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\begin{document}

\title{Projective Geometry}
\author{David Pierce}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}
\date{Last edited January 2, 2019}

\maketitle

\addchap*{Preface}

This document began as a record of a two-week course 
called \emph{Geometriler}
at the Nesin Mathematics Village, \c Sirince,
Sel\c cuk, Izmir, Turkey,
September 12--25, 2016.

I gave a similar course the next summer,
July 31--August 13, 2017,
though I have for this course only a handwritten record,
in a coil-bound yellow A4 notebook from the Village.

In December of 2017,
I published a relevant article,
``Thales and the Nine-point Conic'' \cite{Pierce-Thales-9}.

The first week of my \c Sirince course was on projective geometry,
with Pappus \cite{Pappus-Pierce} as a text;
the second, hyperbolic, with Lobachevsky \cite{Lobachevski}.
The present document covers only the former week.

Otherwise,
I have spelled out many details in my course notes,
sometimes after further consultation with Hilbert \cite{Hilbert-10}
or Coxeter \cite{Coxeter-Geometry}.
I have mostly kept the original ordering of topics,
and the chapters are still titled with the days of the week.

I now draw diagrams with the \url{pst-eucl},
which has commands for drawing parallels
and for naming the intersections of straight lines
that themselves pass through named points.

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

\chapter{Monday}

\section{Quadrangle Theorem}\label{sect:quad}

Suppose five points, $A$ through $E$, 
fall on a straight line,
and $F$ is a random point not on the straight line.
Join $FA$, $FB$, and $FD$,
as in \figref{fig:4a}.
%%%%% Parameters for the next three diagrams
\newcommand{\PB}{2,0}%
\newcommand{\PF}{2.2,4}%
\newcommand{\coC}{1.4}%
\newcommand{\coD}{2}%
\newcommand{\coE}{2.4}%
\begin{figure}[hb]
  \centering
\subfloat[Point $F$ is chosen]{\label{fig:4a}
  \begin{pspicture}(-0.2,-0.6)(5,4.6)
%    \psgrid
    \pstGeonode[PosAngle={-90,-90,90}]A(\PB)B(\PF)F
    \ncline AF\ncline BF
    \pstHomO[HomCoef=\coC,PosAngle=-90]AB[C]
    \pstHomO[HomCoef=\coD,PosAngle=-90]AB[D]
    \ncline DF
    \pstHomO[HomCoef=\coE,PosAngle=-90]AB[E]
    \ncline AE
  \end{pspicture}
}
\hfill
\subfloat[Point $G$ is chosen]{\label{fig:4b}
  \begin{pspicture}(-0.2,-0.6)(5,4.6)
%    \psgrid
    \pstGeonode[PosAngle={-90,-90,90}]A(\PB)B(\PF)F
    \ncline AF\ncline BF
    \pstHomO[HomCoef=\coC,PosAngle=-90]AB[C]
    \pstHomO[HomCoef=\coD,PosAngle=-90]AB[D]
    \ncline DF
    \pstHomO[HomCoef=\coE,PosAngle=-90]AB[E]
    \ncline AE
    \pstHomO[HomCoef=0.6,PosAngle=135]AF[G]
    \ncline CG\ncline EG
%    \pstInterLL[PosAngle=180] BFCGH
%    \pstInterLL[PosAngle=45] DFEGK
  \end{pspicture}
}
\caption{Quadrangle Theorem set up}
\end{figure}
Now let $G$ be a random point on $FA$,
and join $GC$ and $GE$,
as in \figref{fig:4b}.
Supposing these two straight lines
cross $FB$ and $FD$ at $H$ and $K$ respectively,
join $HK$ as in \figref{fig:4c}.
\begin{figure}[ht]
  \centering
\begin{pspicture}(-0.2,-0.6)(7,4.6)
    %\psgrid
    \pstGeonode[PosAngle={-90,-90,90}]A(\PB)B(\PF)F
    \ncline AF\ncline BF
    \pstHomO[HomCoef=\coC,PosAngle=-90]AB[C]
    \pstHomO[HomCoef=\coD,PosAngle=-90]AB[D]
    \ncline DF
    \pstHomO[HomCoef=\coE,PosAngle=-90]AB[E]
    %\ncline AE
    \pstHomO[HomCoef=0.6,PosAngle=135]AF[G]
    \ncline CG\ncline EG
    \pstInterLL[PosAngle=180] BFCGH
    \pstInterLL[PosAngle=45] DFEGK
    \pstInterLL[PosAngle=-90]ABHKL
    \ncline AL\ncline HL
  \end{pspicture}
\caption{Quadrangle Theorem}\label{fig:4c}
\end{figure}
If this straight line crosses the original straight line $AB$ at $L$,
we shall show that
$L$ depends only on the original five points,
not on $F$ or $G$.
Let us call this 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 \figref{fig:quadrangle},
\begin{figure}[ht]
%  \psset{unit=7mm}
\mbox{}\hfill
\subfloat[]{
  \begin{pspicture}(5,3)
    \pstGeonode[PointName=none](0,2)A(2,3)B(5,2.5)C(3.5,0)D
    \ncline AB\ncline AC\ncline AD\ncline BC\ncline BD\ncline CD
  \end{pspicture}
  }
\hfill
\subfloat[]{
  \begin{pspicture}(5,3)
    \pstGeonode[PointName=none](3,1)A(0,1.5)B(5,3)C(3.5,0)D
    \ncline AB\ncline AC\ncline AD\ncline BC\ncline BD\ncline CD
  \end{pspicture}
}
\hfill\mbox{}
\caption{Complete quadrangles}\label{fig:quadrangle}
\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 \figref{fig:quadrilateral},
is a \textbf{complete quadrilateral}
(\emph{tam d\"ortkenar}).
\begin{figure}[ht]
\centering
  \begin{pspicture}(5,3)
    \psset{PointName=none}
    \pstGeonode A(3,3)B(5,0)C
    \pstGeonode[PointSymbol=none](0,4)D(5,2.5)E
    \pstInterLL AECDF
    \pstInterLL AEBCG
    \pstInterLL CDABH
    \ncline AB\ncline BC\ncline AG\ncline CH
  \end{pspicture}
\caption{Complete quadrilateral}\label{fig:quadrilateral}
\end{figure}

As stated,
the Quadrangle Theorem
is a consequence of what we shall call Lemma IV of Pappus.%
\label{IV}%
\index{Lemmas of Pappus!4@IV}
Pappus was the last great mathematician of antiquity,
and Lemma IV
is one of the lemmas in Book VII
of his \emph{Collection} \cite{Pappus,Jones,Jones-2,Pappus-Pierce}
that are intended for use with Euclid's now-lost three books of
\emph{Porisms.}
We shall prove Lemma IV in
\secref{sect:Quad-Pap}.
Lemmas I, II, V, VI, and VII (not proved in these notes) treat other cases,
such as when $HK$ in \figref{fig:4c} is parallel to $AE$.
We shall give a second proof of the Quadrangle Theorem in
\secref{sect:Quad-Des}.

\section{Thales's Theorem}\label{sect:Thales}

\subsection{Proportion Theorem}

Pappus's proofs of results such as Lemma IV
rely heavily on what for now I shall call the \textbf{Proportion Theorem.}
This is Proposition 2 of Book VI of Euclid's \emph{Elements}
\cite{MR17:814b,Oklid-2015}:
\begin{quote}
  If a straight line be drawn parallel to one of the sides of a triangle,
  it will cut the sides of the triangle proportionally, [and conversely].
\end{quote}
Symbolically, if the triangle is $ABC$ as in \figref{fig:VI.2},
\begin{figure}[ht]
%  \psset{unit=1.5cm}
  \subfloat[Internal division]{
    \begin{pspicture}(-0.2,-0.6)(3.2,3.6)
      \pstGeonode[PosAngle={-90,-90,90}] B(4,0)C(1,3)A
      \pstHomO[PosAngle={180,0},HomCoef=0.33]A{B,C}[D,E]
      \pstTranslation[PosAngle=-90]ECD[F]
      \ncline AB\ncline BC\ncline CA\ncline DE\ncline[linestyle=dotted]DF
    \end{pspicture}
    }
  \hfill
    \subfloat[External division]{
    \begin{pspicture}(-0.2,-0.6)(5.5,4.6)
      \pstGeonode[PosAngle={-90,-90,180}] B(4,0)C(2,3)A
      \pstHomO[PosAngle={90,90},HomCoef=-0.33]A{B,C}[D,E]
      \pstTranslation[PosAngle=-90]ECD[F]
      \ncline DB\ncline BC\ncline CE\ncline DE
      \ncline[linestyle=dotted]DF\ncline[linestyle=dotted]CF
    \end{pspicture}
    }
\caption{Proportion Theorem}\label{fig:VI.2}
\end{figure}
and $D$ and $E$ are on $AB$ and $AC$ respectively,
or possibly on the extensions of these bounded straight lines,
then, according to the Proportion Theorem,
\begin{equation}\label{eqn:proportion}
    DE\parallel BC
    \iff AD\rat DB\prop AE\rat EC.
\end{equation}
This result is known in some countries
as \emph{Thales's Theorem} \cite{Thales-Theorem},
but for now I want to reserve this name for a related result, as follows.

\subsection{Thales's Theorem}

In \eqref{eqn:proportion} we can read the proportion
\begin{equation}\label{eqn:ADDB}
  AD\rat DB\prop AE\rat EC
\end{equation}
in any of several ways:
\begin{compactenum}
\item
  $AD$, $DB$, $AE$, and $EC$ are proportional.
\item
  $AD$ is to $DB$ as $AE$ is to $EC$.
\item
  $AD$ has to $DB$ the same ratio that $AE$ has to $EC$.
\end{compactenum}
In particular,
the proportion expresses
not the \emph{equality} of the ratios $AD\rat DB$ and $AE\rat EC$,
but their \textbf{sameness.}

Having the same ratio is an \emph{equivalence relation.}
In particular, it is transitive.
See \secref{sect:eq-rel}.
Thus,
if \eqref{eqn:ADDB} holds, and also
\begin{equation*}
  AE\rat EC\prop AH\rat HG
\end{equation*}
as in \figref{fig:Thales},
\begin{figure}[ht]
  \centering
  \psset{xunit=15mm}
    \begin{pspicture}(5,5)
      \pstGeonode[PosAngle={-90,180,0,90}](2.6,0.6)C(0.2,1.2)B(4.4,2)G(2.2,4.4)A
%      \psgrid
      \pstHomO[HomCoef=0.4,PosAngle={-135,165,45}]A{C,B,G}[E,D,H]
      \ncline CB\ncline[linestyle=dotted] BG\ncline GC
      \ncline ED\ncline[linestyle=dotted] DH\ncline HE
%      \psset{linestyle=dashed}
      \ncline AC\ncline AB\ncline AG
    \end{pspicture}
    \caption{Thales's Theorem}\label{fig:Thales}
  
\end{figure}
then
\begin{equation*}
  AD\rat DB\prop AH\rat HG.
\end{equation*}
By the Proportion Theorem,
\begin{equation}\label{eqn:Thales}
  DE\parallel BC\And EH\parallel CG\implies DH\parallel BG.
\end{equation}
Let us call this \textbf{Thales's Theorem.}
We can count \eqref{eqn:Thales} as true,
even if $G$ lies on $AB$,
since then $DH$ and $BG$ lie on the same straight line.

Historical notes on Thales are in Appendix \ref{app:Thales}.

\subsection{Affine plane}

Truth of Thales's Theorem in the sense just defined
is a fundamental property of an \emph{affine plane.}
By definition, an \textbf{affine plane}
is a collection of points and straight lines
of which the following axioms are true.%
\newcounter{ax}
\begin{compactenum}
\item
  There exist at least three points, not all on the same straight line.
\item\label{ax:post-1}
  Any two distinct points lie on a unique straight line.
\item\label{ax:I-27-29-31}
  To a given straight line, through a given point not on the line,
  there is a unique parallel straight line.
\item
  Thales's Theorem holds.%
  \setcounter{ax}{\value{enumi}}
\end{compactenum}
We can understand axiom \ref{ax:post-1} here
as the first of Euclid's five postulates.
In axiom \ref{ax:I-27-29-31},
\begin{compactitem}
  \item
existence of the parallel is a consequence of Proposition 31
of Book I of the \emph{Elements};
\item
  uniqueness, Propositions 27 and 29,
  the latter relying on the fifth postulate.
\end{compactitem}

\subsection{Ratios}

In an affine plane,
the relation
of having the same ratio is indeed an equivalence relation,
if we take the Proportion Theorem as a \emph{definition} of proportion.
We shall do this.
Then for any triangle $ABC$ and point $D$ on $AB$,
there is unique point $E$ on $AC$ such that \eqref{eqn:ADDB} holds.
The ratio $AD:DB$ is now the equivalence class
consisting of all ordered triples $(A,E,C)$
such that $E$ lies on $AC$ and
\begin{compactitem}
  \item
    if $C$ is not on $AB$, then
    \begin{equation*}
  DE\parallel BC;
    \end{equation*}
    \item
if $C$ is on $AB$,
then for some $G$ not on $AB$ and some $H$ on $AG$,
\begin{equation*}
  DH\parallel BG\And HE\parallel GC.
\end{equation*}
\end{compactitem}

\subsection{Product of ratios}

We can now define the product of two ratios.
In \figref{fig:assoc},
  \begin{figure}[ht]
    \centering
%    \psset{unit=8mm}
    \begin{pspicture}(-2.8,-1.3)(6,3.9)
%      \psgrid
      \pstGeonode[PosAngle={45,0,90,90}](5.4,3.4)O(5.4,1)C(4,1.5)B(3,2.9)A
          \pstHomO[HomCoef=1.5,PosAngle=90]OA[A']
          \pstHomO[HomCoef=1.7,PosAngle=-60]OB[B']
          \pstHomO[HomCoef=1.3]OC[C']
          \pstTranslation[PointName=none,PointSymbol=none] B{A'}{B'}[X]
          \pstInterLL[PosAngle=90]{B'}XOA{A''}
          \pstTranslation[PointName=none,PointSymbol=none] C{B'}{C'}[X]
          \pstInterLL[PosAngle=-60]{C'}XOB{B''}
          \pstTranslation[PointName=none,PointSymbol=none] {B'}{A''}{B''}[X]
          \pstInterLL[PosAngle=90]{B''}XOA{A'''}
          \ncline O{A'''}\ncline O{B''}\ncline O{C'}
          \ncline B{A'}\ncline{B'}{A''}
          \ncline C{B'}\ncline{C'}{B''}
          \ncline{B'}D\ncline{B''}{A'''}
          \psset{linestyle=dotted}
          \ncline C{A''}\ncline{C'}{A'''}
    \end{pspicture}
    \caption{Multiplication of ratios}\label{fig:assoc}
    
  \end{figure}
  with
  \begin{equation}\label{eqn:A'B}
    A'B\parallel A''B',
  \end{equation}
%  then, denoting multiplication by $\&$,
  we let
  \begin{equation}\label{eqn:OAAA'}
    \compratio{OA}{AA'}{OB}{BB'}\prop OA\rat AA''.
  \end{equation}
  Since ratios are equivalence classes,
  we have to confirm that this is a valid definition;
  but it is, by Thales's Theorem.
  As a special case, from which we can now derive \eqref{eqn:OAAA'}, we have
  \begin{equation}\label{eqn:comprat}
    \compratio{OA}{AA'}{OA'}{A'A''}\prop OA\rat AA''.
  \end{equation}

  \subsection{Associativity of multiplication}
  
  Thales's Theorem gives us also associativity of multiplication,
  since, still in \figref{fig:assoc}, if, in addition to \eqref{eqn:A'B}, also
  \begin{align*}
    A''B'&\parallel A'''B'',&
    B'C&\parallel B''C',
  \end{align*}
then
  \begin{gather*}
    \compratio{OB}{BB'}{OC}{CC'}\prop OB\rat BB'',\\
    \compratio{OA}{AA'}{OB}{BB''}\prop OA\rat AA'''.
  \end{gather*}
  By this and \eqref{eqn:comprat},
  \begin{multline*}
    (OA\rat AA')\bigl(\compratio{OB}{BB'}{OC}{CC'}\bigr)\\
    \prop\bigl(\compratio{OA}{AA'}{OB}{BB'}\bigr)(OC\rat CC')\\
    \iff A'''C'\parallel A''C;
  \end{multline*}
  but the parallelism holds by Thales's Theorem.
  
  The commutativity of multiplication of ratios
  will need Pappus's Lemma VIII,
  which is one case of Pappus's Theorem,
  to be defined in \secref{sect:Pap}.

\section{Equivalence relations}\label{sect:eq-rel}

\subsection{Equality}

In Euclid,
two bounded straight lines may be \emph{equal}
without being the same.
For example, in an isosceles triangle,
two of the sides are equal.
Equality of bounded straight lines is an \textbf{equivalence relation.}
This means equality is \emph{transitive, symmetric,} and \emph{reflexive.}
\begin{compactenum}
  \item
Equality is transitive,
by the first of the Common Notions in Euclid's \emph{Elements}:
\begin{quote}\centering
  Equals to the same thing are equal to one another.
\end{quote}
\item
That equality is symmetric
is implicit in the same common notion,
as well as in one of the definitions at the head of the \emph{Elements}
that we have just alluded to:
\begin{quote}
  Of trilateral figures,
  \begin{compactitem}
    \item
      an \textbf{equilateral triangle}
      is that which has its three sides equal;
      \item
        an \textbf{isosceles triangle,}
        that which has two of its sides alone equal; and
      \item
        a \textbf{scalene triangle,} that which has its three sides unequal.
  \end{compactitem}
\end{quote}
If the sides $AB$ and $AC$ of a triangle are equal,
we can write this indifferently as $AB=AC$ or $AC=AB$.
\item
  That equality is reflexive in the \emph{Elements}
  is seen in how Proposition 4 of Book I is applied.
This proposition is the theorem about triangles
that we now call Side-Angle-Side, or SAS.
In triangles $ABC$ and $DEF$ of \figref{fig:I.4},
\begin{figure}[ht]
  \psset{PointSymbol=none}
  \subfloat[I.4]{\label{fig:I.4}
  \begin{pspicture}(3,4)
    \pstTriangle[PosAngleA=-90,PosAngleB=-90,PosAngleC=90]
  (0.2,0.6)B(2.8,0.6)C(2,3.4)A
  \end{pspicture}
  \begin{pspicture}(3,4)
    \pstTriangle[PosAngleA=-90,PosAngleB=-90,PosAngleC=90]
    (0.2,0.6)E(2.8,0.6)F(2,3.4)D
  \end{pspicture}
  }
  \hfill
  \subfloat[I.5]{\label{fig:I.5}
    \begin{pspicture}(4,4)
      \pstGeonode[PosAngle={-90,-90,90}](0.6,0.6)D(3.4,0.6)E(2,3.4)A
      \pstHomO[HomCoef=0.6,PosAngle={180,0}]A{D,E}[B,C]
      \pstHomO[HomCoef=0.85,PosAngle={180,0}]A{D,E}[F,G]
      \ncline DA\ncline AE\ncline BC\ncline FC\ncline BG
    \end{pspicture}
    }
  \caption{\emph{Elements} Propositions I.4 and 5}
\end{figure}
\begin{equation}\label{eqn:I.4}
  \left.
  \begin{gathered}
    BA=ED\\
    \angle BAC=\angle EDF\\
    AC=DF
  \end{gathered}
  \right\}
  \implies
  \left\{
  \begin{gathered}
    \angle ABC=\angle DEF\\
    AC=DF\\
    \angle ACB=\angle DFE\\
    \triangle ABC=\triangle ABC.
  \end{gathered}
  \right.
\end{equation}
The equality of triangles here
is in the sense to be discussed in \secref{sect:=}.
The point now is that,
when \eqref{eqn:I.4} is applied in Proposition 5,
to prove that the base angles of isosceles triangle $ABC$
in \figref{fig:I.5} are equal,
we have $AB=AC$, and we make also $AF=AG$,
and therefore angles $ABG$ and $ACF$ are equal,
since also angles $BAG$ and $CAF$ are equal,
being the same angle.
\end{compactenum}
Equality being an equivalence relation,
we may say that equal straight lines have the same \textbf{length.}
Length is an abstraction, which we cannot draw in a diagram.
We can define the length of a line $AB$ 
as the equivalence class,
denoted by
\begin{equation*}
  \size{AB},
\end{equation*}
consisting of all of the straight lines $XY$
such that $AB=XY$.

Equality has a criterion in
the fourth of Euclid's Common Notions
that his editor Heiberg \cite{Euclid-Heiberg} accepts as genuine:
\begin{quote}\centering
  Things congruent with one another\\
  are equal to one another.
\end{quote}
I discuss this in ``On Commensurability and Symmetry'' \cite{MR3680853}.
Two equal straight lines can have different directions and endpoints:
this is seen in
\begin{compactitem}
  \item
Euclid's third postulate, that a circle can be drawn with any center
and passing through any other point;
\item
Proposition I.3,
whereby, from any straight line,
we can cut off a part that is equal to any shorter straight line.
\end{compactitem}
We shall be interested in seeing how far we can go,
treating only opposite sides of a parallelogram as equal.
This is the only kind of equality that we can talk about in an affine plane.
It is also the equality of which our sign $=$ of equality is an icon;
see the paper \cite{MR3680853} just mentioned.


\subsection{Sameness of ratio}

Being an equivalence relation is even more fundamental to \emph{sameness}
than to equality.
In ancient Greek mathematics at least,
any definition of proportion
should make it obvious
that sameness of ratio is indeed an equivalence relation.
There are two theories of proportion in the \emph{Elements}:
\begin{compactenum}[1)]
  \item
for \emph{magnitudes,}
such as bounded straight lines,
in Books V and VI;
\item
for \emph{numbers,}
in Books VII, VIII, and IX.
\end{compactenum}
By the definition at the head of Book VII,
\begin{quote}
  [Four] numbers are proportional when the first is the same multiple,
  or the same part, or the same parts,
  of the second that the third is of the fourth.
\end{quote}
If we take seriously the use of the word ``same'' here,
then,
in the context of the whole of Book VII of the \emph{Elements,}
the definition of proportion of numbers must mean that,
for counting numbers $k$, $\ell$, $m$, and $n$,
we have $k:\ell\prop m:n$
precisely when the Euclidean algorithm has the same steps,
whether applied to $(k,\ell)$ or $(m,n)$.
Thus for example
$32:14\prop 48:21$, because of the common sequence $(2,3,2)$
of multipliers in the computations
\begin{align*}
  &
  \begin{gathered}
    32=14\cdot2+4,\\
    14=4\cdot3+2,\\
    4=2\cdot2,
  \end{gathered},&
  &
  \begin{gathered}
    48=21\cdot2+6,\\
    21=6\cdot3+3,\\
    6=3\cdot2.
  \end{gathered}  
\end{align*}
I have written about this elsewhere
\cite{Pierce-geo-numbers}.

We shall look at Euclid's definition
(and Hilbert's definition) of proportion of bounded straight lines
in \secref{sect:prop-def}.

As we cannot draw lengths as such in a diagram,
so can we not draw ratios.

\subsection{Parallelism}

Parallelism is transitive by Proposition I.30 of the \emph{Elements.}
Since it is obviously symmetric,
it is an equivalence relation,
provided we understand a straight line to be parallel to itself.

That parallelism is transitive is also a theorem about affine planes.
If $AB\parallel CD$ as in \figref{fig:para},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(5,4)
    \psset{PointSymbol=none}
    \pstGeonode
        [PosAngle={180,180,0,30},PointName={default,none,default,default}]
        (0.6,0.2)A(1.6,3.8)F(2,0.2)C(4.6,1)E
        \pstTranslation[PointName=none] AFC[G]
        \pstInterLL[PosAngle=30] CGEFD
        \pstTranslation[PosAngle=180] CDA[B]
        \ncline AB\ncline CD\ncline ED
        \psset{linestyle=dotted}
        \ncline BF\ncline DF
  \end{pspicture}
  \caption{Parallelism in an affine plane}
  \label{fig:para}
\end{figure}
but a third line $ED$ meets $CD$ at $D$,
then, $CD$ being the only parallel to $AB$ that passes through $C$,
$ED$ must meet $AB$ somewhere.

\subsection{Sameness of direction and length}

Since parallelism is an equivalence relation,
we may say that parallel straight lines have the same \textbf{direction.}
Hence having the same direction and length is an equivalence relation.

In Euclid and Pappus,
expressions such as $AB$ and $BA$
for bounded straight lines are interchangeable.
We may however distinguish them,
considering $AB$ as the ordered pair $(A,B)$.
As in \secref{sect:Thales}
we assigned to an ordered triple $(A,D,B)$ the ratio $AD:DB$,
so now we assign to $AB$, considered as $(A,B)$,
the \textbf{directed length,}
or \textbf{vector,}
denoted by
\begin{equation*}
  \vec AB.
\end{equation*}
This is the equivalence class consisting of all ordered pairs
$(C,D)$ of points such that
\begin{align*}
  AB&\parallel CD,&AB&=CD,
\end{align*}
\emph{and} $A$ is on the same side of $B$ that $C$ is of $D$.
However, this last condition is imprecise.

Propositions 33 and 34
of Book I of Euclid's \emph{Elements}
constitute what we shall call the
\textbf{Equality Theorem:}
two bounded straight lines that are not part of one straight line,
but are parallel,
are equal if and only if they are the sides of a parallelogram.
Now we can say that $\vec AB$ consists of those $(C,D)$ 
such that
\begin{compactitem}
\item
  if $A$ and $B$ are the same, then so are $C$ and $D$;
\item
  if $C$ does not lie on $AB$, then $ABDC$ is a parallelogram;
\item
  if $C$ does lie on $AB$, then for some $E$ and $F$,
  both $ABEF$ and $EFCD$ are parallelograms.
\end{compactitem}
If $(A,B)$ and $(C,D)$ represent the same vector,
we may write
\begin{equation*}
  \vec AB::\vec CD.
\end{equation*}
If $\vec AB::\vec AC$, then $B$ and $C$ must be the same point.
The Side-Angle-Side Theorem,
discussed in \secref{sect:eq-rel},
now takes the form that, in two triangles $ABC$ and $DEF$,
\begin{equation*}
  \vec AB::\vec DE\And\vec BC::\vec EF\implies\vec AC::\vec DF.
\end{equation*}
Let us call this the \textbf{Prism Theorem,}
even though a prism is normally a solid figure,
and we are working in a plane.

