congruent
if they have two angles equal respectively to two angles
and the common sides equal.
(This is the so-called Angle-Side-Angle or ASA Theorem.)
According to Proclus,
who wrote a commentary on the Book I of Euclid's \emph{Elements}
\cite{MR1200456},
Thales also knew the following three theorems found in that book:
\begin{compactenum}[1)]
\item
the diameter of a circle divides the circle into two equal parts,
\item
vertical angles formed by intersecting straight lines are equal to one another;
\item
the base angles of an isosceles triangle are equal to one another.
\end{compactenum}
According to Diogenes La\"ertius \cite[i.24--5]{Diogenes-I}
(who wrote biographies of philosophers,
starting with Thales),
Thales also knew that
\begin{compactenum}[1)]\setcounter{enumi}3
\item
the angle inscribed in a semicircle is right.
\end{compactenum}
Dates of activity for the persons we have named are roughly as in Figure \ref{fig:dates}.
\begin{figure}\centering
\begin{tabular}{ll}
585 \textsc{b.c.e.}&Thales\\
300     &Euclid\\
250 \textsc{c.e.}    &Diogenes La\"ertius\\
340     &Pappus\\
450     &Proclus
\end{tabular}
\caption{Dates of some ancient writers and thinkers}\label{fig:dates}
\end{figure}
All four of the listed theorems can be understood to be true by 
\emph{symmetry.}
For example,
the equation
\begin{equation*}
\angle ABC=\angle CBA
\end{equation*}
basically establishes the equality of vertical angles.
Also, 
suppose we complete the diagram of an angle 
inscribed in a semicircle as in Figure \ref{fig:circ}.
\begin{figure}
\centering
\begin{pspicture}(-3,-3)(3,3)
\pscircle(0,0){2.5}
\psline(-2.5,0)(2.5,0)
\psline(0.7,2.4)(-0.7,-2.4)
\pspolygon(2.5,0)(0.7,2.4)(-2.5,0)(-0.7,-2.4)
\uput[ul](0,0){$A$}
\uput[r](2.5,0){$B$}
\uput[u](0.7,2.4){$C$}
\uput[l](-2.5,0){$D$}
\uput[d](-0.7,-2.4){$E$}
\end{pspicture}
\caption{The angle in a semicircle}\label{fig:circ}
\end{figure}
Here the quadrilateral $BCDE$ has four equal angles.
If it follows that those angles must be right,
then the theorem of the semicircle is proved.

Those four equal angles are right in Euclidean geometry.
Here, by Euclid's fifth postulate,
if the angles at $DCB$ and $CBE$ 
are together less then two right angles,
then $CD$ and $BE$ must intersect when extended.
In that case, for the same reason,
they intersect when extended in the other direction;
but this would be absurd.

\addsec{Non-Euclidean planes}

In Lobachevski's geometry,
Euclid's fifth postulate is denied.
In this case, 
if $AB$ is perpendicular to $AC$
as in Figure \ref{fig:hyp},
\begin{figure}
\centering
\psset{xunit=2cm}
\begin{pspicture}(-2,-0.5)(2,2.72)
\psplot{-1}{2}{2.718 x neg exp}
\psplot{-2}{1}{2.718 x exp}
\psline(-2,0)(2,0)
\psline(0,0)(0,1)
%\psdots[dotstyle=o](0,1)(0,0)
\uput[d](0,0){$A$}
\uput[u](0,1){$B$}
\uput[d](1,0){$C$}
\uput[u](! 1 1 2.718 div){$D$}
\uput[u](! -1 1 2.718 div){$E$}
\end{pspicture}
\caption{Straight lines in the hyperbolic plane}\label{fig:hyp}
\end{figure}
there will be an angle $ABD$,
less than a right angle,
such that $BD$ never intersects $AC$,
no matter how far extended,
though straight lines through $B$
making with $BA$ an angle less than $ABD$
will meet $AC$.
If $EBA=ABD$, then $BE$ too will never meet $AC$,
though $BE$ and $BD$ are different straight lines.

Lobachevski's geometry
may be understood as taking place in the \textbf{hyperbolic plane.}
We can understand Pappus's geometry
as taking place in the \textbf{projective plane.}
Here the several cases of (for example)
the Hexagon Theorem
can be given a single expression,
because formerly parallel straight lines
are now allowed to intersect ``at infinity.''
We obtain the projective plane from the Euclidean plane by adding,
\begin{compactenum}[1)]
\item
for each family of parallel straight lines,
a new point common to all of them, namely their
\textbf{point at infinity,} and
\item
a new straight line, the
\textbf{straight line at infinity,}
which consists precisely of all of the points at infinity.
\end{compactenum}
In the projective plane,
\begin{compactenum}[1)]
\item
any two distinct points lie on a single straight line,
as in Euclidean geometry; but now also
\item
any two distinct straight lines intersect at a single point
(which could be at infinity).
\end{compactenum}

\addsec{Proportions}

We said that Pappus's proof of the Hexagon Theorem
used Thales's Theorem.
How can we prove Thales's Theorem?
We first have to define proportions.
One approach is to say that,
by definition, in Figure \ref{fig:prop},
\begin{figure}
\centering
\psset{unit=7mm}
\begin{pspicture}(0,-0.625)(13,5.625)
\pspolygon(0,0)(10,0)(10,5)
\pspolygon(6,0)(6,3)(7.8,2.4)
\psline(10,0)(13,4)(10,5)
\psline[linestyle=dotted](0,0)(13,4)
\uput[d](0,0){$A$}
\uput[d](6,0){$B$}
\uput[u](6,3){$C$}
\uput[d](10,0){$D$}
\uput[u](10,5){$E$}
\uput[dr](7.8,2.4){$F$}
\uput[dr](13,4){$G$}
\end{pspicture}
\caption{For a definition of proportion}\label{fig:prop}
\end{figure}
where angles $ABC$ and $ADE$ are right,
we have
\begin{equation*}
AB:BC::AD:DE.
\end{equation*}
But then, if $BF=BC$ and $DG=DE$,
and $BF$ is parallel to $DG$,
we want to show that $AFG$ is a straight line.
This is a consequence of Desargues's Theorem.
Indeed, in place of the assumption that $DG=DE$,
assume that $AFG$ is straight.
Since $BC\parallel DE$ and $BF\parallel DG$,
it follows by Desargues that $CF\parallel EG$.
In that case, since in triangle $BCF$
the angles at $C$ and $F$ are equal to one another,
the same is true of the angles at $E$ and $G$
in triangle $DEG$.
(This uses the consequence of Euclid's fifth postulate 
that parallel straight lines
make equal angles with the same straight line.)
As a result, $DE=DG$.
Thus, if this equation is true by construction,
then $AFG$ must be straight.

In the end, we shall have the implications in Figure \ref{fig:imp},
\begin{figure}
\begin{equation*}
\xymatrix{&\text{Desargues}\ar@{=>}[dd]\\
\text{Pappus}\ar@{=>}[ur]&\\
&\text{Thales}\ar@{=>}[ul]}
\end{equation*}
\caption{A circle of implications}\label{fig:imp}
\end{figure}
at least with the assumption of 
a certain part of Book I of Euclid's 
\emph{Elements}.

In fact the theorems of Desargues, Pappus, and Thales
are all simply true in the full geometry of Book I of Euclid's 
\emph{Elements}.
We shall show this for Thales's Theorem on Wednesday.

\addchap[Tuesday]{Tuesday, September 13, 2016}%\label{ch:Tues}

\addsec[Pappus's Theorem: parallel case]{The parallel case of Pappus's Theorem}

We can understand \textbf{Pappus's Theorem}
(or the Hexagon Theorem)
as having the six cases depicted in Figure \ref{fig:Pappus-cases}.
\begin{figure}
\centering
\psset{unit=1.2cm}
\begin{pspicture}(-0.5,-0.5)(3.5,3.5)
\psline(0,0)(0,3)
\psline(3,0)(3,3)
\pspolygon(0,0)(3,0)(0,1)(3,3)(0,3)(3,2)
\uput[l](0,0){$A$}
\uput[r](3,0){$B$}
\uput[l](0,1){$C$}
\uput[r](3,3){$D$}
\uput[l](0,3){$E$}
\uput[r](3,2){$F$}
\end{pspicture}\hfill
\begin{pspicture}(-0.5,-0.5)(4.5,3.5)
\uput[dr](0,3.5){Lemma VIII}
\psline(0,0)(0,3)
\psline(4,0)(3,3)
\pspolygon(0,0)(4,0)(0,1.2)(3,3)(0,3)(3.33,2)
\uput[l](0,0){$A$}
\uput[r](4,0){$B$}
\uput[l](0,1.2){$C$}
\uput[r](3,3){$D$}
\uput[l](0,3){$E$}
\uput[r](3.33,2){$F$}
\end{pspicture}

\begin{pspicture}(-0.5,-0.5)(3.5,3.5)
\psline(0,0)(0,3)
\psline(3,0)(3,3)
\pspolygon(0,0)(3,0)(0,1)(3,3)(0,3)(3,1)
\uput[l](0,0){$A$}
\uput[r](3,0){$B$}
\uput[l](0,1){$C$}
\uput[r](3,3){$D$}
\uput[l](0,3){$E$}
\uput[r](3,1){$F$}
\end{pspicture}\hfill
\begin{pspicture}(-0.5,-0.5)(4.5,3.5)
\psline(0,0)(0,3)
\psline(4,0)(3,3)
\pspolygon(0,0)(4,0)(0,1.2)(3,3)(0,3)(3.67,1)
\uput[l](0,0){$A$}
\uput[r](4,0){$B$}
\uput[l](0,1.2){$C$}
\uput[r](3,3){$D$}
\uput[l](0,3){$E$}
\uput[r](3.67,1){$F$}
\end{pspicture}

\begin{pspicture}(-0.5,-0.5)(3.5,3.5)
\uput[dr](0,3.5){Lemma XII}
\psline(0,0)(0,2.5)
\psline(3,0)(3,3)
\pspolygon(0,0)(3,0)(0,1)(3,3)(0,2.5)(3,1)
\uput[l](0,0){$A$}
\uput[r](3,0){$B$}
\uput[l](0,1){$C$}
\uput[r](3,3){$D$}
\uput[l](0,2.5){$E$}
\uput[r](3,1){$F$}
\end{pspicture}\hfill
\begin{pspicture}(-0.5,-0.5)(4.5,3.5)
\uput[dr](0,3.5){Lemma XIII}
\psline(0,0)(0,2.5)
\psline(4,0)(3,3)
\pspolygon(0,0)(4,0)(0,1.2)(3,3)(0,2.5)(3.67,1)
\uput[l](0,0){$A$}
\uput[r](4,0){$B$}
\uput[l](0,1.2){$C$}
\uput[r](3,3){$D$}
\uput[l](0,2.5){$E$}
\uput[r](3.67,1){$F$}
\end{pspicture}
\caption{Cases of Pappus's Theorem}\label{fig:Pappus-cases}
\end{figure}
In each case, the vertices of hexagon $ABCDEF$
lie alternately on two straight lines;
but these may be parallel or not,
and of the pairs $(AB,DE)$ and $(BC,EF)$
of opposite sides of the hexagon,
two, one, or none may be parallel.

