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{scrartcl}

\usepackage{pstricks}
\usepackage{pst-eucl,pst-plot}
\usepackage{mxedruli}

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\usepackage{amsmath,amsthm,amssymb,url}
%\allowdisplaybreaks
\newcommand{\lto}{\Rightarrow}
\newcommand{\liff}{\Leftrightarrow}

\newcommand{\rat}[2]{#1\mathbin:#2}
\newcommand{\prop}{\mathrel{:\,:}}

\newcommand{\Apex}{A} % Apex of axial triangle
\newcommand{\Beat}{B} % Base endpoint of axial triangle
\newcommand{\Com}{C}  % Complementary endpoint of base of axial triangle
\newcommand{\mbat}{m} % midpoint of base of axial triangle (not shown)
\newcommand{\Dp}{D}   % Distant [end] point of chord
\newcommand{\Dpo}{D'} % Distant point (opposite) of chord
\newcommand{\Edp}{E}  % Extended diameter point
\newcommand{\Endp}{E^*} % Extended new diameter point
\newcommand{\Fp}{F}   % Four-way crossing
\newcommand{\Jp}{J}   % [pro-] Jected point in base
\newcommand{\Jop}{J'} % [pro-] Jected opposite point
\newcommand{\Koc}{O}  % Kentrum of conic
\newcommand{\Lk}{L}% L point next to K
\newcommand{\Lok}{L'}%  L point (opposite) next to K
\newcommand{\Lkn}{L^*}% new L point next to K
\newcommand{\Mp}{M}   % Midpoint of chord
\newcommand{\Mpn}{M^*}% Midpoint of new chord
\newcommand{\Nmp}{N}  % New [projected] midpoint
\newcommand{\Vern}{\Dp}% Vertex new
\newcommand{\Pnt}{P}  % Point on curve
\newcommand{\Pnto}{P'}% Point on curve (opposite)
\newcommand{\Qxr}{Q}  % Q in line with X and R new axial triangle base
\newcommand{\Rxq}{R}  % R in line with X and Q
\newcommand{\Ver}{V}  % vertex of section
\newcommand{\Vasn}{V^*}% Vertex as such (new)
\newcommand{\Wv}{W}   % Way-out (opposite) vertex
\newcommand{\Wnv}{W^*}   % Way-out (opposite) new vertex
\newcommand{\Xp}{X}   % X point (foot of ordinate}
\newcommand{\Xpn}{X^*}% X point (new foot of ordinate)
\newcommand{\Yp}{Y}   % Y point (lined up with P and X*)
\newcommand{\Ypn}{Y^*}% Y point (lined up with P and X)
\newcommand{\hide}{ee}  % hidden variables for constructions
\newcommand{\hida}{a}
\newcommand{\hidb}{b}
\newcommand{\hidc}{c}
\newcommand{\hidd}{d}

% The following are for Desargues's Theorem
\newcommand{\TA}{A}
\newcommand{\TB}{B}
\newcommand{\TC}{C}
\newcommand{\TD}{D}
\newcommand{\TE}{E}
\newcommand{\TF}{F}

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\begin{document}
%\titlehead{\centering}
\title{Affine Planes with Polygons}
\author{David Pierce\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
%  Istanbul\\
  \url{mat.msgsu.edu.tr/~dpierce},
\url{polytropy.com}}
\date{Batumi ({\mxedr batumi}), September, 2019}
\publishers{\smaller{\mxedr sakartvelos matema.ti.kosta .kav+siris}\\
  X {\mxedr qovel.cliuri saerta+soriso .konperencia}\\
  X Annual International Conference\\
  of the Georgian Mathematical Union}