More precisely,
if,
as in Figure \ref{fig:prism},
 $ABC$ is a triangle,
$ABED$ is a parallelogram,
and $AD$ does not lie along $AC$,
but $DF$ is drawn parallel to $AC$,
the Prism Theorem is
\begin{equation}\label{eqn:Prism}
  AD\parallel CF\implies BC\parallel EF.
\end{equation}
This is an easy consequence of the Equality Theorem.
For suppose now, in \figref{fig:prism},
in addition to the conditions already stated,
\begin{equation}\label{eqn:ADCF}
  AD\parallel CF.
\end{equation}
Then
\begin{align*}
  BE&=AD,&AD&=CF,
\end{align*}
by the Equality Theorem.
Since equality is transitive,
\begin{equation*}
  BE=CF.
\end{equation*}
Again by the Equality Theorem,
\begin{equation}\label{eqn:BCEF}
  BC\parallel EF.
\end{equation}
This gives \eqref{eqn:Prism}.
\begin{comment}
  The converse follows easily by contradiction.
  For, supposing \eqref{eqn:BCEF} holds,
  but \eqref{eqn:ADCF} fails,
  since $AC$ and $DF$ are parallel,
  we conclude that they cannot be equal.
  Along $DF$, extended if necessary,
  we can find, as in \figref{fig:prism-Euclid},
  $F'$ so that $DF'=AC$,
  and then $AD\parallel CF'$,
  so $BC\parallel EF'$ by the first direction of the Prism Theorem.
  Both $EF'$ and $EF$ cannot be parallel to $BC$
  unless they are the same line.
  In this way, $F'$ and $F$ must coincide after all.
\end{comment}

Euclid proves the Equality Theorem,
already having equality as an equivalence relation,
in the sense discussed above.
We can use the Theorem as a \emph{definition}
of equality of parallel bounded straight lines,
provided we know that equality so defined will be transitive;
but then the Prism Theorem guarantees this transitivity,
just as Thales's Theorem guarantees that sameness of ratio,
as given by the Proportion Theorem,
is transitive.




\section{Desargues's Theorem}\label{sect:Des}

The Prism Theorem and Thales's Theorem are specials case of
\textbf{Desargues's Theorem.}
We shall use this result for
the second proof of the Quadrangle Theorem,
mentioned at the end of \secref{sect:quad}.
Desargues was a contemporary of Des\-cartes,
and the theorem named for him concerns two triangles.
If these are $ABC$ and $DEF$,
we assume that
the lines $AD$, $BE$, and $CF$ that connect corresponding vertices
either
\begin{compactitem}
  \item
    have a common point $G$, or
  \item
    are parallel to one another.
\end{compactitem}
There are only the following three possibilities
for the pairs $\{BC,EF\}$, $\{AC,DF\}$, and $\{AB,DE\}$
  of corresponding sides.
  \begin{compactdesc}
    \item[Parallelism:]
      each pair are parallel.
    \item[Intersecting:]
      each pair intersect,
      and the three intersection points lie along a common straight line.
    \item[Mixing:]
      Two pairs intersect,
      and the line that the two intersection points determine
      is parallel to each line in the third pair.
  \end{compactdesc}
  That is Desargues's Theorem,
  and there are six cases in all.
  The two cases of parallelism are, again,
  The other cases are shown in \figref{fig:Des-mix}
  \begin{figure}
  \subfloat[]{
    \begin{pspicture}(4.5,5)
      \pstGeonode[PosAngle={-90,90,90,-90}]
      (1,0.6)D(3.5,4.4)B(0.2,4.4)A(2.5,1.5)C
      \pstTranslation[PosAngle=-45] ABD[E]
      \pstTranslation[PointSymbol=none,PointName=none] DAC
      \pstHomO[HomCoef=0.55,PosAngle=90]C{C'}[F]
      \pstInterLL[PosAngle=180] ACDFK
      \pstInterLL BCEFL
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline EF\ncline FD
      \psset{linestyle=dashed}
      \ncline AD\ncline BE\ncline CF
      \ncline[linestyle=dotted]KL
    \end{pspicture}
    }
  \hfill
    \subfloat[]{
    \begin{pspicture}(6,6)
      \pstGeonode[PosAngle={90,90,-165,-90}]
      (5.8,5.4)B(-0.8,5.4)A(2.8,2.4)C(2.5,0.6)G
      \pstHomO[HomCoef=0.3,PosAngle={-135,-45}]G{A,B}[D,E]
      \pstHomO[HomCoef=2.3]GC[F]
      \pstInterLL[PosAngle=-165] ACDFK
      \pstInterLL[PosAngle=-30] BCEFL
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline EF\ncline FD
      \psset{linestyle=dashed}
      \ncline AG\ncline BG\ncline GF
      \ncline[linestyle=dotted]KL
    \end{pspicture}
    }
\caption{Desargues's Theorem, mixed cases}\label{fig:Des-mix}
  
\end{figure}
and \figref{fig:Des-int}.
\begin{figure}
\subfloat[]{\label{fig:Des-proj}
  \begin{pspicture}(5,5)
%    \psgrid[subgriddiv=1]
    \pstGeonode[PosAngle={90,-90,90,90}]
    (0.6,4.4)A(0.6,0.6)B(3.5,3.5)C(1.5,3)G
    \pstHomO[HomCoef=-1.5,PosAngle=-90]GA[D]
    \pstHomO[HomCoef=0.5,PosAngle=-75]GB[E]
    \pstHomO[HomCoef=1.8,PosAngle=90]GC[F]
      \pstInterLL[PosAngle=180] ABDEH
      \pstInterLL[PosAngle=-45] ACDFK
      \pstInterLL[PosAngle=-45] BCEFL
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline EF\ncline FD
      \psset{linestyle=dashed}
      \ncline EH\ncline CK
      \ncline AD\ncline BG\ncline GF
      \ncline[linestyle=dotted]KH
  \end{pspicture}
}
\hfill
\subfloat[]{
    \begin{pspicture}(5,5)
%      \psgrid[subgriddiv=1]
      \psset{PointName=none}
    \pstGeonode(4.8,1.5)A(3,3.5)B(2,0.6)C(-1,1.5)D
    \pstTranslation[PointSymbol=none] ADB
    \pstHomO[HomCoef=0.3]B{B'}[E]
    \pstTranslation[PointSymbol=none] ADC
    \pstHomO[HomCoef=-0.3]C{C'}[F]
    \pstInterLL ABDEL
    \pstInterLL ACDFK
    \pstInterLL BCEFH
    \ncline[linestyle=dotted]KL
    \ncline AB\ncline BC\ncline CA
    \ncline DE\ncline EF\ncline FD
    \psset{linestyle=dashed}
    \ncline EL\ncline BL
    \ncline FK\ncline CK
    \ncline FH\ncline CH
    \ncline AD\ncline BE\ncline CF
    \nput{-90}A{$A$}
    \nput{45}B{$B$}
    \nput{-90}C{$C$}
    \nput{-90}D{$D$}
    \nput{135}E{$E$}
    \nput{-90}F{$F$}
    \nput{195}H{$H$}
    \nput{60}K{$K$}
    \nput{90}L{$L$}
  \end{pspicture}
}
  
  \caption{Desargues's Theorem, intersecting cases}\label{fig:Des-int}
  
\end{figure}

\subsection{Parallel cases in an affine plane}

By definition, in every affine plane,
Thales's Theorem is true.
The Prism Theorem is also true in every affine plane.
To prove this, we first establish a converse of Thales's Theorem.
In \figref{fig:Thales-conv},
\begin{figure}[ht]
  \centering
  \psset{xunit=15mm}
    \begin{pspicture}(5,5)
      \pstGeonode[PosAngle={-90,180,0,90}](2.6,0.6)C(0.2,1.2)B(4.4,2)A(2.2,4.4)G
%      \psgrid
      \pstHomO[HomCoef=0.4,PosAngle={-135,165,45}]G{C,B,A}[F',E,D]
      \pstHomO[HomCoef=1.4,PosAngle=-90]E{F'}[F]
      \ncline CB\ncline BA\ncline AC
      \ncline FE\ncline ED\ncline DF
      \ncline GB\ncline GA
      \psset{linestyle=dotted}
      \ncline GC\ncline D{F'}
    \end{pspicture}
    \caption{Converse of Thales's Theorem}\label{fig:Thales-conv}
  
\end{figure}
assuming that
$DE$ is drawn parallel to $AB$
in the triangle $GAB$,
and the sides $AC$ and $DF$
of the triangles $ABC$ and $DEF$
are also parallel,
we have
\begin{equation*}
  BC\parallel EF\implies\text{$F$ lies on $GC$}.
\end{equation*}
For, if $BC\parallel EF$, but $F$ is not on $GC$,
this intersects $EF$ at some other point $F'$.
Then Thales's Theorem applies, yielding $DF'\parallel AC$.
Thus $F'$ lies on the straight line through $D$ that is parallel to $AC$.
This line being $DF$,
$F'$ must lie on this, as well as on $EF$.
Only one point can do this, and that point is $F$.
So $F'$ and $F$ must be the same point after all,
and $F$ lies on $GC$.

Next we establish a converse of the Prism Theorem.
In \figref{fig:prism-Thales},
\begin{figure}[ht]
  \subfloat[Statement]{\label{fig:prism}
    \begin{pspicture}(0.2,0)(3.6,5)
%      \psgrid
      \pstGeonode[PosAngle={-90,-90,-120,90}]
      (1.8,0.6)A(3.4,1)B(0.6,1.5)C(0.6,4.4)F
      \pstTranslation[PosAngle=-45] CFA[D]
      \pstTranslation[PosAngle=90] CFB[E]
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline[linestyle=dotted] EF\ncline FD
      \psset{linestyle=dashed}
      \ncline DA\ncline EB\ncline[linestyle=dotted] FC
    \end{pspicture}
  }
  \hfill
    \subfloat[Proof of converse]{\label{fig:prism-Thales}
    \begin{pspicture}(0.2,0)(3.6,5)
%      \psgrid
      \pstGeonode[PosAngle={-90,-90,-120,90}]
      (1.8,0.6)A(3.4,1)B(0.6,1.5)C(3.4,3)E
      \pstTranslation[PosAngle=-45] BEA[D]
      \pstHomO[PosAngle=180,HomCoef=2.1]AD[G]
      \pstTranslation[PointSymbol=none,PointName=none]ACD
      \pstInterLL[PosAngle=180] CGD{D'}F
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline[linestyle=dashed] EF\ncline FD
%      \psset{linestyle=dashed}
      \ncline GA\ncline EB\ncline[linestyle=dashed] GC
    \end{pspicture}
    }
    \hfill
  \subfloat[Proof from converse]{\label{fig:prism-Euclid}
    \begin{pspicture}(0.2,0)(3.6,5)
%      \psgrid
      \pstGeonode[PosAngle={-90,-90,-120,90}]
      (1.8,0.6)A(3.4,1)B(0.6,1.5)C(0.6,4.4)F
      \pstTranslation[PosAngle=-45] CFA[D]
      \pstTranslation[PosAngle=90] CFB[E]
      \ncline AB\ncline BC\ncline CA
      \ncline DE\ncline[linestyle=dashed] EF\ncline FD
%      \psset{linestyle=dashed}
      \ncline DA\ncline EB\ncline[linestyle=dashed] FC
      \pstHomO[HomCoef=0.6,PosAngle=-75]DF[F']
      \ncline[linestyle=dotted]{F'}C
      \ncline[linestyle=dotted]{F'}E
    \end{pspicture}
  }
  \caption{Prism Theorem}
  
\end{figure}
$ABED$ is a parallelogram
and $AC\parallel DF$.
If $CF\nparallel AD$,
then they meet at a point $G$.
Since $E$ does not lie on $BG$,
we conclude by the converse of Thales's Theorem
that $BC\nparallel EF$.
By contraposition, the converse of \eqref{eqn:Prism} holds.

Finally we prove the Prism Theorem itself in an affine plane.
In \figref{fig:prism-Euclid},
$ABED$ is a parallelogram and
$AC\parallel DF$.
If $BC\nparallel EF$,
then $BC\parallel EF'$ for some $F'$ on $DF$,
and therefore $AD\parallel CF'$ by what we have just proved,
so $CF\nparallel AD$.

\subsection{Equality in proportions}

In an affine plane,
if some bounded straight line appears in a proportion,
we may now replace it with a bounded straight line
representing the same vector.
For if,
as in Figure \ref{fig:prop-prism},
\begin{figure}
  \centering
  \begin{pspicture}(8,5.3)
%    \psgrid
    \pstGeonode[PosAngle={90,-90,0,180}](2,3)A(3,0.6)B(7.4,2.2){B'}(0.6,1.1)G
    \pstHomO[HomCoef=0.6,PosAngle=-120]AB[C]
    \pstTranslation[PointName=none,PointSymbol=none] BGC[X]
    \pstInterLL[PosAngle=150] CXAGH
    \pstTranslation[PosAngle={90,30,120,165}]
    B{B'}{A,C,H,G}[{A'},{C'},{H'},{G'}]
    \ncline AB\ncline AG\ncline BG
    \ncline {A'}{B'}\ncline {A'}{G'}\ncline {B'}{G'}
    \psset{linestyle=dashed}
    \ncline CH
    \ncline {C'}{H'}
    \psset{linestyle=dotted}
    \ncline A{A'}\ncline B{B'}\ncline C{C'}\ncline G{G'}\ncline H{H'}
  \end{pspicture}
  \caption{Proportions from the Prism Theorem}\label{fig:prop-prism}
  
\end{figure}
\begin{equation}\label{eqn:ACCB}
  AC\rat CB\prop AH\rat HG,
\end{equation}
so that $HC\parallel GB$,
and if all of the straight lines $XX'$ are parallel to one another,
and also
\begin{align*}
  AB&\parallel A'B',&AG&\parallel A'G',
\end{align*}
so that the $XX'$ all represent the same vector,
then
\begin{equation*}
  H'C'\parallel HC\parallel GB\parallel G'B',
\end{equation*}
so
\begin{equation*}
  A'C':C'B'\prop A'H':H'G'.
\end{equation*}
Therefore, in \eqref{eqn:ACCB},
we can replace any particular $XY$ with $X'Y'$.
Thus, back in \figref{fig:VI.2},
if $DF\parallel AC$, then
\begin{equation*}
  BD\rat DA\prop BF\rat FC\prop BF\rat DE.
\end{equation*}
We can conclude from this,
as an alternative form of the Proportion Theorem,
still in \figref{fig:VI.2},
\begin{equation*}
  BA\rat DA\prop BC\rat DE.
\end{equation*}
For, if we augment \figref{fig:VI.2} as in \figref{fig:componendo},
\begin{figure}[ht]
%  \psset{unit=8mm}
  \subfloat[Composition]{
    \begin{pspicture}(-0.2,-0.6)(4.8,4.6)
%      \psgrid
      \pstGeonode[PosAngle={-90,-90,180}] B(3.5,0)C(1,3)A
      \pstHomO[PosAngle={180,0},HomCoef=0.33]A{B,C}[D,E]
      \pstTranslation[PosAngle=-90]ECD[F]
      \ncline AB\ncline BC\ncline CA\ncline DE\ncline[linestyle=dotted]DF
      \pstTranslation[PosAngle=-90] FCC[F']
      \pstTranslation[PosAngle=90] DAA[D']
      \psset{linestyle=dashed}
      \ncline A{D'}\ncline{D'}{F'}\ncline{F'}C
      \pstTranslation DEA[G]
      \psset{linestyle=dotted}
      \ncline AG\ncline GE      
    \end{pspicture}
    }
  \hfill
    \subfloat[Separation]{
      \begin{pspicture}(-0.2,-0.6)(4.8,4.6)
%        \psgrid
      \pstGeonode[PosAngle={-90,-90,0}] B(3.5,0)C(2,3)A
      \pstHomO[PosAngle={90,90},HomCoef=-0.33]A{B,C}[D,E]
      \pstTranslation[PosAngle=-90]ECD[F]
      \ncline DB\ncline BC\ncline CE\ncline DE
      \ncline[linestyle=dotted]DF\ncline[linestyle=dotted]CF
      \pstTranslation[PosAngle=-90] FCC[F']
      \pstTranslation[PosAngle=180] DAA[D']
      \psset{linestyle=dashed}
      \ncline{F'}{D'}
      \pstTranslation[PosAngle=180] DEA[G]
      \psset{linestyle=dotted}
      \ncline AG\ncline GE\ncline G{D'}     
    \end{pspicture}
    }
\caption{Composition and separation of ratios}\label{fig:componendo}
\end{figure}
where now $ADEG$ is a parallelogram,
and the straight line through $G$ parallel to $AC$
cuts $AB$ and $AC$ at $D'$ and $F'$ respectively, then
\begin{equation*}
  BA\rat A{D'}\prop BC\rat C{F'},
\end{equation*}
but also
\begin{align*}
  AD'&=DA,&CF'&=FC=DE.
\end{align*}
We can therefore write the rule \eqref{eqn:comprat}
for multiplication of ratios as
\begin{equation*}
  \compratio{OA}{OA'}{OA'}{OA''}\prop OA\rat OA''.
\end{equation*}
This becomes more succinct when we write $a$ for $OA$, and so forth:
  \begin{equation*}
    \compratio{a}{a'}{a'}{a''}\prop a\rat a''.
  \end{equation*}


\section{Pappus's Theorem}\label{sect:Pap}

We shall prove Desargues's Theorem in general
in \secref{sect:Des-proof},
by means of
\textbf{Pappus's Hexagon Theorem.}%
\index{Theorem!Pappus's Hexagon ---}
This concerns a hexagon, such as $ABCDEF$,
whose vertices
lie alternately on two straight lines.
Thus $AC$ contains $E$, and $BD$ contains $F$,
and the two straight lines either
\begin{compactitem}
  \item
    have a common point $G$, or
  \item
    are parallel to one another.
\end{compactitem}
The following are the only three possibilities
for the pairs $\{AB,DE\}$, $\{BC,EF\}$, and $\{CD,FA\}$
  of opposite sides of the hexagon.
  \begin{compactdesc}
    \item[Parallelism:]
      each pair are parallel.
    \item[Intersecting:]
      each pair intersect,
      and the three intersection points lie along a common straight line.
    \item[Mixing:]
      Two pairs intersect,
      and the line that the two intersection points determine
      is parallel to each line in the third pair.
  \end{compactdesc}
  That is Pappus's Theorem,
  and just as for Desargues's Theorem,
  there are six cases in all.

Pascal generalized the Hexagon Theorem,
though without proof,
to allow the vertices of the hexagon to lie on an arbitrary conic section
\cite{Pascal-Conics,Pascal-Coniques}.

Pappus himself proves three cases of the Hexagon Theorem:
\begin{compactenum}[1)]
\item
  the parallel case with vertices on intersecting lines,
  as Lemma VIII;
\item
  the intersecting case
   with vertices on parallel lines,
  as Lemma XII;
\item
  the intersecting case
   with vertices on intersecting lines,
  as Lemma XIII.
\end{compactenum}
The proofs of the last two lemmas will use Lemmas III, X, and XI,
concerning the \emph{cross ratio} of four straight lines.
We shall take up
\begin{compactitem}
\item
  Lemma VIII, and the other parallel case,
  in \secref{sect:VIII};
\item
  Lemmas III and X in \secref{sect:III};
\item
  Lemmas XI, XII, and XIII
  in \secref{sect:XI};
\item
the mixed cases in \secref{sect:Pap-proj}.
\end{compactitem}
Lemma VIII is illustrated in Figure \ref{fig:VIII-statement},
  \begin{figure}
    \centering
        \begin{pspicture}(-1.2,-0.5)(6,4.5)
%      \psgrid
    \pstGeonode[PosAngle={150,90,0}](-1,0)E(3,4)G(5.8,0)D
    \pstHomO[HomCoef=0.3,PosAngle=150]GE[A]
    \pstHomO[HomCoef=0.45,PosAngle=150]GE[C]
    \pstTranslation[PointSymbol=none,PointName=none] EDA
    \pstInterLL A{A'}GDB
    \pstTranslation[PointSymbol=none,PointName=none] CDA
    \pstInterLL A{A'}GDF
    \ncline BC\ncline CD\ncline EF\ncline FA
    \psset{linestyle=dashed}
    \ncline DE
    \ncline BA
    %    \psset{linestyle=dotted}
%    \ncline AD\ncline CF\ncline EB
    \psset{linewidth=2pt,linestyle=solid}
    \ncline EG\ncline GD
  \end{pspicture}

        \caption{Commutativity of multiplication of ratios}
        \label{fig:VIII-statement}
    
  \end{figure}
  where
  \begin{equation*}
    BC\parallel EF\And CD\parallel FA\implies AB\parallel DE.
  \end{equation*}
  This just means multiplication of ratios is commutative,
  since we have now
  \begin{multline*}
    \compratio{GA}{AC}{GB}{BF}\prop GA\rat AE\\
    \prop GB\rat BD
    \prop \compratio{GB}{BF}{GA}{AC}.
  \end{multline*}
  
Pappus's proof of Lemma VIII
uses only Propositions I.37 and 39 of Euclid's \emph{Elements,}
whereby,
under the hypothesis that two triangles have the same base,
as in \figref{fig:tri},
\begin{figure}[ht]
  \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}
the straight line joining the apices of the triangles
is parallel to the common base
just in case
the triangles are \emph{equal} to one another.

\section{Equality of polygons}\label{sect:=}

\textbf{Equality} of triangles means not congruence of the triangles,
but sameness of their areas.
We have seen in \secref{sect:Thales}
that congruence is \emph{one} way to establish this sameness.
Euclid's proof of his Proposition I.35,
%\label{I.35}
that \emph{parallelograms}
on the same base and in the same parallels are equal,
is by
\begin{compactenum}[1)]
\item
  adding congruent pieces to the parallelograms, then
\item
  dividing the resulting polygons into congruent pieces.
\end{compactenum}
Thus in \figref{fig:I.35},
\begin{figure}[ht]
  \psset{PointName=none,PointSymbol=none,unit=7mm}
  \subfloat[Euclid's case]{\label{fig:I.35}
    \begin{pspicture}(-1,0)(6.5,3)
%      \psgrid
      \pstGeonode A(3,0)B(-1,3)C
      \pstTranslation ABC[D]
      \pstHomO[HomCoef=1.5]CD[E]
      \pstTranslation ABE[F]
      \ncline AB\ncline BD\ncline AC\ncline AE\ncline BF\ncline CF
      \pstInterLL AEBDG
      \rput(0.4,2){$\alpha$}
      \rput(1.8,0.6){$\beta$}
      \rput(2.5,2.6){$\gamma$}
      \rput(4,2){$\delta$}
    \end{pspicture}
  }
  \hfill
%      \psset{unit=5mm}
  \subfloat[Simpler case]{\label{fig:I.35-2}
  \begin{pspicture}(-1,0)(6.5,3)
%      \psgrid
      \pstGeonode A(4,0)B(-1,3)C
      \pstTranslation ABC[D]
      \pstHomO[HomCoef=0.875]CD[E]
      \pstTranslation ABE[F]
      \ncline AB\ncline BD\ncline AC\ncline AE\ncline BF\ncline CF
      \rput(0.4,2){$\alpha$}
      \rput(2.2,1.3){$\beta$}
      \rput(4.2,2){$\delta$}
    \end{pspicture}
  }
  \caption{Euclid's Proposition I.35}
\end{figure}
\begin{equation*}
  \alpha+\gamma=\gamma+\delta
\end{equation*}
because the two sums are congruent triangles.
The third of the Common Notions of Euclid is,
``If equals be subtracted from equals,
the remainders are equal'';
thus
\begin{equation*}
  \alpha=\delta.
\end{equation*}
The second of the Common Notions is,
``If equals be added to equals,
the wholes are equal,''
and so
\begin{equation*}
  \alpha+\beta=\beta+\delta,
\end{equation*}
which is the desired equation of parallelograms.

There is a simpler case, in \figref{fig:I.35-2},
where the parallelograms themselves are composed of congruent parts.
The same is actually true in \figref{fig:I.35},
where we can analyze the parallelograms further
as in \figref{fig:I.35-3}.
\begin{figure}[ht]
  \centering
  \psset{PointName=none,PointSymbol=none}
    \begin{pspicture}(-1,0)(6.5,3)
%      \psgrid
      \pstGeonode A(3,0)B(-1,3)C
      \pstTranslation ABC[D]
      \pstHomO[HomCoef=1.5]CD[E]
      \pstTranslation ABE[F]
      \ncline AB\ncline BD\ncline AC\ncline AE\ncline BF\ncline CF
      \pstInterLL AEBDG
      \pstTranslation DCE[H]
      \pstTranslation DCG[K]
      \ncline HK
      \pstTranslation CDD[L]
      \pstTranslation CDG[M]
      \ncline LM
%      \pstTranslation
      \rput(-0.3,2.67){$\alpha$}
      \rput(0.8,2){$\beta$}
      \rput(1.8,0.6){$\gamma$}
      \rput(2.7,2.67){$\delta$}
      \rput(3.8,2){$\epsilon$}
      \rput(5.7,2.67){$\zeta$}
    \end{pspicture}
  \caption{Alternative proof of I.35}\label{fig:I.35-3}
\end{figure}
Here, by congruence,
\begin{align*}
  a&=\delta=\zeta,&
  \beta&=\epsilon,
\end{align*}
and therefore
\begin{equation*}
  \alpha+\beta+\gamma=\gamma+\epsilon+\zeta.
\end{equation*}
However, as \figref{fig:I.35-4} suggests,
\begin{figure}[ht]
  \centering
  \psset{xunit=15mm}
  \psset{PointName=none,PointSymbol=none}
    \begin{pspicture}(-1.125,0)(4.375,4.5)
%            \psgrid
      \pstGeonode[PosAngle=90](-1.125,4.5){B0}(-0.625,4.5){C-4}
      \multido{\i=-4+1}{10}{    \pstGeonode[PosAngle=-90](\i,0){A\i}}
      \multido{\i= 1+1}{ 5}{\pstTranslation[PosAngle= 90]{A0}{B0}{A\i}[B\i]}
      \multido{\i=-3+1}{ 5}{\pstTranslation[PosAngle=-90]{A-4}{C-4}{A\i}[C\i]}
      \multido{\i=-4+1}{ 5}{\pstInterLL{A\i}{C\i}{A0}{B0}{D\i}
%                            \pstInterLL{A\i}{C\i}{A1}{B1}{F\i}
      \ncline{D\i}{C\i}}
      \multido{\i= 1+1}{ 5}{\pstInterLL{A\i}{B\i}{A1}{C1}{E\i}
        \ncline{E\i}{B\i}}
      \ncline{A0}{A1}\ncline{A1}{C1}\ncline{C1}{B0}\ncline{B0}{D0}
      \rput(0.25,1){$\alpha_1$}
      \rput(0,2){$\alpha_2$}
      \rput(-0.25,3){$\alpha_3$}
      \rput(-0.5,4){$\alpha_4$}
      \rput(1.25,1){$\beta_1$}
      \rput(2,2){$\beta_2$}
      \rput(2.75,3){$\beta_3$}
      \rput(3.5,4){$\beta_4$}
    \end{pspicture}
  \caption{Complications of I.35}\label{fig:I.35-4}
\end{figure}
there is no bound on the number of congruent parts
that we may have to analyze the parallelograms into,
if we want to avoid \emph{adding} congruent parts.