Pappus's \textbf{Lemma VIII}
is the case where the two straight lines intersect,
and the two pairs of opposite sides of the hexa\-gon are parallel.
In particular,
letting the hexagon be \gr{BGHEDZ} in Figure \ref{fig:VIII},
\begin{figure}
\centering
\subfloat[Lemma VIII itself]{\label{fig:VIII}
\psset{unit=1.5cm}
\begin{pspicture}(4,4)
\pspolygon(0,0)(4,0)(3,3)(3,1)(1,1)(0,4)
\psset{linestyle=dotted}
\psline(0,0)(3,1)
\psline(4,0)(1,1)
\psline(0,4)(3,3)
\psset{linewidth=2pt,linestyle=solid}
\psline(0,0)(3,3)
\psline(4,0)(0,4)
\uput[d](2,2){\gr A}
\uput[l](0,0){\gr B}
\uput[r](4,0){\gr G}
\uput[160](1,1){\gr D}
\uput[20](3,1){\gr E}
\uput[l](0,4){\gr Z}
\uput[ur](3,3){\gr H}
\end{pspicture}}

\subfloat[Another version of Lemma VIII]{\label{fig:VIIIc}
\psset{unit=1.2cm}
\begin{pspicture}(0,-0.625)(5,5.625)
\psline[linewidth=2pt](0,0)(0,5)(5,0)
\pspolygon(0,0)(5,0)(0,1)(2,3)(0,3)(2.5,2.5)
\psset{linestyle=dotted}
\psline(0,3)(5,0)
\psline(2,3)(0,0)
\psline(0,1)(2.5,2.5)
\uput[u](0,5){\gr A}
\uput[l](0,3){\gr H}
\uput[ur](2,3){\gr E}
\uput[l](0,1){\gr D}
\uput[dr](5,0){\gr Z}
\uput[dl](0,0){\gr B}
\uput[ur](2.5,2.5){\gr G}
%\uput[u](0,5){\gr H}
\end{pspicture}}
\caption{A case of Pappus's Theorem}\label{fig:VIII-2}
\end{figure}
we suppose
\begin{align*}
	\grm{BG}&\parallel\grm{DE},&
	\grm{HE}&\parallel\grm{ZB}.
\end{align*}
We prove $\grm{GH}\parallel\grm{DZ}$
using several equations of triangles:
\begin{align*}
&\text{1)}&	\mytri\grm{DBE}&=\mytri\grm{DGE},&&\text{[\emph{Elements} I.37, since $\grm{BG}\parallel\grm{DE}$]}\\
&\text{2)}&	\mytri\grm{ABE}&=\mytri\grm{GDA},&&\text{[add $\mytri\grm{DAE}$]}\\
&\text{3)}&	\mytri\grm{BZE}&=\mytri\grm{BZH},&&\text{[\emph{Elements} I.37, since $\grm{BZ}\parallel\grm{EH}$]}\\
&\text{4)}&	\mytri\grm{ABE}&=\mytri\grm{AHZ},&&\text{[subtract $\mytri\grm{ABZ}$]}\\
&\text{5)}&	\mytri\grm{AGD}&=\mytri\grm{AHZ},&&\text{[steps 2 and 4]}\\
&\text{6)}&	\mytri\grm{GDH}&=\mytri\grm{GZH},&&\text{[add $\mytri\grm{AGH}$]}\\
&\text{7)}&	\grm{GH}&\parallel\grm{DZ}.&&\text{[\emph{Elements} I.39]}
\end{align*}
In fact, if Pappus's figure were more like those in Figure \ref{fig:Pappus-cases},
it might be as in Figure \ref{fig:VIIIc};
and then the proof would have to be adjusted by, for example,
subtracting \gr{BZE} and \gr{BZH} \emph{from} \gr{ABZ} in step 4.

\addsec[Quadrangle Theorem: Pappus's proof]{Pappus's proof of the Quadrangle Theorem}

\textbf{Lemma IV}%
\index{Lemmas of Pappus!4@IV|(textbf}
is that if, as in Figure \ref{fig:IV},
\begin{figure}
\centering
\psset{unit=6mm}
\begin{pspicture}(0,-1)(16,11)
\pspolygon[linewidth=2pt](10,4)(7,4)(6,6)(10,10)
\pspolygon[linewidth=2pt](0,0)(16,4)
\psline(5.714,1.429)(10,10)
\psline(6,6)(0,0)
\psline[linestyle=dotted](16,4)(10,4)
\psline(8,2)(6,6)(12,3)
\psline[linestyle=dotted](12,3)(11.5,4)
\psline(10,10)(10,2.5)
\psline[linestyle=dotted](6,6)(10,7)
\uput[d](0,0){\gr A}
\uput[d](5.714,1.429){\gr B}
\uput[d](8,2){\gr G}
\uput[d](10,2.5){\gr D}
\uput[d](12,3){\gr E}
\uput[d](16,4){\gr Z}
\uput{2pt}[ur](10,4){\gr H}
\uput[l](7,4){\gr J}
\uput[ul](6,6){\gr K}
\uput[u](10,10){\gr L}
\uput[r](10,7){\gr M}
\uput{2pt}[dr](8.286,6.571){\gr N}
\uput{2pt}[ur](11.5,4){\gr{\myXi}}
\end{pspicture}
\caption{Lemma IV}\label{fig:IV}
\end{figure}
the given points \gr{A, B, G, D, E,} and \gr Z along a straight line
satisfy the proportion
\begin{equation}\label{eqn:IV}
\grm{AZ}\cdot\grm{BG}:\grm{AB}\cdot\grm{GZ}
::\grm{AZ}\cdot\grm{DE}:\grm{AD}\cdot\grm{EZ},
\end{equation}
then \gr J, \gr H, and \gr Z are in a straight line.
To prove this,
by alternation of \eqref{eqn:IV}, we obtain
\begin{equation}\label{eqn:IV-2}
\grm{AZ}\cdot\grm{BG}:\grm{AZ}\cdot\grm{DE}
::\grm{AB}\cdot\grm{GZ}:\grm{AD}\cdot\grm{EZ}.
\end{equation}
Considering first the left-hand member, we have by simplification
\begin{equation*}
\grm{AZ}\cdot\grm{BG}:\grm{AZ}\cdot\grm{DE}
::\grm{BG}:\grm{DE},
\end{equation*}
and we write the latter ratio as a composition of three ratios:
\begin{equation*}
\grm{BG}:\grm{DE}
::\grm{BG}:\grm{KN}\And\grm{KN}:\grm{KM}\And\grm{KM}:\grm{DE}.
\end{equation*}
Now we analyze the right-hand member of \eqref{eqn:IV-2}
as a composite:
\begin{equation*}
\grm{AB}\cdot\grm{GZ}:\grm{AD}\cdot\grm{EZ}
::\grm{BA}:\grm{AD}\And\grm{GZ}:\grm{ZE}.
\end{equation*}
Assuming \gr{KM} is drawn parallel to \gr{AZ},
by Thales's Theorem we have
\begin{equation*}
\grm{NK}:\grm{KM}::\grm{BA}:\grm{AD}.
\end{equation*}
Eliminating this common ratio from either member of \eqref{eqn:IV-2},
and reversing the order of the new members,
we obtain
\begin{equation*}
\grm{GZ}:\grm{ZE}
::\grm{BG}:\grm{KN}\And\grm{KM}:\grm{DE},
\end{equation*}
and therefore, by Thales's Theorem
applied to each ratio in the compound,
\begin{equation}\label{eqn:IV-lemma}
\grm{GZ}:\grm{ZE}
::\grm{JG}:\grm{KJ}\And\grm{KH}:\grm{HE}.
\end{equation}
Pappus says now that \gr{JHZ} is indeed straight.
Although he provides a reminder,
he may expect his readers to know,
as some students today know from high school,
the generalization of Thales's Theorem
known as \textbf{Menelaus's Theorem}
whose diagram is in Figure \ref{fig:IV-lemma}.%%%%%
\footnote{%The result is what is now called Menelaus's Theorem.
Menelaus's \emph{Sphaerica} survives in Arabic translation
\cite[p.\ 261]{MR654680};
but we also have Menelaus's Theorem in Ptolemy,
where I read it as a student at St John's College,
just before Toomer's 1984 translation \cite{Ptolemy-Toomer} came out;
we used the translation that Taliaferro had made for the College
\cite[I.13, p.\ 26]{Ptolemy}.
Thomas also puts Menelaus's Theorem in his anthology
\cite[pp. 445 ff.]{MR13:419b}.
In the commentary for their translation of Desargues,
Field and Gray remark that Pappus's Lemma IV is proved by
``chasing ratios much in the fashion Desargues was later to use. 
In this case collinearity could have been established 
by appealing to the converse of Menelaus' theorem, 
but when Pappus reached that point he missed that trick 
and continued to chase ratios 
until the conclusion was established---%
in effect, proving the converse of Menelaus' theorem without saying so''
\cite[pp.\ 10--1]{MR869422}.
I would add that the similarity of ``fashion'' in Pappus and Desargues
is probably due to the latter's having studied the former.
Moreover, Pappus seems \emph{not} to have ``missed the trick,''
since he asserts the desired collinearity at a point when it can be recognized
only by somebody who knows Menelaus's Theorem.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{figure}
\centering
\psset{unit=0.75cm}
\begin{pspicture}(6,1.33)(16,6.67)
\uput[d](8,2){\gr G}
\uput[d](12,3){\gr E}
\uput[d](16,4){\gr Z}
\uput[ur](10,4){\gr H}
\uput[l](7,4){\gr J}
\uput[ul](6,6){\gr K}
\uput[u](11.5,4){\gr{\myXi}}
\psline(16,4)(8,2)(6,6)(12,3)
\psline(7,4)(10,4)
\psline[linestyle=dashed](10,4)(11.5,4)(12,3)(16,4)
\psline[linestyle=dotted](11.5,4)(16,4)
\end{pspicture}
\begin{equation*}
\grm{GZ}:\grm{ZE}
::\grm{GJ}:\grm{JK}\And\grm{KH}:\grm{HE}
\iff\grm{JHZ} \text{ is straight}
\end{equation*}
\caption{Lemma from Lemma IV}\label{fig:IV-lemma}
\end{figure}
Here $\grm{E\myXi}\parallel\grm{JG}$,
and \gr{JH} is extended to \gr{\myXi}.
Then from \eqref{eqn:IV-lemma} we have
\begin{align*}
\grm{GZ}:\grm{ZE}
%&::\grm{JG}:\grm{KJ}\And\grm{KH}:\grm{HE}\\
&::\grm{JG}:\grm{KJ}\And\grm{KJ}:\grm{E\myXi}\\
&::\grm{JG}:\grm{E\myXi}.
\end{align*}
By Thales's Theorem again,
the points \gr J, \gr{\myXi}, and \gr Z must be collinear,
and therefore the same is true for \gr J, \gr H, and \gr Z.
This completes the proof of Lemma VIII.

The steps of the proof are reversible.
Thus, if we are given the complete quadrangle \gr{HJKL} of Figure \ref{fig:IV}
and the points \gr{A, B, G, D, E,} and \gr Z where its sides cross a given straight line,
the proportion \eqref{eqn:IV} must be satisfied.
Therefore if five sides of another complete quadrangle,
as \gr{PRST} in Figure \ref{fig:IV-CQT},
\begin{figure}
\centering
\psset{unit=8mm}
\begin{pspicture}(0,-4.5)(10,6.5)
\psline(0,0)(10,0)
\psline(6,6)(0,0)
\psline(6,6)(2,0)
\psline(6,6)(6,0)
\psline(4,4)(4,0)
\psline(4,4)(8,0)
\psline(4,3)(10,0)
\psline(4,-4)(0,0)
\psline(4,-4)(2,0)
\psline(4,-4)(6,0)
\psline(2,-2)(4,0)
\psline(2,-2)(8,0)
\psline(! 2 2 3 div add -1 1 3 div sub)(5.6,-0.8)
\psline[linestyle=dotted](5.6,-0.8)(10,0)
\uput[l](0,0){\gr A}
\uput[ul](2,0){\gr B}
\uput[ul](4,0){\gr G}
\uput[ul](6,0){\gr D}
\uput[ur](8,0){\gr E}
\uput[r](10,0){\gr Z}
\uput[u](6,6){\gr L}
\uput[ul](4,4){\gr K}
\uput[l](4,3){\gr J}
\uput[ur](6,2){\gr H}
\uput[d](4,-4){\gr T}
\uput[dl](2,-2){\gr S}
\uput[l](! 2 2 3 div add -1 1 3 div sub){\gr R}
\uput[dr](5.6,-0.8){\gr P}
\end{pspicture}
\caption{The Quadrangle Theorem}\label{fig:IV-CQT}

\end{figure}
should pass through the points
\gr{A, B, G, D,} and \gr E,
then the sixth side must pass through \gr Z.
This is the \textbf{Quadrangle Theorem.}

\addchap[Wednesday]{Wednesday, September 14, 2016}%\label{ch:Wednes}

\addsec[Thales's Theorem: Euclid's proof]{Euclid's proof of Thales's Theorem}

Suppose in Figure \ref{fig:TT},
\begin{figure}
\centering\psset{unit=1.5cm}
\begin{pspicture}(0,-0.33)(3,2.33)
\pspolygon(0,0)(3,0)(1,2)
\psline(0.6,1.2)(1.8,1.2)
\uput[u](1,2){$A$}
\uput[d](0,0){$B$}
\uput[d](3,0){$C$}
\uput[ul](0.6,1.2){$D$}
\uput[ur](1.8,1.2){$E$}
\psset{linestyle=dotted}
\psline(0,0)(1.8,1.2)
\psline(3,0)(0.6,1.2)
\end{pspicture}
\caption{Thales's Theorem}\label{fig:TT}
\end{figure}
$DE\parallel BC$.
Then $\triangle DEB=\triangle DEC$, so
\begin{align*}
	AD:DB
&	::\triangle ADE:\triangle DBE\\
&::\triangle ADE:\triangle ECD
::AE:EC.
\end{align*}
Conversely, if $AD:DB::AE:EC$,
then
\begin{align*}
\triangle ADE:\triangle DBC
&::	AD:DB\\
&::AE:EC
::\triangle ADE:\triangle BCE,
\end{align*}
and so $\triangle BCD=\triangle BCE$,
which gives us $DE\parallel BC$.
This is Euclid's proof of Thales's Theorem.