{\relscale{0.2}\maketitle}
%\maketitle

    \newpage
    \thispagestyle{empty}
    \mbox{}\vfill
    \begin{center}
      \psset{dotsize=8pt,PointName=none}
      \newcommand{\myxrad}{6}\newcommand{\myyrad}{4.5}
      \begin{pspicture}(-5,-3)(5.5,3.5)
        \parametricplot[linewidth=2.4pt]0{360}
              {t cos \myxrad\space mul t sin \myyrad\space mul}
           \pstGeonode[PosAngle={180,-12,78,35}]
%              PointSymbol={default,default,default,default,o}]
        (0,0){\Koc}
        (! -12 cos \myxrad\space mul -12 sin \myyrad\space mul){\Ver}
(! 90 -12 add cos \myxrad\space mul 90 -12 add sin \myyrad\space mul)
                   {\Lk}
     (!  35 cos \myxrad\space mul  35 sin \myyrad\space mul){\Vasn}
%           (! -85 cos \myxrad\space mul -85 sin \myyrad\space mul){\Pnt}
                   %                   \ncline{\Koc}{\Ver}\ncline{\Koc}{\Vasn}
                   \uput[r](\Koc){${\Koc}$}
                   \uput[-10](\Ver){${\Ver}$}
                   \uput[80](\Lk){${\Lk}$}
                   \uput[35](\Vasn){${\Vasn}$}
                   \rput(0,2.3){\begin{minipage}{\textwidth}\centering
                       Let
                       \begin{equation*}
             {\Vasn}={\Koc}+a\cdot\vec{\Koc}{\Ver}+b\cdot\vec{\Koc}{\Lk}
                       \end{equation*}
          arbitrarily on the locus of ${\Pnt}$, where
              \begin{gather*}
          {\Pnt}={\Koc}+x\cdot\vec{\Koc}{\Ver}+y\cdot\vec{\Koc}{\Lk},\qquad
            x^2\pm y^2=1.
              \end{gather*}
        \end{minipage}}
                   \rput(0,-2.4){\begin{minipage}{\textwidth}\centering
                       \textbf{Apollonius's Theorem.}
          The affinity
            fixing ${\Koc}$ and\\
            interchanging ${\Ver}$ and ${\Vasn}$
            fixes the locus.\\
            \mbox{}\\
            \textbf{Modern proof.}
            The affinity corresponds to
            \begin{equation*}
              \begin{pmatrix}
                x\\y
              \end{pmatrix}\mapsto
              \begin{pmatrix}
                a&\pm b\\b&-a
              \end{pmatrix}
              \begin{pmatrix}
                x\\y
              \end{pmatrix}.
            \end{equation*}
        \end{minipage}}
      \end{pspicture}
    \end{center}
    \vfill\mbox{}
    \newpage

    

    \textbf{Apollonius's Proof.}
    \hfill
    The locus is given by
      \fbox{${\Xp}{\Pnt}{\Yp}
        ={\Ver}{\Xp}{\Ypn}{\Endp}$,}

      \vfill\hfill
      because
      this is true when ${\Pnt}$ is ${\Vasn}$, and
      
      \vfill
\hfill
            ${\Xp}{\Pnt}{\Yp}
            \propto{\Xp}{\Pnt}^2
            \propto{\Xp}{\Ver}\cdot{\Xp}{\Wv}
            \propto{\Ver}{\Xp}{\Ypn}{\Endp}$.

        \vfill
          \newcommand{\myxrad}{3.5}\newcommand{\myyrad}{2.3}
%\psset{unit=9mm}
      \begin{pspicture}(-4,-1.1)(3,2.7)
%\psgrid
        \psset{PointSymbol=none}
        \parametricplot[linewidth=2.4pt]0{360}{t cos \myxrad\space mul t sin \myyrad\space mul}
        \pstGeonode[PosAngle={105,-60,78,75,0,0,-85},
          PointName={default,default,default,default,none,none,default}]
        (0,0){\Koc}
        (! -12 cos \myxrad\space mul -12 sin \myyrad\space mul){\Ver}
        (! 90 -12 add cos \myxrad\space mul 90 -12 add sin \myyrad\space mul){\Lk}
        (!  35 cos \myxrad\space mul  35 sin \myyrad\space mul){\Vasn}
        (! 0 47 cos)x
        (0,1)t
        (! -85 cos \myxrad\space mul -85 sin \myyrad\space mul){\Pnt}
        \pstTranslation[PointName=none] t{\Ver}x
        \pstInterLL[PosAngle=-105]{x'}x{\Koc}{\Ver}{\Mp}
        \pstTranslation[PointName=none] x{\Ver}t
        \pstInterLL[PosAngle=-60]{t'}t{\Koc}{\Ver}{\Edp}
        \pstTranslation[PointName=none]{\Mp}{\Vasn}{\Ver,\Pnt}[u,v]
        \pstInterLL[PosAngle=90]{\Ver}u{\Koc}{\Vasn}{\Endp}
        \pstInterLL[PosAngle=-60]{\Pnt}v{\Koc}{\Ver}{\Xp}
        \pstInterLL[PosAngle=90]{\Pnt}{\Xp}{\Koc}{\Vasn}{\Ypn}
        \pstTranslation[PointName=none]{\Edp}{\Vasn}{\Pnt}[w]
        \pstInterLL[PosAngle=75]{\Pnt}w{\Koc}{\Ver}{\Yp}
        \pstInterLL[PosAngle=-105]{\Pnt}{\Yp}{\Koc}{\Vasn}{\Xpn}
        \pstTranslation[PosAngle=150]{\Ver}{\Koc}{\Koc}[\Wv]
        \pstTranslation[PosAngle=-150,PointName=none]{\Vasn}{\Koc}{\Koc}[\Wnv]
%        \pstInterLL[PosAngle=10,PointName=none]{\Vasn}{\Edp}{\Ver}{\Endp}{\Fp}
        \ncline[linewidth=2.4pt]{\Wv}{\Edp}
        \ncline{\Mp}{\Vasn}
        \ncline{\Ver}{\Endp}
        \ncline{\Pnt}{\Ypn}
        \ncline{\Koc}{\Lk}
        \psset{linestyle=dashed}
        \ncline{\Vasn}{\Edp}
        \ncline{\Pnt}{\Yp}
        \psset{linestyle=dotted}        
        \ncline{\Wnv}{\Endp}
      \end{pspicture}
      