\section{Projective plane}\label{sect:proj-plane}

The first four books of Euclid's \emph{Elements}
concern a \textbf{Euclidean plane.}
Provided we can,
as we shall in \secref{sect:Thales-Hil},
prove Thales's Theorem using only those books,
and not Book VI, which we mentioned in \secref{sect:Thales},
and which we shall use in \secref{sect:Thales-Eucl},
a Euclidean plane
is a special case of an affine plane,
as defined in \secref{sect:Thales}.
Books XI--XIII of Euclid concern Euclidean space.

We can understand Pappus's geometry
as concerning a \emph{projective plane.}
Here, for either of Desargues's Theorem and Pappus's Theorem,
the six cases can be given a single expression,
because formerly parallel straight lines
are now allowed to intersect ``at infinity.''
In Chapter \ref{ch:proj},
we shall obtain a projective plane from an affine plane
%such as the Euclidean plane
by adding
\begin{compactenum}[1)]
\item
  new points, called \textbf{points at infinity,}
  one for each family of parallel straight lines,
  to which the point is considered to be common;
%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 is common to all of the points at infinity.
\end{compactenum}
Meanwhile,
by definition, a \textbf{projective plane} satisfies the following axioms.
%\newcounter{proj}
\begin{compactenum}
\item
Any two distinct points lie on a single straight line.
\item
  Any two distinct straight lines intersect at a single point.
\item
  There is a complete quadrangle, in the sense of \secref{sect:quad}.
\end{compactenum}
In not every projective plane is Pappus's Theorem true.
A \textbf{Pappian plane}
is a projective plane in which Pappus's Theorem \emph{is} true,
in the sense illustrated below by \figref{fig:XIII}
in \secref{sect:XI}:
when the vertices of a hexagon
lie alternately on two straight lines, which now necessarily intersect,
then the intersection points, which now always exist, of opposite sides
lie on a straight line.
This is Pappus's Lemma XIII,
which Pappus proves for a Euclidean plane.

A Pappian plane has no specified line at infinity.
When we remove any straight line and its points,
what remains is an affine plane,
for which the line removed may be conceived as a line at infinity.

We shall show in \secref{sect:Des-proof} that Desargues's Theorem,
in the projective sense illustrated by \figref{fig:Des-proj},
is true in every Pappian plane.

\chapter{Tuesday}%\label{ch:Tues}

\section{Pappus's Theorem, parallel cases}\label{sect:VIII}

In a Euclidean plane,
we prove now the parallel cases of Pappus's Theorem,
stated in \secref{sect:Pap}.
One of them, Lemma VIII,
will give us commutativity of multiplication of ratios in an affine plane,
as we said.

In Lemma VIII,
two pairs of opposite sides of the hexa\-gon are parallel,
and the two bounding lines intersect.
In particular,
letting the hexagon be \gr{BGHEDZ} in \figref{fig:VIII}
\begin{figure}[ht]
%  \psset{unit=8mm}
  \subfloat[Pappus's figure]{\label{fig:VIII}
    \begin{pspicture}(-0.2,-0.5)(5,4.5)
%      \psgrid
    \psset{PointName=none,PointSymbol=none}
  \pstGeonode B(4.8,0)G(1,4)Z(3.6,3.5)H
  \ncline BG\ncline BZ
  \pstInterLL BHGZA
  \pstTranslation ZBH
  \pstInterLL H{H'}GZE
  \ncline HE
  \pstTranslation GBE
  \pstInterLL E{E'}BHD
  \ncline ED
  \psset{linestyle=dotted}
  \ncline HZ\ncline BE\ncline GD
    \psset{linestyle=dashed}
    \ncline GH\ncline DZ
\psset{linewidth=2pt,linestyle=solid}
\ncline BH\ncline GZ
\uput[d](A){\gr A}
\uput[d](B){\gr B}
\uput[d](G){\gr G}
\uput[160](D){\gr D}
\uput[20](E){\gr E}
\uput[u](Z){\gr Z}
\uput[ur](H){\gr H}
  \end{pspicture}
  }
  \hfill
  \subfloat[Alternative figure]{\label{fig:VIII-alt}
    \begin{pspicture}(-0.2,-0.5)(5,4.5)
%      \psgrid
    \psset{PointName=none}
    \pstGeonode B(3,4)A(4.8,0)Z
    \nput dB{\gr B}\nput uA{\gr A}\nput dZ{\gr Z}
    \pstHomO[HomCoef=0.3]AB[H]
    \nput lH{\gr H}
    \pstHomO[HomCoef=0.45]AB[D]
    \nput lD{\gr D}
    \pstTranslation[PointSymbol=none] BZH
    \pstInterLL H{H'}AZE
    \nput{r}E{\gr E}
    \pstTranslation[PointSymbol=none] DZH
    \pstInterLL H{H'}AZG
    \nput{r}G{\gr G}
    \ncline HE\ncline ED\ncline ZB\ncline BG
    \psset{linestyle=dashed}
    \ncline GH\ncline DZ
    \psset{linestyle=dotted}
    \ncline HZ\ncline DG\ncline BE
    \psset{linewidth=2pt,linestyle=solid}
    \ncline BA\ncline AZ
  \end{pspicture}
  }
  \caption{Lemma VIII}
\end{figure}
(which is close to the figure in Hultsch's text of Pappus \cite{Pappus},
except that the lines there all look the same),
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*}
If the diagram is as in \figref{fig:VIII-alt},
then we must adjust the proof
by \emph{subtracting} \gr{BZE} and \gr{BZH} from \gr{ABZ} in step 4,
and subtracting \gr{AGH} from \gr{AGD} and \gr{AHZ} in step 6.
Note then that the proof does not make sense in an abstract affine plane,
where there is no ordering of points on a straight line.

Possibly the intersection point \gr A does not exist,
because $\grm{BD}\parallel\grm{GE}$.
This is the situation of \figref{fig:par},
\begin{figure}[ht]
\subfloat[]{\label{fig:para-dif}
    \begin{pspicture}(5,4)
%      \psgrid
    \psset{PointName=none,PointSymbol=none}
    \pstGeonode(0.2,0.6)B(4.8,3.4)E(4.2,0.6)G
    \pstTranslation GEB[D]
    \pstHomO[HomCoef=0.7]BD[H]
    \pstTranslation HGD[Z]
  \ncline BG\ncline BZ
  \ncline HE
  \ncline ED
%  \psset{linestyle=dotted}
%  \ncline HZ\ncline BE\ncline GD
      \psset{linestyle=dashed}
    \ncline GH\ncline DZ
\psset{linewidth=2pt,linestyle=solid}
\ncline BD\ncline GE
\uput[d](B){\gr B}
\uput[d](G){\gr G}
\uput[u](D){\gr D}
\uput[u](E){\gr E}
\uput[r](Z){\gr Z}
\uput[l](H){\gr H}
    \end{pspicture}
}
\hfill
\subfloat[]{\label{fig:para-sum}
    \begin{pspicture}(5,4)
%      \psgrid
    \psset{PointName=none,PointSymbol=none}
    \pstGeonode(0.2,0.6)D(4.8,3.4)G(4.2,0.6)Z
    \pstTranslation ZGD[H]
    \pstHomO[HomCoef=0.7]DH[B]
    \pstTranslation BZH[E]
  \ncline BG\ncline BZ
  \ncline HE
  \ncline ED
%  \psset{linestyle=dotted}
%  \ncline HZ\ncline BE\ncline GD
      \psset{linestyle=dashed}
    \ncline GH\ncline DZ
\psset{linewidth=2pt,linestyle=solid}
\ncline HD\ncline GZ
\uput[l](B){\gr B}
\uput[u](G){\gr G}
\uput[d](D){\gr D}
\uput[r](E){\gr E}
\uput[d](Z){\gr Z}
\uput[u](H){\gr H}
    \end{pspicture}
}
    \caption{Parallel Theorem}\label{fig:par}
\end{figure}
where,
by the Equality Theorem of \secref{sect:Des},
being opposite sides of parallelograms,
\begin{align*}
  \grm{BD}&=\grm{GE},&\grm{BH}&=\grm{ZE}.
\end{align*}
Therefore
\begin{compactitem}
  \item
the differences are equal in \figref{fig:para-dif},
by the third common notion, mentioned in \secref{sect:=};
\item
the sums are equal in \figref{fig:para-sum},
by the second common notion.
\end{compactitem}
That is,
\begin{equation*}
  \grm{DH}=\grm{GZ}.
\end{equation*}
These too must be the opposite sides of a parallelogram,
by the Equality Theorem again;
in particular,
\begin{equation}\label{eqn:GHZD}.
  \grm{GH}\parallel\grm{ZD}.
\end{equation}
Let us call this case of Pappus's Theorem the \textbf{Parallel Theorem.}
We shall use it in \secref{sect:+}.
Again the proof is in a Euclidean plane;
but now there is a proof for an affine plane.
In \figref{fig:par-aff},
\begin{figure}[ht]
\centering
    \begin{pspicture}(8,5)
%      \psgrid
    \psset{PointName=none,PointSymbol=none}
    \pstGeonode(0.6,0.2)B(7.4,4.4)E(6,0.2)G
    \pstTranslation GEB[D]
    \pstHomO[HomCoef=0.6]BD[H]
    \pstTranslation HGD[Z]
  \ncline BG\ncline BZ
  \ncline HE
  \ncline ED
  \psset{linestyle=dashed}
  \ncline HZ\ncline BE\ncline GD
      \psset{linestyle=dotted}
    \ncline GH\ncline DZ
\psset{linewidth=2pt,linestyle=solid}
\ncline BD\ncline GE
\uput[l](B){\gr B}
\uput[r](G){\gr G}
\uput[l](D){\gr D}
\uput[r](E){\gr E}
\uput[r](Z){\gr Z}
\uput[l](H){\gr H}
\pstInterLL BEGDA
\uput[d](A){\gr A}
    \end{pspicture}

    \caption{Parallel Theorem in an affine plane}\label{fig:par-aff}
\end{figure}
the triangles \gr{BGZ} and \gr{EDH} having corresponding sides parallel,
the straight lines \gr{BE}, \gr{GD}, and \gr{ZH}
must have a common point \gr A,
by the converse of Thales's Theorem.
Applied then to the triangles \gr{BGH} and \gr{EDZ},
Thales's Theorem yields \eqref{eqn:GHZD}.

\section{Quadrangle Theorem proved by Pappus}\label{sect:Quad-Pap}

Pappus's \textbf{Lemma IV}%
\index{Lemmas of Pappus!4@IV|(textbf}
is that, in \figref{fig:IV},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(0,-0.5)(10.2,5.5)
%    \psgrid
    \psset{PointSymbol=none,PointName=none}
    \pstGeonode A(3,0)B(7,5)L
    \uput[d](A){\gr A}
    \uput[d](B){\gr B}
    \uput[u](L){\gr L}
    \pstHomO[HomCoef=1.4,PosAngle=-90]AB[G]
    \uput[d](G){\gr G}
    \pstHomO[HomCoef=2,PosAngle=-90]AB[D]
    \uput[d](D){\gr D}
    \pstHomO[HomCoef=2.4,PosAngle=-90]AB[E]
    \uput[d](E){\gr E}
    \pstHomO[HomCoef=0.65,PosAngle=135]AL[K]
    \uput[ul](K){\gr K}    
    \pstTranslation ABK
    \pstInterLL K{K'}DLM
    \uput[r](M){\gr M}    
    \pstInterLL K{K'}BLN
    \uput[-60](N){\gr N}
    \pstInterLL[PosAngle=180] BLGKJ
    \uput[l](J){\gr J}
    \pstInterLL[PosAngle=45] DLEKH
    \uput[30](H){\gr H}
    \pstInterLL[PosAngle=-90]ABHJZ
    \uput[d](Z){\gr Z}
    \pstTranslation GKE
    \pstInterLL E{E'}JZX
%    \uput[u](X){\gr X}
    \ncline AK\ncline BL\ncline GJ\ncline DH\ncline EK
    \psset{linestyle=dotted}
    \ncline HZ\ncline KM
    %\ncline EX
    \psset{linestyle=solid,linewidth=2pt}
    \ncline AZ\ncline HJ\ncline JK\ncline KL\ncline LH
  \end{pspicture}

  \caption{Lemma IV}\label{fig:IV}
\end{figure}
where the solid lines are straight,
if the proportion
\begin{equation}\label{eqn:IV}
  \compratio{\grm{AZ}}{\grm{AB}}{\grm{BG}}{\grm{GZ}}
  \prop
  \compratio{\grm{AZ}}{\grm{AD}}{\grm{DE}}{\grm{EZ}}
\end{equation}
holds for the points \gr{A, B, G, D, E,} and \gr Z
one one of the straight lines,
then \gr J, \gr H, and \gr Z are in a straight line.
Proving this will involve various manipulations.
Pappus writes the products of ratios in \eqref{eqn:IV} as
ratios of products:
\begin{equation}\label{eqn:IV-Pap}
\grm{AZ}\cdot\grm{BG}:\grm{AB}\cdot\grm{GZ}
\prop \grm{AZ}\cdot\grm{DE}:\grm{AD}\cdot\grm{EZ}.
\end{equation}
Now we apply \textbf{alternation,}
which is the rule
\begin{equation}\label{eqn:alt}
  a\rat b\prop c\rat d\implies a\rat c\prop b\rat d.
\end{equation}
We prove this using commutativity of multiplication.
From the hypothesis $a\rat b\prop c\rat d$, we compute
\begin{equation*}
  a\rat c
  \prop\compratio{a}{b}{b}{c}
  \prop\compratio{c}{d}{b}{c}
  \prop\compratio{b}{c}{c}{d}
  \prop b\rat d.
\end{equation*}
Now \eqref{eqn:IV-Pap} is equivalent to
\begin{equation}\label{eqn:IV-2}
\grm{AZ}\cdot\grm{BG}\rat\grm{AZ}\cdot\grm{DE}
\prop \grm{AB}\cdot\grm{GZ}\rat\grm{AD}\cdot\grm{EZ}.
\end{equation}
The left-hand member simplifies,
and then we expand it as a product:
\begin{multline}\label{eqn:IV-3}
\grm{AZ}\cdot\grm{BG}\rat\grm{AZ}\cdot\grm{DE}
\prop\grm{BG}\rat\grm{DE}\\
\prop(\grm{BG}\rat\grm{NK})(\grm{NK}\rat\grm{KM})(\grm{KM}\rat\grm{DE}).
\end{multline}
Pappus has \gr{KN} for \gr{NK} here, and similar variants elsewhere.
He analyzes the right-hand member of \eqref{eqn:IV-2}
as a product of ratios:
\begin{equation}\label{eqn:IV-4}
\grm{AB}\cdot\grm{GZ}\rat\grm{AD}\cdot\grm{EZ}
\prop\compratio{\grm{BA}}{\grm{AD}}{\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}\rat\grm{KM}\prop \grm{BA}\rat\grm{AD}.
\end{equation*}
Eliminating this common ratio from the members of \eqref{eqn:IV-2}
given in \eqref{eqn:IV-3} and \eqref{eqn:IV-4},
then reversing the order of the new members,
we obtain
\begin{equation*}
\grm{GZ}\rat\grm{ZE}
\prop\compratio{\grm{BG}}{\grm{NK}}{\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}\rat\grm{ZE}
\prop\compratio{\grm{JG}}{\grm{JK}}{\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 \figref{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}[ht]
  \centering
  \begin{pspicture}(0,-0.5)(6,3.5)
    \psset{PointName=none}
    \pstGeonode G(6,0)Z(1,3)K
    \pstHomO[HomCoef=0.6]GK[J]
    \pstHomO[HomCoef=0.6]GZ[E]
    \pstInterLL JZKEH
    \pstTranslation[PointSymbol=none] GKE
    \pstInterLL E{E'}JZX
    \ncline GZ\ncline GK\ncline KE\ncline JH
    \psset{linestyle=dashed}
    \ncline HX\ncline XE
    \ncline[linestyle=dotted]XZ
    \nput dG{\gr G}
    \nput dE{\gr E}
    \nput dZ{\gr Z}
    \nput lJ{\gr J}
    \nput uK{\gr K}
    \nput{ur}H{\gr H}
    \nput uX{\gr{\myXi}}
  \end{pspicture}
  \caption{Menelaus's Theorem}\label{fig:IV-lemma}
\end{figure}
Rewriting \eqref{eqn:IV-lemma},
we have the hypothesis
\begin{equation}\label{eqn:IV-lemma'}
\grm{GZ}:\grm{ZE}
\prop\compratio{\grm{GJ}}{\grm{JK}}{\grm{KH}}{\grm{HE}}.  
\end{equation}
We extend \gr{JH}
and let it be met at \gr{\myXi}
by the parallel to \gr{GK} through \gr E.
By Thales's Theorem then,
from \eqref{eqn:IV-lemma'} we have
\begin{equation*}
\grm{GZ}\rat\grm{ZE}
\prop\compratio{\grm{JG}}{\grm{KJ}}{\grm{KJ}}{\grm{E\myXi}}
\prop \grm{JG}\rat\grm{E\myXi}.  
\end{equation*}
By the same theorem in the other direction,
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 proofs of
\begin{compactitem}
  \item
one direction of Menelaus's Theorem,
\item
  Lemma VIII.
\end{compactitem}
The steps of the proof are reversible.
Thus, if we are given the complete quadrangle \gr{HJKL} of \figref{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 \figref{fig:IV-CQT},
\begin{figure}[ht]
\centering
  \begin{pspicture}(0,-4.5)(10.2,6.5)
%    \psgrid
    \psset{PointSymbol=none,PointName=none}
    \pstGeonode A(3,0)B(7,6)L(4,-4)T
    \uput[l](A){\gr A}
    \uput[ul](B){\gr B}
    \uput[u](L){\gr L}
    \uput[d](T){\gr T}
    \pstHomO[HomCoef=1.4]AB[G]
    \uput[ur](G){\gr G}
    \pstHomO[HomCoef=2]AB[D]
    \uput[ul](D){\gr D}
    \pstHomO[HomCoef=2.4]AB[E]
    \uput[ur](E){\gr E}
    \pstHomO[HomCoef=0.65]AL[K]
    \pstHomO[HomCoef=0.55]AT[S]
    \uput[ul](K){\gr K}    
    \uput[dl](S){\gr S}    
%    \pstTranslation ABK
%    \pstInterLL K{K'}DLM
%    \uput[r](M){\gr M}    
%    \pstInterLL K{K'}BLN
%    \uput[-60](N){\gr N}
    \pstInterLL[PosAngle=180] BLGKJ
    \uput[l](J){\gr J}
    \pstInterLL[PosAngle=180] BTGSR
    \uput[l](R){\gr R}
    \pstInterLL[PosAngle=45] DLEKH
    \uput[30](H){\gr H}
    \pstInterLL[PosAngle=45] DTESP
    \uput[dr](P){\gr P}
    \pstInterLL[PosAngle=-90]ABHJZ
    \uput[r](Z){\gr Z}
    \pstTranslation GKE
    \pstInterLL E{E'}JZX
    \ncline AZ
    \ncline AL\ncline BL\ncline GK\ncline DL\ncline EK\ncline ZJ
    \ncline AT\ncline BT\ncline GS\ncline DT\ncline ES
    \ncline[linestyle=dotted] ZR
  \end{pspicture}
\caption{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 would pass through \gr Z.
This is the \textbf{Quadrangle Theorem.}

\chapter{Wednesday}%\label{ch:Wednes}

\section{Thales's Theorem proved by Euclid}\label{sect:Thales-Eucl}

Proposition VI.1 of Euclid's \emph{Elements}
is that triangles and parallelograms under the same height
are to one another in the ratio of their bases.
Thus, in \figref{fig:TT},
\begin{figure}[ht]
\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}
\begin{equation}\label{eqn:tri-rat}
  \left.\qquad
  \begin{gathered}
AD\rat DB\prop \mytri ADE:\mytri DBE,\\
AE\rat EC\prop \mytri ADE:\mytri EDC.
  \end{gathered}
\qquad  \right\}
\end{equation}
By V.7 and 8,
the ratios of $ADE$ to $DBE$
and $EDC$ are the same,
just in case these two triangles are equal
in the sense of \secref{sect:=};
symbolically,
\begin{equation}\label{eqn:ADE}
  \mytri ADE:\mytri DBE\prop \mytri ADE:\mytri EDC\\
  \iff\mytri DBE=\mytri EDC.
\end{equation}
Finally,
by I.35 and 37,
as discussed in \secref{sect:=},
\begin{equation}\label{eqn:DBE}
  \mytri DBE=\mytri EDC\iff DE\parallel BC.
\end{equation}
Combining \eqref{eqn:tri-rat}, \eqref{eqn:ADE}, and \eqref{eqn:DBE},
using the transitivity of sameness of ratio,
we conclude
\begin{equation*}
  AD\rat DB\prop AE\rat EC\iff\mytri DBE=\mytri EDC.
\end{equation*}
That is Euclid's proof of Thales's Theorem,
or more precisely the Proportion Theorem,
which is Proposition VI.2 of the \emph{Elements},
as discusses in \secref{sect:Thales}.
The proof takes place in a Euclidean plane,
in the sense of \secref{sect:proj-plane},
but also using a theory of proportion.
We have developed such a theory
only in an affine plane,
in the sense of \secref{sect:Thales}.

\section{Proportion in a Euclidean plane}\label{sect:prop-def}

For Euclid,
a proportion is a relation of four \textbf{magnitudes.}
Magnitudes come in three kinds:
lines, surfaces, or solids
(all bounded).
The magnitudes in the proportion are considered in two pairs,
the magnitudes in each pair \textbf{having a ratio} to one another;
and these ratios composing the proportion are the same.
Only magnitudes of the same kind can have a ratio.

In any proportion,
we expect to be able to replace a magnitude by an equal magnitude
in the sense of \secref{sect:=}.
We may then confuse a magnitude with its
\textbf{size}---its length, area, or volume---,
understood as the class of magnitudes equal to the original one.

Given three lengths $a$, $b$, and $c$,
we can form the products
\begin{align*}
  &ab,&&abc,
\end{align*}
which are respectively
\begin{compactitem}
\item
the area of a rectangle having dimensions $a$ and $b$,
\item
the volume of a rectangular parallelepiped
  having dimensions $a$, $b$, and $c$.
\end{compactitem}
The order in which dimensions are given is irrelevant.
Suppose we have
\begin{align}\label{eqn:ay}
  ay&=bx,&cy&=dx.
\end{align}
Then
\begin{equation}\label{eqn:bcx}
  bcx=acy=adx.
\end{equation}
The fifth of Euclid's Common Notions is,
``the whole is greater than the part.''
If $bc\neq ad$,
then we may assume $bc<ad$,
and thus $bc$ is the area of a part of a surface that has area $ad$.
In this case,
$bcx$ is the volume of a part of a solid having volume $adx$;
in particular, \eqref{eqn:bcx} fails.
Thus from \eqref{eqn:bcx},
and therefore from \eqref{eqn:ay}, we conclude
\begin{equation}\label{eqn:bc}
  bc=ad.
\end{equation}
As a consequence,
the relation of ratios of segments
given by the rule
\begin{multline}\label{eqn:prop}
  AB\rat CD\prop EF\rat GH\\
  \iff\size{AB}\cdot\size{GH}=\size{CD}\cdot\size{EF}
\end{multline}
is transitive.
It now makes some sense to take
\eqref{eqn:prop} for a definition of proportion of lengths.
A convenience of this definition is that
alternation, as in \eqref{eqn:alt}, is immediate.
There remain two problems, one more serious than the other.
\begin{compactenum}
  \item
    As observed in \secref{sect:Thales},
    one way to read the proportion in \eqref{eqn:prop} is as
    ``$AB$ has the same ratio to $CD$ that $EF$ has to $GH$'';
    and as argued in \secref{sect:eq-rel},
    transitivity of sameness of ratio ought to be obvious,
    not needing such a proof as we have given.
\item
  Euclid's proof of Thales's Theorem uses also ratios of \emph{areas.}
\end{compactenum}

\subsection{Hilbert's definition}

We can avoid these problems
by using the method of Descartes \cite{Descartes-Geometry,Descartes-Geometrie}
and fixing a unit length,
so that we can obtain products of lengths as lengths.
However, Descartes assumes Euclid's theory of proportion to begin with.
Hilbert does not,
but works only in a Euclidean plane.
Following Hilbert's idea \cite[pp.\ 29--32]{MR0116216},
in \figref{fig:Hilbert},
\begin{figure}[ht]
\subfloat[]{\label{fig:Hilbert}
  \begin{pspicture}(-0.2,-0.6)(5,4.6)
%    \psgrid
\psset{PointSymbol=none}
\pstGeonode[PosAngle={-90,-90,90}]A(3.5,0)B(3.5,4)C
\pstHomO[HomCoef=0.6,PosAngle={-90,0}]B{A,C}[D,E]
\ncline DE
%\psset{RightAngleSize=0.21} % default is 0.28 unit
\pstRightAngle ABC
\psset{LabelSep=0.7}
\pstMarkAngle BDE{$\alpha$}
\pstMarkAngle BAC{$\alpha$}
\ncline DB\nbput{$1$}
\ncline BE\nbput{$\tan\alpha$}
\ncline AD\nbput{$a$}
\ncline EC\nbput{$\tan\alpha\cdot a$}
\ncline AC
\end{pspicture}}
\hfill
\subfloat[Hilbert's diagram]{\label{fig:Hil-own}
\begin{pspicture}(-0.6,-0.6)(4.2,3.2)
\psset{PointName=none,PointSymbol=none}
\pstTriangle[PosAngleB=-90,PosAngleC=180](0,0)O(4,0)A(0,3)B
\pstHomO[HomCoef=0.6]O{A,B}[C,D]
\nput{-90}C{$1$}
\nput{-90}A{$b$}
\nput{180}D{$a$}
\nput{180}B{$a\hilmul b$}
\nput{-135}O{$0$}
\ncline CD
%\psset{RightAngleSize=0.21} % default is 0.28 unit
%\pstRightAngle AOB
\end{pspicture}
  }
\caption{Hilbert's multiplication}
\end{figure}
we suppose
$DB$ has unit length.
If the size of angle $EDB$ is $\alpha$,
then by definition
\begin{equation*}
  \size{BE}=\tan\alpha.
\end{equation*}
Thus every length is the tangent of the size of \emph{some} angle.
If $AC\parallel DE$, we define
\begin{equation*}
  \size{BC}=\tan\alpha\cdot\size{AB}.
\end{equation*}
Hence also
\begin{equation*}
  \size{EC}=\tan\alpha\cdot\size{AD}.
\end{equation*}
Hilbert does not mention tangents of angles,
but just defines multiplication using a figure like \ref{fig:Hil-own}.

