\documentclass[%
version=last,%
%a4paper,%
%25pt,%
a6paper,%
12pt,%
landscape,%
%twocolumn,%
titlepage,%
%headings=small,%
bibliography=totoc,%
twoside=false,%
reqno,%
parskip=half,%
draft=true,%
%DIV=classic,%
DIV=12,%
headinclude=false,%
pagesize]
{scrartcl}

\usepackage[margin=1cm]{geometry}
\pagestyle{empty}
\usepackage{pstricks,subfig,pst-eucl,pst-plot}
%\psset{linewidth=3pt,labelsep=8pt}
%\psset{unit=5mm}

\setlength{\footskip}{1\baselineskip}

\usepackage{graphicx}

\usepackage{multicol}

\setlength{\columnseprule}{0.4pt}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\usepackage{relsize}

%\usepackage[polutonikogreek,english]{babel}

%\usepackage{gfsneohellenic} % From the Greek Text Society

%\newcommand{\gk}[1]{\foreignlanguage{polutonikogreek}{%
%\relscale{0.9}%
%\textneohellenic{#1}}}%  Greek text
%\newcommand{\gkm}[1]{\text{\gk{#1}}}

\renewenvironment{quote}{\begin{list}{}
{\relscale{.9}\setlength{\leftmargin}{0.05\textwidth}
\setlength{\rightmargin}{\leftmargin}}
\item[]}
{\end{list}}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\usepackage{hfoldsty,verbatim}

\newenvironment{optional}{}{}

%\let\optional\comment
%\let\endoptional\endcomment

\usepackage{paralist}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\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}

\newcommand{\abs}[1]{\lvert#1\rvert}
\newcommand{\as}{\mathrel{:\;:}}

\renewcommand{\vec}[2]{\overrightarrow{#1#2}}
\renewcommand{\theta}{\vartheta}

\renewcommand{\theequation}{\fnsymbol{equation}}

\usepackage{bm}

\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\newcommand{\bce}{\textsc{b.c.e.}}
\newcommand{\ce}{\textsc{c.e.}}

\raggedright

\begin{document}
\title{Apollonian Proof}
\author{David Pierce}
\date{Logic Colloquium 2019, Prague, August}
\publishers{Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
  \url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}

{\relscale{0.2}\maketitle}

%\setcounter{page}0
    \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){${\Koc}
                     \begin{pmatrix}
                       0\\0
                     \end{pmatrix}$}
                   \uput[r](\Ver){${\Ver}
                     \begin{pmatrix}
                       1\\0
                     \end{pmatrix}$}
                   \uput[30](\Lk){${\Lk}
                     \begin{pmatrix}
                       0\\1
                     \end{pmatrix}$}
                   \uput[r](\Vasn){${\Vasn}
                     \begin{pmatrix}
                       a\\b
                     \end{pmatrix}$}
          \rput(0,2.3){\begin{minipage}{\textwidth}\centering
          In an \textbf{affine plane,} the\\
          locus of ${\Koc}+x\cdot\vec{\Koc}{\Ver}+y\cdot\vec{\Koc}{\Lk}$,
          where
          \begin{equation*}
            x^2\pm y^2=1,
          \end{equation*}
          is (for any ${\Vasn}$,
          namely
          ${\Koc}+a\cdot\vec{\Koc}{\Ver}+b\cdot\vec{\Koc}{\Lk}$,
          on the locus)\\ fixed by
          the \emph{affinity}
            fixing ${\Koc}$ and
            interchanging ${\Ver}$ and ${\Vasn}$.\\
        \end{minipage}}
          \rput(0,-2.5){\begin{minipage}{\textwidth}\centering
            \emph{Modern proof.}
            The affinity is $
            \begin{pmatrix}
              x\\y
            \end{pmatrix}\mapsto
            \begin{pmatrix}
              a&\pm b\\b&-a
            \end{pmatrix}
            \begin{pmatrix}
              x\\y
            \end{pmatrix}$. \qedsymbol\\\mbox{}\\
              1. Apollonius's proof uses \textbf{areas.}
              2. So does an\\ \textbf{axiomatization} of affine planes
              in which the\\
              \emph{theorems} of \textbf{Pappus} and \textbf{Desargues}\\
              are just that.
        \end{minipage}}
      \end{pspicture}
    \end{center}
    \vfill\mbox{}
    \newpage

    

    \textbf{Proof of Apollonius.}
    The locus of $P$ is given by
    \begin{equation*}
      {\Xp}{\Pnt}{\Yp}
      ={\Ver}{\Xp}{\Ypn}{\Endp},
    \end{equation*}
    because\hfill
    $
    \begin{aligned}[t]
      {\Xp}{\Pnt}{\Yp}
      &\propto{\Xp}{\Pnt}^2\\
    &\propto{\Xp}{\Ver}\cdot{\Xp}{\Wv}\\
      &\propto{\Ver}{\Xp}{\Ypn}{\Endp},\\
{\Mp}{\Vasn}{\Edp}&={\Ver}{\Mp}{\Vasn}{\Endp}.
    \end{aligned}$