In any proportion, any of the four terms,
be it a straight line, a plane figure, or a solid,
can be replaced by a term of the same kind.
Beyond this,
the foregoing proof relies on two properties of proportions:
\begin{compactenum}
\item
If $a$ and $b$ are bounded straight lines, and $c$ and $d$ are plane figures,
then the proportion
\begin{equation*}
a:b::c:d
\end{equation*}
is true if and only if there is a rectangle as in Figure \ref{fig:rect}.
\begin{figure}
\centering
\begin{pspicture}(0,-0.5)(6,3)
\pspolygon(0,0)(6,0)(6,3)(0,3)
\psline(2,0)(2,3)
\uput[d](1,0){$a$}
\uput[d](4,0){$b$}
\rput(1,1.5){$c$}
\rput(4,1.5){$d$}
\end{pspicture}
\caption{A proportion of lengths and areas}\label{fig:rect}
\end{figure}
\item
Sameness of ratio is indeed an equivalence relation.
In particular, it is transitive.
\end{compactenum}
These properties follow from Euclid's definition of proportion,
found in Book V of the \emph{Elements.}
But this definition relies on the \textbf{Archimedean Axiom,}
that for any two magnitudes that have a ratio,
some multiple of either of them must exceed the other.

\addsec{A simple definition of proportion}

We can avoid using the Archimedean Axiom by, first of all,
letting Property 1 above be a definition.
Moreover, if $a$, $b$, $c$, and $d$ are all straight lines,
we define
\begin{equation*}
a:b::c:d
\end{equation*}
if, for some plane figures $e$ and $f$,
\begin{align*}
	a:b&::e:f,&c:d&::e:f.
\end{align*}
But we must check the following.
Suppose this condition is met,
and also $g$ and $h$ are plane figures
such that
\begin{equation*}
a:b::g:h.
\end{equation*}
Does it follow that $c:d::g:h\;$?
It does, as is shown in Figure \ref{fig:prop-def-a},
\begin{figure}
\centering
\subfloat[]{\label{fig:prop-def-a}
\begin{pspicture}(-2,-1.5)(7,3.5)
\psline(0,2)(7,2)(7,0)(0,0)(0,3.5)(4,3.5)(4,-1)
\psline(0,0)(0,-1)(7,-1)(7,0)
\uput[d](2,-1){$a$}
\uput[d](5.5,-1){$b$}
\uput[l](0,1){$c$}
\uput[l](0,2.75){$d$}
\rput(2,1){$e$}
\rput(5.5,1){$f$}
\rput(2,2.75){$f$}
\rput(2,-0.5){$g$}
\rput(5.5,-0.5){$h$}
\end{pspicture}}

\subfloat[]{\label{fig:prop-def-b}
\begin{pspicture}(-2,-1.5)(7,3.5)
\psline(0,2)(7,2)(7,0)(0,0)(0,3.5)(4,3.5)(4,-1)
\psline(0,0)(0,-1)(7,-1)(7,0)
\uput[d](2,-1){$a$}
\uput[d](5.5,-1){$b$}
\uput[l](-2,1){$c$}
\uput[l](-2,2.75){$d$}
\uput[ul](2,1){$e$}
\rput(5.5,1){$f$}
\rput(2,2.75){$f$}
\rput(2,-0.5){$g$}
\rput(5.5,-0.5){$h$}
\rput(-1,1){$g$}
\rput(-1,2.75){$h$}
\psset{linestyle=dotted}
\psline(-2,-1)(7,3.5)(4,3.5)
\psline(7,3.5)(7,2)
\psline(0,3.5)(-2,3.5)(-2,-1)(0,-1)
\psline(-2,2)(0,2)
\psline(-2,0)(0,0)
\end{pspicture}}

\caption{A definition of proportion}%\label{fig:prop-def}
\end{figure}
which can be completed as in Figure \ref{fig:prop-def-b}.
Here the line across the big parallelogram
(which may not be a rectangle) really is straight,
so that the the smaller parallelograms are equal in pairs,
as shown.

\addsec{Cross ratio in Pappus}

In \textbf{Lemma III,}%
\index{Lemmas of Pappus!3@III}
the straight lines \gr{JE} and \gr{JD}
cut the straight lines \gr{AB}, \gr{GA}, and \gr{DA} as in Figure \ref{fig:III}.
\begin{figure}
\centering
\psset{unit=1.2cm}
\begin{pspicture}(-2,-1.4)(4,2.4)
%\psgrid
  \psline[linewidth=2pt](1,0)(2,1)(4,0)
\psline[linewidth=2pt](2,0)(2,1)
\psline(2.667,0.667)(0,0)(4,0)
\psset{linestyle=dotted}
\psline(2,1)(0,2)(0,-1)
\psline(1,0)(0,-1)(-2,0)(0,0)
%\pspolygon(0,0)(-2,0)(0,-1)
\psdot(0,0)
\uput[u](2,1){\gr A}
\uput[ul](1.333,0.333){\gr B}
\uput[dr](2,0.5){\gr G}
\uput[ur](2.667,0.667){\gr D}
\uput[-67.5](1,0){\gr E}
\uput[d](2,0){\gr Z}
\uput[r](4,0){\gr H}
\uput[ul](0,0){\gr J}
\uput[u](0,2){\gr K}
\uput[d](0,-1){\gr L}
\uput[u](-2,0){\gr M}
\end{pspicture}
\caption{Lemma III}\label{fig:III}
\end{figure}
The diagram is completed by making
\begin{align*}
  \grm{KL}&\parallel\grm{ZGA},&
\grm{LM}&\parallel\grm{DA}.
\end{align*}
We first have
\begin{equation}\label{eqn:III}
  \begin{aligned}
\grm{EZ}:\grm{ZA}&::\grm{EJ}:\grm{JL},\\
\grm{AZ}:\grm{ZH}&::\grm{JL}:\grm{JM};
\end{aligned}
\end{equation}
so \emph{ex aequali,}
\begin{gather*}
  \grm{EZ}:\grm{ZH}::\grm{EJ}:\grm{JM},\\
\grm{JE}\cdot\grm{HZ}=\grm{EZ}\cdot\grm{JM}.
\end{gather*}
Being equal,
these areas have the same ratio to $\grm{EZ}\cdot\grm{JH}$, and so
\begin{align}
\grm{JE}\cdot\grm{HZ}:\grm{EZ}\cdot\grm{JH}
&::\grm{EZ}\cdot\grm{JM}:\grm{EZ}\cdot\grm{JH}\notag\\
  &::\grm{JM}:\grm{JH}\notag\\
	&::\grm{LJ}:\grm{JK}.\label{eqn:III-conc}
\end{align}
The ratio $\grm{LJ}:\grm{JK}$ 
is independent of the choice of the straight line through \gr J
that cuts the three straight lines that pass through \gr A.
In particular, we can conclude immediately
\begin{equation*}
\grm{JE}\cdot\grm{HZ}:\grm{EZ}\cdot\grm{JH}
::\grm{JB}\cdot\grm{DG}:\grm{BG}\cdot\grm{JD},
\end{equation*}
which is the theorem.

The ratio $\grm{JE}\cdot\grm{HZ}:\grm{EZ}\cdot\grm{JH}$
can be called the \textbf{cross ratio} (\emph{\c capraz oran})
of the points \gr{J, E, Z,} and \gr H.
By the lemma,
if four straight lines in a plane intersect at a point,
then the cross ratio of the four points where some other straight line crosses the lines
is always the same.
We can see this using Figure \ref{fig:CR}.
\begin{figure}
\centering
\begin{pspicture}(-1.5,-6)(3,0)
\psline(0,0)(-1.5,-6)
\psline(0,0)(0,-6)
\psline(0,0)(1.8,-6)
\psline(0,0)(3,-6)
\psline(2,0)(-0.5,-2)(2,-4)(-2,-5)
\end{pspicture}
\caption{Invariance of cross ratio in Pappus}\label{fig:CR}
\end{figure}