To prove that this multiplication is associative and commutative,
Hartshorne,
in \emph{Geometry: Euclid and Beyond}
\cite[pp.\ 168--72]{MR1761093},
presents the streamlined method
found in later editions of Hilbert \cite[pp.\ 203--6]{Hilbert-10}
and attributed to Enriques.
In \figref{fig:Hartshorne},
\begin{figure}%[ht]
  \centering
%  \psset{unit=8mm}
  \begin{pspicture}(-4,-4.6)(4,4.6)
%    \psgrid
\psset{PointSymbol=none}
%\pstGeonode[PosAngle=135](-0.5,-0.25)A
\pstGeonode[PosAngle=45](-0.5,-1)A
{%
\psset{PointName=none,PointSymbol=none}
\pstGeonode(0,0)O(4,0)X(0,1)Y
\pstTranslation OXA[F]
\pstTranslation OYA[G]
}
\pstCircleOA OX
\pstInterLC[PosAngleA=180] AFOXBC
\pstInterLC[PosAngleA=-90,PosAngleB=90] AGOXDE
\ncline BD\ncline DC\ncline CE\ncline EB
\ncline AB\taput{$c$}
\ncline AE\tlput{$a\hilmul c$}
\ncline AD\tlput{$b\hilmul c$}
\ncline AC\taput{$b\hilmul(a\hilmul c)$}\tbput{$a\hilmul(b\hilmul c)$}
\pstMarkAngle DBA{}
\pstMarkAngle AEC{}
\psset{MarkAngleRadius=0.5}
\pstMarkAngle ABE{$\alpha$}
\pstMarkAngle CDA{$\alpha$}
\psset{MarkAngleRadius=0.6}
\pstMarkAngle DBA{$\beta$}
\pstMarkAngle AEC{$\beta$}
\pstRightAngle BAD
\end{pspicture}
\caption{Associativity and commutativity}\label{fig:Hartshorne}
\end{figure}
using that,
``In a circle, the angles in the same segment
are equal to one another'' (\emph{Elements} III.21),
if
\begin{align*}
  \tan\alpha&=a,&\tan\beta&=b,
\end{align*}
then $AC$ has the two lengths indicated,
so these are the same;
that is,
\begin{equation}\label{eqn:abc}
a\hilmul(b\hilmul c)=b\hilmul(a\hilmul c).
\end{equation}
Letting $c=1$ gives commutativity;
then this with \eqref{eqn:abc} gives
\begin{equation*}
a\hilmul(c\hilmul b)=(a\hilmul c)\hilmul b
\end{equation*}
and thus associativity.
It is clear how to add lengths,
but see also \secref{sect:+}.
Distributivity of multiplication over addition
follows from \figref{fig:dist},
\begin{figure}%[ht]
\centering
\begin{pspicture}(0,-0.5)(5,4)
\psset{PointSymbol=none,PointName=none}
\pstGeonode (0,0)A(5,0)B(5,4)C
\pstHomO[HomCoef=0.4]CA[D]
\pstProjection ABD[E]
\ncline AE\tbput{$b$}
\ncline ED\trput{$a\hilmul b$}
\pstProjection BCD[F]
\ncline DF\tbput{$c$}
\ncline FC\trput{$a\hilmul c$}
\ncline AC
\pstRightAngle AED
\pstRightAngle DFC
\psset{linestyle=dotted}
\ncline EB\ncline BF
\end{pspicture}
\caption{Distributivity}\label{fig:dist}
\end{figure}
where
\begin{equation*}
a\hilmul b+a\hilmul c=a\hilmul(b+c).
\end{equation*}
Every product of lengths is now a length.
Moreover, with respect to this multiplication,
every length has an inverse:
in \figref{fig:Hil-own}, we can let $ab=1$.
Thus multiplication admits cancellation:
\begin{equation*}
  ax=bx\implies a=b.
\end{equation*}
Finally,
\begin{equation*}
  a<b\implies ax<bx.
\end{equation*}

In algebraic terms,
lengths are now the positive elements of an ordered field $\ordfield$.
We obtain the whole ordered field $\ordfield$ by selecting points $O$ and $U$
on an infinite straight line that are the unit distance apart;
then the points on the line correspond to
all of the elements of $\ordfield$,
with $O$ as $0$ and $U$ as $1$.
We can now define the ratio $AB\rat CD$ to be the quotient
\begin{equation*}
  \frac{\size{AB}}{\size{CD}},
\end{equation*}
in $\ordfield$, of the lengths of $AB$ and $CD$.
This makes
\begin{compactitem}
\item
  sameness of ratio transitive;
\item
  alternation, \eqref{eqn:alt}, an easy theorem;
  \item
\eqref{eqn:prop} also an easy theorem.
\end{compactitem}
Now we can \emph{state} the Proportion Theorem,
which implies Thales's Theorem.

\subsection{Euclid's definition}

It remains to \emph{prove}
the Proportion Theorem,
and thus Thales's Theorem, under Hilbert's definition of ratios.
Meanwhile, let us note that, for Euclid,
the ratio $a:b$ is effectively what we now call a \textbf{Dedekind cut,}
because of its use in Dedekind's development of the real numbers
\cite{MR0159773}.
We can understand a Dedekind cut
as a partition of the positive rational numbers
into two nonempty sets $A$ and $B$,
where every element of $A$ is less than every element of $B$.
In the cut corresponding to $a:b$,
\begin{align*}
  A&=\left\{\frac xy:bx\leq ay\right\},&
  B&=\left\{\frac xy:bx> ay\right\}.
\end{align*}
Here $x$ and $y$ are counting numbers.
We can replace $\leq$ with $<$, and $>$ with $\geq$:
this moves at most one element from $A$ to $B$.
We assume that $A$ and $B$ are indeed both nonempty,
when $a$ and $b$ are magnitudes of the same kind:
this is the \textbf{Archimedean Axiom.}
In this case, the ordered pair $(a,b)$ defines the given cut;
Dedekind's innovation was to recognize that
he could define cuts without reference to magnitudes.


\section{Thales's Theorem proved by Hilbert}\label{sect:Thales-Hil}

As Hilbert shows,
we can prove the Proportion Theorem,
under the definition of ratios as quotients of lengths,
without using the Archimedean Axiom.
\begin{figure}
  \centering
  \begin{pspicture}(-0.5,-0.5)(10.5,13.5)
    \psset{PointSymbol=none}
    \pstTriangle A(10,0)B(3,13)C
    \pstBissectBAC[PointName=none,linestyle=none] BACH
    \pstBissectBAC[PointName=none,linestyle=none] ACBK
    \pstInterLL[PosAngle=65] AHCKI
    \psset{HomCoef=0.7}
    \pstHomO[PosAngle=-80] IA[D]
    \pstHomO[PosAngle=80] IB[E]
    \pstHomO[PosAngle=180] IC[F]
    \ncline AI\ncline BI\ncline CI
    \ncline DE\ncline EF\ncline FD
    \psset{PointName=none}
    \pstProjection ABI[K]
    \pstProjection BCI[G]
    \pstProjection CAI[H]
    \ncline IG\ncline IH\ncline IK
%    \psset{RightAngleSize=0.14} % default is 0.28 unit
    \pstRightAngle BKI
    \pstRightAngle CGI
    \pstRightAngle AHI
    \psset{linestyle=dotted}
    \pstCircleOA IG
    \pstInterLL IGEFL
    \pstCircleOA IL
    \pstRotation[RotAngle=105] IG[M]
    \ncline[arrows=->]IM\naput{$r$}
    \pstRotation[RotAngle=-75] IL[N]
    \ncline[arrows=->]IN\nbput{$r'$}
    \psset{linestyle=none}
    \ncline AH\naput{$a$}\ncline AK\nbput{$a$}
    \ncline BK\naput{$b$}\ncline BG\nbput{$b$}
    \ncline CG\naput{$c$}\ncline CH\nbput{$c$}
    \pstInterLL IHFDM
    \pstInterLL IKDEN
    \ncline DM\naput{$a'$}\ncline DN\nbput{$a'$}
    \ncline EN\naput{$b'$}\ncline EL\nbput{$b'$}
    \ncline FL\naput{$c'$}\ncline FM\nbput{$c'$}
  \end{pspicture}
  \caption{Proof of Thales's Theorem}\label{fig:incenters}
  
\end{figure}
Given triangles $ABC$ and $DEF$
that are \textbf{similar,} in the sense that
the angles at $A$ and $B$
are respectively equal to the angles at $D$ and $E$,
and therefore also the angles at $C$ and $F$ are equal
by Euclid's I.32,
we want to show
\begin{equation}\label{eqn:Thales-AB}
  AB\rat DE\prop AC\rat DF.
\end{equation}
If the angles at $B$ and $E$ are right angles,
then \eqref{eqn:Thales-AB}
follows from our definition of multiplication of lengths.
In the general case,
we can arrange the triangles as in \figref{fig:incenters},
so that their corresponding sides are parallel
and their incenters coincide at a point $I$.
Thus $AI$ and $BI$ bisect the angles at $A$ and $B$,
and therefore $CI$ does the same for the angle at $C$.
Then because of the right angles,
\begin{align*}
  ar'&=a'r,& br'&=b'r,&cr'&=c'r,  
\end{align*}
so that
\begin{equation*}
  (a+b)r'=ar'+br'=a'r+b'r=(a'+b')r,
\end{equation*}
and likewise
\begin{equation*}
  (a+c)r'=(a'+c')r.
\end{equation*}
Therefore
\begin{equation*}
  \frac{a+b}{a'+b'}=\frac r{r'}=\frac{a+c}{a'+c'},
\end{equation*}
which yields \eqref{eqn:Thales-AB}.


\section{Cross ratio}\label{sect:III}

In Pappus's \textbf{Lemma III,}%
\index{Lemmas of Pappus!3@III}
straight lines \gr{JD} and \gr{JH}
cut the straight lines \gr{AB}, \gr{AG}, and \gr{AD}
as in \figref{fig:III}.
\begin{figure}[ht]
\subfloat[Pappus's figure]{
\begin{pspicture}(-0.5,-2)(4,4.3)
%  \psgrid[subgriddiv=2]
    \psset{PointName=none}
    \pstGeonode J(1.7,-0.2)Z(1.7,2)A
    \pstHomO[HomCoef=2.4]JZ[H]
    \pstHomO[HomCoef=0.7]AZ[G]
    \pstHomO[HomCoef=0.4]JZ[E]
    \pstTranslation[PointSymbol=none] AZJ
    \pstInterLL J{J'}AHK
    \pstInterLL AEKJL
    \pstTranslation[PointSymbol=none] AHL
%    \pstInterLL L{L'}JZM
    \pstInterLL JGAHD
    \pstInterLL JDAEB
    \ncline JH\ncline JD
    \psset{linestyle=dotted}
    %\ncline JM\ncline ML
    \ncline LE\ncline LK\ncline KA
    \psset{linewidth=2pt,linestyle=solid}
    \ncline AE\ncline AZ\ncline AH
    \nput uA{\gr A}
    \nput{ul}B{\gr B}
    \nput{-45}G{\gr G}
    \nput{ur}D{\gr D}
    \nput{-65}E{\gr E}
    \nput{d}Z{\gr Z}
    \nput{ur}H{\gr H}
    \nput{ul}J{\gr J}
    \nput{u}K{\gr K}
    \nput{d}L{\gr L}
%    \nput{u}M{\gr M}
\end{pspicture}
}
\hfill
\subfloat[alternative figure]{
\begin{pspicture}(-2.5,-0.5)(2.6,5.4)
%  \psgrid[subgriddiv=2]
    \psset{PointName=none}
    \pstGeonode(2.2,1.5)J(0,3.2)Z(0,2.5)A
    \pstHomO[HomCoef=0.82]JZ[H]
    \pstHomO[HomCoef=-2.5]AZ[G]
    \pstHomO[HomCoef=2.2]JZ[E]
    \pstTranslation[PointSymbol=none] AZJ
    \pstInterLL J{J'}AHK
    \pstInterLL AEKJL
    \pstTranslation[PointSymbol=none] AHL
%    \pstInterLL L{L'}JZM
    \pstInterLL JGAHD
    \pstInterLL JDAEB
    \ncline JE\ncline JD
    \psset{linestyle=dotted}
    %\ncline JM\ncline ML
    \ncline LE\ncline LK\ncline KA
    \psset{linewidth=2pt,linestyle=solid}
    \ncline BE\ncline GZ\ncline DH
    \nput lA{\gr A}
    \nput{d}B{\gr B}
    \nput{d}G{\gr G}
    \nput{d}D{\gr D}
    \nput{u}E{\gr E}
    \nput{u}Z{\gr Z}
    \nput{r}H{\gr H}
    \nput{ur}J{\gr J}
    \nput{u}K{\gr K}
    \nput{d}L{\gr L}
    %\nput{r}M{\gr M}
\end{pspicture}
}
\caption{Lemma III}\label{fig:III}
\end{figure}
We are going to show
\begin{equation*}
       \compratio{\grm{JE}}{\grm{EZ}}{\grm{ZH}}{\grm{HJ}}
  \prop\compratio{\grm{JB}}{\grm{BG}}{\grm{GD}}{\grm{DJ}}.
\end{equation*}
We complete the diagram by making
\begin{align*}
  \grm{KL}&\parallel\grm{AZ},&
\grm{LM}&\parallel\grm{AH}.
\end{align*}
By Thales,
\begin{equation}\label{eqn:JEEZ}
  \compratio{\grm{JE}}{\grm{EZ}}{\grm{ZH}}{\grm{HJ}}
  \prop\compratio{\grm{JL}}{\grm{AZ}}{\grm{AZ}}{\grm{JK}}
  \prop\grm{JL}\rat\grm{JK}.
\end{equation}
Since the last ratio is independent of the choice of \gr H along \gr{AD},
we are done.

In \eqref{eqn:JEEZ},
the first product of ratios,
written also as
\begin{equation*}
  \grm{JE}\cdot\grm{ZH}:\grm{EZ}\cdot\grm{HJ},
\end{equation*}
and which we shall denote by
\begin{equation*}
  \crossratio{\grm J}{\grm E}{\grm Z}{\grm H},
\end{equation*}
is the \textbf{cross ratio} (\emph{\c capraz oran})
of the ordered quadruple $(\grm J, \grm E, \grm Z, \grm H)$.
Some permutations of the points do not change the cross ratio.
For example, we can reverse their order:
\begin{align*}
  \crossratio{\grm H}{\grm Z}{\grm E}{\grm J}
  &\prop\grm{HZ}\cdot\grm{EJ}:\grm{ZE}\cdot\grm{JH}\\
  &\prop\grm{JE}\cdot\grm{ZH}:\grm{EZ}\cdot\grm{HJ}
  \prop\crossratio{\grm J}{\grm E}{\grm Z}{\grm H}.
\end{align*}
Thus,
by Lemma III,
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 \figref{fig:CR}.
\begin{figure}[ht]
  \centering
  \psset{yunit=5mm}
\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}\label{fig:CR}
\end{figure}

\textbf{Lemma X} is a converse to Lemma III.
The hypothesis is that, in \figref{fig:X},
\begin{figure}
\centering
  \begin{pspicture}(-6.2,-5.4)(4.2,1.8)
%    \psgrid
    \psset{PointName=none}
    \pstGeonode J(1.7,-2)Z(0,-2)A
    \pstHomO[HomCoef=1.9]JZ[E]
    \pstHomO[HomCoef=-1.4]AZ[G]
    \pstHomO[HomCoef=0.7]JZ[H]
    \pstTranslation[PointSymbol=none] AZJ
    \pstInterLL J{J'}AHK
    \pstInterLL AEKJL
    \pstTranslation[PointSymbol=none] AHL
    \pstInterLL L{L'}JZM
    \pstInterLL JGAHD
    \pstInterLL JDAEB
    \pstTranslation[PointSymbol=none] AEK
%    \pstInterLL GJK{K'}N
    \ncline DH\ncline BE
    \ncline[linestyle=dotted] GZ
    \psset{linestyle=dashed}
    \ncline LK\ncline LB\ncline LM\ncline MJ%\ncline JN\ncline NK
    \ncline KH
    \psset{linewidth=2pt,linestyle=solid}
    \ncline JE\ncline JD
    \nput uA{\gr A}
    \nput{u}B{\gr B}
    \nput{ul}G{\gr G}
    \nput{dl}D{\gr D}
    \nput{-45}E{\gr E}
    \nput{ur}Z{\gr Z}
    \nput{u}H{\gr H}
    \nput{u}J{\gr J}
    \nput{u}K{\gr K}
    \nput{l}L{\gr L}
    \nput{u}M{\gr M}
 %   \nput{u}N{\gr N}
  \end{pspicture}

  \caption{Lemma X}\label{fig:X}
\end{figure}
\begin{equation}\label{eqn:X-hyp}
  [\grm J,\grm H,\grm Z,\grm E]
  \prop[\grm J,\grm D,\grm G,\grm B].
\end{equation}
We shall show that \gr G, \gr A, and \gr Z are collinear.
Through \gr J we draw a line parallel to \gr{GA},
and this line is cut at \gr L and \gr K
by the extensions of \gr{EB} and \gr{DH}.
We let \gr{LM} be parallel to \gr{DH}
and meet the extension of \gr{EJ} at \gr M.
We compute
\begin{align*}
  [\grm J,\grm D,\grm G,\grm B]
  &\prop\grm{JK}\rat\grm{JL}&&\text{[Lemma III proof]}\\
  &\prop\grm{JH}\rat\grm{JM}&&\text{[Thales]}\\
  &\prop\grm{JH}\cdot\grm{EZ}\rat\grm{JM}\cdot\grm{EZ}.&&
\end{align*}
By definition,
\begin{equation*}
  [\grm J,\grm H,\grm Z,\grm E]
  \prop\grm{JH}\cdot\grm{ZE}\rat\grm{HZ}\cdot\grm{EJ}.
\end{equation*}
Our hypothesis \eqref{eqn:X-hyp} then yields
\begin{align*}
  \grm{ZH}\cdot\grm{EJ}&=\grm{JM}\cdot\grm{EZ},&&\\
  \grm{EJ}\rat\grm{JM}&\prop\grm{EZ}\rat\grm{ZH},&&\\
  \grm{EJ}\rat\grm{EM}
  &\prop\grm{EZ}\rat\grm{EH},&&\text{[\emph{componendo}]}\\
    \grm{EJ}\rat\grm{EZ}
    &\prop\grm{EM}\rat\grm{EH}&&\text{[alternation]}\\
    &\prop\grm{EL}\rat\grm{EA}.&&\text{[Thales]}
  \end{align*}
By the converse of Thales's Theorem,
\begin{equation*}
  \grm{AZ}\parallel\grm{LJ},
\end{equation*}
and this yields the claim.%
\index{Lemmas of Pappus!10@X|)textbf} 

\chapter{Friday}\label{ch:planes}

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.

In \secref{sect:prop-def},
we saw how to understand the points of a line
in a Euclidean plane
as the elements of a commutative field.
Now we are going to do the same thing
in an arbitrary affine plane,
except that the field may not be commutative:
it may be a \textbf{skew field,}
usually called today a non-commutative division ring.
To ensure that the field of ratios \emph{is} commutative,
to the axioms for an affine plane
we can add one axiom:
\begin{compactenum}\setcounter{enumi}{\value{ax}}
  \item
  Pappus's Lemma VIII, proved in \secref{sect:VIII}
  on the basis of Book I of the \emph{Elements.}
\end{compactenum}
The commutative field in this case still need not be ordered.

Here is where we are going:
\begin{compactitem}
\item
  Today, after obtaining the field of ratios,
we shall independently prove the remaining cases of Pappus's Theorem,
stated in \secref{sect:Pap},
of which Lemma VIII is a special case.
\item
  Tomorrow we shall look at geometry over fields, possibly skew.
\item
  The day after that,
we shall prove Desargues's Theorem in general,
in \secref{sect:Des-proof}.
\end{compactitem}

\section{Arithmetic in an affine plane}\label{sect:+}

\subsection{Multiplication}

We have seen in \secref{sect:Thales} and \secref{sect:Des}
that we can define ratios in an affine plane
to meet the following conditions.
\begin{compactenum}
  \item
In \figref{fig:ratios-a},
\begin{figure}[ht]
    \subfloat[]{\label{fig:ratios-a}
    \begin{pspicture}(3.4,5)
      \pstGeonode[PosAngle={90,180,-90}](3,4.4)A(0.6,0.2)B(3.4,2)D
      \pstHomO[HomCoef=0.6,PosAngle=180]AB[C]
      \pstHomO[HomCoef=0.75,PosAngle=180]AB[F]
      \pstTranslation[PointSymbol=none,PointName=none] BDC
      \pstInterLL[PosAngle=0] C{C'}ADE
      \pstTranslation[PointSymbol=none,PointName=none] FDE
      \pstInterLL[PosAngle=180] E{E'}ABG
      \ncline AB\ncline AD\ncline BD\ncline CE
      \ncline FD\ncline GE
    \end{pspicture}
    }
    \hfill
      \subfloat[]{  \label{fig:ratios-b}
    \begin{pspicture}(0.6,0)(3.4,5)
      \pstGeonode[PosAngle={90,180,0}](3,4.4)A(0.6,0.2)B(3.4,1)H
      \pstHomO[HomCoef=0.6,PosAngle=180]AB[C]
      \pstTranslation[PointSymbol=none,PointName=none] BHC
      \pstInterLL[PosAngle=0] C{C'}AHK
      \ncline AB\ncline AH\ncline BH\ncline CK
    \end{pspicture}
    }
    \hfill
      \subfloat[]{  \label{fig:ratios-c}
    \begin{pspicture}(4,5)
      \pstGeonode[PosAngle={90,180,0}](2,4.4)A(0.6,0.2)B(3,4)L(3.4,1)M
      \pstHomO[HomCoef=0.6,PosAngle=180]AB[C]
      \pstTranslation[PosAngle=-90,PointName=none]LAM
      \pstTranslation[PointSymbol=none,PointName=none] B{M'}C
      \pstInterLL[PosAngle=0,PointName=none] C{C'}A{M'}P
      \pstTranslation ALP[N]
      \ncline AB\ncline A{M'}\ncline B{M'}\ncline CP
      \ncline LM\ncline AL\ncline PN\ncline{M'}M
    \end{pspicture}
    }
\caption{Ratios}
\end{figure}
\begin{compactenum}
  \item
if $C$ is on $AB$, and $E$ is on $AD$,
but $D$ is not on $AB$, then
\begin{equation*}
  AC\rat AB\prop AE\rat AD\iff CE\parallel BD;
\end{equation*}
\item
if also $F$ and $G$ lie on $AB$, then
\begin{multline*}
  AC\rat AB\prop AE\rat AD\And AE\rat AD\prop AG\rat AF\\
  \implies AC\rat AB\prop AE\rat AD.
\end{multline*}
\end{compactenum}
\item
For any ratio $AC\rat AB$ and any point $H$,
there is a unique point $K$ such that, as in \figref{fig:ratios-b},
\begin{equation*}
  AK\rat AH\prop AC\rat AB.
\end{equation*}
\item
  For any ratio $AC\rat AB$ and any points $L$ and $M$,
there is a unique point $N$ such that, as in \figref{fig:ratios-c},
\begin{equation*}
  LN\rat LM\prop AC\rat AB.
\end{equation*}
\end{compactenum}
Given a ratio $r$ and two points $A$ and $B$,
we can now define
\begin{equation*}
  r\cdot_AB=C,
\end{equation*}
where $C$ lies on $AB$ and
\begin{equation*}
  AC\rat AB\prop r.
\end{equation*}
If $s$ is another ratio,
we have defined the product $sr$;
associativity of the multiplication here gives us
\begin{equation*}
  (sr)\cdot_AB=s\cdot_A(r\cdot_AB).
\end{equation*}

\subsection{Addition}

With respect to a point $A$,
we can form the sum of two points $B$ and $C$ by completing the parallelogram,
if there is one.
That is, assuming $C$ does not lie on $AB$,
we define
\begin{equation*}
  B+_AC=D,
\end{equation*}
where $ABDC$ is a parallelogram as in \figref{fig:+}.
\begin{figure}[ht]
  \centering
  \begin{pspicture}(0,-2.1)(6.8,2.6)
    \pstGeonode[PosAngle={180,105,90,-90}] A(2,0)B(1,2)C(1.5,-1.5)G
    \pstTranslation[PosAngle=90] ABC[D]
    \pstTranslation[PosAngle=-90] ABG[H]
    \pstHomO[HomCoef=2.4,PosAngle=75]AB[E]
    \pstTranslation CDE[F]
    \ncline AF\ncline CD\ncline GH
    \ncline AC\ncline BD\ncline CE\ncline DF
    \psset{linestyle=dashed}
    \ncline AG\ncline BH\ncline GE\ncline HF
    \psset{linestyle=dotted}
    \ncline CG\ncline DH
  \end{pspicture}
  \caption{Addition}\label{fig:+}
  
\end{figure}
This means
\begin{align*}
  AB&\parallel CD,&AC&\parallel BD.
\end{align*}
This definition of the sum is symmetric in $B$ and $C$.
If however $E$ lies on $AB$,
then, using the point $C$, which does not lie on $AB$,
and using the sum $B+_AC$, which is $D$,
we define
\begin{equation*}
  E+_AB=F,
\end{equation*}
where, as in \figref{fig:+} again,
\begin{equation*}
  F=E+_CD.
\end{equation*}
Thus
\begin{equation*}
  E+_AB=E+_C(B+_AC).
\end{equation*}
This definition is independent of the choice of $C$,
by the Prism Theorem.
Indeed, if also $G$ does not lie on $AB$, and
\begin{equation*}
  B+_AG=H,
\end{equation*}
so that,
as in \figref{fig:+},
\begin{align*}
  CD&\parallel AB\parallel GH,&
  AC&\parallel BD,&
  AG&\parallel BH,
\end{align*}
then
\begin{equation*}
  CG\parallel DH.
\end{equation*}
Likewise, since also $CE\parallel DF$,
we have $GE\parallel HF$.