        \vfill
          \newcommand{\myxrad}{3.5}\newcommand{\myyrad}{2.3}
%\psset{unit=9mm}
      \begin{pspicture}(-4,-0.8)(4,-0.4)
                          %\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\mbox{}
\hfill
        Adding ${\Xp}{\Yp}{\Xpn}{\Ypn}$
        yields\\
\hfill                  ${\Ypn}{\Pnt}{\Xpn}
          ={\Ver}{\Yp}{\Xpn}{\Endp}$, then
          \begin{equation*}
            {\Ypn}{\Pnt}{\Xpn}
          ={\Edp}{\Yp}{\Xpn}{\Vasn}.
          \end{equation*}
      
  
\newpage

\begin{multicols}{2}
  Fundamental to the geometry of areas is
%Each \textbf{axiom} will justify a step in the proof of
\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}
  \begin{enumerate}
  \item
    Assuming $AF\parallel CD$,
    let
    \begin{align*}
      AC&\parallel BD,
      &CE&\parallel DF.
    \end{align*}
  \item
    By translation,
    \begin{equation*}
    ACE=BDF.      
    \end{equation*}
  \item
    By polygon algebra,
    \begin{equation*}
      ACDB=ECDF.
    \end{equation*}
    \begin{comment}
      \begin{multline*}
        ACDB
        =ACGB+CDG\\
        =ACE-BGE+CDG\\
        =BDF-BGE+CDG\\
        =ECDF.
      \end{multline*}
    \end{comment}
  \item
    By bisection,
    \begin{gather*}
      ACDB=2ACD,\\
      ECDF=2FCD.
    \end{gather*}
  \item
    By halving,
    \begin{equation}\label{eqn:ACD}
      ACD=FCD.
    \end{equation}
  \item
    If $AF\nparallel CD$,
    let $AH\parallel CD$.
    \begin{align*}
      ACD&=HCD,\\
      FCD&=HCD+FCH,
    \end{align*}
    so \eqref{eqn:ACD} fails.
  \end{enumerate}
\end{multicols}
    \newpage

    \begin{multicols}{2}
      In order 3, 6, 1, 5, 4, 2,
      the steps
      are justified by:
      
      \begin{asparadesc}
      \item[Axiom 1.]
        The polygons compose an abelian group $\Pi$ where,
        $*$ and $\dag$ being strings of vertices,
        \begin{gather*}
          A*{}= A* A={}* A,\\
          A* B+B\mathrel{\dag} A= A* B\mathrel{\dag},\\
          -ABC\cdots= \cdots CBA.
        \end{gather*}
      \item[Axiom 2.]
        $ABC=0$ means that $A$, $B$, and $C$ are collinear.
      \item[Axiom 3.]
        Playfair's Axiom.
      \item[Axiom 4.]
        All nonzero elements of $\Pi$
        have the same order,
        not $2$.
        
        \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[Axiom 5, 6.]
        If
        $ABCG$,
        $ABED$, and $BCFE$ are parallelograms,
        then
        \begin{equation*}
          CGA=ABC=DEF.
        \end{equation*}
      \end{asparadesc}

    \end{multicols}

    \newpage

    \begin{minipage}{5cm}
      \begin{pspicture}(5,4)
        \psset{PointSymbol=none}
%              \psgrid
        \pstGeonode[PosAngle={180,0,180,0}]
        (0.6,0.2)D(4.4,0.2)C(2,3.8)F(4,3.8)A
        \pstHomO[HomCoef=0.45,PosAngle=150]FD[B]
        \pstTranslation[PointSymbol=none,PointName=none] BCF
        \pstInterLL F{F'}ACE
        \ncline DF\ncline AC
        \psset{linestyle=dotted}
        \ncline AD\ncline BE\ncline FC
        \psset{linewidth=2pt,linestyle=solid}
        \ncline AB\ncline BC\ncline DE\ncline EF
        \psset{linestyle=dashed}
        \ncline CD
        \ncline AF
        %    \psset{linestyle=dotted}
        %    \ncline AC\ncline BE\ncline DA
      \end{pspicture}
    \end{minipage}\hfill
    \begin{minipage}{7cm}
\centerline{\emph{From Euclid \textsc i.37, 39:}}
      