\textbf{Lemma X} is a converse to Lemma III.
The diagram is as in Figure \ref{fig:X},
\begin{figure}
\centering
\psset{unit=5mm}
\begin{pspicture}(0,-0.7)(12,8.7)
\pspolygon[linestyle=dotted](0,6)(12,6)(8,8)(6,6)(4,8)(0,6)
\psline[linestyle=dotted](12,6)(8,4)
\psline[linestyle=dotted](0,6)(4,4)
\psline[linestyle=dashed](3,3)(9,3)
\psline(0,0)(8,4)
\psline(12,0)(4,4)
\psline[linewidth=2pt](0,0)(6,6)(12,0)
\psdot(6,3)
\uput[d](6,3){\gr A}
\uput[u](4,4){\gr B}
\uput[ul](3,3){\gr G}
\uput[dl](0,0){\gr D}
\uput[dr](12,0){\gr E}
\uput[ur](9,3){\gr Z}
\uput[u](8,4){\gr H}
\uput[u](6,6){\gr J}
\uput[r](12,6){\gr K}
\uput[l](0,6){\gr L}
\uput[u](4,8){\gr M}
\uput[u](8,8){\gr N}
\end{pspicture}
\caption{Lemma X}\label{fig:X}
\end{figure}
where the hypothesis is
\begin{equation}\label{eqn:X}
\grm{JD}\cdot\grm{BG}:\grm{DG}\cdot\grm{BJ}::\grm{JH}\cdot\grm{ZE}:\grm{JE}\cdot\grm{ZH}.
\end{equation}
We make \gr{KL} parallel to \gr{GA},
and then extend \gr{AB} and \gr{AD} to meet \gr{KL} 
at two points, which Pappus writes as \gr K and \gr L,
though today we might say this is the wrong order.
We ensure $\grm{LM}\parallel\grm{AD}$ and $\grm{KN}\parallel\grm{AB}$,
with \gr{EJ} extended to \gr M, and \gr{DJ} to \gr N.
Now we repeat part of the proof of in Lemma III, 
at least in the configuration of Figure \ref{fig:III-alt},
\begin{figure}
\centering
\psset{unit=1.0cm}
\begin{pspicture}(-2,-1.5)(4,2.5)
%\psgrid
  \psline[linewidth=2pt](1,0)(2,1)(4,0)
\psline[linewidth=2pt](2,0)(2,1)
\psline(0,0)(4,0)
\psset{linestyle=dotted}
\psline(2,1)(0,2)(0,-1)(1,0)
\psline(0,2)(-2,0)(0,0)
\psdot(0,0)
\uput[u](2,1){\gr A}
\uput[-67.5](1,0){\gr H}
\uput[d](2,0){\gr Z}
\uput[r](4,0){\gr E}
\uput[ul](0,0){\gr J}
\uput[u](0,2){\gr L}
\uput[d](0,-1){\gr K}
\uput[u](-2,0){\gr M}
\end{pspicture}
\caption{Lemma III reconfigured}\label{fig:III-alt}
\end{figure}
with two of the three concurrent straight lines interchanged.
That is,
using only the part of the diagram shown in Figure \ref{fig:X-1},
\begin{figure}
%\mbox{}\hfill
\psset{unit=4.2mm}
\subfloat[]{\label{fig:X-1}
\begin{pspicture}(-0.2,-0.7)(12,8.7)
\psline(6,6)(0,0)
\psline[linewidth=2pt](6,3)(4,4)
\psline[linewidth=2pt](6,3)(3,3)
\psline[linewidth=2pt](6,3)(0,0)
\psset{linestyle=dotted}
\psline(6,3)(12,6)(0,6)(4,4)
\psline(12,6)(8,8)(6,6)
\psdot(6,6)
\uput[d](6,3){\gr A}
\uput[u](4,4){\gr B}
\uput[l](3,3){\gr G}
\uput[d](0,0){\gr D}
\uput[u](6,6){\gr J}
\uput[d](12,6){\gr K}
\uput[u](0,6){\gr L}
\uput[u](8,8){\gr N}
\end{pspicture}
}
\hfill
\subfloat[]{\label{fig:X-2}
\begin{pspicture}(0,-0.7)(12.2,8.7)
\psline(6,6)(12,0)
\psline[linewidth=2pt](6,3)(12,0)
\psline[linewidth=2pt](6,3)(9,3)
\psline[linewidth=2pt](6,3)(8,4)
\psset{linestyle=dotted}
\psline(6,3)(0,6)(12,6)(8,4)
\psline(0,6)(4,8)(6,6)
\psdot(6,6)
\uput[d](6,3){\gr A}
\uput[d](12,0){\gr E}
\uput[r](9,3){\gr Z}
\uput[u](8,4){\gr H}
\uput[u](6,6){\gr J}
\uput[u](12,6){\gr K}
\uput[d](0,6){\gr L}
\uput[u](4,8){\gr M}
\end{pspicture}
}
%\hfill\mbox{}
\caption{Lemma X: two halves of the proof}
\end{figure}
we have
\begin{gather*}
\grm{DJ}:\grm{JN}::\grm{DG}:\grm{GB},\\
\grm{DJ}\cdot\grm{GB}=\grm{DG}\cdot\grm{JN},\\
\begin{aligned}
	\grm{DJ}\cdot\grm{BG}:\grm{DG}\cdot\grm{BJ}
	&::\grm{GD}\cdot\grm{JN}:\grm{DG}\cdot\grm{BJ}\\
	&::\grm{JN}:\grm{JB}\\
	&::\grm{KJ}:\grm{JL}.
	\end{aligned}
	\end{gather*}
Now we move to the part of the diagram shown in Figure \ref{fig:X-2},
where we have proportions corresponding to the last two:
\begin{align*}
\grm{KJ}:\grm{JL}
&::\grm{HJ}:\grm{JM}\\
&::\grm{JH}\cdot\grm{ZE}:\grm{JM}\cdot\grm{ZE}.	
\end{align*}
In sum, we have shown
\begin{equation*}
\grm{DJ}\cdot\grm{BG}:\grm{DG}\cdot\grm{BJ}
::\grm{JH}\cdot\grm{ZE}:\grm{JM}\cdot\grm{ZE}.
\end{equation*}
But the left member already appears in \eqref{eqn:X}.
Hence the right members of the two proportions are the same,
that is,
\begin{gather*}
\grm{JH}\cdot\grm{ZE}:\grm{JE}\cdot\grm{ZH}::\grm{JH}\cdot\grm{ZE}:\grm{JM}
\cdot\grm{ZE},\\
\grm{JE}\cdot\grm{ZH}=\grm{JM}\cdot\grm{ZE},\\
\grm{JM}:\grm{JE}::\grm{HZ}:\grm{ZE}.
\end{gather*}
Thus what we did in Figure \ref{fig:X-1},
we have done in reverse in Figure \ref{fig:X-2}.
It remains to draw the conclusion
$\grm{AZ}\parallel\grm{KL}$,
corresponding to the hypothesis
$\grm{AG}\parallel\grm{KL}$.
Pappus argues as follows
(the bracketed proportions replaced with a reference to addition and alternation):
\begin{gather*}
\left[\begin{gathered}
\grm{JM}+\grm{JE}:\grm{JE}::\grm{HZ}+\grm{ZE}:\grm{ZE},\\
\grm{ME}:\grm{JE}::\grm{HE}:\grm{ZE},
\end{gathered}\right]\\
\grm{ME}:\grm{EH}::\grm{JE}:\grm{EZ},\\
\grm{LE}:\grm{EA}::\grm{JE}:\grm{EZ},
\end{gather*}
and so 
$\grm{AZ}\parallel\grm{KL}$,
which means
\gr{GAZ} must be straight.%
\index{Lemmas of Pappus!10@X|)textbf} 

\addchap[Friday]{Friday, September 16, 2016}

\addsec{Addition and multiplication}

In Euclid, a \textbf{bounded straight line}
(\emph{s\i n\i rlanm\i\c s do\u gru \c cizgi})
is called more simply a \textbf{straight line}
(\emph{do\u gru \c cizgi}),
and more simply still, a \textbf{``straight''} (\emph{do\u gru}).
In English, this is usually called a \textbf{line segment}
(\emph{do\u gru par\c cas\i}),
although for Euclid, and sometimes in English too,
a \textbf{line} (\emph{\c cizgi}) may be curved.
For example, a circle is a certain kind of line.
For us though, henceforth lines will always be straight and unbounded.

It is clear how to add the lengths of two line segments:
place the segments end to end in one straight line.
Thus in Figure \ref{fig:ABC},
\begin{figure}
\centering
\begin{pspicture}(5,0.5)
\psline(0,0.5)(5,0.5)
\psdots(0,0.5)(3,0.5)(5,0.5)
\uput[d](0,0.5){$A$}
\uput[d](3,0.5){$B$}
\uput[d](5,0.5){$C$}
\end{pspicture}
\caption{Addition of line segments}\label{fig:ABC}
\end{figure}
\begin{equation*}
AB+BC=AC.
\end{equation*}
It is implicit in Euclid and Pappus,
and we are making it explicit now,
that $AB$, for example, 
can be reversed and placed on itself
so that $A$ is on $B$ and $B$ is on $A$:
briefly, $AB=BA$.
Then
\begin{equation*}
BC+AB=CB+BA=CA=AC,
\end{equation*}
so addition is commutative.

We can consider addition in another way.
We select a point $A$ 
on a Euclidean unbounded straight line.
If $B$ and $C$ are also points on the line,
distinct from one another and from $A$,
then there are two points $D$ on the line
such that
\begin{equation*}
BD=AC,
\end{equation*}
but for only one of them do we also have
\begin{equation*}
CD=AB.
\end{equation*}
See Figure \ref{fig:B+C}.
\begin{figure}
\centering
\begin{pspicture}(5,1.5)
\psline(0,0.5)(5,0.5)
\psline(0,1.5)(5,1.5)
\psdots(0,0.5)(2,0.5)(3,0.5)(5,0.5)
\psdots(0,1.5)(2,1.5)(3,1.5)(5,1.5)
\uput[d](0,1.5){$C$}
\uput[d](2,1.5){$A$}
\uput[d](3,1.5){$D$}
\uput[d](5,1.5){$B$}
\uput[d](0,0.5){$A$}
\uput[d](2,0.5){$C$}
\uput[d](3,0.5){$B$}
\uput[d](5,0.5){$D$}
\end{pspicture}
\caption{Sum of two points with respect to a third}\label{fig:B+C}
\end{figure}
We may refer to this $D$ as the \textbf{sum of $B$ and $C$ with respect to $A$,}
writing
\begin{equation*}
D=B+_AC=C+_AB.
\end{equation*}
If $B$ and $C$ are the same point,
then $B+_AC$ is the unique point $D$ different from $A$
such that $AB=BD$.
Finally, if one of $B$ and $C$ is the same as $A$,
then $B+_AC$ is the same as the other.
Thus defined,
the operation $+_A$ makes the points of the line
into an abelian group
with neutral element $A$.

Can we define a multiplication that makes this group into a field?
We have already treated the product of two line segments as a rectangle
having sides equal to those two segments.
But now we should like the product of line segments to be another segment.

If $a$ and $b$ are points on the real number line as studied in calculus,
then $a+b$ and $a\cdot b$ are also points on that line.
The \emph{directed} segment from $a$ to $a+b$ 
is equal to that from $0$ to $b$.
But how is $a\cdot b$ defined?

Descartes showed the way \cite{Descartes-Geometry}.
The position of $1$ on the real line must be known.
More generally, on our Euclidean straight line with point $A$ chosen,
let us select another point $B$.
We can then denote by $1$ the length of $AB$.
If $a$ and $b$ are two lengths,
then by the proportion
\begin{equation*}
1:a::b:a\cdot b,
\end{equation*}
we can define the length $a\cdot b$.
To be more precise, 
since this length depends on the segment $AB$,
we might write the length as
\begin{equation*}
a\cdot_{AB}b.
\end{equation*}
We can construct a line segment having this length
by means of Thales's Theorem,
as in Figure \ref{fig:Descartes}.
\begin{figure}
\centering
\psset{unit=6.7mm}
\begin{pspicture}(0,-0.75)(5,4)
\pspolygon(0,0)(5,0)(4,4)
\psline(3,0)(2.4,2.4)
\uput[d](3,0){$1$}
\uput[d](5,0){$b$}
\uput[ul](2.4,2.4){$a$}
\uput[ul](4,4){$a\cdot b$}
\end{pspicture}
\caption{Descartes's definition of multiplication}\label{fig:Descartes}
\end{figure}

Why is the multiplication $\cdot_{AB}$ commutative?
We do know from Thales's Theorem that if
$A$, $B$, $C$, and $D$ are now arbitrary line segments,
and $A:B::C:D$, 
then $A\cdot D=B\cdot C$,
where the products are rectangles as before.
Since then $B\cdot C=C\cdot B$, we obtain
$A:C::B:D$.
Therefore
\begin{equation*}
1:b::a:a\cdot b,
\end{equation*}
and so, by interchanging $a$ and $b$,
\begin{gather*}
1:a::b:b\cdot a,\\
b\cdot a=a\cdot b.
\end{gather*}
For an alternative proof of commutativity of multiplication,
we can apply the case of Pappus's Theorem
established in Lemma VIII,
as in Figure \ref{fig:ab=ba}.
\begin{figure}
\centering
\psset{xunit=5mm}
\begin{pspicture}(-5,0)(5,5)
%\psgrid
\pspolygon(-5,0)(5,0)(0,5)
\psline(-5,0)(3,2)(-2,3)(2,3)(-3.33,1.67)(5,0)
\uput[l](-2,3){$1$}
\uput[r](2,3){$1$}
\uput[l](-3.33,1.67){$a$}
\uput[r](3,2){$b$}
\uput[l](-5,0){$ab$}
\uput[r](5,0){$ba$}
\end{pspicture}
\caption{Commutativity of Descartes's multiplication}\label{fig:ab=ba}
\end{figure}

We can do Euclidean geometry in the set $\R^2$ of ordered pairs of real numbers.
We can do some geometry in $\K^2$ for any field $\K$.
We can even allow $\K$ to be a non-commutative field
(also known as a division ring);
but then Pappus's Theorem will not be true in this geometry.
We shall see this in more detail tomorrow.