\subsubsection{Commutativity}

By the symmetry of the definition,
as we have noted,
addition of points not collinear with the reference point is commutative.
For points that \emph{are} collinear with the reference point,
the definition of their sum is not symmetric,
but commutativity is equivalent to the case of Pappus's Theorem
called the Parallel Theorem,
and proved for affine planes, in \secref{sect:VIII}.
In \figref{fig:weak-2} now,
\begin{figure}
  \centering
  \begin{pspicture}(10,3)
%      \psgrid
    \pstGeonode[PosAngle={-90,-90,90}](0.2,0.6)A(4,0.6)B(2.2,2.4)D
    \pstHomO[HomCoef=1.5,PosAngle=-90]AB[C]
    \pstTranslation[PosAngle={-90,90}] AC{B,D}[F,E]
    \pstTranslation[PosAngle=90] ADB[G]
%    \pstInterLL[PosAngle=60] CGBEH
    \ncline AD
    \ncline AF\ncline FE\ncline ED
    \ncline DB\ncline CE\ncline BG
    \psset{linestyle=dotted}
    \ncline GF\ncline DC
%    \psset{linestyle=dashed}
%    \ncline DF\ncline BE\ncline CG
  \end{pspicture}
  \caption{Commutativity of addition}\label{fig:weak-2}
  
\end{figure}
\begin{gather*}
  B+_AC=B+_D(D+_AC)=B+_DE=F,\\
  C+_AB=C+_D(D+_AB)=C+_DG,
\end{gather*}
and therefore
\begin{equation*}
  B+_AC=C+_AB\iff DC\parallel GF.
\end{equation*}
We have the last parallelism by the Parallel Theorem,
applied to the hexagon $DBGFEC$, since
\begin{align*}
  BF&\parallel DE,&DB&\parallel FE,&BG&\parallel EC.
\end{align*}


\subsubsection{Associativity}

To prove associativity of addition,
we have three cases to consider.
\begin{asparaenum}
  \item
When $A$ is not collinear with any two of $B$, $C$, and $D$,
as in \figref{fig:ass},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-1,-0.6)(7,4.5)
%    \psgrid
    \pstGeonode[PosAngle={-90,-90,45,180}] A(5,0)B(2,1.5)C(-1,2.5)D
    \pstTranslation ABC[E]
    \pstTranslation[PosAngle=90] ADE[F]
    \pstTranslation[PosAngle=90] ACD[G]
    \ncline AB\ncline AC\ncline AD
    \psset{linestyle=dashed}
    \ncline BE\ncline EC
    \ncline AE\ncline EF\ncline FD
    \ncline CG\ncline GD\ncline AG
    \ncline[linestyle=dotted] FG
    \ncline[linestyle=dotted] BF
  \end{pspicture}

  \caption{Associativity of addition: easy case}\label{fig:ass}
\end{figure}
then
\begin{align*}
  B+_AC&=E,&E+_AD&=F,&C+_AD&=G,
\end{align*}
and then
\begin{equation*}
  B+_AG=F\iff AG\parallel BF\And GF\parallel AB.
\end{equation*}
They \emph{are} parallel, by the Prism Theorem.
Indeed,
\begin{align*}
  CG&\parallel AD\parallel EF,&AC&\parallel DG,&AE&\parallel DF,
\end{align*}
and therefore $CE\parallel GF$; but also $AB\parallel CE$, so
\begin{equation*}
  AB\parallel GF.
\end{equation*}
Now we can derive $AG\parallel BF$ similarly.
Thus
\begin{equation}\label{eqn:+ass}
  (B+_AC)+_AD=B+_A(C+_AD).
\end{equation}
\item
When $AB$ contains $C$, but not $D$,
then \eqref{eqn:+ass} still holds,
since in \figref{fig:+as},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-0.2,-0.6)(9.2,3.6)
%      \psgrid
    \pstGeonode[PosAngle={-90,-90,90}] A(3,0)B(1,3)D
    \pstHomO[HomCoef=1.5,PosAngle=-90]AB[C]
    \pstTranslation[PosAngle={-90,90}] AC{B,D}[F,E]
    \pstTranslation[PosAngle=90] ADB[G]
    \pstTranslation[PosAngle=90] ADF[K]
    \ncline AF\ncline FK\ncline KD\ncline DA
    \ncline DB\ncline BG\ncline CE\ncline EF
    \psset{linestyle=dotted}
    \ncline AE\ncline BK
%    \psset{linestyle=dashed}
%    \ncline DF\ncline BE\ncline CG
  \end{pspicture}
  \caption{Associativity of addition: less easy case}\label{fig:+as}
  
\end{figure}
\begin{align*}
  (B+_AC)&=F,&F+_AD&=K,&
  C+_AD&=E,
\end{align*}
so that
\begin{equation*}
  K=B+_AE\iff AE\parallel BK.
\end{equation*}
Now apply the Parallel Theorem to $ADBKFE$.
\item
Finally, when $AB$ contains both $C$ and $D$,
then, making use of commutativity,
in \figref{fig:+ass}
\begin{figure}[ht]
  \centering
  \begin{pspicture}(0,-0.6)(9.6,4.5)
%    \psgrid
    \pstGeonode[PosAngle={-90,-90,90}] A(3,0)B(1.5,4)E
    \psset{PosAngle=-90}
    \pstHomO[HomCoef=0.67]AB[C]
    \pstHomO[HomCoef=1.5]AB[D]
    \pstTranslation[PosAngle=90] AEB[F]
    \ncline BF\ncline EB
    \pstTranslation ECF[G]
    \ncline FG\ncline EG
    \pstTranslation[PosAngle=90] AED[H]
    \pstTranslation EGH[K]
    \ncline HK
    \pstTranslation ECH[L]
    \ncline HL
    \pstTranslation[PosAngle=90] AEL[M]
    \ncline LM
    \ncline EM\ncline BK
    \psset{linestyle=dashed}
    \ncline MK
    \ncline BH\ncline GM
    \psset{linestyle=dotted}
    \ncline AB\ncline EC\ncline AE\ncline DH
%    \psset{linestyle=dashed}\ncline BM\ncline EK\ncline GH
  \end{pspicture}
  \caption{Associativity of addition: hardest case}\label{fig:+ass}
  
\end{figure}
we have
\begin{align*}
  C+_AB&=G,&
G+_AD&=K,&
C+_AD&=L,
\end{align*}
and
\begin{equation*}
  B+_AL=K\iff EB\parallel MK.
\end{equation*}
By the Parallel Theorem
applied to $BFGMLH$,
\begin{equation*}
  BH\parallel GM;
\end{equation*}
then, applied to $KHBEGM$,
$BE\parallel MK$.
\end{asparaenum}

\subsubsection{Negatives}

Having defined vectors as in \secref{sect:eq-rel},
we can define
\begin{equation*}
  \vec AB+\vec CD=\vec AE,
\end{equation*}
where,
as in \figref{fig:vec-add},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-0.2,-1)(7,3.2)
%    \psgrid
    \pstGeonode[PointSymbol=none,PosAngle={180,180,-90,90}]
    A(1,3)B(3,1)C(7,0.5)D
    \pstTranslation[PosAngle=-45] CAD[F]
    \pstTranslation[PointSymbol=none] ABF[E]
    \psset{linestyle=dotted}
    \ncline AC\ncline CD\ncline DF\ncline FA
    \ncline BE\ncline EF
    \ncline[arrows=->,linestyle=dashed]AE
    \psset{linewidth=2pt,linestyle=solid}
    \ncline[arrows=->]AB
    \ncline[arrows=->]CD
  \end{pspicture}
  \caption{Addition of vectors}\label{fig:vec-add}
  
\end{figure}
\begin{align*}
  E&=B+_AF,&F&=(A+_CD)
\end{align*}
By what we have shown,
the vectors compose an abelian group with respect to this addition.
In particular, the $0$ of the group is $\vec AA$, and
\begin{equation*}
  -\vec AB=\vec BA.
\end{equation*}

\subsection{Distributivity}

By Thales's Theorem,
\begin{equation*}
  r\cdot_A(B+_AC)=r\cdot_AB+_Ar\cdot_AC.
\end{equation*}
For any additional ratio $s$,
there is a ratio $t$ such that,
for all $A$ and $B$,
\begin{equation*}
r\cdot_AB+s\cdot_AB=  t\cdot_AB.
\end{equation*}
Then we can define
\begin{equation*}
  r+s=t.
\end{equation*}
We can also define
\begin{equation*}
  r\cdot\vec AB=r\cdot_AB.
\end{equation*}
We are not writing out all details of the proofs here.
Ratios now compose a field,
possibly a skew field,
and the vectors themselves
compose a vector space over this field.


\section{A skew field}\label{sect:H}

An \emph{example} of a skew field
is the field $\Ham$ of \textbf{quaternions,}
discovered by Hamilton.
We can obtain this field from the field $\C$ of complex numbers
as we can obtain $\C$ from $\R$.

\subsection{Complex numbers}

We define
\begin{equation*}
\C=\left\{\begin{pmatrix}x&y\\-y&x\end{pmatrix}\colon(x,y)\in\R^2\right\}.
\end{equation*}
This is a subspace of the
space of all $2\times 2$ matrices over $\R$.
Using the abbreviations
\begin{align*}
  \begin{pmatrix}x&0\\0&x\end{pmatrix}&=x,&
    \begin{pmatrix}0&1\\-1&0\end{pmatrix}&=\mathbf i,
\end{align*}
we have
\begin{equation*}
  \begin{pmatrix}x&y\\-y&x\end{pmatrix}
    = \begin{pmatrix}x&0\\0&x\end{pmatrix}
      + \begin{pmatrix}0&y\\-y&0\end{pmatrix}
    =x+y\mathbf i.
\end{equation*}
In particular, $(1,\mathbf i)$ is a basis of $\C$ over $\R$.
Moreover, since
\begin{equation*}
  \mathbf i^2=-1,
\end{equation*}
$\C$ is closed under multiplication,
and multiplication is commutative on $\C$.
Finally,
\begin{equation*}
  (x+y\mathbf i)(x-y\mathbf i)=x^2+y^2,
\end{equation*}
which is in $\R$, and therefore $\C$ is a field.
We define
\begin{equation*}
  \overline{x+y\mathbf i}=x-y\mathbf i.
\end{equation*}

\subsection{Quaternions}

By analogy, we define
\begin{equation*}
  \Ham=\left\{\begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix}
  \colon(z,w)\in\C^2\right\}.
\end{equation*}
This is \emph{not}
a subspace of the space of $2\times 2$ matrices over $\C$.
It is however a \emph{real} subspace.
Moreover, using the abbreviations
\begin{align*}
  \begin{pmatrix}z&0\\0&\overline z\end{pmatrix}&=z,&
    \begin{pmatrix}0&1\\-1&0\end{pmatrix}&=\mathbf j,&
\end{align*}
we have
\begin{equation*}
  \begin{pmatrix}z&w\\-\overline w&\overline z\end{pmatrix}
    = \begin{pmatrix}z&0\\0&\overline z\end{pmatrix}
      + \begin{pmatrix}0&w\\-\overline w&0\end{pmatrix}
        =z+w\mathbf j,
\end{equation*}
but also
\begin{align*}
  \overline w\mathbf j&=\mathbf jw,&
  \mathbf j^2&=-1.
\end{align*}
Thus $\Ham$ is a sub-ring of the ring of $2\times 2$ matrices over $\C$,
and moreover, $\Ham$ in turn has the sub-ring
\begin{equation*}
  \left\{
  \begin{pmatrix}
    z&0\\0&\overline z
  \end{pmatrix}\colon z\in\C\right\},
\end{equation*}
which is isomorphic to $\C$,
and $\Ham$ is both a \emph{left}
and a \emph{right} vector-space over this,
with basis $(1,\mathbf j)$ in each case.
Also
\begin{equation*}
  (z+w\mathbf j)(\overline z-w\mathbf j)
  =z\overline z+w\mathbf j\overline z-zw\mathbf j-w\mathbf jw\mathbf j
  =z\overline z+w\overline w,
\end{equation*}
which is in $\R$, and therefore $\Ham$ is a field,
albeit a skew field.

$\Ham$ is also a two-sided real vector space,
with basis $(1,\mathbf i,\mathbf j,\mathbf k)$, where
\begin{equation*}
  \mathbf k=\mathbf i\mathbf j.
\end{equation*}
In analytic geometry, in $\R^2$, we define lines by equations
\begin{equation*}
  ax+by=c,
\end{equation*}
where not both of $a$ and $b$ are $0$.
We  can do the same in $\Ham^2$, or else we can use equations
\begin{equation*}
  xa+yb=c,
\end{equation*}
but we cannot do both.
For example, the solution sets of
\begin{align*}
  \mathbf ix+y&=0,&x\mathbf i+y&=0
\end{align*}
have $(0,0)$ and $(\mathbf i,1)$ in common,
but are not identical:
$(\mathbf j,-\mathbf k)$ solves the former equation, not the latter;
$(\mathbf j,\mathbf k)$, the latter, not the former.
We shall continue with these ideas tomorrow.

\section{Pappus's Theorem, intersecting cases}\label{sect:XI}

Pappus's \textbf{Lemma XI} is that, 
in \figref{fig:XI}
or \ref{fig:XI-alt2} (Pappus only alludes to the latter),
\begin{figure}[ht]
%  \centering
  \subfloat[]{\label{fig:XI}
    \begin{pspicture}(0,-2.5)(5.2,3.5)
%    \psgrid
    \psset{PointName=none}
    \pstGeonode E(5,0)G(3,3)A
    \pstHomO[HomCoef=0.3]AG[H]
    \pstHomO[HomCoef=0.45]EG[B]
    \pstTranslation[PointSymbol=none] EGA
    \pstInterLL A{A'}EHD
    \pstTranslation[PointSymbol=none] EHG
    \pstInterLL G{G'}ABJ
    \pstInterLL ABEHZ
    \ncline DE\ncline EG
    \psset{linestyle=dotted}
    \ncline BJ\ncline JG
    \psset{linewidth=2pt,linestyle=solid}
    \ncline AB\ncline AG\ncline AD
    \nput uA{\gr A}
    \nput{-45}B{\gr B}
    \nput{-45}G{\gr G}
    \nput{u}D{\gr D}
    \nput{d}E{\gr E}
    \nput{ul}Z{\gr Z}
    \nput{-20}H{\gr H}
    \nput{d}J{\gr J}
    \end{pspicture}
  }
  \hfill
  \subfloat[]{\label{fig:XI-alt2}
    \begin{pspicture}(-1.4,-3.5)(3.2,2.5)
%    \psgrid
    \psset{PointName=none}
    \pstGeonode E(2.5,0)G(3,2)A
    \pstHomO[HomCoef=2.5]AG[H]
    \pstHomO[HomCoef=0.4]EG[B]
    \pstTranslation[PointSymbol=none] EGA
    \pstInterLL A{A'}EHD
    \pstTranslation[PointSymbol=none] EHG
    \pstInterLL G{G'}ABJ
    \pstInterLL ABEHZ
    \ncline DH\ncline EG
    \psset{linestyle=dotted}
    \ncline BJ\ncline JG
    \psset{linewidth=2pt,linestyle=solid}
    \ncline AZ\ncline AH\ncline AD
    \nput uA{\gr A}
    \nput{ul}B{\gr B}
    \nput{-45}G{\gr G}
    \nput{u}D{\gr D}
    \nput{l}E{\gr E}
    \nput{-135}Z{\gr Z}
    \nput{d}H{\gr H}
    \nput{ul}J{\gr J}
    \end{pspicture}
  }
  \caption{Lemma XI}
\end{figure}
\begin{equation*}
\grm{DE}\cdot\grm{ZH}:\grm{EZ}\cdot\grm{HD}\prop \grm{GB}:\grm{BE}.
\end{equation*}
This is,
in the notation for cross ratios of \secref{sect:III},
with a correction of the ordering of points,
\begin{equation}\label{eqn:DEZH}
  \crossratio{\grm D}{\grm E}{\grm Z}{\grm H}\prop\grm{GB}\rat\grm{EB}.
\end{equation}
By Lemma III, if \gr{AD} and \gr{EG} met at a point \gr K,
then
\begin{equation*}
  \crossratio{\grm D}{\grm E}{\grm Z}{\grm H}
  \prop\crossratio{\grm K}{\grm E}{\grm B}{\grm G}
  \prop(\grm{EK}\rat\grm{GK})(\grm{GB}:\grm{EB}).
\end{equation*}
This yields \eqref{eqn:DEZH},
if $\grm{EK}\rat\grm{GK}$ becomes identity when $K$ is at infinity.
Pappus does not argue this way,
but, drawing \gr{GJ} parallel to \gr{DE},
he has by Thales
\begin{gather*}
	\grm{GJ}:\grm{ZH}\prop
	\grm{GA}:\grm{AH}\prop \grm{ED}:\grm{DH},\\
\grm{ED}\cdot\grm{ZH}=\grm{GJ}\cdot\grm{DH},\\
\begin{aligned}
	\grm{ED}\cdot\grm{ZH}:\grm{DH}\cdot\grm{EZ}
	&\prop \grm{GJ}\cdot\grm{DH}:\grm{DH}\cdot\grm{EZ}\\
	&\prop \grm{GJ}:\grm{EZ}\\
	&\prop \grm{GB}:\grm{EB},
\end{aligned}
\end{gather*}
which is \eqref{eqn:DEZH}.
Alternatively,
from the proof of Lemma III, we know
\begin{equation*}%\label{eqn:EHDZ}
  \crossratio{\grm E}{\grm H}{\grm D}{\grm Z}\prop\grm{EG}\rat\grm{EB}.
\end{equation*}
From this we obtain \eqref{eqn:DEZH}
in modern notation as follows.
\begin{multline*}
  \crossratio{\grm D}{\grm E}{\grm Z}{\grm H}
  =\frac{\grm{DE}\cdot\grm{ZH}}{\grm{EZ}\cdot\grm{HD}}
  =\frac{(\grm{DH}+\grm{HE})\cdot\grm{ZH}}{\grm{EZ}\cdot\grm{HD}}\\
  =\frac{\grm{DH}\cdot\grm{ZH}+\grm{HE}\cdot\grm{ZH}}
  {\grm{EZ}\cdot\grm{HD}}\\
  =\frac{\grm{DH}\cdot(\grm{ZE}+\grm{EH})+\grm{HE}\cdot\grm{ZH}}
  {\grm{EZ}\cdot\grm{HD}}\\
  =1+\frac{\grm{DH}\cdot\grm{EH}+\grm{HE}\cdot\grm{ZH}}
  {\grm{EZ}\cdot\grm{HD}}\\
  =1-\frac{\grm{HD}\cdot\grm{EH}+\grm{EH}\cdot\grm{ZH}}
  {\grm{EZ}\cdot\grm{HD}}\\
  =1-\frac{\grm{ZD}\cdot\grm{EH}}
  {\grm{EZ}\cdot\grm{HD}}\\
  =1-  [\grm E,\grm H,\grm D,\grm Z]
  =1-\frac{\grm{EG}}{\grm{EB}}=\frac{\grm{GB}}{\grm{EB}}.
\end{multline*}


\textbf{Lemma XII} is that,
in \figref{fig:XII}
\begin{figure}[ht]
\centering
\begin{pspicture}(0,-0.5)(5,4.5)
    \psset{PointName=none}
    \pstGeonode[PosAngle={-90,-90,90,90}]G(5,0)D(0.3,4)A(4.8,4)B
    \pstHomO[PosAngle=90,HomCoef=0.61]AB[E]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \ncline AD\ncline DE\ncline EG\ncline GB\ncline BZ\ncline ZA
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \ncline[linestyle=dotted]KH
    \psset{linestyle=solid,linewidth=2pt}
    \ncline AB\ncline DG
    \nput uA{\gr A}
    \nput uB{\gr B}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput dZ{\gr Z}
    \nput lH{\gr H}
    \nput uJ{\gr J}
    \nput rK{\gr K}
    \nput uL{\gr L}
    \nput dM{\gr M}
\end{pspicture}

    \caption{Lemma XII}\label{fig:XII}
\end{figure}
and \figref{fig:XII-alt},
\begin{figure}
  \subfloat[]{
  \begin{pspicture}(-0.2,-1.7)(5,5)
%    \psgrid
    \psset{PointName=none}
    \pstGeonode[PosAngle={-90,-90,90,90}]G(5,0)D(1.3,3)B(3.5,3)A
    \pstHomO[PosAngle=90,HomCoef=0.55]AB[E]
    \pstHomO[PosAngle=-90,HomCoef=0.45]GD[Z]
    \ncline AD\ncline DE\ncline EG\ncline GB\ncline BZ\ncline ZA
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \ncline[linestyle=dotted]KH
    \psset{linestyle=dashed}
    \ncline BK\ncline KE\ncline EH\ncline HA\ncline AM\ncline MB
    \psset{linestyle=solid,linewidth=2pt}
    \ncline AB\ncline DG
    \nput rA{\gr A}
    \nput lB{\gr B}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput dZ{\gr Z}
    \nput uH{\gr H}
    \nput uJ{\gr J}
    \nput uK{\gr K}
    \nput uL{\gr L}
    \nput uM{\gr M}
\end{pspicture}
}
  \hfill
  \subfloat[]{
    \begin{pspicture}(0,-2.7)(4.2,6)
%      \psgrid
    \psset{PointName=none}
    \pstGeonode[PosAngle={-90,-90,90,90}](0,3)A(3.5,3)B(0.2,0)D(4,0)Z
    \pstHomO[PosAngle=90,HomCoef=0.5]AB[E]
    \pstHomO[PosAngle=-90,HomCoef=0.2]DZ[G]
    \ncline AD\ncline DE\ncline EG\ncline GB\ncline BZ\ncline ZA
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \ncline[linestyle=dotted]KM
    \psset{linestyle=dashed}
    \ncline EK\ncline BK\ncline DJ\ncline GL\ncline DL\ncline GJ
    \psset{linestyle=solid,linewidth=2pt}
    \ncline AB\ncline DZ
    \nput uA{\gr A}
    \nput rB{\gr B}
    \nput{-45}G{\gr G}
    \nput lD{\gr D}
    \nput{135}E{\gr E}
    \nput dZ{\gr Z}
    \nput rH{\gr H}
    \nput dJ{\gr J}
    \nput uK{\gr K}
    \nput dL{\gr L}
    \nput{-105}M{\gr M}
  \end{pspicture}
  }
  \caption{Lemma XII, alternative figures}\label{fig:XII-alt}
\end{figure}
where $\grm{AB}\parallel\grm{GD}$,
the points \gr{H, M,} and \gr K
are on a straight line.
The proof considers the parts of the diagram 
shown in \figref{fig:XII-steps}.
\begin{figure}
    \psset{PointName=none,PointSymbol=none,unit=6mm}
    \subfloat[]{
  \begin{pspicture}(0,-0.8)(5,4.8)
      \pstGeonode G(5,0)D(0.3,4)A(4.8,4)B
    \pstHomO[HomCoef=0.61]AB[E]
    \pstHomO[HomCoef=0.4]GD[Z]
    \ncline AD\ncline EG\ncline ZA
    \pstInterLL AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL ADGBM
    \ncline AE\ncline DG
    \nput uA{\gr A}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput dZ{\gr Z}
    \nput lH{\gr H}
    \nput uJ{\gr J}
\end{pspicture}
    }
    \hfill
    \subfloat[]{
  \begin{pspicture}(0,-0.8)(5,4.8)
      \pstGeonode[PosAngle={-90,-90,90,90}]G(5,0)D(0.3,4)A(4.8,4)B
    \pstHomO[PosAngle=90,HomCoef=0.61]AB[E]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \ncline DE\ncline GB\ncline BZ
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \ncline EB\ncline DG
    \nput uB{\gr B}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput dZ{\gr Z}
    \nput rK{\gr K}
    \nput lL{\gr L}
\end{pspicture}
    }
    \hfill
        \subfloat[]{
  \begin{pspicture}(0,-0.8)(5,4.8)
      \pstGeonode[PosAngle={-90,-90,90,90}]G(5,0)D(0.3,4)A(4.8,4)B
    \pstHomO[PosAngle=90,HomCoef=0.61]AB[E]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \ncline GE\ncline ED
    \pstInterLL[PointSymbol=*] AZGEH
    \pstInterLL[PointSymbol=*] EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \ncline JD\ncline LG
    \pstInterLL[PosAngle=-90] ADGBM
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput lH{\gr H}
    \nput{ul}J{\gr J}
    \nput rK{\gr K}
    \nput{ur}L{\gr L}
    \nput dM{\gr M}
\end{pspicture}
    }

        \caption{Steps of Lemma XII}\label{fig:XII-steps}
\end{figure}
Applying Lemma XI to the first two parts yields
\begin{equation*}
  [\grm E,\grm G,\grm H,\grm J]
  \prop\grm{DZ}\rat\grm{GZ}
  \prop[\grm E,\grm L,\grm K,\grm D]
\end{equation*}
By Lemma X then, \gr{HMK} is straight.%
\index{Lemmas of Pappus!12@XII|)textbf} 

\textbf{Lemma XIII} is the same,
except that \gr{AB} and \gr{GD} meet at a point \gr N,
as in \figref{fig:XIII}
\begin{figure}[ht]
  \centering
    \begin{pspicture}(-5,-0.5)(5,4.5)
    \psset{PointName=none}
    \pstGeonode[PosAngle={-90,-90,90}](-5,0)N(5,0)D(4,4)B
    \pstHomO[HomCoef=0.5,PosAngle=90]NB[A]
    \pstHomO[PosAngle=90,HomCoef=0.5]AB[E]
    \pstHomO[HomCoef=0.5,PosAngle=-90]ND[G]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \ncline AD\ncline DE\ncline EG\ncline GB\ncline BZ\ncline ZA
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \ncline[linestyle=dotted]KH
    \psset{linestyle=solid,linewidth=2pt}
    \ncline NB\ncline DN
    \nput uA{\gr A}
    \nput uB{\gr B}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uE{\gr E}
    \nput dZ{\gr Z}
    \nput lH{\gr H}
    \nput uJ{\gr J}
    \nput rK{\gr K}
    \nput uL{\gr L}
    \nput dM{\gr M}
    \nput dN{\gr N}
    \end{pspicture}

\caption{Lemma XIII: Pappus's figure}\label{fig:XIII}
\end{figure}
and \figref{fig:XIII-alt},
\begin{figure}
  \centering
    \begin{pspicture}(-3.4,-0.5)(5,4.5)
%      \psgrid
    \psset{PointName=none}
    \pstGeonode[PosAngle={-90,-90,90,90}]G(5,0)D(0.3,3.8)K(4.8,4)H
    \pstHomO[PosAngle=90,HomCoef=0.61]KH[M]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \pstInterLL[PosAngle=180] KZGMB
    \pstInterLL MDZHA
    \pstInterLL KDMGL
    \pstInterLL MDGHJ
    \pstInterLL[PosAngle=-90] KDGHE
    \pstInterLL ABDGN
    \ncline GB\ncline BZ\ncline ZA\ncline AD\ncline DE\ncline EG
    \ncline[linestyle=dotted]KH
    \psset{linestyle=dashed}
    \ncline BK\ncline KE
    \ncline EH\ncline HA
    \ncline AM\ncline MB
    \psset{linestyle=solid,linewidth=2pt}
    \ncline NA\ncline ND
    \nput uK{\gr K}
    \nput uH{\gr H}
    \nput dG{\gr G}
    \nput dD{\gr D}
    \nput uM{\gr M}
    \nput dZ{\gr Z}
    \nput{165}B{\gr B}
    \nput uJ{\gr J}
    \nput rA{\gr A}
    \nput uL{\gr L}
    \nput dE{\gr E}
    \nput dN{\gr N}
  \end{pspicture}

\caption{Lemma XIII: alternative figure}\label{fig:XIII-alt}
\end{figure}
so that Lemma III is used in place of Lemma XI.
This gives us the intersecting cases of Pappus's Theorem.