$\bullet$ \textbf{Pappus's Hexagon Theorem,}
by his proof:
      In hexagon  $ABCDEF$, if
      \begin{align*}
        AB&\parallel DE,&
        BC&\parallel EF,
      \end{align*}
      then $FAD=FAEB=FAC$, so
      \begin{equation*}
        CD\parallel FA.
      \end{equation*}
    \end{minipage}

    \begin{minipage}{6cm}
      $\bullet$ \textbf{Euclid \textsc i.43 plus.}
      \begin{align*}
        OGL=0
        &\iff \alpha=\beta\\
        &\iff BD\parallel AC.
      \end{align*}
$\bullet$ \emph{Desargues's Theorem.}
    \end{minipage}\hfill
    \begin{minipage}{6cm}
      \begin{pspicture}(6,4)
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={180,180,0}](0.6,3.4)B(1,0.6)O(5.4,0.6)D
        \pstHomO[HomCoef=0.45,PosAngle=-90]OD[C]
        \pstHomO[HomCoef=0.35,PosAngle=180]OB[A]
        \pstTranslation[PosAngle=0] ODB[L]
        \pstTranslation[PosAngle=0] ODA[M]
        \pstTranslation[PosAngle=90] OBC[N]
        \pstInterLL[PosAngle=140] AMCNG
        \psset{PointName=none}
        \pstTranslation[PosAngle=30]ACB[P]
        \pstTranslation[PosAngle=75] CAD[Q]
        {\psset{linewidth=2.4pt}
        \pspolygon(O)(B)(L)(D)
        \ncline AM\ncline CN
                \psset{linestyle=dashed}
                \ncline AC\ncline BP\ncline QD}
        \ncline OG
        \ncline PQ
                \psset{linestyle=dashed}
        \ncline GL
        \pstMiddleAB AN{Ma}
        \rput(Ma){$\alpha$}
        \pstMiddleAB CM{Mb}
        \rput(Mb){\psframebox[fillstyle=solid,fillcolor=white,linestyle=none]{$\beta$}}
      \end{pspicture}
    \end{minipage}

    \newpage
    \begin{minipage}{8cm}
      \textbf{Desargues's Theorem.}
      If
      \begin{equation*}
        AB\parallel DE\And AC\parallel DF,
      \end{equation*}
      then
      $BC\parallel EF$,
      so $ABC\sim DEF$.
      
      \emph{Proof.}
        \begin{compactitem}
        \item
          True when
          $AB\parallel OC$,
          by \textsc i.43+.
        \item
          Enough now that, since
          \begin{equation*}
            BAG\sim EDH,
          \end{equation*}
          for all $X$ (not shown) on $OA$,
          \begin{equation*}
            BXG\sim EYH
          \end{equation*}
          for some $Y$ on $OA$.
        \end{compactitem}
\makebox[0pt][l]{Note  $BAE\sim GKH$
      by Pappus,
where $BA\parallel GK$.}
    \end{minipage}
    \hfill
    \begin{minipage}{4cm}
      \begin{pspicture}(1.5,0)(5.5,8.5)
        %    \psgrid
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={0,180,180,0}]
        (5,8.3)A(0.6,8)O(0.6,2.5)C
        \pstHomO[HomCoef=0.35,PosAngle=90]OA[D]
        \pstTranslation[PointName=none]ACD[x]
        \pstInterLL[PosAngle=180] DxOCF
        \pstTranslation[PointName=none]OCA[P]
        \pstHomO[HomCoef=0.48]AP[B]
        \pstTranslation[PointName=none]OBC[Q]
        \pstTranslation[PointName=none]OCD[K]
        \pstInterLL[PosAngle=-65] DKOBE
        \pstInterLL[PosAngle=115] ACOBG
        \pstInterLL[PosAngle=-170] DFOBH
        {\psset{PointName=none}
        \pstInterLL DKCQL
        \pstTranslation CPF[M]
        \pstTranslation CQF[N]
        \pstTranslation AEH[x]}
        \pstInterLL[PosAngle=90] HxOAK
        \ncline CP\ncline BQ\ncline QC\ncline EL
        \ncline FM\ncline FN
                {\psset{linestyle=dotted}
                  \ncline HK\ncline KG\ncline AE}
        \psset{linewidth=2.4pt}
        \ncline OA\ncline OC\ncline OB
        \ncline AB\ncline DE
        \psset{linestyle=dashed}
        \ncline AC\ncline DF
        \psset{linestyle=dotted}
        \ncline CB\ncline FE
      \end{pspicture}
    \end{minipage}