      \vfill\hfill
      Adding ${\Xp}{\Yp}{\Xpn}{\Ypn}$ yields

      \vfill\hfill
          ${\Ypn}{\Pnt}{\Xpn}
          ={\Ver}{\Yp}{\Xpn}{\Endp}$,

          \vfill\hfill
        then
        \fbox{${\Ypn}{\Pnt}{\Xpn}
          ={\Edp}{\Yp}{\Xpn}{\Vasn}$.}
      
      \newpage

    \begin{multicols}{2}
      The foregoing happens in an \textbf{affine plane,}
      satisfying
      \begin{asparaenum}[1)]
      \item
        two points determine a line;
      \item
        through a point not on a line, a single parallel passes;
      \item
        there is a proper triangle.
      \end{asparaenum}
      \begin{pspicture}(3,3.6)
%    \psgrid
  \psset{PointSymbol=none}
  \pstGeonode[PosAngle={-90,-90,-45,90}]
  (2,0.6){\TE}(0.2,1.2){\TD}(2.8,1.5){\TF}(2.6,3){\TC}
  %      \psgrid
  \pstHomO[HomCoef=3,PointName=none]{\TF}{\TC}[O]
  \pstTranslation[PointName=none]{\TF}{\TD}{\TC}[x]
  \pstInterLL[PosAngle=135]{\TC}x{\TD}O{\TA}
  \pstTranslation[PointName=none]{\TF}{\TE}{\TC}[x]
  \pstInterLL[PosAngle=-135]{\TC}x{\TE}O{\TB}
%      \psset{linestyle=dashed}
      \ncline{\TB}{\TE}\ncline{\TA}{\TD}\ncline{\TC}{\TF}
      \psset{linewidth=2.4pt}
%      \psset{linestyle=solid}
  \ncline {\TD}{\TE}\ncline {\TD}{\TF}
  \ncline {\TB}{\TA}\ncline {\TA}{\TC}
      \psset{linestyle=dashed}
  \ncline{\TF}{\TE}
  \ncline{\TC}{\TB}      
    \end{pspicture}
  \hfill
      \begin{pspicture}(2.6,3.6)
        \psset{PointSymbol=none}
%              \psgrid
        \pstGeonode[PosAngle={-90,-90,90,90}]
        (0,0.6)D(2.6,0.6){\TE}(1.2,3){\TB}(2,3)A
        \pstHomO[HomCoef=0.45,PosAngle=150]{\TB}D[{\TC}]
        \pstTranslation[PointSymbol=none,PointName=none] {\TC}{\TE}{\TB}[x]
        \pstInterLL {\TB}xA{\TE}{\TF}
        \ncline D{\TB}\ncline A{\TE}
      \ncline{\TB}{\TD}\ncline{\TA}{\TE}
      \psset{linewidth=2.4pt}
  \ncline {\TD}{\TE}\ncline {\TD}{\TF}
  \ncline {\TB}{\TA}\ncline {\TA}{\TC}
      \psset{linestyle=dashed}
  \ncline{\TF}{\TB}
  \ncline{\TC}{\TE}      
      \end{pspicture}