\section{Pappus's Theorem by projection}\label{sect:Pap-proj}

The mixed cases of Pappus's Theorem are as in \figref{fig:mixed},
\begin{figure}
\subfloat[Intersecting bounding lines]{\label{fig:Pap-n1}
      \begin{pspicture}(-0.2,-0.5)(5,5.4)
%      \psgrid
    \psset{PointName=none}
    \pstGeonode B(3,4)A(4.8,0)Z
    \nput dB{\gr B}\nput lA{\gr A}\nput dZ{\gr Z}
    \pstHomO[HomCoef=-0.3]AB[H]
    \nput rH{\gr H}
    \pstHomO[HomCoef=0.5]AB[D]
    \nput lD{\gr D}
    \pstTranslation[PointSymbol=none] BZH
    \pstInterLL H{H'}AZE
    \nput{l}E{\gr E}
    \pstHomO[HomCoef=0.4]AZ[G]
    \nput{r}G{\gr G}
    \ncline HE\ncline ED\ncline ZB\ncline BG
    \ncline GH\ncline DZ
    \pstInterLL ZDGHL
    \pstInterLL BGDEK
   \nput dL{\gr L}
    \nput dK{\gr K}
    \psset{linestyle=dashed}
    \ncline DL\ncline LH\ncline GK\ncline KE
    \psset{linestyle=dotted}
    \ncline KL
    \psset{linewidth=2pt,linestyle=solid}
    \ncline BH\ncline EZ
  \end{pspicture}
}
\hfill
\subfloat[Parallel bounding lines]{\label{fig:Pap-p1}
        \begin{pspicture}(-1.2,-0.5)(3.2,5.2)
%      \psgrid
    \psset{PointName=none}
    \pstGeonode B(-0.3,2)H(3,0)Z
    \pstTranslation BHZ[E]
    \nput dB{\gr B}\nput lH{\gr H}\nput dZ{\gr Z}\nput{r}E{\gr E}
    \pstHomO[HomCoef=-1.5]HB[D]
    \nput lD{\gr D}
    \pstHomO[HomCoef=-0.9]EZ[G]
    \nput{r}G{\gr G}
    \ncline HE\ncline ED\ncline ZB\ncline BG
    \ncline GH\ncline DZ
    \pstInterLL ZDGHL
    \pstInterLL BGDEK
   \nput lL{\gr L}
    \nput rK{\gr K}
     \psset{linestyle=dashed}
    \ncline DL\ncline LH\ncline GK\ncline KE
    \psset{linestyle=dotted}
    \ncline KL
    \psset{linewidth=2pt,linestyle=solid}
    \ncline BD\ncline GZ
  \end{pspicture}
}
  \caption{Pappus's Theorem: mixed cases}\label{fig:mixed}
  
\end{figure}
where
\begin{compactitem}
  \item
    the hexagon is \gr{BGHEDZ},
  \item
    \gr{BG} and \gr{ED} meet at \gr K,
  \item
    \gr{GH} and \gr{DZ} meet at \gr L.
\end{compactitem}
The theorem is,
\begin{equation*}
  \grm{BZ}\parallel\grm{HE}\implies
  \grm{HE}\parallel\grm{KL}.
\end{equation*}
We can prove this,
and all other cases of Pappus's Theorem,
except for the Parallel Theorem,
using only Lemma VIII
and \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.

For example, adding to \figref{fig:VIII},
%or \figref{fig:VIII-alt},
we draw \gr{MN} through \gr A, parallel to \gr{BZ},
and we conceive of the diagram
as lying in a vertical plane in \figref{fig:Pap-proj}
\begin{figure}[ht]
  \centering
  \psset{unit=5mm}
  \begin{pspicture}(-15,-4)(7,21.5)
%    \psgrid[subgriddiv=1]
    \psset{PointName=none,PointSymbol=none,Alpha=-65,Beta=15}
%      \pstThreeDCoor[xMax=5,yMax=5,zMax=5]
    \pstThreeDPut(4,0,10){\pnode A}
    \pstThreeDPut(4,2,7.5){\pnode H} % height was 5
    \pstThreeDPut(4,-1,7.5){\pnode E} % height was 5
    \pstHomO[HomCoef=-4]AH[B] % coef was -2
    \pstHomO[HomCoef=-1.5]AH[D]  % coef was -1.2
    \pstTranslation[PointSymbol=none] HEB[BB]
    \pstInterLL B{BB}AEZ
    \pstTranslation BZA[MM]
    \pstHomO[HomCoef=0.75]A{MM}[M]
    \pstHomO[HomCoef=2]MA[N]
    \pstThreeDPut(8,0,10){\pnode I}\nput uI{\gr I}
    \pstThreeDDot(8,0,10)
    \pstHomO[HomCoef=4] IH
    \pstHomO[HomCoef=4] IE
    \pstHomO[HomCoef=-1] IB
    \pstHomO[HomCoef=-1] IZ
    \pstTranslation[PointSymbol=none] DZH[HH]
    \pstInterLL H{HH}AEG
    \pstInterLL ID{B'}{H'}{D'}
    \pstInterLL IG{Z'}{E'}{G'}
    \pstInterLL{E'}{D'}{G'}{B'}K
    \pstInterLL{H'}{G'}{D'}{Z'}L
    \nput{0}B{\gr B}
    \nput{135}A{\gr A}
    \nput{180}Z{\gr Z}
    \nput lH{\gr H}
    \nput lD{\gr D}
    \nput{45}E{\gr E}
    \nput{0}G{\gr G}
    \nput{-120}{B'}{$\grm B'$}
    \nput{105}{Z'}{$\grm Z'$}
    \nput d{H'}{$\grm H'$}
    \nput{90}{D'}{$\grm D'$}
    \nput d{E'}{$\grm E'$}
    \nput{60}{G'}{$\grm G'$}
    \nput d{K}{\gr K}
    \nput u{L}{\gr L}
    \nput u{M}{\gr M}
    \nput u{N}{\gr N}
    \ncline[linestyle=dotted] MN
    \ncline HE\ncline ED\ncline DZ\ncline ZB\ncline BG\ncline GH
    \ncline[linewidth=2pt] BH\ncline[linewidth=2pt] GZ
    \ncline{B'}{Z'}
    \ncline[linewidth=2pt]{Z'}{E'}
    \ncline{E'}{H'}
    \ncline[linewidth=2pt]{D'}{H'}
    \ncline{B'}{G'}
    \ncline{G'}{H'}
    \ncline{E'}{D'}
    \ncline{D'}{Z'}
    \psset{linestyle=dashed}
    \ncline{D'}K\ncline{B'}K
    \ncline{G'}L\ncline{Z'}L
    \ncline[linestyle=dotted]KL
  \end{pspicture}
  \caption{Mixed case by projection}\label{fig:Pap-proj}
  
\end{figure}
\begin{comment}
  or \figref{fig:Pap-proj-2},
  \begin{figure}
    \centering
    \psset{unit=4.5mm}
    \begin{pspicture}(-10,-5.2)(14,16.5)
      %    \psgrid[subgriddiv=1]
      \psset{PointName=none,PointSymbol=none,Alpha=-75,Beta=20}
      %      \pstThreeDCoor[xMax=5,yMax=5,zMax=5]
      \pstThreeDPut(4,0,12){\pnode A}
      \pstThreeDPut(4,0.2,8){\pnode H}
      \pstThreeDPut(4,-1,8){\pnode E}
      \pstHomO[HomCoef=2]AH[B]
      \pstHomO[HomCoef=1.6]AH[D]
      \pstTranslation[PointSymbol=none] HEB[BB]
      \pstInterLL B{BB}AEZ
      \pstTranslation BZA[M]
      \pstTranslation ZBA[N]
      \pstThreeDPut(12,0,12){\pnode I}\nput uI{\gr I}
      \pstThreeDDot(12,0,12)
      \pstHomO[HomCoef=3] IH
      \pstHomO[HomCoef=3] IE
      \pstHomO[HomCoef=1.5] IB
      \pstHomO[HomCoef=1.5] IZ
      \pstTranslation[PointSymbol=none] DZH[HH]
      \pstInterLL H{HH}AEG
      \pstInterLL ID{B'}{H'}{D'}
      \pstInterLL IG{Z'}{E'}{G'}
      \pstInterLL{E'}{D'}{G'}{B'}K
      \pstInterLL{H'}{G'}{D'}{Z'}L
      \nput{180}B{\gr B}
      \nput uA{\gr A}
      \nput{0}Z{\gr Z}
      \nput lH{\gr H}
      \nput lD{\gr D}
      \nput{45}E{\gr E}
      \nput{r}G{\gr G}
      \nput u{B'}{$\grm B'$}
      \nput u{Z'}{$\grm Z'$}
      \nput d{H'}{$\grm H'$}
      \nput{-135}{D'}{$\grm D'$}
      \nput d{E'}{$\grm E'$}
      \nput{0}{G'}{$\grm G'$}
      \nput u{K}{\gr K}
      \nput u{L}{\gr L}
      \nput u{M}{\gr M}
      \nput u{N}{\gr N}
      \ncline[linestyle=dotted] MN
      \ncline HE\ncline ED\ncline DZ\ncline ZB\ncline BG\ncline GH
      \ncline BA\ncline AZ
      \ncline{B'}{Z'}
      \ncline{Z'}{E'}
      \ncline{E'}{H'}
      \ncline{B'}{H'}
      \ncline{B'}{G'}
      \ncline{G'}{H'}
      \ncline{E'}{D'}
      \ncline{D'}{Z'}
      \psset{linestyle=dashed}
      \ncline{D'}K\ncline{B'}K
      \ncline{G'}L\ncline{Z'}L
      \ncline[linestyle=dotted]KL
    \end{pspicture}
    \caption{Pappus's Theorem by projection}\label{fig:Pap-proj-2}
    
  \end{figure}
\end{comment}
where \gr{MN} is horizontal.
We let \gr I lie not on \gr{MN},
but in a horizontal plane that contains \gr{MN},
and we project the diagram from \gr I onto another horizontal plane,
so that \gr B becomes $\grm B'$, and \gr G becomes $\grm G'$, and so on.
All straight lines, such as \gr{BZ} and \gr{HE},
that were parallel to \gr{MN}
remain parallel to one another in the new diagram.
Lines such as \gr{BG} and \gr{DE} that were parallel to one another,
but not to \gr{MN}, now intersect;
but their intersection points all lie on a single line, \gr{KL},
which represents the line at infinity of the old diagram.  
Lines such as \gr{AB} and \gr{AZ} that intersected on \gr{MN}
become parallel in the new diagram.
Thus we obtain the mixed cases of Pappus's Theorem
with parallel bounding lines (\figref{fig:Pap-p1}).
Similarly,
\begin{compactitem}
  \item
  if \gr{MN} is parallel to \gr{BZ},
  but does not contain \gr A,
  we obtain the mixed case with intersecting bounding lines
  (\figref{fig:Pap-n1});
  \item
    if \gr{MN} is not parallel to any other lines of the figure, and
    \begin{compactitem}
    \item
      does not contain \gr A,
  we obtain Lemma XIII (\figref{fig:XIII});
  \item
  but does contain \gr A,
  we obtain Lemma XII (\figref{fig:XII}).
    \end{compactitem}
\end{compactitem}
Tomorrow
we shall obtain a third proof of Pappus's Theorem
by coordinatizing a projective plane.
 
\chapter{Saturday}\label{ch:proj}

\section{Cartesian coordinates}

In \secref{sect:+},
we have obtained a field $\field$ of ratios
in an affine plane.
Given a triangle $ABC$ in such a plane,
we can consider the plane as a left vector space over $\field$
with neutral point $C$ and basis $(A,B)$.
In particular, for any point $M$ of the plane,
there is a unique ordered pair $(s,t)$ of ratios,
meaning $(s,t)\in\field^2$, such that
\begin{equation*}
M=s\cdot_CA+_Ct\cdot_CB,
\end{equation*}
or equivalently
\begin{equation}\label{eqn:M-vec}
  \vec CM=s\cdot\vec CA+t\cdot\vec CB.
\end{equation}
Here $(s,t)$ is the ordered pair of
\textbf{Cartesian coordinates} of $M$
with respect to $ABC$.
Conversely, every element of $\field^2$
corresponds to a point of the plane in this way.

\section{Barycentric coordinates}

Since $\vec CC$ is the neutral or zero vector,
we can rewrite \eqref{eqn:M-vec} as
\begin{equation}\label{eqn:M-vec-2}
  \vec CM=s\cdot\vec CA+t\cdot\vec CB+(1-s-t)\cdot\vec CC.
\end{equation}
The point of doing this is that now the coefficients on the right add up to $1$.
Since, for all points $X$ and $D$,
\begin{equation*}
  \vec CX=\vec CD+\vec DX,
\end{equation*}
substitution into \eqref{eqn:M-vec-2} yields
\begin{equation*}
  \vec DM=s\cdot\vec DA+t\cdot\vec DB+(1-s-t)\cdot\vec DC,
\end{equation*}
or equivalently
\begin{equation*}
  M=s\cdot_DA+_Dt\cdot_DB+_D(1-s-t)\cdot_DC.
\end{equation*}
Since $D$ is arbitrary, we may write simply
\begin{equation*}
M=sA+tB+(1-s-t)C.
\end{equation*}
Conversely, 
if $p$, $q$, and $r$ are three ratios (elements of $\field$) for which
\begin{equation}\label{eqn:pqr1}
  p+q+r=1,
\end{equation}
then the linear combination
\begin{equation*}
pA+qB+rC
\end{equation*}
is unambiguous, defining the point
\begin{equation*}
  p\cdot_CA+_Cq\cdot_CB.
\end{equation*}
We may write the same point as
\begin{equation*}
(p:q:r)
\end{equation*}
when we consider $ABC$ as fixed.
But now we can allow
\begin{equation}\label{eqn:pqr}
(p:q:r)=(pt:qt:rt)
\end{equation}
for any nonzero $t$ in $\field$.
It will be important that the multiplier $t$ is on the right.
Given arbitrary ratios $p$, $q$, and $r$ for which the equation
\begin{equation}\label{eqn:=0}
  p+q+r=0
\end{equation}
\emph{fails,}
we have
\begin{equation}\label{eqn:p+q+r}
  (p:q:r)
  =ptA+qtB+rtC=pt\cdot_CA+qt\cdot_CB,
\end{equation}
where
\begin{equation*}%\label{eqn:t}
  t=(p+q+r)\inv.
\end{equation*}
The point $(p:q:r)$
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, from \grp{bar'us, -e~ia, -'u} ``heavy'')
of the system with weights $p$, $q$, and $r$
at $A$, $B$, and $C$ respectively.

\section{Ceva's Theorem}

Given $(p,q,r)$ satisfying \eqref{eqn:pqr1},
if we define
\begin{equation}\label{eqn:qBrC}
D=(q+r)\inv qB+(q+r)\inv rC=(q+r)\inv q\cdot_CB,
\end{equation}
then $D$ is a point on $BC$,
and since $t$ in \eqref{eqn:p+q+r} is $1$ we have
\begin{equation*}
  (p:q:r)  =pA+(q+r)D.
\end{equation*}
Thus $(p:q:r)$ is a point on $AD$.
Similarly, when we define
\begin{gather*}
  E=(p+r)\inv pA+(p+r)\inv rC,\\
  F=(p+q)\inv pA+(p+q)\inv qC,
\end{gather*}
these points are on $AC$ and $AB$ respectively,
so $(p:q:r)$ is on $BE$ and $CF$.
We have for example
\begin{equation*}
  BD\rat DC\prop\compratio{BD}{BC}{BC}{DC},
\end{equation*}
which from \eqref{eqn:qBrC} is
\begin{equation*}
  (q+r)\inv qr\inv(q+r).
\end{equation*}
This ratio is just $qr\inv$,
if $\field$ is commutative,
and in this case we have \textbf{Ceva's Theorem:}
in \figref{fig:Ceva},
\begin{figure}[ht]
  \centering
  \begin{pspicture}(0,-0.6)(4,3.6)
    \pstTriangle A(4,0)B(3,3)C
    \pstHomO[HomCoef=0.3]BC[D]
    \pstHomO[HomCoef=0.4,PosAngle=135]CA[E]
    \pstInterLL[PointName=none] ADBEG
    \pstInterLL[PosAngle=-90] CGABF
    \ncline AB\ncline BC\ncline CA
    \psset{linestyle=dotted}
    \ncline AD\ncline BE\ncline CF
  \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\rat DC\And CE\rat EA\And AF\rat FB=1.
\end{equation*}

\section{Projective coordinates}

We have given geometric meaning to $(p:q:r)$
whenever the equation \eqref{eqn:=0}
fails.
We can still understand $(p:q:r)$
to be the equivalence class defined by \eqref{eqn:pqr},
even when \eqref{eqn:=0} holds.
In this case, we shall give geometric meaning to $(p:q:r)$,
if at least one of $p$, $q$, and $r$ is not $0$.

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*}
Here it will be important that the coefficients are on the left.
If the same point $(s,t)$ has barycentric coordinates $(p:q:r)$,
where \eqref{eqn:pqr1} holds, then $(s,t)=(p,q)$, and so
\begin{align*}
0&=ap+bq+c(p+q+r)\\
&=(a+c)p+(b+c)q+cr.
\end{align*}
Thus the same line is given by
\begin{equation*}
ax+by+c=0
\end{equation*}
in Cartesian coordinates and
\begin{equation}\label{eqn:cx}
(a+c)x+(b+c)y+cz=0
\end{equation}
in barycentric coordinates.
The straight lines parallel to this one
are obtained by changing $c$ alone.
Since at least one of $a$ and $b$ is not $0$,
the coefficients in \eqref{eqn:cx} are not all the same.
We obtain all parallel lines
by adding the same ratio to each coefficient.

Relabelling, we now have that
every straight line is given by an equation
\begin{equation}\label{eqn:ax}
ax+by+cz=0
\end{equation}
in barycentric coordinates,
where one of the coefficients $a$, $b$, and $c$
is different from the others.
As $(p:q:r)$ and $(pt:qt:rt)$ are the same point if $t\neq0$,
so then \eqref{eqn:ax} and
\begin{equation}\label{eqn:tax}
  tax+tby+tcz=0
\end{equation}
define the same line.
If $(p:q:r)$ satisfies \eqref{eqn:ax} and therefore \eqref{eqn:tax},
and also \eqref{eqn:=0} holds,
although $(p,q,r)\neq(0,0,0)$,
then $(p:q:r)$ satisfies the equation
of every straight line parallel to the one defined by \eqref{eqn:ax},
and no other straight line.
Thus we can understand $(p:q:r)$
as the point at infinity 
of the straight lines parallel to \eqref{eqn:ax}.
The line at infinity is then defined by
\begin{equation*}
  x+y+z=0.
\end{equation*}

A 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.
The definition in \secref{sect:proj-plane}
of a projective plane is indeed satisfied
by such points
and the lines defined by the equations \eqref{eqn:ax}.
For, given two such equations,
the coefficients in one
not being the same multiples of the coefficients of the other,
by Gaussian elimination we can find a nonzero solution,
unique up to scaling.
(If $\field$ is not commuting, we cannot use Cramer's Rule.)
In the same way, two distinct points determine uniquely,
again up to scaling,
the coefficients in the equation of the line that contains them.
Finally, no three of the points $(1:0:0)$, $(0:1:0)$, $(0:0:1)$
and $(1:1:1)$ are collinear.

\section{Change of coordinates}

Any triangle $ABC$
determines a system of barycentric coordinates
for the points of the affine plane of the triangle,
hence a system of projective coordinates for the projective plane.
However, suppose a fourth point $D$ in this plane
does not lie on any of the three sides of $ABC$.
Then $D$ has projective coordinates $(\mu:\nu:\rho)$,
with $\mu\nu\rho\neq0$.
There is now a bijection
\begin{equation*}
(x:y:z)\mapsto(\mu\inv x:\nu\inv y:\rho\inv z)
\end{equation*}
from the set of points of the projective plane to itself.
This bijection preserves linearity;
in particular,
it takes the line given by \eqref{eqn:ax} to the line given by
\begin{equation*}
  a\mu x+b\nu y+c\rho z=0.
\end{equation*}
The bijection fixes $A$, $B$, and $C$,
which are $(1:0:0)$,
$(0:1:0)$, and $(0:0:1)$ respectively,
but takes $D$ to $(1:1:1)$,
the barycenter of $ABC$.
In particular,
the points that used to be on the line at infinity,
defined by \eqref{eqn:=0},
are sent to the line given by
\begin{equation}\label{eqn:mx}
\mu x+\nu y+\rho z=0,
\end{equation}
while of course the points now at infinity satisfy \eqref{eqn:=0}.

If $\mu$, $\nu$, and $\rho$ are not all equal to one another,
that is, $D$ is not the barycenter of $ABC$,
then, as noted in the previous section,
there is a unique point satisfying both \eqref{eqn:mx}
and \eqref{eqn:=0}.

\section{Pappus's Theorem, third proof}\label{sect:Pap-3}

We show that Pappus's Hexagon Theorem holds a projective plane
if and only if the field $\field$ of ratios is commutative.
For convenience, let us write
\begin{equation*}
  \var{ax+by+cz}
\end{equation*}
for the line given by \eqref{eqn:ax}.

We have a hexagon $ABCDEF$,
vertices lying alternately on two lines,
opposite sides intersecting at $G$, $H$, and $K$ respectively,
as in \figref{fig:Pap-coord}.
\begin{figure}[ht]
  \centering
  \begin{pspicture}(10,7)
    \psset{PointSymbol=none}
    \pstTriangle[linestyle=none](0.6,0.6)K(9.4,0.6)E(9,6.4)B
    \pstCGravABC[PosAngle=135] KEBA
    \pstHomO[HomCoef=0.3,PosAngle=-120]AE[C] % should be 0.3
    \pstHomO[HomCoef=0.4,PosAngle=135]AB[G] % should be 0.4
    \pstInterLL[PosAngle=-105] EGKCD
    \pstInterLL KADBF
    \pstInterLL[PosAngle=95] CBEFH
    \ncline AB\ncline BC\ncline CD\ncline DE\ncline EF\ncline FA
    \psset{linestyle=dashed}
    \ncline DG\ncline FH\ncline CK\ncline KA
    \psset{linestyle=dotted}
    \ncline KH
    \psset{linewidth=2pt,linestyle=solid}
    \ncline AE\ncline DB
  \end{pspicture}
\caption{Hexagon Theorem in projective coordinates}\label{fig:Pap-coord}
\end{figure}
The three lines through $A$
pass respectively through $B$, $E$, and $K$.
Assuming these last three points are not collinear,
we may let
\begin{align*}
	A&=(1:1:1),&B&=(1:0:0),\\
	E&=(0:1:0),&K&=(0:0:1).
\end{align*}
In particular, $K$ is the barycenter of $ABC$.
For the lines through $A$ we have equations as follows:
\begin{align*}
	AB&=\var{y-z},&
	AE&=\var{x-z},&
	AK&=\var{x-y}.
\end{align*}
For some $p$, $q$, and $r$ then,
\begin{align*}
	G&=(p:1:1),&C&=(1:q:1),&F&=(1:1:r).
\end{align*}
Consequently,
\begin{align*}
BF&=\var{z-ry},&
EG&=\var{x-pz},&
KC&=\var{y-qx}.
\end{align*}
Since these three lines have a common point,
namely $D$,
this is each of
\begin{align*}
  &(1:q:rq),&&(pr:1:r),&&(p:qp:1).
\end{align*}
Hence for example
\begin{equation*}
  (1,q,rq)=(qpr,q,qr),
\end{equation*}
and therefore
\begin{equation}\label{eqn:1=qpr}
  1=qpr.
\end{equation}
In the same way, the products $prq$ and $rqp$ are $1$.
But we have also
\begin{align*}
	BC&=\var{y-qz},&
	EF&=\var{z-rx},&
	KG&=\var{x-py}.
\end{align*}
The intersection of the first two of these is $H$,
so
\begin{equation*}
H=(1:qr:r).
\end{equation*}
This lies on $KG$ if and only if $pqr=1$.
Comparison with \eqref{eqn:1=qpr} yields the claim.