\addsec{A non-commutative field}

Meanwhile,
an example of a non-commutative field is
$\Ham$, the field, or rather skew field, of \textbf{quaternions,}
discovered by Hamilton.
We can obtain this field from the field $\C$ of complex numbers
roughly as $\C$ is obtained from $\R$.
Indeed, by one definition,
\begin{equation*}
\C=\left\{\begin{pmatrix}x&y\\-y&x\end{pmatrix}\colon x\in\R\And y\in\R\right\}.
\end{equation*}
We can write $x$ for $\left(\begin{smallmatrix}x&0\\0&x\end{smallmatrix}\right)$
and $\mathbf i$ for $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$,
so
\begin{equation*}
\begin{pmatrix}x&y\\-y&x\end{pmatrix}=x+y\mathbf i.
\end{equation*}
Now $\C$ is a commutative field, and it has the automorphism $z\mapsto\overline z$
given by
\begin{equation*}
\overline{x+y\mathbf i}=x-y\mathbf i.
\end{equation*}
We define
\begin{equation*}
\Ham=\left\{\begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix}\colon z\in\C\And w\in\C\right\}.
\end{equation*}
One shows that this is an additive abelian group
and is closed under multiplication,
and nonzero elements have multiplicative inverses:
it is a division ring.
We can write $z$ for $\left(\begin{smallmatrix}z&0\\0&\overline z\end{smallmatrix}\right)$
and $\mathbf j$ for $\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$,
so
\begin{equation*}
\begin{pmatrix}z&w\\-w&z\end{pmatrix}=z+w\mathbf j.
\end{equation*}
Then
\begin{align*}
	\mathbf{ij}&=
	\begin{pmatrix}\mathbf i&0\\0&-\mathbf i\end{pmatrix}
	\begin{pmatrix}0&1\\-1&0\end{pmatrix}
		=	\begin{pmatrix}0&\mathbf i\\\mathbf i&0\end{pmatrix},\\
	\mathbf{ji}&=
	\begin{pmatrix}0&1\\-1&0\end{pmatrix}
	\begin{pmatrix}\mathbf i&0\\0&-\mathbf i\end{pmatrix}
		=	\begin{pmatrix}0&-\mathbf i\\-\mathbf i&0\end{pmatrix}.
\end{align*}

\addsec[Pappus's Theorem: intersecting case]{Intersecting case of Pappus's Theorem}

Pappus's \textbf{Lemma XI} is that, 
in Figure \ref{fig:XI},
\begin{figure}
\centering
\psset{unit=0.5mm}
\begin{pspicture}(-45,-46)(75,66)
\psline[linewidth=2pt](0,0)(30,60)(75,60)
\psline[linewidth=2pt](30,60)(60,0)
\psdot(-45,0)
\psline(75,60)(-45,0)(60,0)
\psline[linestyle=dotted](0,0)(-20,-40)(60,0)
%\pspolygon(30,60)(-20,-40)(60,0)
%\psline(30,60)(75,60)(-45,0)(60,0)
\uput[u](30,60){\gr A}
\uput[dr](0,0){\gr B}
\uput[dr](60,0){\gr G}
\uput[u](75,60){\gr D}
\uput[d](-45,0){\gr E}
\uput[ul](15,30){\gr Z}
\uput[-25](39,42){\gr H}
\uput[d](-20,-40){\gr J}
\end{pspicture}
\caption{Lemma XI}\label{fig:XI}
\end{figure}
\begin{equation*}
\grm{DE}\cdot\grm{ZH}:\grm{EZ}\cdot\grm{HD}::\grm{GB}:\grm{BE}.
\end{equation*}
Since \gr{GJ} is drawn parallel to \gr{AE},
we have the two proportions
\begin{gather*}
	\grm{GA}:\grm{AH}::\grm{GJ}:\grm{ZH},\\
	\grm{GA}:\grm{AH}::\grm{ED}:\grm{DH},
\end{gather*}
and therefore
\begin{gather*}
	\grm{ED}:\grm{DH}::\grm{JG}:\grm{ZH},\\	
\grm{ED}\cdot\grm{ZH}=\grm{GJ}\cdot\grm{DH},\\
\begin{aligned}
	\grm{ED}\cdot\grm{ZH}:\grm{DH}\cdot\grm{EZ}
	&::\grm{GJ}\cdot\grm{DH}:\grm{DH}\cdot\grm{EZ}\\
	&::\grm{GJ}:\grm{EZ}\\
	&::\grm{GB}:\grm{BE},
\end{aligned}
\end{gather*}
as desired.
By Lemma III,
if \gr{EG} meets \gr{AD} at a point \gr K,
then
\begin{align*}
\grm{ED}\cdot\grm{ZH}:\grm{EZ}\cdot\grm{HD}
&::\grm{EK}\cdot\grm{GB}:\grm{EB}\cdot\grm{GK}\\
&::\grm{GB}:\grm{EB}\And\grm{EK}:\grm{GK}.
\end{align*}
In Lemma XI, the point $K$ has been sent off to infinity,
which means $\grm{EK}:\grm{GK}$ has become the unit ratio.

Pappus alludes to another case,
presumably as in Figure \ref{fig:XI-alt2}.
\begin{figure}
\centering
\psset{unit=6.25mm}
  \begin{pspicture}(0,-1.2)(6,6.8)
    \psline[linewidth=2pt](0,6)(6,6)(6,0)
\psline[linewidth=2pt](3,3)(6,6)
    \psline(0,6)(6,0)
\psline(2,4)(6,4)
\psline[linestyle=dotted](6,4)(5,5)
\uput[u](6,6){\gr A}
\uput[-67.5](4,4){\gr B}
\uput[r](6,4){\gr G}
\uput[u](0,6){\gr D}
\psdot(2,4)
\uput[dl](2,4){\gr E}
\uput[dl](3,3){\gr Z}
\uput[d](6,0){\gr H}
\uput[ul](5,5){\gr J}
  \end{pspicture}
  \caption{Lemma XI variant}\label{fig:XI-alt2}
 
\end{figure}
There is no change in the proof.

\textbf{Lemma XII} is that,
in Figure \ref{fig:XII},
\begin{figure}
\centering
\psset{unit=3.3mm}
\begin{pspicture}(0,-1.5)(12,13.5)
\psline[linewidth=2pt](0,0)(12,0)
\psline[linewidth=2pt](0,12)(12,12)
\pspolygon(0,0)(6,12)(12,0)(0,12)(6,0)(12,12)
%\psline(0,0)(6,12)(12,0)
%\psline(0,12)(6,0)(12,12)
\psline[linestyle=dotted](3,6)(9,6)
\uput[u](0,12){\gr A}
\uput[u](12,12){\gr B}
\uput[d](0,0){\gr G}
\uput[d](12,0){\gr D}
\uput[u](6,12){\gr E}
\uput[d](6,0){\gr Z}
\uput[l](3,6){\gr H}
\uput[100](4,8){\gr J}
\uput[r](9,6){\gr K}
\uput[80](8,8){\gr L}
\uput[d](6,6){\gr M}
\end{pspicture}
\caption{Lemma XII}\label{fig:XII}
\end{figure}
the points \gr{H, M,} and \gr K
are on a straight line.
The proof considers the parts of the diagram 
shown in Figure \ref{fig:XII-steps}.
\begin{figure}
\psset{unit=2.5mm}
\subfloat[]{
\begin{pspicture}(-0.4,-2)(12.4,14)
\uput[u](0,12){\gr A}
\uput[d](0,0){\gr G}
\uput[d](12,0){\gr D}
\uput[u](6,12){\gr E}
\uput[d](6,0){\gr Z}
\uput[l](3,6){\gr H}
\uput[r](4,8){\gr J}
\psline(6,0)(0,12)(12,0)(0,0)(6,12)(0,12)
\end{pspicture}
}
\hfill
\subfloat[]{
\begin{pspicture}(0,-2)(12,14)
\uput[u](12,12){\gr B}
\uput[d](0,0){\gr G}
\uput[d](12,0){\gr D}
\uput[u](6,12){\gr E}
\uput[d](6,0){\gr Z}
\uput[r](9,6){\gr K}
\uput[l](8,8){\gr L}
\psline(6,0)(12,12)(0,0)(12,0)(6,12)(12,12)
\end{pspicture}
}
\hfill
\subfloat[]{
\begin{pspicture}(-0.4,-2)(12.4,14)
\uput[d](0,0){\gr G}
\uput[d](12,0){\gr D}
\uput[u](6,12){\gr E}
\uput[l](3,6){\gr H}
\uput[ul](4,8){\gr J}
\uput[r](9,6){\gr K}
\uput[ur](8,8){\gr L}
\uput[d](6,6){\gr M}
\psdots(3,6)(9,6)
\psline(4,8)(12,0)(6,12)(0,0)(8,8)
\end{pspicture}
}
\caption{Steps of Lemma XII}\label{fig:XII-steps}
\end{figure}
Pappus's argument is:
\begin{compactenum}[a)]
\item
By Lemma XI,
\begin{equation}\label{eqn:DZZG}
\grm{DZ}:\grm{ZG}::\grm{GE}\cdot\grm{HJ}:\grm{GH}\cdot\grm{JE}.
\end{equation}
\item
By Lemma XI again, inversion, and \eqref{eqn:DZZG},
\begin{align*}
	\grm{GZ}:\grm{ZD}&::\grm{DE}\cdot\grm{LK}:\grm{DK}\cdot\grm{LE},\\
	\grm{DZ}:\grm{ZG}&::\grm{DK}\cdot\grm{LE}:\grm{DE}\cdot\grm{LK},\\
\grm{GE}\cdot\grm{HJ}:\grm{GH}\cdot\grm{JE}
&::\grm{DK}\cdot\grm{LE}:\grm{DE}\cdot\grm{KL}.
\end{align*}
\item
By Lemma X then, \gr{HMK} is straight.
\end{compactenum}%
\index{Lemmas of Pappus!12@XII|)textbf} 

\textbf{Lemma XIII} is similar,
with Lemma III used in place of Lemma XI.
This gives us Pappus's Theorem as proved by Pappus,
in the three cases labelled in Figure \ref{fig:Pappus-cases}
on page \pageref{fig:Pappus-cases}.

\addsec[Pappus's Theorem: a second proof]{A second proof of Pappus's Theorem}

Alternatively,
from Lemma VIII,
except for the easier case where everything that might be parallel is parallel,
all other cases of Pappus's Theorem 
can be derived by \emph{projection.}

If a diagram is drawn on a transparent notebook cover,
and the cover is raised at an angle to the first page,
and a shadow of the diagram is cast on that page,
all straight lines will remain straight,
but some parallel lines will cease to be so,
and some intersecting lines will become parallel.

Specifically, we can choose one straight line in a diagram 
that will become the line at infinity for a new diagram.
Call this line in the original diagram $A$.
All straight lines parallel to $A$ will remain parallel in the new diagram.
Also, one line parallel to these in the new diagram
will represent the line at infinity of the old diagram.  
Call this line in the new diagram $B$.
Lines that intersected on $A$
become parallel in the new diagram;
lines that were parallel to one another, but not to $A$,
will intersect on $B$.
See for example Figure \ref{fig:Pap-proj}.
\begin{figure}
\psset{xunit=3.3mm}
\subfloat[Lemma VIII]{%\label{fig:Pap-proj}
\begin{pspicture}(-6.5,-0.5)(6.5,5.5)
%\psgrid
\pspolygon(-5,0)(5,0)(0,5)
\psline(-5,0)(3,2)(-2,3)(2,3)(-3.33,1.67)(5,0)
\psline[linestyle=dotted](-5,5)(5,5)
\uput[u](0,5){new line at $\infty$}
\uput[l](-2,3){$A$}
\uput[r](2,3){$B$}
\uput[l](-3.33,1.67){$C$}
\uput[r](3,2){$F$}
\uput[l](-5,0){$E$}
\uput[r](5,0){$D$}
\end{pspicture}}
\hfill
\psset{unit=3.3mm}
\subfloat[A new case]{%\label{fig:Pap-2-per}
  \begin{pspicture}(0,-1.5)(16.5,16.5)
\pspolygon(6,6)(9,10)(6,15)(9,15)(6,9)(9,6)
\psline(6,6)(6,15)
\psline(9,6)(9,15)
\psset{linestyle=dashed}
\psline(6,6)(1.5,0)(6,9)
\psline(9,6)(15,0)(9,10)
\psline[linestyle=dotted](1.5,0)(15,0)
\uput[l](6,15){$A$}
\uput[r](9,15){$B$}
\uput[l](6,9){$C$}
\uput[d](9,6){$D$}
\uput[d](6,6){$E$}
\uput[r](9,10){$F$}
\uput[l](1.5,0){$G$}
\uput[r](15,0){$H$}
\uput[u](8.25,0){old line at $\infty$}
\end{pspicture}}
  \caption{Pappus's Theorem by projection}\label{fig:Pap-proj}
  
\end{figure}

Tomorrow
we shall obtain a third proof of Pappus's Theorem
by coordinatizing the projective plane.
 