      The plane is $K^2$ for some field $K$,
      if also,
      assuming
      \begin{equation*}
   {\TA}{\TB}\parallel{\TD}{\TE}\And{\TA}{\TC}\parallel{\TD}{\TF},
      \end{equation*}
      %we require 
      \begin{asparaenum}[1)]\setcounter{enumi}{3}
      \item
        \textbf{Desargues's Theorem:}
        \begin{equation*}
  {\TB}{\TC}\parallel {\TE}{\TF},
        \end{equation*}
  if
  ${\TA}{\TD}$,
  ${\TB}{\TE}$,  
  and ${\TC}{\TF}$ either
  \begin{compactenum}
    \item
      are mutually parallel
  or
\item
  have a common point;
  \end{compactenum}
\item
  \textbf{Pappus's Theorem:}
      \begin{equation*}
  {\TB}{\TF}\parallel {\TC}{\TE},
      \end{equation*}
      if \parbox{5cm}{%
      \begin{compactitem}
        \item
      ${\TD}$ lies on ${\TB}{\TC}$
          and
        \item
          ${\TA}$ lies on ${\TE}{\TF}$.
      \end{compactitem}}
      \end{asparaenum}


%      Now we need a theory of \emph{areas.}
    \end{multicols}

    \newpage

    \begin{multicols}{2}
      Of Desargues, case (a),
      \textbf{``Prism,''}
      lets us define,
      for non-collinear
      directed segments,
%      $AD$ and $BE$,
      \begin{multline*}
        \vec AD=\vec BE
        \iff\\ ABED \text{ is a parallelogram;}
      \end{multline*}
      \begin{center}
        \begin{pspicture}(5,3.6)
%          \psgrid
          \psset{PointSymbol=none}
          \pstGeonode[PosAngle={-30,180,0,0}]
          (3,0.3){\TD}(0.6,1){\TE}(4.4,1.7){\TF}(4.4,3.4){\TC}
%          \psgrid
        \pstTranslation[PosAngle={180,-30}] {\TF}{\TC}{{\TE},{\TD}}[{\TB},{\TA}]
        \ncline {\TE}{\TD}\ncline[linestyle=dotted] {\TE}{\TF}\ncline {\TF}{\TD}
        \ncline {\TA}{\TB}\ncline[linestyle=dotted] {\TB}{\TC}\ncline {\TC}{\TA}
        {\psset{linewidth=2.4pt}
          \ncline {\TA}{\TD}\ncline {\TB}{\TE}}
          \psset{linestyle=dashed}
          \ncline {\TC}{\TF}
        \end{pspicture}
      \end{center}
      case (b),
      \textbf{``Pyramid,''}
      for non-parallel pairs of parallel vectors,
      \begin{multline*}
        \rat{\vec OA}{\vec OD}
        \prop\rat{\vec OB}{\vec OE}\\
        \iff AB\parallel DE.
      \end{multline*}


      \begin{center}
        \begin{pspicture}(5,3.6)
%          \psgrid
          \psset{PointSymbol=none}
          \pstGeonode[PosAngle={-30,180,0,90}]
          (3,0.3){\TD}(0.6,1){\TE}(4.4,1.7){\TF}(2.2,3)O
          \pstHomO[HomCoef=0.4,PosAngle={-135,165,45}]O{{\TD},{\TE},{\TF}}[{\TA},{\TB},{\TC}]
          \ncline {\TE}{\TD}\ncline[linestyle=dotted] {\TE}{\TF}\ncline {\TF}{\TD}
          \ncline {\TA}{\TB}\ncline[linestyle=dotted] {\TB}{\TC}\ncline {\TC}{\TA}
          {\psset{linewidth=2.4pt}
          \ncline O{\TD}\ncline O{\TE}}
          \psset{linestyle=dashed}
          \ncline O{\TF}
        \end{pspicture}
      \end{center}
      On the plane, ratios of vectors act as a field or skew-field.
      
      With Pappus, it is a field.
    \end{multicols}
    
      \newpage

    \begin{multicols}{2}
      For Pappus and Desargues to be \emph{Theorems,}
            I propose axioms:

      
      \begin{asparadesc}
      \item[Addition.]\label{ax:group}
        The polygons compose an abelian group where
        \begin{gather*}
          -ABC\cdots= \cdots CBA,\\
          AAB=0,\qquad
          A*{}={}* A,\\
          A* B+B\mathrel{\dag} A= A* B\mathrel{\dag},
        \end{gather*}
        $*$ and $\dag$ being strings of vertices.
      \item[Linearity.]\label{ax:col}
%        $ABC=0$ means $A$, $B$, and $C$ are collinear:
        \begin{multline*}
          ABC\neq0\land BCD=0\\
          {}\land C\neq D\lto ACD\neq0.
        \end{multline*}
      \item[Parallels\lips]\label{ax:Playfair}
%        Playfair's Axiom.
      \item[Translation.]\label{ax:2-par}
        $ABED$ and $BCFE$ being parallelograms,
        \begin{equation*}
          ABC=DEF.
        \end{equation*}
        \begin{pspicture}(4.8,3.2)
%                    \psgrid
          \psset{PointSymbol=none}
          \pstGeonode[PosAngle={180,90,-90}]
          (0.6,1.1)G(1,2.6)A(1.5,0.6)C
          \pstGeonode[PointName=none](3,1.6)O
          \pstSymO[PosAngle={-90,0,90}] O{A,G,C}[F,E,D]
          \pstTranslation[PosAngle=-45] DAE[B]
          \ncline AD\ncline CF\ncline BE
          \psset{linestyle=dotted}
          \psset{linestyle=dashed}
          %          \ncline FG\ncline BG
          %          \ncline DA\ncline DE
          \psset{linestyle=solid,linewidth=2.4pt}
          \ncline AC\ncline CB\ncline BA
          \ncline DF\ncline FE\ncline ED
          \ncline AG\ncline CG
        \end{pspicture}
      \item[Bisection.]\label{ax:1-par}
        $ABCG$ being a parallelogram,
        \begin{equation*}
          CGA=ABC.
        \end{equation*}
      \item[Halving.]\label{ax:not-2}
        All nonzero polygons
        have the same order,
        not $2$.
      \end{asparadesc}
    \end{multicols}

\newpage

\begin{multicols}{2}

Hence
\textbf{Euclid \textsc i.37, 39:}

  \begin{pspicture}(4.8,4)
    %      \psgrid
    \psset{PointSymbol=none}
    \pstGeonode[PosAngle={-90,-90,90,90}](1.1,0.6)C(3.5,0.6)D
    (0.2,3.4)A
    \pstTranslation[PosAngle=90]CDA[B]
    \pstHomO[HomCoef=1.2,PosAngle=90]AB[E]
    \pstTranslation[PosAngle=90]CDE[F]
    \pstInterLL[PosAngle=-95] CEDBG
               {\psset{linewidth=2.4pt}
                 \ncline AD\ncline DC\ncline CA
                 \ncline CF\ncline FD\ncline DC}
               \ncline AF
               %    \psset{linestyle=dashed}
               \ncline BD\ncline CE
               \pstHomO[HomCoef=0.8,PosAngle=-30]DF[H]
               \psset{linestyle=dotted}
               \ncline AH\ncline HC
               %        \pstHomO[HomCoef=1.8,PosAngle=-90]CD[K]
               %        \ncline[linestyle=dashed]DK
  \end{pspicture}

$\bullet$    Assuming $AF\parallel CD$,    
    by \textbf{Parallels} we may let
    \begin{align*}
      AC&\parallel BD,
      &CE&\parallel DF;
    \end{align*}
    by \textbf{Translation,}
    \begin{equation*}
    ACE=BDF;
    \end{equation*}
    by \textbf{Addition,}
    \begin{comment}
      \begin{equation*}
        ACDB=ECDF,
      \end{equation*}
      since
    \end{comment}
    \begin{multline*}
        ACDB
        =ACGB+CDG={}\\
        ACE-BGE+CDG={}\\
        BDF-BGE+CDG
        =ECDF;
      \end{multline*}
    by \textbf{Bisection,}
    \begin{gather*}
      ACDB=2ACD,\\
      ECDF=2FCD;
    \end{gather*}
    by \textbf{Halving,}
    \begin{equation*}%\label{eqn:ACD}
      ACD=FCD.
    \end{equation*}
$\bullet$
    If $AF\nparallel CD$,
    let $AH\parallel CD$.
    \begin{align*}
      ACD&=HCD,\\
      FCD&=HCD+FCH,
    \end{align*}
    so $ACD\neq FCD$
    %\eqref{eqn:ACD} fails
    by \textbf{Linearity.}
\end{multicols}