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

\section{Projective plane}

Given an affine plane, we have obtained a field $\field$ of ratios,
possibly not commutative.
Using this, we have extended the affine plane to a projective plane.
This plane has points and lines.

\subsection{Points}

We have obtained the set
\begin{equation*}
  \bigl\{(x:y:z)\colon
  (x,y,z)\in\field^3\setminus\{(0,0,0)\}\bigr\}
\end{equation*}
of points of the projective plane,
where
 \begin{equation*}
(p:q:r)=\bigl\{(pt,qt,rt)\colon t\in\field\setminus\{0\}\bigr\}.
 \end{equation*}
 Thus the set of points of the projective plane
 is the quotient of $\field^3\setminus\{(0,0,0)\}$
 by the equivalence relation $L$ given by
 \begin{equation*}
   (p,q,r)\mathrel R(p',q',r')\iff(p:q:r)=(p':q':r').
 \end{equation*}

 \subsection{Lines}
 
 We have shown that the set of lines in the projective plane is
 \begin{equation*}
   \bigl\{\var{ax+by+cz}\colon(a,b,c)\in\K^3\setminus\{(0,0,0\}\bigr\},
 \end{equation*}
 whose elements are well-defined by the rule
 \begin{equation*}
   \var{ax+by+cz}=\{(p:q:r):ap+bq+cr=0\}.
 \end{equation*}
 There is an equivalence relation $L$ on $\field^3\setminus\{(0,0,0)\}$
 such that
 \begin{multline*}
   (a,b,c)\mathrel L(a',b',c')\\
   \iff
   \var{ax+by+cz}=\var{a'x+b'y+c'z}.
 \end{multline*}
The equivalence class of $(a,b,c)$ with respect to $L$ is
 \begin{equation*}
   \bigl\{(ta,tb,tc)\colon t\in\field\setminus\{0\}\bigr\}.
 \end{equation*}

 \subsection{Duality}

 There is no particular reason not to let  $\var{ax+by+cz}$
 simply \emph{be} the $L$-class of $(a,b,c)$.
 There is then no reason why points should be $R$-classes,
 and lines, $L$-classes,
 and not the other way around.
 It just depends on the side we want to write coefficients on.
We have chosen the left in the equation \eqref{eqn:ax} for a line.

The field $\field$ gives us the sets
\begin{align*}
  &\Mat31,&&\Mat13,
\end{align*}
of $3\times1$ matrices, or column vectors,
and of $1\times3$ matrices, or row vectors,
respectively.
The sets are isomorphic an abelian groups with respect to addition.
 If we write column vectors as $\bm x$,
 we can write row vectors as $\bm x\trans$.
 We are interested in three matrix multiplications.
 \begin{compactitem}
 \item
   With respect to
   \begin{equation*}
     (\bm x,t)\mapsto\bm xt
   \end{equation*}
   from $\Mat31\times\field$ to $\Mat31$,
   the latter is a right vector space over $\field$.
 \item
   With respect to
   \begin{equation*}
     (t,\bm x\trans)\mapsto t\bm x\trans
   \end{equation*}
   from $\field\times\Mat13$ to $\Mat31$,
   the latter is a left vector space over $\field$.
 \item
   Using
   \begin{equation*}
     (\bm x\trans,\bm y)\mapsto\bm x\trans\bm y
   \end{equation*}
   from $\Mat13\times\Mat31$ to $\field$,
   we shall define the relation of a line to a point,
   whereby the point is on the line,
   or the line contains or passes through the point.
 \end{compactitem}
 These multiplications are associative,
 in the sense that
 \begin{align*}
   (t\bm x\trans)\bm y
   &=t(\bm x\trans\bm y),&
   (\bm x\trans\bm y)t
   &=\bm x\trans(\bm yt).
 \end{align*}
We have also
 \begin{equation*}
   t\bm x\trans=(\bm x t)\trans.
 \end{equation*}
 For all $\bm a$ and $\bm b$, the condition
 \begin{equation*}
   \bm a\trans\bm b=0
 \end{equation*}
is equivalent to either of the following:
 \begin{compactenum}[(i)]
 \item
   for all nonzero $t$ in $\field$,
   \begin{equation*}
        \bm a\trans(\bm bt)=0;
   \end{equation*}
 \item
   for all nonzero $t$ in $\field$,
   \begin{equation*}
        (t\bm a\trans)\bm b=0.
   \end{equation*}
 \end{compactenum}
We can understand the relations $R$ and $L$ defined above
as being on $\Mat31\setminus\{\bm0\}$ and $\Mat13\setminus\{\bm0\trans\}$
respectively.
Thus, if $\bm a$ and $\bm b$ are nonzero, the two conditions
\begin{align*}
  \bm a&\mathrel R\bm b,&
  \bm a\trans&\mathrel L\bm b\trans
\end{align*}
are equivalent to one another and to the existence of a nonzero $t$ such that
\begin{equation*}
  \bm at=\bm b.
\end{equation*}
Therefore,
for all nonzero $\bm a$, $\bm b$, and $\bm c$,
\begin{equation*}
  \bm a\trans\bm b=0\And \bm b\mathrel R\bm c\implies\bm a\trans\bm c=0,
\end{equation*}
and equivalently
\begin{equation*}
  \bm b\trans\bm a=0\And \bm b\trans\mathrel L\bm c\trans
  \implies\bm c\trans\bm a=0.
\end{equation*}
If the $R$-class of $\bm a$ is $[\bm a]$,
and the $L$-class of $\bm b\trans$ is $[\bm b\trans]$,
we can understand $[\bm a]$ as a point,
and $[\bm b\trans]$ as a line.
The point is on the line
if and only if
\begin{equation*}
  \bm b\trans\bm a=0.
\end{equation*}

\subsection{Plane}

Given a field $\field$,
we can denote by
\begin{equation*}
  \Proj^2(\field)
\end{equation*}
the projective plane that we have just described.
The points and lines compose the quotients
\begin{align*}
  &(\Mat31\setminus\{\bm0\})/R,&&(\Mat13\setminus\{\bm0\trans\})/L
\end{align*}
respectively.
Typical elements are respectively
\begin{align*}
  &\bigl\{\bm at\colon t\in\field\setminus\{0\}\bigr\},&
  &\bigl\{t\bm a\trans\colon t\in\field\setminus\{0\}\bigr\},
\end{align*}
denoted by
\begin{align*}
  &[\bm a],&&[\bm a\trans].
\end{align*}
We already conceive of $\field^3$
as consisting of points in a three-dimensional space.
Then $\Mat31$ and $\Mat13$ are somehow the same space,
and each of $[\bm a]$ and $[\bm a\trans]$ consists of the nonzero points
on a line that passes through zero and the same nonzero point.
However, when $\field$ is non-commutative, the lines may be different,
as we noted at the end of \secref{sect:H}.
Returning to our earlier notation,
we have the one-to-one correspondence
\begin{equation*}
  (p:q:r)\leftrightsquigarrow\var{p^*x+q^*y+r^*z}
\end{equation*}
between
$(\Mat31\setminus\{\bm0\})/R$ and $(\Mat13\setminus\{\bm0\trans\})/L$,
where
\begin{equation*}
  t^*=
  \begin{cases}
    t\inv,&\text{ if }t\neq0,\\
    0,&\text{ if }t=0.
  \end{cases}
\end{equation*}

\section{Fano Plane}

For the simplest example of a projective plane,
we may let $\field$ 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 $\Proj^2(\field)$:
\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*}
\var{x}&=BCG,&\var{y-z}&=ADG,\\
\var{y}&=ACF,&\var{x-z}&=BDF,\\
\var{z}&=ABE,&\var{x-y}&=CDE,
\end{align*}
and finally,
\begin{equation*}
  \var{x+y+z}=EFG,
\end{equation*}
 the line at infinity.
All of this can be depicted as in \figref{fig:Fano}.
\begin{figure}[ht]
\centering
\begin{pspicture}(-4,-2.3)(4.3,4)
%  \psgrid
\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{Fano Plane}\label{fig:Fano}
\end{figure}

\section{Desargues's Theorem proved}\label{sect:Des-proof}

\subsection{In the projective plane over a field}

We now prove Desargues's Theorem in $\Proj^2(\field)$
for arbitrary $\field$.
First we prove it in $\field^3$.
We suppose triangles $ABC$ and $DEF$
lie in two different planes,
and
\begin{compactitem}
\item
  $AB$ and $DE$ intersect at $H$,
\item
  $BC$ and $EF$ intersect at $K$,
\item
  $CA$ and $FD$ intersect at $L$.
\end{compactitem}
In particular,
the two lines in each of the three respective pairs
are not identical.
There are two conclusions.
\begin{compactenum}
\item
  The two lines in each of the three respective pairs
  lie in a common plane,
  and the three resulting planes have a common point $G$
  (possibly at infinity),
  which is where $AD$, $BE$, and $CF$ intersect.
  \item
Each of $H$, $K$, and $L$
lies in both the plane of $ABC$ and the plane of $DEF$,
and therefore in the intersection of those planes,
which is a straight line.
\end{compactenum}
If the planes of $ABC$ and $DEF$ are parallel, 
then they meet at their common line at infinity.

If now $ABC$ and $DEF$ lie in the same plane,
and $AD$, $BE$, and $CF$ intersect at $G$,
the triangles are projections of triangles in different planes
meeting the conditions above.
Thus Desargues's Theorem holds in $\Proj^2(\field)$.

\subsection{In a Pappian plane}

  \begin{figure}[ht]
  \centering
  \begin{pspicture}(-1.4,-2.2)(6,6.5)
    %\psgrid
    \psset{PointSymbol=none}
    \pstGeonode[PosAngle={-90,-90,90,90,-90}]
    (1.5,0)D(3,0)F(0,6)G(6,6)K(5.5,0)L
    \ncline DG\ncline DL\ncline FG\ncline FK
    \pstHomO[HomCoef=0.4,PosAngle=225]DG[A]
    \ncline AL
    \pstInterLL[PosAngle=190] ALFGC
    \ncline CK
    \pstHomO[HomCoef=0.5]KL[H]
    \ncline AH\ncline DH
    \pstInterLL[PosAngle=-30] DHFKE
    \pstInterLL[PosAngle=-90] GEAHB
    \ncline EG
    \psset{linestyle=dashed}
    \pstInterLL[PosAngle=180] BCDEM
    \ncline MC\ncline MD
    \pstInterLL[PosAngle=90] CDGEN
    \ncline ND
    \pstInterLL[PosAngle=180] ACGMP
    \ncline PA\ncline GM
    \ncline[linestyle=dotted]HP
    \pstInterLL[PosAngle=180] DFGMQ
    \ncline DQ
    \ncline[linestyle=dotted]KQ
    \ncline[linestyle=dotted]KL
  \end{pspicture}
\caption{Hessenberg's proof}\label{fig:Des-Q}
  \end{figure}
If $\field$ is not commutative,
then Pappus's Theorem does not hold in $\Proj^2(\field)$.
However,
in a Pappian plane,
as defined in \secref{sect:proj-plane},
we can prove Desargues's Theorem
as follows.
There will be two cases.
First we assume
\begin{compactitem}
\item
  $C$ is not on $DE$,
\item
  $D$ is not on $BC$,
\end{compactitem}
as in \figref{fig:Des-Q}.
  This is the general case proved by Hessenberg in 1905,
  in a paper \cite{Hessenberg} using diagrams.
\begin{compactenum}
\item
  Let $BC$ and $DE$ intersect at $M$.
\item
  Because $ABC$ and $DEF$ are proper triangles, not lines,
  $M$ cannot be $A$ or $F$.
\item
  Because $BC$ and $EF$ are not the same line,
  so that $BC$ does not contain $G$,
  also $M$ cannot be $G$.
\item
  Because of our additional assumptions for the present case,
  $M$ cannot be any of the points $B$, $C$, $D$, or $E$.
\item
  Thus the vertices of the hexagon $ABGMDC$ are distinct.
  \item
Let  the intersection of
  \begin{compactitem}
    \item
      $DC$ and $BE$ (which is $BG$) be $N$,
    \item
      $CA$ and $GM$ be $P$.
  \end{compactitem}
\item
  Since the vertices of hexagon $ABGMDC$
  lie alternately on two distinct straight lines,
  by Pappus's Theorem the intersections
  $H$, $N$, and $P$ of the pairs of opposite sides are collinear.
\item
Let $DF$ and $GM$ intersect at $Q$.
\item
  Since the vertices of hexagon $CDFEGM$
  are distinct and lie alternately on two distinct straight lines,
  the intersections $K$, $N$, and $Q$ are collinear.
\item
  Since $CD$ does not contain $B$,
  it does not contain $M$ either.
\item
  Since $AD$ is not identical with $CF$,
  the point $P$, which lies on $AC$,
  can lie also on $CD$ only by being the point $C$;
  but then $MC$ would contain both $B$ and $G$,
  so $BC$ would contain $G$, and it doesn't.
  Thus $CD$ does not contain $P$.
\item
  Likewise,
  $Q$ can lie on $CD$ only by being $D$,
  but then $DE$ would contain $G$;
  so $CD$ does not contain $Q$.
  \item
Now Pappus applies to the hexagon $CMDQNP$,
and the points $H$, $K$, and $L$ are collinear.
\end{compactenum}
In the other case,
as in \figref{fig:Cron},
\begin{figure}[ht]
  \centering
  \psset{yunit=7mm}
  \begin{pspicture}(-4,-8)(6.5,6.5)
%    \psgrid
    \psset{PointSymbol=none}
    \pstGeonode[PosAngle={30},PointNameSep=0.6em] G
    \pstGeonode[PosAngle={-135,45,0}](-0.5,-1.6)D(-0.7,4)E(6,-1.6)F
    \pstInterLL[PosAngle=45] DGEFA
    \pstInterLL[PosAngle=-45] EGDFB
    \pstInterLL[PosAngle=210,PointNameSep=0.7em] FGDEC
    \pstInterLL[PosAngle=180] FDCAL
    \pstInterLL[PosAngle=-45] DEABH
    \pstInterLL[PosAngle=180] EFBCK
    \ncline AH\ncline EH
    \ncline BK\ncline FK
    \ncline AL\ncline FL
    \ncline CF\ncline BE\ncline AD
    \pstInterLL[PosAngle=180] ELAGS
    \pstInterLL[PosAngle=180] ELCGR
    \psset{linestyle=dashed}
    \ncline ES\ncline DS\ncline CR
    \psset{linestyle=dotted}
    \ncline FS\ncline DK\ncline HK
  \end{pspicture}
  \caption{Cronheim's proof}
\label{fig:Cron}  
\end{figure}
\begin{compactitem}
\item
  $C$ lies on $DE$,
\item
  $B$ lies on $DF$,
\item
  $A$ lies on $EF$.
\end{compactitem}
By applying Pappus's Theorem in turn to hexagons 
$GCELBA$, $GAELBC$, and $SRDCAF$,
we have that $FHS$, $DKR$ and finally $LKH$ are straight.
This proof is by Cronheim,
in a 1953 paper \cite{MR0053531} that uses no diagrams.

\section{Duality}

The \textbf{dual} of a statement about a 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 contain two points,
then the straight lines containing pairs of opposite vertices
have 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 \figref{fig:Pappus-conv}.
\begin{figure}[ht]
  \centering
  \begin{pspicture}(10,7)
    \psset{PointSymbol=none}
    \pstGeonode[PosAngle={90,90,-90}](0.2,6.4)G(9.8,6.4)H(4,0.6)A
    \pstHomO[HomCoef=0.3,PosAngle=-150]AG[B]
    \pstHomO[HomCoef=0.3,PosAngle=-45]BH[C]
    \pstHomO[HomCoef=0.4,PosAngle=-150]CG[D]
    \pstHomO[HomCoef=0.2,PosAngle=90]DH[E]
    \pstInterLL[PosAngle=-30] GEAHF
    \pstInterLL[PosAngle=180] ADBEK
    \ncline BG\ncline GD\ncline GE\ncline EH\ncline HC\ncline HF
    \psset{linestyle=dashed}
    \ncline BE\ncline AD
    \psset{linestyle=dotted}
    \ncline FK
    \psset{linewidth=2pt,linestyle=solid}
    \ncline AB\ncline BC\ncline CD\ncline DE\ncline EF\ncline FA
  \end{pspicture}
  \caption{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 that the dual of Desargues's Theorem is true
in a Pappian plane.
But the dual is precisely the converse.

\section{Quadrangle Theorem proved by Desargues}\label{sect:Quad-Des}

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 \figref{fig:CQT},
\begin{figure}
  \centering
  \begin{pspicture}(0,-10)(10,5)
    \psset{PointSymbol=none}
    \pstGeonode[PosAngle={180,-60,-60,0,90,-90}]
        (0.6,0)A(4.5,0)C(7.5,0)D(9.4,0)F(7,4.4)G(3,-9.4)R
    \pstHomO[HomCoef=0.4,PosAngle=135]AG[H]
    \pstHomO[HomCoef=0.6,PosAngle=175]CH[K]
    \pstInterLL[PosAngle=-120]AFKGB
    \pstInterLL[PosAngle=45]KFDGL
    \pstInterLL[PosAngle=-60]HLAFE
    \pstHomO[HomCoef=0.8,PosAngle=-165]GR[M]
    \pstInterLL[PosAngle=-150]AMHRN
    \pstInterLL[PosAngle=50]MBNCP
    \pstInterLL[PosAngle=-60]NEPFQ
    \ncline AF
    \ncline AG\ncline BG\ncline DG
    \ncline HC\ncline HE
    \ncline KF
    \ncline AM\ncline BM\ncline DM
    \ncline NC\ncline NE
    \ncline PQ
    \psset{linestyle=dotted}
    \ncline QF
    \ncline KR
    \psset{linestyle=dashed}
    \ncline GR\ncline HR\ncline LR
  \end{pspicture}
  \caption{Quadrangle Theorem from Desargues}
  \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 $R$.
\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
Since now $HN$, $KP$, and $LQ$ intersect at $R$,
the respective sides of triangles $HKL$ and $NPQ$
intersect along a straight line,
by Desargues's Theorem.
But
\begin{compactitem}
  \item
    $HK$ and $NP$ meet at $C$,
  \item
    $HL$ and $NQ$ meet at $E$, and
  \item
    $KL$ and $CE$ meet at $F$;
\end{compactitem}
therefore $PQ$ must also meet $CE$ at $F$.
\end{compactenum}

\appendix


\chapter{Thales himself}\label{app:Thales}

There is little evidence
that Thales knew, in full generality,
the theorem named for him.
I learned this while preparing for the Thales Meeting
held on Saturday, September 24, 2016,
in Thales's home town of Miletus \cite{Pierce-Thales}.
See also my ``Thales and the Nine-point Conic''
\cite{Pierce-Thales-9}.
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).

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 the commentary by Proclus
on 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
%who wrote biographies of philosophers, starting with Thales
\cite[i.24--5]{Diogenes-I},
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 \figref{fig:dates}.
\begin{figure}[ht]\centering
\begin{tabular}{|ll|}\hline
585 \textsc{b.c.e.}&Thales\\
300     &Euclid\\
250 \textsc{c.e.}    &Diogenes La\"ertius\\
340     &Pappus\\
450     &Proclus\\\hline
\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 \figref{fig:circ}.
\begin{figure}[ht]
\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{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.


\chapter{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.
We students were taught to write proofs in the two-column,
statement--reason format.
I understood the purpose of the class,
not as learning geometry as such,
but as learning proof.
That was good,
but I did not much care for our textbook, by Weeks and Adkins.
Their example of congruence
was a machine in a photograph,
stamping out foil trays for TV dinners
\cite[p.\ 13]{Weeks-Adkins}.

Weeks and Adkins confuse equality with sameness,
as I mention in ``On Commensurability and Symmetry''
\cite{MR3680853}.
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 $A\rat B\prop C\rat D$,
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
at St John's College \cite{Pierce-SJC},
along with Homer, Aeschylus, and Plato,
and Apollonius, Ptolemy, Newton, and Lobachevski.

In 2008,
the first course I taught at the Nesin Mathematics Village
was an opportunity to review some of that reading.
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,
while the course was great for me,
I don't think it meant much for the students
who sat and watched me at the board.
One has to engage with the mathematics for oneself,
especially when it is something so unusual 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 \textsc{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 \textsc{Metu}
I loved the content of the course called Fundamentals of Mathematics,
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 traditional mathematics.
When I oved to Mimar Sinan in 2011,
my colleagues and I were able
to develop a course in which first-year students 
read and presented the proofs 
that taught mathematics to practically all mathematicians 
until the twentieth century.
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 here 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
\grp{kr'atistos, -h, -on} (``most excellent''); but
``in Commandino's Latin translation her name vanishes, 
leaving the absurdity of the polite epithet
\grp{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 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.
For Hilbert,
the lemma is Pascal's Theorem;
he never mentions Pappus \cite{Hilbert-10}.

%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.
I offered it in the fall of 2015;
it had last been taught in the fall of 2010.
For use in the first half of the course,
from Hultsch's text \cite{Pappus} 
I translated the first 19 of Pappus's 38 lemmas for Euclid's \emph{Porisms}
\cite{Pappus-Pierce};
in the second half, we used the existing English translation of Lobachevski
\cite{Lobachevski}.
I was able to go over the same material
in the following summer at the Nesin Mathematics Village in \c Sirince.

I had not thought there was an English version of the Pappus;
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} (\grp{>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.


\chapter{Involutions}

\newcommand{\dZ}{O}%
\newcommand{\dA}{A}%
\newcommand{\dB}{B}%
\newcommand{\dC}{C}%
\newcommand{\dCp}{C'}%
\newcommand{\dBp}{B'}%
\newcommand{\dAp}{A'}%
\newcommand{\dD}{D}%
\newcommand{\dE}{E}%
\newcommand{\dF}{F}%
\newcommand{\dDp}{D'}%
\newcommand{\dEp}{E'}%
\newcommand{\dFp}{F'}%
\newcommand{\dG}{G}%
\newcommand{\dH}{H}%
\newcommand{\dK}{K}%
\newcommand{\dQ}{O}%
Pappus's Lemma IV is that if the points ${\dA}{\dB}{\dC}{\dCp}{\dBp}{\dAp}$ 
in \figref{fig:4c-again}
\begin{figure}[ht]
  \centering
\begin{pspicture}(10,5)
  %\psgrid
  \psset{PointSymbol=none,PointName=none}
  \pstGeonode[PosAngle={-90,-90,-90,-90,45,-90}]
  (0.2,0.5)A(2.2,0.5)B(4.2,0.5)C(5,0.5)D(6,4.6)F(9.8,0.5)L
  \pstHomO[HomCoef=0.6,PosAngle=135]AF[G]
    \pstInterLL[PosAngle=165] BFCGH
    \pstInterLL[PosAngle=30] DFHLK
    \pstInterLL[PosAngle=-90]ABGKE
    \ncline AF\ncline BF
    \ncline DF
    \ncline CG\ncline EG
    \ncline AL\ncline HL
    \uput[d](A){${\dA}$}
    \uput[d](B){${\dB}$}
    \uput[d](C){${\dC}$}
    \uput[d](D){${\dCp}$}
    \uput[d](E){${\dBp}$}
    \uput[ur](F){${\dQ}$}
    \uput[ul](G){${\dD}$}
    \uput[165](H){${\dE}$}
    \uput[30](K){${\dF}$}
    \uput[d](L){${\dAp}$}
  \end{pspicture}
\caption{Lemma IV (Quadrangle Theorem)}\label{fig:4c-again}
\end{figure}
satisfy the proportion
\begin{equation}\label{eqn:Pap-IV}
  ({\dA}{\dB}\rat{\dB}{\dC})({\dC}{\dAp}\rat {\dAp}{\dA})
  \prop
  ({\dAp}{\dBp}\rat{\dBp}{\dCp})({\dCp}{\dA}\rat {\dA}{\dAp}),
\end{equation}
then ${\dE}{\dF}{\dAp}$ is straight.
Eliminating ${\dA}{\dAp}$,
Chasles
\cite[p.\ 102]{Chasles}
 rewrites \eqref{eqn:Pap-IV} in the form
\begin{equation*}
  {\dB}{\dC}\cdot {\dCp}{\dA}\cdot {\dAp}{\dBp}
  ={\dA}{\dB}\cdot {\dBp}{\dCp}\cdot {\dC}{\dAp}.
\end{equation*}
We can write this as
\begin{equation*}
  {\dB}{\dC}\cdot {\dCp}{\dA}\rat {\dA}{\dB}\cdot {\dCp}{\dBp}
  \prop {\dC}{\dAp}\rat {\dBp}{\dAp},
\end{equation*}
which makes it easier to see that,
when ${\dA}{\dB}{\dC}{\dCp}{\dBp}$ are given,
then some unique ${\dAp}$ exists so as to satisfy \eqref{eqn:Pap-IV}.
As ${\dAp}$ varies, the ratio ${\dC}{\dAp}\rat {\dBp}{\dAp}$
takes on all possible values but unity.
If ${\dB}{\dC}\cdot {\dCp}{\dA}={\dA}{\dB}\cdot {\dCp}{\dBp}$,
that is,
\begin{equation*}
{\dA}{\dB}\rat {\dB}{\dC}\prop {\dCp}{\dA}\rat {\dCp}{\dBp},
\end{equation*}
then ${\dE}{\dF}\parallel {\dA}{\dBp}$ by Lemma I.
Otherwise,
by Lemma IV, ${\dE}{\dF}$ must pass through the ${\dAp}$ 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},
\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,}
as in \figrefp{fig:quadrangle}.
The proposition to which Thomas refers
is apparently the
\textbf{Involution Theorem,}
which Desargues proves 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 understand the Involution Theorem is to observe that, 
in \figref{fig:4c-again},
if the points ${\dB}{\dC}{\dCp}{\dBp}$ are conceived of as fixed,
then ${\dA}$ determines ${\dAp}$.
Moreover, ${\dAp}$ determines ${\dA}$ in the same way,
as in \figref{fig:inv}.
\begin{figure}[ht]
  \centering
\begin{pspicture}(0,-3.9)(10,5)
  %\psgrid
  \psset{PointSymbol=none,PointName=none}
  \pstGeonode[PosAngle={180,120,-90,-60,90,0}]
  (0.2,0.5)A(2.2,0.5)B(4,0.5)C(5,0.5)D(6,4.4)F(9.8,0.5)L
  \pstHomO[HomCoef=0.6,PosAngle=135]AF[G]
    \pstInterLL[PosAngle=165] BFCGH
    \pstInterLL[PosAngle=30] DFHLK
    \pstInterLL[PosAngle=-75]ABGKE
    \ncline AF\ncline BF
    \ncline DF
    \ncline CG\ncline EG
    \ncline AL\ncline HL
    \uput[l](A){${\dA}$}
    \uput[ul](B){${\dB}$}
    \uput[d](C){${\dC}$}
    \uput[dr](D){${\dCp}$}
    \uput[dr](E){${\dBp}$}
    \uput[u](F){${\dQ}$}
    \uput[ul](G){${\dD}$}
    \uput[l](H){${\dE}$}
    \uput[30](K){${\dF}$}
    \uput[r](L){${\dAp}$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \pstHomO[HomCoef=0.4,PosAngle=45]LF[G']
    \pstInterLL[PosAngle=-135] BFC{G'}{H'}
    \pstInterLL[PosAngle=-90] DF{H'}A{K'}
    \psset{linestyle=dotted}
    \ncline LF\ncline{H'}F
    \ncline D{K'}
    \ncline{H'}{G'}\ncline{K'}{G'}
    \ncline{K'}A
    \uput[ur](G'){${\dDp}$}
    \uput[dl](H'){${\dEp}$}
    \uput[d](K'){${\dFp}$}
  \end{pspicture}
\caption{Involution}\label{fig:inv}
\end{figure}
Thus we have an operation that \emph{transposes} $A$ and ${\dAp}$,
and so it is an \emph{involution} of the straight line ${\dB}{\dBp}$.