    \newpage

    \begin{minipage}{6cm}
      \begin{pspicture}(6,8.5)
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={180,180,0,0}]
        (0.6,6)A(0.6,0.2)D(5.4,1)H(3,8.3)E
        \pstHomO[HomCoef=0.5,PosAngle=180]AD[C]
        \pstTranslation[PointName=none] CED[x]
        \pstInterLL[PosAngle=0] DxEHF
        \pstTranslation[PointName=none] EAF[x]
        \pstInterLL[PosAngle=180] FxADB
        \pstTranslation[PointName=none] EAC[x]
        \pstInterLL[PosAngle=0,PointSymbol=default] CxEHL
        \pstTranslation[PointName=none] DHB[x]
        \pstInterLL[PosAngle=0] BxEHG
        \pstTranslation[PointName=none] HDE[x]
        \pstInterLL[PosAngle=180] ExADK
        %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
        \ncline KD\ncline EH
        \ncline AE\ncline BF
        \psset{linestyle=none,hatchangle=0}
        \pspolygon[fillstyle=hlines]
        (K)(E)(A)(G)(B)(F)
        %        \psset{hatchsep=8pt}
        \pspolygon[fillstyle=vlines]
        (K)(E)(C)(H)(D)(F)
        \psset{linestyle=solid,linewidth=2.4pt}
        \ncline KF\ncline AG\ncline CH
        \psset{linestyle=dashed}
        \ncline CE\ncline DF
        \psset{linestyle=dotted}
        \ncline KE\ncline BG\ncline DH
      \end{pspicture}
    \end{minipage}
    \hfill
    \begin{minipage}{6cm}
      \textbf{Lemma.}
      Given
      \begin{equation*}
        AEC\sim BFD,
      \end{equation*}
      we noted
      \begin{equation*}
        AEB\sim CLD
      \end{equation*}
            for some $L$ on $EF$.
      Now let
      \begin{equation*}
        KF\parallel AG\parallel CH.
      \end{equation*}
      By Pappus twice,
      \begin{equation*}
        BG\parallel KE\parallel DH,
      \end{equation*}
      whence
      \begin{equation*}
        AGB\sim CHD.
      \end{equation*}
    \end{minipage}

    \begin{comment}
      \newpage
      \noindent
      \begin{pspicture}(0,4.4)(12,7)
        %      \psgrid%[subgriddiv=1]
        \psset{PointSymbol=none}
        \pstGeonode[PosAngle={90,-90,0,90}]
        (0.2,6.4)O(11,0.6)A(12,3.1)B(11.5,6.4)C
        \pstHomO[HomCoef=0.47,PosAngle={-135,60,90},
          PointName={default,none,default}]
        O{A,B,C}[D,E,F]
        \pstTranslation[PointName=none] OC{B,E,D}[x,y,z]
        \pstInterLL[PosAngle=-135] BxOAG
        \pstInterLL[PosAngle=-135] EyOAH
        \pstInterLL[PointName=none] DzOBK
        \pstTranslation[PointName=none] OA{K,E}[x,y]
        \pstInterLL[PosAngle=90] KxOCL
        \pstInterLL[PosAngle=90] EyOCM
        \pstTranslation[PointName=none] FDM[x]
        \pstInterLL[PosAngle=-135] MxOAN
        \pspolygon[fillstyle=solid,fillcolor=gray]
        (A)(K)(D)(E)(G)(B)
        \ncline OB
        \ncline EG\ncline KA
        \ncline AL\ncline GM
        \ncline MH\ncline LD
        \ncline FH\ncline NL\ncline CG
        \ncline BC\ncline EF
        \psset{linestyle=dashed}
        \psset{linestyle=dotted}
        \psset{linewidth=2.4pt}
        \psset{linestyle=solid}
        \ncline AB\ncline DE
        \ncline AC\ncline DF\ncline NM
        \ncline OA\ncline KL\ncline EM
        \ncline OC\ncline BG\ncline DK\ncline EH
        \psset{linestyle=dashed}
        \psset{linestyle=dotted}
        \psset{linestyle=none,hatchangle=0}
        \pspolygon[fillstyle=hlines]
        (H)(M)(N)(L)(D)(F)
        \pspolygon[fillstyle=vlines]
        (G)(M)(N)(L)(A)(C)
        \uput*[30](E){$E$}
        \uput*[60](K){$K$}
      \end{pspicture}

      \begin{math}
        \begin{gathered}[b]
          AB\parallel DE,\\
          OC\parallel EH\parallel DK\parallel BG,\\
          OA\parallel EM\parallel KL,\\
          NM\parallel DF\parallel AC
        \end{gathered}
      \end{math}

      implies
      \begin{math}
        \begin{aligned}[t]
          HM&\parallel DL&&\text{[Desargues case]}\\
          GE&\parallel AK&&\text{[Pappus]}\\
          MG&\parallel LA&&\text{[Desargues case]}\\
          HF&\parallel NL\parallel GC&&\text{[Pappus]}\\
          FE&\parallel CB&&\text{[Desargues case]}
        \end{aligned}
      \end{math}
    \end{comment}

\end{document}