    \newpage

    \begin{minipage}{9cm}
      We can now prove:
      \begin{compactitem}
      \item 
        \textbf{Prism,}
        by Translation, \textsc i.39, and Parallels.
        \item
        \textbf{Pappus.}
        From ${\TA}{\TB}\parallel{\TD}{\TE}$
        and ${\TA}{\TC}\parallel{\TD}{\TF}$,
        we obtain                  ${\TB}{\TF}\parallel{\TC}{\TE}$,
        since,
        by \textsc i.37,
        \begin{equation*}
          {\TB}{\TF}{\TC}
          ={\TB}{\TF}{\TA}{\TD}
          ={\TB}{\TF}{\TE}.
        \end{equation*}
      \end{compactitem}
    \end{minipage}
\hfill    \begin{minipage}{3cm}
            \begin{pspicture}(3,3)
        \psset{PointSymbol=none}
         %     \psgrid
        \pstGeonode[PosAngle={-150,-30,180,0}]
        (0,0){\TC}(2.6,0){\TE}(1,2.8){\TB}(2,2.8){\TF}
        \pstHomO[HomCoef=0.45,PosAngle=150]{\TB}{\TC}[{\TD}]
        \pstTranslation[PointSymbol=none,PointName=none] {\TD}{\TE}{\TB}[x]
        \pstInterLL {\TB}x{\TF}{\TE}{\TA}
        \ncline {\TC}{\TB}\ncline {\TF}{\TE}
        \ncline{\TB}{\TC}\ncline{\TF}{\TE}
        %\psset{linestyle=dotted}
        \ncline{\TF}{\TC}\ncline{\TB}{\TE}\ncline{\TD}{\TA}
        \psset{linestyle=solid}
      \psset{linewidth=2.4pt}
  \ncline{\TA}{\TB}
  \ncline {\TC}{\TA}
  \ncline{\TD}{\TE}      
  \ncline {\TF}{\TD}
      \psset{linestyle=dashed}
  \ncline {\TC}{\TE}
  \ncline {\TB}{\TF}
      \end{pspicture}
    \end{minipage}

    \begin{minipage}{4.5cm}
      \begin{pspicture}(4.5,5)
%        \psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={0,180,180}]
        (3.9,4.8){\TE}(0.6,4)O(0.6,0.3){\TF}
        \pstHomO[HomCoef=0.45,PosAngle=180]O{\TF}[{\TC}]
        \pstHomO[HomCoef=0.35,PosAngle=135]O{\TE}[{\TB}]
        \pstTranslation[PosAngle=0] O{\TF}{\TE}[{\TD}]
        \pstTranslation[PosAngle=-45] O{\TF}{\TB}[M]
        \pstTranslation[PosAngle=0] O{\TE}{\TC}[N]
        \pstInterLL[PosAngle=135] {\TB}M{\TC}N{\TA}
        \psset{PointName=none}
        \pstTranslation[PosAngle=30]{\TB}{\TC}{\TE}[P]
        \pstTranslation[PosAngle=75] {\TC}{\TB}{\TF}[Q]
        \pspolygon[fillstyle=solid,fillcolor=lightgray]
        ({\TB})({\TE})(N)({\TA})
        \pspolygon[fillstyle=solid,fillcolor=lightgray]
        ({\TC})({\TF})(M)({\TA})
        {\psset{linewidth=2.4pt}
        \pspolygon(O)({\TE})({\TD})({\TF})
        \ncline {\TB}M\ncline {\TC}N
                \psset{linestyle=dashed}
                \ncline {\TB}{\TC}\ncline {\TE}P\ncline Q{\TF}}
 %        \ncline O{\TA}
%        \ncline PQ
%                \psset{linestyle=dashed}
        \ncline {\TA}{\TD}
        \pstMiddleAB {\TB}N{Ma}
        \rput*(Ma){$\beta$}
        \pstMiddleAB {\TC}M{Mb}
        \rput*(Mb){$\gamma$}
      \end{pspicture}
    \end{minipage}
    \hfill
        \begin{minipage}{7.5cm}
          $\bullet$ \textbf{Pyramid, special case.}
          Assume
          \begin{compactitem}[--]
            \item
          $BE$ and $CF$ meet at $O$,
\item
  $AB\parallel DE$ and $AC\parallel DF$,
\item
  ${\TA}{\TB}\parallel O{\TC}$
      and ${\TA}{\TC}\parallel O{\TB}$.
          \end{compactitem}
          Then
          \begin{align*}
            {\TA}{\TD} \text{ contains }O
            &\iff\beta=\gamma\\
            &\iff BC\parallel DF.
          \end{align*}
          \begin{comment}
            \begin{gather}\label{eqn:cont}
              {\TA}{\TD} \text{ contains }O,\\\label{eqn:par}
              AB\parallel DE\And AC\parallel DF
            \end{gather}
            imply
            \begin{equation}\label{eqn:BC||DF}
              BC\parallel DF.
            \end{equation}
            If \eqref{eqn:par},
            \textbf{in case} ${\TA}{\TB}\parallel O{\TC}$
            and ${\TA}{\TC}\parallel O{\TB}$,
            \begin{equation*}
              \eqref{eqn:cont}
              \iff \beta=\gamma
              \iff \eqref{eqn:BC||DF}.
            \end{equation*}
          \end{comment}