Desargues proceeds towards the Involution Theorem
by first observing that if seven points ${\dZ}{\dA}{\dB}{\dC}{\dCp}{\dBp}{\dAp}$ are arranged,
as in \figref{fig:Des-invo},
\begin{figure}[ht]
\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){${\dZ}$}
\uput[d](0.5,0){${\dA}$}
\uput[d](0.63,0){${\dB}$}
\uput[d](0.80,0){${\dC}$}
%\uput[d](1,0){${\dBp}$}
\uput[d](1.26,0){${\dCp}$}
\uput[d](1.59,0){${\dBp}$}
\uput[d](2,0){${\dAp}$}
  \end{pspicture}
  \caption{Points in involution}\label{fig:Des-invo}
  
\end{figure}
on a straight line so that
\begin{equation}\label{eqn:3}
{\dZ}{\dA}\cdot {\dZ}{\dAp}={\dZ}{\dB}\cdot {\dZ}{\dBp}
={\dZ}{\dC}\cdot {\dZ}{\dCp},
\end{equation}
then, without reference to ${\dZ}$, we have
\cite[p.\ 48]{MR869422}
\begin{equation}\label{eqn:Des-invo}
{\dC}{\dBp}\cdot {\dCp}{\dBp}\rat {\dB}{\dC}\cdot {\dB}{\dCp}
\prop {\dA}{\dBp}\cdot {\dBp}{\dAp}\rat {\dA}{\dB}\cdot {\dB}{\dAp}.
\end{equation}
To prove this,
by second equation in \eqref{eqn:3},
we have
\begin{equation}\label{eqn:AGAFADAC}
{\dZ}{\dBp}\rat {\dZ}{\dCp}\prop {\dZ}{\dC}\rat {\dZ}{\dB},
\end{equation}
so by subtracting the terms of the first ratio from the terms of the second,
we obtain
\begin{equation}\label{eqn:AGAF}
{\dZ}{\dBp}\rat {\dZ}{\dCp}\prop {\dC}{\dBp}\rat {\dB}{\dCp}.
\end{equation}
Similarly,
after alternating \eqref{eqn:AGAFADAC},
we obtain
\begin{gather*}
  {\dZ}{\dCp}\rat {\dZ}{\dB}\prop {\dZ}{\dBp}\rat {\dZ}{\dC},\\
{\dZ}{\dCp}\rat {\dZ}{\dB}\prop {\dCp}{\dBp}\rat {\dB}{\dC}.
\end{gather*}
Composing the latter with \eqref{eqn:AGAF} yields
\begin{equation}\label{eqn:AGAC}
{\dZ}{\dBp}\rat {\dZ}{\dB}\prop {\dC}{\dBp}\cdot {\dCp}{\dBp}\rat {\dB}{\dC}\cdot {\dB}{\dCp}.
\end{equation}
Replacing ${\dC}$ with ${\dA}$ and ${\dCp}$ with ${\dAp}$
in the second equation of \eqref{eqn:3}
yields the first equation.
Hence we can do the same in \eqref{eqn:AGAC},
obtaining
\begin{equation*}
{\dZ}{\dBp}\rat {\dZ}{\dB}\prop {\dA}{\dBp}\cdot {\dBp}{\dAp}\rat {\dA}{\dB}\cdot {\dB}{\dAp}.
\end{equation*}
Eliminating the common ratio from the last two proportions yields
\eqref{eqn:Des-invo}.
For meeting this condition,
the three pairs ${\dA}{\dAp}$, ${\dB}{\dBp}$ and ${\dC}{\dCp}$ of points
are said to be in \textbf{involution,}
by Desargues's definition.


Now suppose instead that the pairs ${\dA}{\dAp}$, ${\dB}{\dBp}$ and ${\dC}{\dCp}$
consist of the points 
where a transversal cuts the pairs of opposite sides 
of a complete quadrangle,
as in Thomas's description
and as in \figref{fig:4c-again}.
Then the converse of Pappus's Lemma IV,
expressed in \eqref{eqn:Pap-IV}, gives us now
\begin{equation}\label{eqn:IV-Des}
  {\dA}{\dAp}\cdot {\dB}{\dC}\rat {\dA}{\dB}\cdot {\dC}{\dAp}\prop {\dA}{\dAp}\cdot {\dCp}{\dBp}\rat {\dA}{\dCp}\cdot {\dBp}{\dAp}.
\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,}
as we did in \secref{sect:III},
though apparently Desargues does not do this \cite[p.\ 52]{MR869422}.
Despite the misgivings expressed on 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}=\crossratio 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}
  \crossratio{\dA}{\dAp}{\dC}{\dB}
  =\crossratio{\dA}{\dAp}{\dBp}{\dCp},
\end{equation}
while, since \eqref{eqn:Des-invo} is, in modern notation,
\begin{equation*}
  \frac{{\dC}{\dBp}\cdot {\dCp}{\dBp}}{{\dB}{\dC}\cdot {\dB}{\dCp}}=\frac{{\dA}{\dBp}\cdot {\dBp}{\dAp}}{{\dA}{\dB}\cdot {\dB}{\dAp}},
\end{equation*}
we obtain from this
\begin{equation*}
  \frac{{\dC}{\dBp}\cdot {\dA}{\dB}}{{\dB}{\dC}\cdot {\dA}{\dBp}}=\frac{{\dB}{\dCp}\cdot {\dBp}{\dAp}}{{\dCp}{\dBp}\cdot {\dB}{\dAp}},
\end{equation*}
that is,
\begin{equation}\label{eqn:DGBC}
  \crossratio{\dC}{\dBp}{\dA}{\dB}
  =\crossratio{\dB}{\dCp}{\dBp}{\dAp}.
\end{equation}
But \eqref{eqn:IV-cross}, obtained from Pappus, 
must still hold 
if we permute the pairs ${\dA}{\dAp}$, ${\dB}{\dBp}$, and ${\dC}{\dCp}$.
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}
  \crossratio{\dB}{\dBp}{\dA}{\dC}=
  \crossratio{\dB}{\dBp}{\dCp}{\dAp}.
\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*}
  \crossratio CDAB=\crossratio ABCD,\\
\crossratio BADC=\crossratio ABCD.
\end{gather*}
Moreover, 
from \eqref{eqn:cross} we can read off
\begin{equation*}
  \crossratio ADCB=\frac1{\crossratio ABCD},
\end{equation*}
while
\begin{align*}
  \crossratio 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}-\crossratio ABCD\\
&=1-\crossratio 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{multline*}
  \crossratio{\dB}{\dBp}{\dA}{\dC}=\crossratio{\dB}{\dBp}{\dCp}{\dAp}\\
  \begin{aligned}
&\iff\crossratio{\dB}{\dA}{\dBp}{\dC}=\crossratio{\dB}{\dCp}{\dBp}{\dAp}\\
&\iff\crossratio{\dA}{\dB}{\dC}{\dBp}=\crossratio{\dB}{\dCp}{\dBp}{\dAp}\\
&\iff\crossratio{\dC}{\dBp}{\dA}{\dB}=\crossratio{\dB}{\dCp}{\dBp}{\dAp},
  \end{aligned}
\end{multline*}
that is, \eqref{eqn:CGDB} and \eqref{eqn:DGBC} are equivalent,
as desired.
\begin{figure}
  \centering
    \psset{PointSymbol=none}
  \begin{pspicture}(0,-0.6)(10,3)
    %    \psgrid
    \pstGeonode[PosAngle={-90,-90,-90,90},
      PointName={default,none,default,default}]
    (0.4,0)A(9.6,-1.4)K(6,0)B(8,2.4)O
    \pstInterLL[PosAngle=-90] ABOK{A'}
    \pstHomO[HomCoef=0.7,PosAngle=90]AO[D]
        \pstInterLL[PosAngle=-90] DKAB{B'}
    \pstInterLL[PosAngle=-90] D{B'}BOH
      \pstInterLL[PosAngle=90] D{A'}BOG
      \ncline A{A'}\ncline OB
      \psset{linestyle=dotted}
      \ncline OA\ncline DB\ncline D{B'}\ncline D{A'}
      \psdot(D)
  \end{pspicture}

  \begin{pspicture}(5.4,-2)(10,3)
%        \psgrid
    \pstGeonode[PosAngle={-90,-90,-130,90},
      PointName={none,default,default,default}]
    (0.4,0)A(9.6,-1.4)K(6,0)B(8,2.4)O
    \pstInterLL ABOK{A'}
    \pstHomO[HomCoef=0.7,PosAngle=90]AO[D]
        \pstInterLL[PosAngle=-135] DKAB{B'}
    \pstInterLL[PosAngle=-90] D{B'}BOH
      \pstInterLL[PosAngle=90] D{A'}BOG
      \ncline OB\ncline DK
      \psset{linestyle=dotted}
      \ncline{A'}B\ncline{A'}H\ncline{A'}D\ncline OK
      \psdot(A')
  \end{pspicture}

  \begin{pspicture}(0,-2)(10,3)
    %    \psgrid
    \pstGeonode[PosAngle={-90,-90,-130,90}]
    (0.4,0)A(9.6,-1.4)K(6,0)B(8,2.4)O
    \pstInterLL ABOK{A'}
    \pstHomO[HomCoef=0.7,PosAngle=135]AO[D]
        \pstInterLL[PosAngle=-135] DKAB{B'}
    \pstInterLL[PosAngle=90] D{B'}BOH
%      \pstInterLL[PosAngle=0] D{A'}BOG
      \ncline A{A'}\ncline DK
      \psset{linestyle=dotted}
      \ncline OA\ncline OB\ncline O{B'}\ncline OK
            \psdot(O)
  \end{pspicture}
  \caption{Involution}\label{fig:invo}
  
\end{figure}

The involution of the straight line ${\dB}{\dBp}$
in \figref{fig:4c-again}
that transposes ${\dA}$ and ${\dAp}$
will transpose ${\dB}$ and ${\dBp}$,
and also ${\dC}$ and ${\dCp}$.
Following Coxeter \cite[\textbf{14.56}, p.\ 246]{Coxeter-Geometry},
we obtain this transposition as in \figref{fig:invo},
as a composition of three projections of one straight line onto another.
In Coxeter's notation \cite[p.\ 242]{Coxeter-Geometry},
the projections are
\begin{gather*}
  {\dA}{\dB}{\dBp}{\dAp}
  \overset {\dD}{\underset{\wedge}=}{\dQ}{\dB}{\dH}{\dG},\\
  {\dQ}{\dB}{\dH}{\dG}
  \underset{\wedge}{\overset {\dAp}=}{\dK}{\dBp}{\dH}{\dD},\\
  {\dK}{\dBp}{\dH}{\dD}
  \underset{\wedge}{\overset {\dQ}=}{\dAp}{\dBp}{\dB}{\dA},
\end{gather*}
where for example the first expression
means that
\begin{align*}
  &{\dA}{\dD}{\dQ},& 
  &{\dBp}{\dD}{\dH},&
  &{\dAp}{\dD}{\dG},&
  &{\dA}{\dB}{\dBp}{\dAp},&
  &{\dQ}{\dB}{\dH}{\dG}
\end{align*}
are straight.
It follows from Lemma IV that this transformation is an involution
and is uniquely determined by the pairs ${\dA}{\dAp}$ and ${\dB}{\dBp}$.


\chapter{Locus problems}\label{app:locus}

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,
as in \figref{fig:5-line},
\begin{figure}[ht]
\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}
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 \figref{fig:5-line};
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.
In \figref{fig:5-line-sol},
\begin{figure}[ht]
\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.

\chapter{Ancients and moderns}

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.
In some of his lemmas, as Jones observes \cite[p.\ 456]{Jones-2},
\begin{quote}
  Pappus does not
  take the trouble to define his figure
  (the reader with Euclid's \emph{Porisms} before him
  perhaps would not have needed such help).
\end{quote}
For example, in Lemma I of Pappus,
there is no \textbf{enunciation}
in the sense of Proclus \cite[203, p.\ 159]{MR1200456},
but the \textbf{exposition} reads,
\begin{quote}
\grp{>'Estw katagraf`h}\hfill Let the diagram be\\
\grp{<h ABGDEZH,}\hfill \gr{ABGDEZH},\\
\grp{ka`i >'estw}\hfill and let it be that\\
\grp{<ws <h AZ pr`os t`hn ZH,}\hfill as \gr{AZ} is to \gr{ZH},\\
\grp{o<'utws <h AD pr`os t`hn DG,}\hfill so is \gr{AD} to \gr{DG};\\
\grp{ka`i >epeze'uqjw <h JK.}\hfill and let \gr{JK} have been joined.
\end{quote}
Thus we are given
\begin{equation}\label{eqn:AZZH}
  \grm{AZ}\rat\grm{ZH}\prop\grm{AD}\rat\grm{DG},
\end{equation}
and we can infer that \gr Z lies on \gr{AH},
and \gr D on \gr{AG}.
If \gr H did not lie on \gr{AG},
then we could immediately apply Thales's Theorem;
so probably all of the points \gr{AGDZH} are on one straight line.
Pappus mentions two more points, \gr B and \gr E;
probably they are not on \gr{AGDZH}.
From the \textbf{specification,}
we can infer that \gr J and \gr K do not lie on \gr{AG}:
\begin{quote}
\grp{<'oti}\hfill [I say] that\\
\grp{par'allhl'os >estin <h JK t~h| AG.}\hfill parallel is \gr{JK} to \gr{AG}.
\end{quote}
Thus we aim to show
\begin{equation}\label{eqn:JKAG}
  \grm{JK}\parallel\grm{AG}.
\end{equation}
The \textbf{construction} gives one more point:
\begin{quote}
  \grp{>'Hqjw di`a to~u Z}\hfill Let have been drawn through \gr Z,\\
  \grp{t~h| BD par'allhlos <h ZL}\hfill to \gr{BD} parallel, \gr{ZL}.\\
\end{quote}
We have now
\begin{equation}\label{eqn:BDZL}
  \grm{BD}\parallel\grm{ZL}.
\end{equation}
Beginning as follows,
the \textbf{demonstration} shows, in terms of \gr B,
where \gr L is along \gr{ZL}:
\begin{quote}
  \grp{>epe`i o>~un}\hfill Since then\\
  \grp{>estin <ws <h AZ pr`os t`hn ZH,}\hfill as \gr{AZ} is to \gr{ZH},\\
  \grp{o<'utws <h AD pr`os t`hn DG,}\hfill so \gr{AD} to \gr{DG},\\
  \grp{>an'apalin}\hfill
  by inversion,\\
  \grp{ka`i sunj'enti ka`i >enall'ax}\hfill \emph{componendo, alternando,}\\
  \grp{>estin <ws <h DA pr`os t`hn AZ,}\hfill as \gr{DA} is to \gr{AZ},\\
  \grp{tout'estin >en parall'hlw|}\hfill that is, in parallel,\\
  \grp{<ws <h BA pr`os t`hn AL,}\hfill as \gr{BA} is to \gr{AL},\\
  \grp{o<'utws <h GA pr`os t`hn AH;}\hfill so \gr{GA} to \gr{AH};\\
    \grp{par'allhlos >'ara}\hfill  parallel then\\
  \grp{>est`in <h LH t~h| BG;}\hfill  is \gr{LH} to \gr{BG}.
\end{quote}
Thus we manipulate \eqref{eqn:AZZH} to get
\begin{equation*}
  \grm{DA}\rat\grm{AZ}\prop\grm{GA}\rat\grm{AH}.
\end{equation*}
Also, by Thales's Theorem, apparently
\begin{equation*}
  \grm{DA}\rat\grm{AZ}\prop\grm{BA}\rat\grm{AL}.
\end{equation*}
This shows \gr L must have lain along \gr{AB}.
Now, with \eqref{eqn:BDZL}, the position of \gr L is determined.
Combining the last two proportions yields
\begin{equation*}
  \grm{BA}\rat\grm{AL}\prop\grm{GA}\rat\grm{AH},
\end{equation*}
so by the converse of Thales,
\begin{equation*}
  \grm{BG}\parallel\grm{LH}.
\end{equation*}
What we know is in \figref{fig:I-part}.
\begin{figure}[ht]
  \subfloat[]{\label{fig:I-part}
  \begin{pspicture}(5,3.7)
    \psset{PointSymbol=none,PointName=none}
    \pstGeonode(0.2,0.6)A(4.8,0.6)H(2,0.6)G(4,2)L
    \pstHomO[HomCoef=0.6]A{G,H}[D,Z]
    \pstTranslation HLG
    \pstInterLL G{G'}ALB
    \ncline AH\ncline AL\ncline DB\ncline BG\ncline ZL\ncline LH
    \uput[d](A){\gr A}
    \uput[u](B){\gr B}
    \uput[d](G){\gr G}
    \uput[d](D){\gr D}
    \uput[d](Z){\gr Z}
    \uput[d](H){\gr H}
    \uput[u](L){\gr L}
  \end{pspicture}
  }
  \hfill
  \subfloat[]{\label{fig:I}
    \begin{pspicture}(5,3.7)
%      \psgrid
    \psset{PointSymbol=none,PointName=none}
    \pstGeonode(0.2,0.6)A(4.8,0.6)H(2,0.6)G(4,2)L
    \pstHomO[HomCoef=0.6]A{G,H}[D,Z]
    \pstTranslation HLG
    \pstInterLL G{G'}ALB
    \pstHomO[HomCoef=0.75]AL[E]
    \pstInterLL EZDBK
    \pstInterLL EHGBJ
    \ncline AH\ncline AL\ncline DB\ncline BG\ncline ZL\ncline LH
    \ncline GJ\ncline HJ\ncline ZK\ncline DK\ncline JK
    \uput[d](A){\gr A}
    \uput[155](B){\gr B}
    \uput[d](G){\gr G}
    \uput[d](D){\gr D}
    \uput[-125](E){\gr E}
    \uput[d](Z){\gr Z}
    \uput[d](H){\gr H}
    \uput[u](J){\gr J}
    \uput[u](K){\gr K}
    \uput[u](L){\gr L}
  \end{pspicture}
}  
  \caption{Pappus's Lemma I}
\end{figure}
The demonstration continues:
  \begin{quote}
  \grp{>'estin >'ara <ws <h EB pr`os t`hn BL,}\hfill
  hence as \gr{EB} is to \gr{BL},\\
  \grp{o<'utws >en parall'hlw|}\hfill so in parallel\\
  \grp{<h EK pr`os t`hn KZ,}\hfill \gr{EK} to \gr{KZ},\\
  \grp{ka`i <h EJ pr`os t`hn JH;}\hfill and \gr{EJ} to \gr{JH};\\
  \grp{ka`i <ws >'ara <h EK pr`os t`hn KZ,}\hfill
  hence also as \gr{EK} to \gr{KZ},\\
  \grp{o<'utws >est`in <h EJ pr`os t`hn JH;}\hfill so is \gr{EJ} to \gr{JH};\\
  \grp{par'allhlos >'ara}\hfill
  parallel then\\
  \grp{>est`in <h JK t~h| AG.}\hfill
  is \gr{JK} to \gr{AG}.
  \end{quote}
  By Thales's Theorem, apparently,
  \begin{equation*}
    \grm{EB}\rat\grm{BL}\prop
    \begin{cases}
      \grm{EK}\rat\grm{KZ},\\
      \grm{EJ}\rat\grm{JH}.
    \end{cases}
  \end{equation*}
  We can infer
  \begin{compactitem}
    \item
  from the ratios,
  \begin{compactitem}
  \item
    \gr E lies on \gr{BL},
  \item
    \gr K lies on \gr{EZ},
  \item
    \gr J lies on \gr{EH};
  \end{compactitem}
\item
  from the proportions,
  \begin{compactitem}
  \item
    \gr K lies on \gr{DB},
  \item
    \gr J lies on \gr{GB}.
  \end{compactitem}
  \end{compactitem}
  Now we can complete the diagram as in \figref{fig:I}.
Since, finally,
  \begin{equation*}
    \grm{EK}\rat\grm{KZ}\prop
      \grm{EJ}\rat\grm{JH},
  \end{equation*}
we obtain \eqref{eqn:JKAG} by the converse of Thales.  

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 \secref{sect:Pap-3}
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, in \secref{sect:Pap-proj},
where we pass to a third dimension and 
\emph{project,} starting from the diagram of Lemma VIII.
Coxeter's diagram for Pappus's Theorem \cite[p.\ 232]{Coxeter-Geometry}
is labelled as in \figref{fig:XIII-Cox};
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-2,-0.5)(5,4.5)
%    \psgrid
    \psset{PointName=none,PointSymbol=none}
    \pstGeonode[PosAngle={-90,-90,90}]
    (-15,0)N(5,0)D(4,4)B
    \pstHomO[HomCoef=0.75,PosAngle=90]NB[A]
    \pstHomO[PosAngle=90,HomCoef=0.5]AB[E]
    \pstHomO[HomCoef=0.65,PosAngle=-90]ND[G]
    \pstHomO[PosAngle=-90,HomCoef=0.4]GD[Z]
    \pstLineAB AD\pstLineAB DE\pstLineAB EG\pstLineAB GB\pstLineAB BZ\pstLineAB ZA
    \pstInterLL[PosAngle=180] AZGEH
    \pstInterLL EDZBK
    \pstInterLL ADEGJ
    \pstInterLL EDGBL
    \pstInterLL[PosAngle=-90] ADGBM
    \pstLineAB KH
%    \psset{linestyle=solid,linewidth=2pt}
    \pstLineAB AB\pstLineAB DG
    \nput uA{$A_1$}
    \nput uB{$C_1$}
    \nput dG{$A_2$}
    \nput dD{$C_2$}
    \nput uE{$B_1$}
    \nput dZ{$B_2$}
    \nput lH{$C_3$}
    \nput rK{$A_3$}
    \nput dM{$B_3$}
    \end{pspicture}

\caption{Pappus's Lemma XIII in modern notation}\label{fig:XIII-Cox}

\end{figure}
it allows us to observe, 
once for all,
that $A_iB_jC_k$ is straight
whenever $\{i,j,k\}=\{1,2,3\}$.
Coxeter can also express Pappus's Theorem with a matrix,
\begin{equation*}
  \begin{pmatrix}
    A_1&B_1&C_1\\
    A_2&B_2&C_2\\
    A_3&B_3&C_3
  \end{pmatrix}.
\end{equation*}
Here each of the first two rows consists of collinear points,
as does each of the six diagonals $A_iB_{i\pm 1}C_{i\pm 2}$,
indices considered \emph{modulo} $3$.
The theorem then is that the bottom row consists of collinear points.

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.
By constrast, projective geometry collapses all cases into one.

Pappus knew, and probably Euclid before him knew \cite[p. 594]{Jones-2},
that all of the conics sections can be given a single expression
as the locus of points,
whose distances
from a given \emph{focus} and a given \emph{directrix}
have a given ratio.
In analytic geometry,
we say that the conic sections
are just the curves given by quadratic equations.
However,
when Omar Khayy\=am used conic sections
to solve what we should call cubic equations,
he considered 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.
Like Euclid's and Pappus's proofs,
but unlike the proofs of analytic geometry,
Apollonius's proofs rely on \emph{areas.}
There are areas of scalene triangles
and non-rectangular parallelograms.
In Appendix \ref{app:locus}
we considered 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.

\chapter{What happened in 2016}

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.
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.


\AfterBibliographyPreamble{\smaller}

%\bibliographystyle{plain}
%\bibliography{../../../references}

\begin{thebibliography}{10}

\bibitem{MR0053831}
{Apollonius of Perga}.
\newblock {\em On conic sections}.
\newblock Great Books of the Western World, no. 11. Encyclopaedia Britannica,
  Chicago, London, Toronto, 1952.

\bibitem{MR1660991}
{Apollonius of Perga}.
\newblock {\em Conics. {B}ooks {I}--{III}}.
\newblock Green Lion Press, Santa Fe, NM, revised edition, 1998.
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\end{thebibliography}

\end{document}