\addchap[Saturday]{Saturday, September 17, 2016}

\addsec{Cartesian coordinates}

We saw yesterday that by choosing a point $A$ on an infinite straight line,
we could make the points of the line into an abelian group with the addition
denoted by $+_A$.  By choosing another point $B$ on the line,
we obtained a multiplication $\cdot_{AB}$ of lengths.
This is not quite a multiplication of points.
However, given points $C$ and $D$ on the line,
we can define
\begin{equation*}
C\cdot_{AB}D=E,
\end{equation*}
where $E$ is a point on the line such that,
if $a$ and $b$ are the lengths of $AC$ and $AD$,
then $a\cdot_{AB}b$ is the length of $AE$,
and also
$A$ lies between $B$ and $E$
just in case it lies between $C$ and $D$.
Equipped with addition and multiplication so defined,
the Euclidean line becomes a field.

We can obtain an isomorphic field
by considering ratios of line segments.
On Wednesday we defined ratios of line segments
and of rectangles.
By this definition,
any two line segments have the ratio of two rectangles.
Conversely, for any two rectangles,
we can find line segments that have their ratio.
For example, in Figure \ref{fig:rect-rat},
\begin{figure}
\centering
\psset{unit=4mm}
\begin{pspicture}(0,-1)(16,9)
\psline(10,0)(10,8)(16,8)(16,0)(0,0)(0,7)(10,7)
\psline(0,3)(16,3)
\psset{linestyle=dotted}
\psline(0,7)(0,8)(10,8)
\psline(0,8)(16,0)
\uput[l](0,7){$A$}
\uput[l](0,3){$B$}
\uput[l](0,0){$C$}
\uput[u](10,8){$D$}
\uput[r](10,7){$E$}
\uput[ur](10,3){$F$}
\uput[d](10,0){$G$}
\uput[r](16,8){$H$}
\uput[r](16,3){$K$}
\uput[r](16,0){$L$}
\end{pspicture}
\caption{Line segments in the ratio of rectangles}\label{fig:rect-rat}
\end{figure}
\begin{equation*}
ABFE:DFKH::AB:BC
\end{equation*}
since $DFKH=BCGF$.
Ratios constitute a group,
the operation being composition,
since
\begin{gather*}
(a:b\And b:c)\And c:d::a:c\And c:d::a:d,\\
	a:b\And(b:c\And c:d)::a:b\And b:d::a:d.
\end{gather*}
Composition is commutative, since if
\begin{equation*}
a:b::c:d,
\end{equation*}
then by Thales's Theorem
\begin{align*}
ad&=bc,&a:c&::b:d
\end{align*}
(see Figure \ref{fig:alt}),
\begin{figure}
\centering
\psset{unit=5mm}
\begin{pspicture}(0,-1)(10,5)
\pspolygon(0,0)(10,0)(10,5)(0,5)
\psline(0,0)(10,5)
\psline(6,0)(6,5)
\psline(0,3)(10,3)
\uput[l](0,1.5){$b$}
\uput[l](0,4){$d$}
\uput[d](3,0){$a$}
\uput[d](8,0){$c$}
\rput(8,1.5){$bc$}
\rput(3,4){$ad$}
\end{pspicture}
\caption{Alternation of ratio}\label{fig:alt}
\end{figure}
so that
\begin{align*}
	a:b\And b:c&::a:c,\\
	b:c\And a:b&::b:c\And c:d::b:d::a:c.	
\end{align*}

We have defined the product of two line segments in two different ways:
\begin{compactenum}[1)]
\item
as the rectangle bounded by them, or
\item
as the straight line that has the same ratio to one of the two segments
that the other one has to some predetermined segment.
\end{compactenum}
In the former case,
commutativity is immediate;
in the latter case,
it follows from the theorems of Thales and Pappus,
as we showed.
In either case,
we can define sums of ratios of line segments,
through the usual process of finding a common denominator:
\begin{equation*}
(a:b)+(c:d)=ad+bc:bd.
\end{equation*}
We get the same result,
no matter which multiplication of segments we use.

If we allow negative lengths and a zero length;
then the ratios of straight lines compose a field,
which we shall call $\K$.

We shall now work in the \textbf{Euclidean plane:}
the plane in which the propositions of Book I of the \emph{Elements} are true.
If we choose a neutral point $A$,
we obtain an additive abelian group,
as we did on a line.
Associativity can be checked as in Figure \ref{fig:ass}.
\begin{figure}
\centering
\psset{unit=6.67mm}
\begin{pspicture}(-1,-1.75)(9,7.75)
\psline(0,0)(7,-1)
\psline(0,0)(2,3)
\psline(-1,4)
\psline[linestyle=dotted](9,2)(2,3)(1,7)(-1,4)(8,6)(7,-1)(9,2)(8,6)(1,7)
\uput[dl](0,0){$A$}
\uput[d](7,-1){$B$}
\uput[ur](2,3){$C$}
\uput[l](-1,4){$D$}
\uput[r](9,2){$B+_AC$}
\uput[u](1,7){$C+_AD$}
\uput[ur](8,6){$B+_AC+_AD$}
\psset{linestyle=dashed}
\psline(0,0)(9,2)
\psline(0,0)(1,7)
\psline(0,0)(8,6)
\end{pspicture}
\caption{Associativity of addition}\label{fig:ass}
\end{figure}
The group can also be understood as a vector space
over the field $\K$ of ratios.
Here if $s$ is such an ratio,
and $B$ is a point of the plane different from $A$,
we define
\begin{equation*}
s\cdot_AB=C,
\end{equation*}
where $C$ is the point on $AB$ such that
\begin{equation*}
AC:AB=s.
\end{equation*}

Now suppose $A$, $B$, and $C$ are vertices of a triangle,
that is, none of the three is on the straight line determined by the other two.
Then $B$ and $C$ compose a basis of the vector space
whose neutral point is $A$.
In particular, for any point $D$ of the plane,
there is a unique ordered pair $(s,t)$ of ratios,
meaning $(s,t)\in\K^2$, such that
\begin{equation*}
D=s\cdot_AB+_At\cdot_AC.
\end{equation*}
Here $(s,t)$ is the pair of
\textbf{Cartesian coordinates} of $D$
with respect to $ABC$.
Conversely, every element of $\K^2$
corresponds to a point of the plane in this way.

\addsec{Barycentric coordinates}

Again, we have chosen triangle $ABC$,
and for some $D$ in the same plane we have
\begin{equation*}
D=s\cdot_AB+_At\cdot_AC.
\end{equation*}
This fails if we replace $A$ with some other point.
However, the equations
\begin{align*}
D&=A+_A s\cdot_A(B-_AA) +_A t\cdot_A(C-_AA)\\
&=(1-s-t)\cdot_AA+_As\cdot_AB+_At\cdot_AC
\end{align*}
are still correct if the subscripts $A$ are replaced
with any other point.
Thus we may write simply
\begin{equation*}
D=(1-s-t)A+sB+tC.
\end{equation*}
Conversely, 
if $p$, $q$, and $r$ are three ratios whose sum is $1$,
then the linear combination
\begin{equation*}
pA+qB+rC
\end{equation*}
is unambiguous.
Considering $ABC$ as fixed,
we may write the same point $pA+qB+rC$
as
\begin{equation*}
(p:q:r).
\end{equation*}
But now we can allow
\begin{equation*}
(p:q:r)=(tp:tq:tr)
\end{equation*}
if $t\neq0$.
Thus,
given arbitrary ratios $p$, $q$, and $r$
whose sum is not $0$, we have
\begin{align*}
(p:q:r)
&=\frac p{p+q+r}A+\frac q{p+q+r}B+\frac r{p+q+r}C\\
&=\frac q{p+q+r}\cdot_AB+_A\frac r{p+q+r}\cdot_AC.
\end{align*}
This point has the \textbf{barycentric coordinates} $p$, $q$, and $r$,
but each must be considered together with the sum $p+q+r$.
The idea is that the point is the center of gravity
(the barycenter) of the system with weights $p$, $q$, and $r$
at $A$, $B$, and $C$ respectively.
If we define
\begin{equation*}
D=\frac q{q+r}B+\frac r{q+r}C,
\end{equation*}
then $D$ is a point on $BC$, and
\begin{equation*}
(p:q:r)=\frac p{p+q+r}A+\frac{q+r}{p+q+r}D,
\end{equation*}
so $(p:q:r)$ is a point on $AD$.
Similarly we can define $E$ and $F$
on $AC$ and $AB$ respectively,
so that $(p:q:r)$ is on $BE$ and $CF$.
Note that
\begin{equation*}
BD:DC::r:q,
\end{equation*}
and so on, so that we have \textbf{Ceva's Theorem:}
in Figure \ref{fig:Ceva},
\begin{figure}
\centering
\psset{unit=6.6mm}
\begin{pspicture}(0,-0.75)(7,5.75)
\pspolygon(0,0)(7,0)(3,5)
\psset{linestyle=dotted}
\psline(0,0)(5.4,2)
\psline(7,0)(2.4,4)
\psline(3,5)(6,0)
\uput[d](0,0){$A$}
\uput[d](6,0){$F$}
\uput[d](7,0){$B$}
\uput[ur](5.4,2){$D$}
\uput[u](3,5){$C$}
\uput[ul](2.4,4){$E$}
\end{pspicture}
\caption{Ceva's Theorem}\label{fig:Ceva}
\end{figure}
the lines $AD$, $BE$, and $CF$ have a common point
if and only if
\begin{equation*}
BD:DC\And CE:EA\And AF:FB::1.
\end{equation*}

We can understand $(p:q:r)$ as the equivalence class
consisting of all ordered triples $(px,qx,rx)$,
where $x$ is a nonzero ratio.
The notation makes sense for all ratios $p$, $q$, and $r$.
We are going to show how to extend the Euclidean plane
to the projective plane
by giving geometric meaning to $(p:q:r)$ when $p+q+r=0$
but at least one of $p$, $q$, and $r$ is not $0$.

\addsec{Projective coordinates}

For every straight line in the plane,
there are ratios $a$, $b$, and $c$,
where at least one of $a$ and $b$ is not $0$, 
such that the straight line consists of the points
such that,
if their Cartesian coordinates are $(s,t)$,
then
\begin{equation*}
as+bt+c=0.
\end{equation*}
If the same point has barycentric coordinates $(p:q:r)$,
where $p+q+r=1$, then $(s,t)=(q,r)$, and so
\begin{align*}
0&=aq+br+c(p+q+r)\\
&=cp+(a+c)q+(b+c)r.
\end{align*}
Thus the same line is given by
\begin{equation*}
ax+by+c=0
\end{equation*}
in Cartesian coordinates and
\begin{equation*}
cx+(a+c)y+(b+c)z=0
\end{equation*}
in barycentric coordinates.
The straight lines parallel to this one
are obtained by changing $c$ alone.
Relabelling, we now have that
every straight line is given by an equation
\begin{equation*}
ax+by+cz=0
\end{equation*}
in barycentric coordinates,
where one of the coefficients $a$, $b$, and $c$
is different from the others
(the coefficients are not all equal).
We obtain parallel lines
by adding the same ratio to each coefficient.

Thus, if $ap+bq+cr=0$,
and $p+q+r=0$,
but $(p,q,r)\neq(0,0,0)$,
then $(p:q:r)$ satisfies the equation
of every straight line parallel to
$ax+by+cz=0$,
and no other straight line.
We can understand $(p:q:r)$
as the point at infinity 
of the straight lines parallel to $ax+by+cz=0$.
The line at infinity is then defined by
$x+y+z=0$.

The projective plane now consists of points $(p:q:r)$,
where $(p,q,r)\neq(0,0,0)$.
The expression $(p:q:r)$ consists of
\textbf{projective coordinates} for the point.

There is now a one-to-one correspondence
between points of the projective plane
and straight lines in that plane:
if the point is $(p:q:r)$ in projective coordinates,
the straight line is given by $px+qy+rz=0$.