        \end{minipage}

    \newpage
    \begin{minipage}{7cm}
      \textbf{Pyramid, less special case.}  If
      \begin{compactitem}[--]
        \item
            ${\TA}{\TD}$, ${\TB}{\TE}$, and
          ${\TC}{\TF}$ meet at $O$,
          \item
            $AB\parallel DE$ and $AC\parallel DF$,
          \item
            $AB\parallel OC$,\hfill
          then $BC\parallel DF$.
      \end{compactitem}
\vfill
          \textbf{General case.}
          Assuming $AC'\parallel DF'$,
          \begin{equation*}
            ABC\sim DEF\implies ABC'\sim DEF',
          \end{equation*}
          provided $CC'\parallel OA$;
          we can remove this;
          putting $C'$ on $O{\TC}$ is enough.
          
          \begin{pspicture}(7,2.5)
            \psset{PointSymbol=none}
%            \psgrid
      \pstGeonode[PosAngle={-90,-90,180,0}]
        (0.2,0.3)O(6.8,0.3){\TE}(2,2.1){F'}(5.8,2.1){\TF}
      \pstHomO[HomCoef=0.7,PosAngle=-90]O{\TE}[{\TD}]
%        \pstTranslation[PosAngle=-90]{F'}O{\TF}[{\TD}]
        \pstHomO[HomCoef=0.4,PosAngle=135]O{\TF}[{\TC}]
        \pstTranslation[PointName=none] {\TF}{F'}{\TC}[x]
        \pstInterLL[PosAngle=180] {\TC}xO{F'}{C'}
        \pstTranslation[PointName=none] {\TF}{\TE}{\TC}[x]
        \pstInterLL[PosAngle=-90] {\TC}xO{\TE}{\TB}
        \pstTranslation[PointName=none]{\TF}{\TD}{\TC}[x]
        \pstInterLL[PosAngle=-90] {\TC}xO{\TE}{\TA}
        %%%%%%%%%%%%%%%%%%%%%%%%
\pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]
        ({\TD})({\TF})({\TE})
\pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]
        ({\TA})({\TC})({\TB})
\pspolygon[linestyle=none,fillstyle=hlines]
({\TD})(F')({\TE})
\pspolygon[linestyle=none,fillstyle=hlines]
({\TA})(C')({\TB})
\ncline O{\TE}
\ncline O{\TF}\ncline O{F'}
\ncline{\TF}{F'}
\ncline{\TC}{C'}
\ncline{\TA}{C'}\ncline{C'}{\TB}
\ncline{\TD}{F'}\ncline{F'}{\TE}
\ncline{\TA}{\TC}\ncline{\TC}{\TB}
\ncline{\TD}{\TF}\ncline{\TF}{\TE}
          \end{pspicture}


    \end{minipage}
    \hfill
    \begin{minipage}{4cm}
      \begin{pspicture}(0.7,-1.5)(4.7,7)
%            \psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={90,90,-90}]
        (4.5,6.5){\TE}(0.2,5.8)O(0.2,1.5){\TF}
        \pstHomO[HomCoef=0.35,PosAngle=90]O{\TE}[{\TB}]
        \pstTranslation[PointName=none]{\TE}{\TF}{\TB}[x]
        \pstInterLL[PosAngle=180] {\TB}xO{\TF}{\TC}
        \pstTranslation[PointName=none]
        O{\TF}{\TE}[P]
        \pstHomO[HomCoef=0.48]{\TE}P[{\TD}]
        \pstTranslation[PointName=none]
        O{\TD}{\TF}[Q]
        \pstTranslation[PointName=none]
        O{\TF}{\TB}[R]
        \pstInterLL[PointName=none] {\TB}RO{\TD}{\TA}
%        \pstInterLL[PosAngle=115] {\TE}{\TF}O{\TD}G
%        \pstInterLL[PosAngle=-170] {\TB}{\TC}O{\TD}H
        {\psset{PointName=none}
        \pstInterLL {\TB}R{\TF}QL
        \pstTranslation {\TF}P{\TC}[M]
        \pstTranslation {\TF}Q{\TC}[N]
        \pstTranslation {\TE}{\TA}H[x]
        %        \pstInterLL[PosAngle=90] HxO{\TE}K
        \pstInterLL{\TB}LCNS
        \pstInterLL{\TC}M{\TB}LT}
        \pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]({\TC})({\TF})(L)(S)
        \pspolygon[linestyle=none,fillstyle=hlines]({\TC})({\TF})(R)(T)
        \pspolygon[linestyle=none,fillstyle=solid,fillcolor=lightgray]({\TA})(S)(N)({\TD})
        \pspolygon[linestyle=none,fillstyle=hlines]({\TB})(T)(M)(E)
        \uput*[30]({\TA}){${\TA}$}
       \ncline {\TF}P\ncline {\TD}Q\ncline Q{\TF}\ncline {\TA}L
        \ncline {\TC}M\ncline {\TC}N
 %               {\psset{linestyle=dotted}
 %                 \ncline HK\ncline KG\ncline {\TE}{\TA}}
        \psset{linewidth=2.4pt}
        \ncline O{\TE}\ncline O{\TF}\ncline O{\TD}
        \ncline {\TE}{\TD}\ncline {\TB}{\TA}
        \ncline {\TF}{\TD}\ncline {\TC}{\TA}
        \psset{linestyle=dashed}
        \ncline {\TE}{\TF}\ncline {\TB}{\TC}
        \psset{linestyle=dotted}
      \end{pspicture}
    \end{minipage}