We may use the notation
\begin{equation*}
\Proj^2(\K)=\Bigl\{(x:y:z)\colon(x,y,z)\in\K^3\setminus\bigl\{(0,0,0)\bigr\}\Bigr\}.
\end{equation*}

Any triangle $ABC$ in the Euclidean plane
determines a system of barycentric coordinates
for the points of the plane,
hence a system of projective coordinates for the projective plane.
However, suppose a fourth point $D$ in the Euclidean plane
does not lie on any of the three sides of $ABC$.
Then $D$ has barycentric coordinates $(\mu:\nu:\rho)$,
with $\mu\nu\rho\neq0$.
There is now a bijection
\begin{equation*}
(x:y:z)\mapsto\left(\frac x{\mu}:\frac y{\nu}:\frac z{\rho}\right)
\end{equation*}
from the set of points of the projective plane to itself.
This bijection takes the line $ax+by+cz=0$ to the line
$\mu ax+\nu by+\rho cz=0$,
while fixing $A$, $B$, and $C$,
which are $(1:0:0)$,
$(0:1:0)$, and $(0:0:1)$ respectively.
The points that used to be on the line at infinity
are sent to the line
\begin{equation*}
\mu x+\nu y+\rho z=0,
\end{equation*}
while of course the points now at infinity are on $x+y+z=0$.
The only former point at infinity that is still at infinity
is $(\nu-\rho:\rho-\mu:\mu-\nu)$.
In particular, if $E$ is a point that used to be at infinity,
we can have chosen $D$ so that $E$ is no longer at infinity.

In this way,
for any quadrangle $ABCD$ in the projective plane,
there is a system of projective coordinates
in which the vertices of the quadrangle are $(1:0:0)$,
$(0:1:0)$, $(0:0:1)$, and $(1:1:1)$ respectively.

\addsec[Pappus's Theorem: third proof]{A third proof of Pappus's Theorem}

To prove Pappus's Hexagon Theorem
in yet another way,
again we suppose the vertices of the hexagon $ABCDEF$
lie alternately on straight lines $ACE$ and $BDF$,
and we let
\begin{compactitem}
\item
$AB$ and $DE$ intersect at $G$,
\item
$BC$ and $EF$ intersect at $H$,
\item
$CD$ and $FA$ intersect at $K$.
\end{compactitem}
We may now suppose further
\begin{align*}
	A&=(1:1:1),&B&=(1:0:0),\\
	E&=(0:1:0),&K&=(0:0:1),
\end{align*}
as in Figure \ref{fig:Pap-coord}.
\begin{figure}
\centering
\psset{xunit=0.4mm,yunit=0.15mm}
\begin{pspicture}(0,-252)(252,252)%36x7
\pspolygon(168,0)(252,252)(189,-63)(216,-72)(252,-252)(224,0)
\psset{linestyle=dashed}
\psline(252,252)(216,-72)
\psline(168,0)(252,-252)
\psset{linestyle=dotted}
\psline(216,-72)(189,63)
\psline(224,0)(216,72)
\psline(189,-63)(0,0)(168,0)
\uput[ul](168,0){$A$}
\uput[r](252,252){$B$}
\uput[dl](189,-63){$C$}
\uput[dl](216,-72){$D$}
\uput[r](252,-252){$E$}
\uput[r](224,0){$F$}
\uput[ul](189,63){$G$}
\uput[ul](216,72){$H$}
\uput[u](0,0){$K$}
\psdots(168,0)(252,252)(189,-63)(216,-72)(252,-252)(224,0)(189,63)(216,72)(0,0)
\end{pspicture}

\caption{Pappus's Hexagon Theorem}\label{fig:Pap-coord}
\end{figure}
The point of the choice is that
no line of the diagram contains two of $B$, $E$, and $K$,
but these are on the lines through $A$.
If $B$, $E$, and $K$ are all on a straight line,
we may replace them with $F$, $C$, and $H$.
For the lines through $A$ we have
\begin{align*}
	AB&=\{y=z\},&
	AE&=\{x=z\},&
	AK&=\{x=y\},
\end{align*}
and so, for some $p$, $q$, and $r$,
\begin{align*}
	G&=(p:1:1),&C&=(1:q:1),&F&=(1:1:r).
\end{align*}
Consequently,
\begin{align*}
BF&=\{z=ry\},&
EG&=\{x=pz\},&
KC&=\{y=qx\}.
\end{align*}
Since these three lines have a common point,
namely $D$, we must have
$D=(1:q:rq)=(pr:1:r)=(p:qp:1)$,
and so
\begin{equation*}
1=prq=qpr=rqp.
\end{equation*}
But we have also
\begin{align*}
	BC&=\{y=qz\},&
	EF&=\{z=rx\},&
	KG&=\{x=py\}.
\end{align*}
In particular,
$H$ being the intersection of the first two of these lines,
we have
\begin{equation*}
H=(1:qr:r),
\end{equation*}
and this lies on $KG$ since $pqr=1$.
Figure \ref{fig:Pap-coord}
shows the situation when $p=2$ and $q=2$, so $r=1/4$.

\addchap[Sunday]{Sunday, September 18, 2016}

\addsec{The projective plane over a field}

Given a field $\K$
(possibly a ``skew'' or non-commutative field),
we have
\begin{equation*}
\Proj^2(\K)=\Bigl(\K^3\setminus\bigl\{(0,0,0)\bigr\}\Bigr)/\mathord{\sim},
\end{equation*}
where
 \begin{equation*}
(p,q,r)\sim(p',q',r')
 \end{equation*}
if and only if, for some $t$ in $\K\setminus\{0\}$,
\begin{align*}
	p'&=tp,&q'&=tq,&r'&=tr.
\end{align*}
We may understand this both as the set of points in the projective plane
and as the set of lines in the projective plane.
The point $(p:q:r)$ sits on the line $(a:b:c)$ if and only if $ap+bq+cr=0$.
The line at infinity is $(1:1:1)$, that is, the line given by the equation
$x+y+z=0$.  So the points at infinity satisfy this.

\addsec{The Fano Plane}

For the simplest example,
we may let $\K$ be the two-element field $\F_2$,
thus obtaining the \textbf{Fano Plane.}
The four points of $\F_2{}^2$ in barycentric coordinates are
\begin{align*}
	&(1:0:0),&
	&(0:1:0),&
	&(0:0:1),&
	&(1:1:1).
\end{align*}
There are three points at infinity in $\Proj^2(\K)$:
\begin{align*}
	&(1:1:0),&
	&(1:0:1),&
	&(0:1:1).
\end{align*}
If we call these $A$, $B$, $C$, $D$, $E$, $F$, and $G$ respectively,
then the seven lines are as follows:
\begin{align*}
\{x=0\}&=BCG,&\{y=z\}&=ADG,\\
\{y=0\}&=ACF,&\{x=z\}&=BDF,\\
\{z=0\}&=ABE,&\{x=y\}&=CDE,
\end{align*}
and the line at infinity, $EFG$.
All of this can be depicted as in Figure \ref{fig:Fano}.
\begin{figure}
\centering
\begin{pspicture}(-4,-3)(4,4)
\pscircle(0,0)2
\pspolygon(4;90)(4;210)(4;330)
\psline(4;90)(2;270)
\psline(4;210)(2;30)
\psline(4;330)(2;150)
\rput*(0,0){$(1:1:1)$}
\rput*(4;210){$(1:0:0)$}
\rput*(4;330){$(0:1:0)$}
\rput*(4;90){$(0:0:1)$}
\rput*(2;30){$(0:1:1)$}
\rput*(2;150){$(1:0:1)$}
\rput*(2;270){$(1:1:0)$}
\end{pspicture}
\caption{The Fano Plane}\label{fig:Fano}
\end{figure}

\addsec[Desargues's Theorem: a proof]{A proof of Desargues's Theorem}

We now prove Desargues's Theorem in $\Proj^2(\K)$
for arbitrary $\K$.
The diagram is as in Figure \ref{fig:Des},
repeated as Figure \ref{fig:Des2}.
\begin{figure}
\centering
\psset{xunit=5.0mm,yunit=5.0mm,dotstyle=o,dotsize=5pt}
\begin{pspicture}(-1.00,-1)(13.00,12)
\psline[linestyle=dotted](12,12)(12,0) % Kappa to Lambda
\psline(0,12)(3,0)   % Eta to Delta
\psline(0,12)(6,0)   % Eta to Zeta
\psline(0,12)(7.5,3) % Eta to Epsilon
\psline(12,6)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Theta to Alpha
\psline(12,6)(3,0)                         % Theta to Delta
\psline(12,12)(! 12 7 19 div mul 
              12 24 7 19 div mul sub) % Kappa to Gamma
\psline(12,12)(6,0)                   % Kappa to Zeta 
\psline(12,0)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Lambda to Alpha
\psline(12,0)(3,0)                         % Lambda to Delta
\psdot(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
\uput[dl](! 12 7 7 36 add div mul
         12 48 7 7 36 add div mul sub){$A$}
\psdot(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
\uput[u](! 12 14 mul 14 14.5 add div 
      12 14.5 14 mul 14 14.5 add div sub){$B$} 
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput[dl](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0)(7.5,3)(6,0) % Delta, Epsilon, Zeta
\uput[d](3,0){$D$}
\uput[dr](7.5,3){$E$}
\uput[d](6,0){$F$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,6)(12,12)(12,0) % Theta, Kappa, Lambda
\uput[r](12,6){$H$}
\uput[r](12,12){$K$}
\uput[dr](12,0){$L$}
\end{pspicture}
\caption{Desargues's Theorem}\label{fig:Des2}
\end{figure}
In $\K^3$,
if triangles $ABC$ and $DEF$
lie in two different planes,
then each of $H$, $K$, and $L$
lies in each of those planes,
and therefore the three points lie in the intersection of the two planes,
which is a straight line.
(If the planes are parallel, 
they meet at their common line at infinity.)
If $ABC$ and $DEF$ lie in the same plane,
the diagram can still be considered as the ``shadow''
or projection of the three-dimensional case.
So Desargues's Theorem holds in $\Proj^2(\K)$.