    \newpage

    \begin{minipage}[c][8cm]{7.5cm}
      Given
      \fbox{${\TA}{\TB}{\TC}\sim {\TD}{\TE}{\TF}$,}
      %for arbitrary $C'$ on ${\TC}{\TF}$
      we
      let
      \begin{gather*}
%        {\TD}{F'}\parallel{\TA}{C'},\\
        {\TB}G\parallel{\TA}{\TC},{\TD}{\TF},\\
        HG,{\TD}{F'}\parallel{\TA}{C'}.
      \end{gather*}
      By Pappus,
      \begin{enumerate}[1)]
      \item
        from
      ${\TA}{\TC}HG{\TB}{C'}$,
        $H{\TC}\parallel {\TB}{C'}$;
\item
      from ${\TB}{\TC}{\TD}{\TF}{\TE}G$,
        ${\TD}{\TC}\parallel{\TE}G$;
\item
      from
      $H{\TC}{\TD}{F'}{\TE}G$,
        $H{\TC}\parallel {\TE}{F'}$.
      \end{enumerate}
      Thus
      \begin{equation*}
        {\TB}C'\parallel{\TE}F',
      \end{equation*}
      \fbox{${\TA}{\TB}{C'}\sim {\TD}{\TE}{F'}$;}
also \fbox{${\TA}{\TD}{\TC}\sim {\TB}{\TE}G$.}


        \end{minipage}
    \hfill    
    \begin{minipage}{4.5cm}
      \begin{pspicture}(0.5,0)(5,8.5)
%        \psgrid[subgriddiv=1]
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={180,0,180,0}]
        (0,8.3){\TA}(4.5,8){\TC}(1.5,0.2){\TE}(4.5,1.5){F'}
        \pstHomO[HomCoef=0.65,PosAngle=180]{\TA}{\TE}[{\TD}]
        \pstTranslation[PointName=none] {\TD}{\TC}{\TE}[x]
        \pstInterLL[PosAngle=0] {\TE}x{\TC}{F'}G
        \pstTranslation[PointName=none] {\TC}{\TA}G[x]
        \pstInterLL[PosAngle=180] Gx{\TA}{\TE}{\TB}
        \pstTranslation[PointName=none] {\TC}{\TA}{\TD}[x]
        \pstInterLL[PosAngle=0] {\TD}x{\TC}{F'}{\TF}
        \pstTranslation[PointName=none] {\TE}{F'}{\TB}[x]
        \pstInterLL[PosAngle=0] {\TB}x{\TC}{F'}{C'}
        \pstTranslation[PointName=none] {F'}{\TE}{\TC}[x]
        \pstInterLL[PosAngle=180] {\TC}x{\TA}{\TE}H
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \psset{fillstyle=solid,fillcolor=lightgray}
        \pspolygon
        (H)({\TC})({\TD})({F'})({\TE})(G)
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \psset{hatchangle=41,fillcolor=gray}
        \pspolygon%[fillstyle=hlines]
        ({\TB})({\TC})({\TD})({\TF})({\TE})(G)
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \psset{hatchangle=!0.3 4.5 atan neg}
        \pspolygon[fillstyle=hlines]
        (H)({\TC})({\TA})({C'})({\TB})(G)
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \psset{linestyle=solid}
        \ncline{\TA}{\TE}\ncline {\TC}{F'}
%        \psset{linestyle=dashed}
%        \ncline {\TD}{\TC}\ncline {\TE}G
%        \psset{linestyle=dotted}
%        \ncline H{\TC}\ncline {\TB}{C'}\ncline {\TE}{F'}
        \psset{linewidth=2.4pt}
        \ncline HG\ncline {\TA}{C'}\ncline {\TD}{F'}
        \ncline{\TA}{\TC}\ncline{\TB}G\ncline{\TD}{\TF}
      \end{pspicture}
    \end{minipage}

\end{document}