\begin{sloppypar}
If $\K$ is not commutative,
then Pappus's Theorem does not hold in $\Proj^2(\K)$.
However,
with three applications of Pappus's Theorem alone,
we can prove Desargues's Theorem as sketched in Figure \ref{fig:Des-2}.
\begin{figure}
\psset{xunit=3.8mm,yunit=4.6mm,dotstyle=o,dotsize=5pt}
%\subfloat[Construction]{\label{fig:Des-Greek}% XI AND PI ADDED
\begin{pspicture}(-1.00,-4)(13.00,12)
\rput(6,12){Construction}
\psline(0,12)(3,0)   % Eta to Delta
\psline(0,12)(6,0)   % Eta to Zeta
\psline(0,12)(7.5,3) % Eta to Epsilon
\psline(12,6)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Theta to Alpha
\psline(12,6)(3,0)                         % Theta to Delta
\psline(12,12)(! 12 7 19 div mul 
              12 24 7 19 div mul sub) % Kappa to Gamma
\psline(12,12)(6,0)                   % Kappa to Zeta 
\psline(12,0)(! 12 7 7 36 add div mul
             12 48 7 7 36 add div mul sub) % Lambda to Alpha
\psline(12,0)(3,0)                         % Lambda to Delta
\psset{linestyle=dashed}
\psline(! 12 7 19 div mul 
      12 24 7 19 div mul sub)(0,-2) % Gamma to Mu
\psline(3,0)(0,-2)                  % Delta to Mu
\psline(3,0)(! 12 5 mul 11 div 
               12 5 mul 11 div) % Delta to Nu
\psline(0,12)(0,-2) % Eta to Mu
\psline(! 12 7 7 36 add div mul
       12 48 7 7 36 add div mul sub)(0,5) % Alpha to Xi
\psline(3,0)(0,0) % Delta to Pi
\psdot(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
\uput[dl](! 12 7 7 36 add div mul
         12 48 7 7 36 add div mul sub){$A$}
\psdot(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
\uput[d](! 12 14 mul 14 14.5 add div 
      12 14.5 14 mul 14 14.5 add div sub){$B$} 
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput{7pt}[15](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0)(7.5,3)(6,0) % Delta, Epsilon, Zeta
\uput[d](3,0){$D$}
\uput{2pt}[dr](7.5,3){$E$}
\uput[d](6,0){$F$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,6)(12,12)(12,0) % Theta, Kappa, Lambda
\uput[r](12,6){$H$}
\uput[r](12,12){$K$}
\uput[dr](12,0){$L$}
\psdot(0,-2) % Mu
\uput[l](0,-2){$M$}
\psdot(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
\uput[u](! 12 5 mul 11 div 12 5 mul 11 div){$N$}
\psdots(0,5)(0,0) % Xi, Pi
\uput[l](0,5){$P$}
\uput[l](0,0){$Q$}
\end{pspicture}
%}
\hfill
%\subfloat[$HNP$ is straight]{% FIRST HEXAGON, ABHMDG, WITH JNX
\begin{pspicture}(-1.00,-4)(13.00,12)
\rput(6,12){$HNP$ is straight}
\psline[linewidth=2pt](0,12)(3,0)   % Eta to Delta
\psline[linewidth=2pt](0,-2)(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Mu to Beta
\pspolygon(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
  (! 12 14 mul 14 14.5 add div 
12 14.5 14 mul 14 14.5 add div sub) % Beta
(0,12)(0,-2)(3,0) % Eta Mu Delta
(! 12 7 19 div mul 
12 24 7 19 div mul sub) % Gamma
\psset{linestyle=dashed}
\psline(! 12 7 7 36 add div mul
       12 48 7 7 36 add div mul sub)(0,5) % Alpha to Xi
\psline(! 12 7 19 div mul 
      12 24 7 19 div mul sub)
			(! 12 5 mul 11 div 12 5 mul 11 div) % Gamma to Nu
\psline(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
		(12,6)(3,0)                         % Theta to Delta
\psset{linestyle=dotted}
\psline(0,5)(12,6)
\psdot(! 12 7 7 36 add div mul
      12 48 7 7 36 add div mul sub) % Alpha
\uput[dl](! 12 7 7 36 add div mul
         12 48 7 7 36 add div mul sub){$A$}
\psdot(! 12 14 mul 14 14.5 add div 
    12 14.5 14 mul 14 14.5 add div sub) % Beta
\uput[dr](! 12 14 mul 14 14.5 add div 
      12 14.5 14 mul 14 14.5 add div sub){$B$} 
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput[dr](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0) % Delta
\uput[d](3,0){$D$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,6) % Theta
\uput[r](12,6){$H$}
\psdot(0,-2) % Mu
\uput[l](0,-2){$M$}
\psdot(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
\uput[u](! 12 5 mul 11 div 12 5 mul 11 div){$N$}
\psdots(0,5) % Xi
\uput[l](0,5){$P$}
\end{pspicture}
%}
%\caption{Proof of Desargues's Theorem: first steps}\label{fig:Des-1}
%\end{figure}
% and \ref{fig:Des-2},
%\begin{figure}
%\psset{xunit=3.8mm,yunit=3.8mm,dotstyle=o,dotsize=5pt}

%\subfloat[$KNQ$ is straight]{% SECOND HEXAGON
\begin{pspicture}(-1.00,-3)(13.00,12)
\rput(6,12){$KNQ$ is straight}
\psline[linewidth=2pt](0,12)(6,0)   % Eta to Zeta
\psline[linewidth=2pt](0,-2)(7.5,3) % Mu to Epsilon
\pspolygon(7.5,3)(6,0)(3,0) % Epsilon Zeta Delta
(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma
			(0,-2)(0,12) % Mu Eta
\psset{linestyle=dashed}
\psline(3,0)(0,0) % Delta to Pi
\psline(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma
			(! 12 5 mul 11 div 
               12 5 mul 11 div) % Nu
\psline(7.5,3)(12,12) % Epsilon Kappa
(! 12 7 19 div mul 
   12 24 7 19 div mul sub)	% Gamma
\psset{linestyle=dotted}
\psline(0,0)(12,12)
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput{7pt}[15](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0)(7.5,3)(6,0) % Delta, Epsilon, Zeta
\uput[d](3,0){$D$}
\uput[dr](7.5,3){$E$}
\uput[d](6,0){$F$}
\psdot(0,12) % Eta
\uput[l](0,12){$G$}
\psdots(12,12) % Kappa
\uput[r](12,12){$K$}
\psdot(0,-2) % Mu
\uput[l](0,-2){$M$}
\psdot(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
\uput[u](! 12 5 mul 11 div 12 5 mul 11 div){$N$}
\psdots(0,0) % Pi
\uput[l](0,0){$Q$}
\end{pspicture}
%}
\hfill
%\subfloat[$HKL$ is straight]{% THIRD HEXAGON
\begin{pspicture}(-1.00,-3)(13.00,12)
\rput(6,12){$HKL$ is straight}
\psline[linewidth=2pt](3,0)(! 12 5 mul 11 div 12 5 mul 11 div) % Delta to Nu
\psline[linewidth=2pt](0,-2)(0,5) % Mu to Xi
\pspolygon(! 12 7 19 div mul 
          12 24 7 19 div mul sub)             % Gamma
					(0,5)                               % Xi
					(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
					(0,0)                               % Pi
					(3,0)                               % Delta
					(0,-2)                              % Mu
\psset{linestyle=dashed}
\psline(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
       (12,12)                             % Kappa
			 (! 12 7 19 div mul 
       12 24 7 19 div mul sub)             % Gamma
\psline(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
       (12,6)                              % Theta
			 (3,0)                               % Delta
\psline(! 12 7 19 div mul 
       12 24 7 19 div mul sub)             % Gamma
			 (12,0)                              % Lambda
       (3,0)                               % Delta
\psset{linestyle=dotted}
\psline(12,0)(12,12)
\psdot(! 12 7 19 div mul 
      12 24 7 19 div mul sub) % Gamma 
\uput{7pt}[15](! 12 7 19 div mul 
        12 24 7 19 div mul sub){$C$}
\psdots(3,0) % Delta
\uput[d](3,0){$D$}
\psdots(12,6)(12,12)(12,0) % Theta, Kappa, Lambda
\uput[r](12,6){$H$}
\uput[r](12,12){$K$}
\uput[dr](12,0){$L$}
\psdot(0,-2) % Mu
\uput[l](0,-2){$M$}
\psdot(! 12 5 mul 11 div 12 5 mul 11 div) % Nu
\uput[u](! 12 5 mul 11 div 12 5 mul 11 div){$N$}
\psdots(0,5)(0,0) % Xi, Pi
\uput[l](0,5){$P$}
\uput[l](0,0){$Q$}
\end{pspicture}
%}
\caption{Proof of Desargues's Theorem%: last steps
}\label{fig:Des-2}
\end{figure}
using in turn the hexagons
$ABGMDC$, $CDFEGM$, and $CMDQNP$.
%\gr{ABHMDG}, \gr{GDZEHM}, and \gr{GMDPN\myXi}.
Strictly,
we have now proved Desargues's Theorem
on the assumption of three axioms:
\begin{compactenum}
\item
Two points determine a line.
\item
Two lines determine a point.
\item
Pappus's Theorem is true.
\end{compactenum}
\end{sloppypar}

\addsec{Duality}

The \textbf{dual} of a statement about the projective plane
is obtained by interchanging points and lines.
Thus the dual of Pappus's Theorem
is that if the sides of a hexagon
alternately meet two points,
then the straight lines met by pairs of opposite vertices
meet a common point.
So,
in the hexagon $ABCDEF$, 
let $AB$, $CD$, and $EF$ intersect at $G$,
and let $BC$, $DE$, and $FA$ intersect at $H$,
as in Figure \ref{fig:Pappus-conv}.
\begin{figure}
\centering
\psset{xunit=3mm,yunit=2.5mm}
\begin{pspicture}(0,-1)(24,25)
\psline(12,24)(0,0)
\psline(12,24)(12,6)
\psline(12,24)(19.2,9.6)
\psline(24,12)(0,0)
\psline(24,12)(6,12)
\psline(24,12)(9.6,19.2)
\psset{linestyle=dashed,dotstyle=o,dotsize=5pt}
\psline(0,0)(12,18)
\psline(6,12)(16,16)
\psline[linestyle=dotted](19.2,9.6)(! 8 8 11 div add
                                     13 1 11 div add)
\psdots(0,0)(6,12)(12,12)(12,18)(16,16)(19.2,9.6)(12,24)(24,12)
(! 8 8 11 div add
  13 1 11 div add)
\uput[d](0,0){$A$}
\uput[l](6,12){$B$}
\uput[dl](12,12){$C$}
\uput[ur](12,18){$D$}
\uput[ur](16,16){$E$}
\uput[dr](19.2,9.6){$F$}
\uput[u](12,24){$G$}
\uput[r](24,12){$H$}
\uput[ul](! 8 8 11 div add
           13 1 11 div add){$K$}
\end{pspicture}
\caption{The dual of Pappus's Theorem}\label{fig:Pappus-conv}
\end{figure}
If the diagonals $AD$ and $BE$ meet at $K$,
then the diagonal $CF$ also passes through $K$.
For we can apply Pappus's Theorem itself
to the hexagon $ADGEBH$,
since $AGB$ and $DEH$ are straight.
Since $AD$ and $EB$ intersect at $K$,
and $DG$ and $BH$ at $C$,
and $GE$ and $HA$ at $F$,
it follows that $KCF$ is straight.

It now follows from the three axioms above
that the dual of Desargues's Theorem is true.
But the dual is precisely the converse.

\addsec[Quadrangle Theorem: second proof]{A second proof of the Quadrangle Theorem}

We now use the converse of Desargues's Theorem twice,
and the original Theorem once, 
to prove the Quadrangle Theorem.%
\index{Theorem!Quadrangle ---}
We shall show that,
in Figure \ref{fig:CQT},
\begin{figure}
\centering
\psset{xunit=8mm,yunit=6mm}
\begin{pspicture}(0,-4.5)(10,6.5)
\psline(0,0)(10,0)
\psline(6,6)(0,0)
\psline(6,6)(2,0)
\psline(6,6)(6,0)
\psline(4,4)(4,0)
\psline(4,4)(8,0)
\psline(4,3)(10,0)
\psline(4,-4)(0,0)
\psline(4,-4)(2,0)
\psline(4,-4)(6,0)
\psline(2,-2)(4,0)
\psline(2,-2)(8,0)
\psline(! 2 2 3 div add -1 1 3 div sub)(5.6,-0.8)
\psline[linestyle=dotted](5.6,-0.8)(10,0)
\uput[l](0,0){$A$}
\uput[ul](2,0){$B$}
\uput[ul](4,0){$C$}
\uput[ul](6,0){$D$}
\uput[ur](8,0){$E$}
\uput[r](10,0){$F$}
\uput[u](6,6){$G$}
\uput[ul](4,4){$H$}
\uput[l](4,3){$K$}
\uput[ur](6,2){$L$}
\uput[d](4,-4){$M$}
\uput[dl](2,-2){$N$}
\uput[l](! 2 2 3 div add -1 1 3 div sub){$P$}
\uput[dr](5.6,-0.8){$Q$}
\end{pspicture}
\caption{The Quadrangle Theorem}\label{fig:CQT}

\end{figure}
the line $PQ$ passes through $F$.
\begin{compactenum}
\item
by the converse of Desargues's Theorem
applied to triangles $GHL$ and $MNQ$,
since 
\begin{compactitem}
\item
$GH$ and $MN$ meet at $A$,
\item
$GL$ and $MQ$ meet at $D$, and
\item
$HL$ and $NQ$ meet at $E$,
\end{compactitem}
and $ADE$ is straight,
it follows that $GM$, $HN$, and $LQ$ intersect at a common point $R$ (not drawn).
\item
Likewise, in triangles $GHK$ and $MNP$,
since 
\begin{compactitem}
\item
$GH$ and $MN$ meet at $A$,
\item
$GK$ and $MP$ meet at $B$, and
\item
$HK$ and $NP$ meet at $C$,
\end{compactitem}
and $ABC$ is straight,
it follows that
$KP$ passes through the intersection point of $GM$ and $HN$,
which is $R$.
\item
So now we know that $HN$, $KP$, and $LQ$ intersect at $R$.
Therefore, by Desargues's Theorem,
the respective sides of triangles $HKL$ and $NPQ$ intersect along a straight line.
But $HK$ and $NP$ intersect at $C$, and $HL$ and $NQ$ intersect at $E$;
and $KL$ intersects $CE$ at $F$;
therefore $PQ$ must also intersect $CE$ at $F$.
\end{compactenum}

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\end{thebibliography}


\end{document}
