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\begin{document}
%\titlehead{\centering}
\title{Analitik Geometri}
\author{David Pierce\\
  Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
  %  Istanbul\\
  Matematik B\"ol\"um\"u\\
  \url{mat.msgsu.edu.tr/~dpierce}\\
\url{polytropy.com}}
\date{20 \c Subat 2020 tasla\u g\i}
\maketitle

\tableofcontents

\chapter{Giri\c s}

\textbf{Analitik Geometri,}
cebirsel y\"ontemleri kullanan geometridir.
Bu y\"ontemler,
bir \emph{orant\i lar} kuram\i na ba\u gl\i d\i r.
Bu kuramda
\begin{compactenum}[1)]
\item
  orant\i l\i l\i k ba\u g\i nt\i s\i,
  bir denklik ba\u g\i nt\i s\i d\i r;
%  \"ozellikle ge\c ci\c slidir;
\item
  \emph{Thales Teoremi} do\u grudur.
\end{compactenum}
Bu notlarda
\begin{compactitem}
\item 
orant\i l\i l\i k kuram\i n\i,
\emph{Desargues Teoremi}'nden;
\item
  Desargues Teoremi'ni,
  \emph{Pappus Teoremi}'nden;
\item
  Pappus Teoremi'ni,
  \"Oklid'in \emph{\"O\u geler}'inin \textsc i.\ kitab\i ndan
\end{compactitem}
elde edece\u giz.

Geometrik tan\i mlar\i ndan birine g\"ore bir \textbf{elips,}
iki verilen \textbf{odak noktas\i ndan}
uzakl\i klar\i n\i n toplam\i\ ayn\i\ olan noktalar\i n bir yeridir.
Analitik geometride,
bir $a$ ve bir $b$ i\c cin,
elipsi tan\i mlayan ko\c sul
iki de\u gi\c skeni olan
\begin{equation*}
  \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
\end{equation*}
cebirsel denklemi taraf\i ndan
ifade edilebilir.

Cebirsel bir denklemde
de\u gi\c skenlerin ve sabitlerin de\u gerleri,
ger\c cel say\i lar\i n olu\c sturdu\u gu
$\R$ gibi bir \emph{cisimden} gelir.
Bir cisimde iki eleman\i n
\emph{toplam\i} ve \emph{\c carp\i m\i} vard\i r.
Toplama ve \c carpma i\c slemine
geometrik bir tan\i m\i\ verece\u giz.

\chapter{Orant\i lar}

\section{Thales Teoremi}

\"Oklid'in \emph{\"O\u geler}'inin birinci kitab\i nda
toplaman\i n geometrik anlam\i\ vard\i r,
ama \c carpma kavram\i\ yoktur
\cite{bones,MR1932864,Oklid-2015}.

\emph{La G\'eometrie}
adl\i\ kitab\i nda
Descartes, ger\c cel say\i lar\i n \c carpmas\i na
geometrik bir anlam\i\ verir
\cite{Descartes-Geometry,Descartes-Geometrie}.
Bunun i\c cin \textbf{Thales Teoremi'ni}
kullan\i r
\cite{Thales-Theorem}.
Bu teoreme g\"ore e\u ger \Sekilde{fig:Thales}ki gibi
\begin{figure}
              \psset{PointSymbol=none}
  \subfloat[]{
    \begin{pspicture}(5,4)
      \pstGeonode[PosAngle={180,180,0}]
        (0.5,0.2)A(0.5,3.8)B(4.5,1.5)O
      \pstHomO[HomCoef=0.6,PosAngle={-90,90}]O{A,B}[C,D]
      \ncline AO\ncline BO
%      \psset{linewidth=2.4pt}
      \ncline AB\ncline CD
  \end{pspicture}}
  \hfill
    \subfloat[]{
    \begin{pspicture}(5,4)
      \pstGeonode[PosAngle={180,180,0,0}]
      (0.5,0.2)A(0.5,3.8)B(4.5,3)C(4.5,1)D
      \pstInterLL[PosAngle=90] ACBDO
      \ncline AC\ncline BD
%      \psset{linewidth=2.4pt}
      \ncline AB\ncline CD
    \end{pspicture}}
\caption{Thales Teoremi}
  \label{fig:Thales}
\end{figure}
bir $OAB$ \"u\c cgeninin $OA$ kenar\i nda $C$
ve $OB$ kenar\i nda $D$ oturursa,
o zaman
\begin{equation*}
  AB\parallel CD\liff\rat{\vec OA}{\vec OC}\prop\rat{\vec OB}{\vec OD}.
\end{equation*}
Burada
\begin{itemize}
\item
  $\vec OA$, $\vec OC$, $\vec OB$, ve $\vec OD$,
  \textbf{y\"onl\"u do\u gru par\c cas\i d\i r};
  \item
    $\rat{\vec OA}{\vec OC}$ ve $\rat{\vec OB}{\vec OD}$,
    \sayfada{oran} tan\i mlayaca\u g\i m\i z \c sekilde
    \emph{orand\i r};
    \item
$\rat{\vec OA}{\vec OC}\prop\rat{\vec OB}{\vec OD}$,
      \textbf{orant\i d\i r.}
\end{itemize}
\.Iki farkl\i\ y\"onl\"u do\u gru par\c cas\i\
\c cizilebilir ve birbirine e\c sit olabilir.
\.Iki y\"onl\"u do\u gru par\c cas\i n\i n oran\i\ olabilir,
ama soyut bir \c sey oldu\u gundan bir oran\i n kendisi \c cizilemez.
Bundan dolay\i\ bir orant\i ya gore iki oran birbirine e\c sit de\u gil,
birbiriyle \emph{ayn\i d\i r.}
Bununla birlikte bir orant\i da
$\prop$ i\c saretinde $=$ i\c sareti kullan\i labilir.

Thales Teoremi, \emph{\"O\u geler}'in alt\i nc\i\ kitab\i n\i n
ikinci \"onermesidir.
\"Onermeyi g\"ostermek i\c cin \"Oklid,
be\c sinci kitapta bulunan orant\i lar kuram\i n\i\ kullan\i r.
Bu kuram,
uzunluklar\i n
\textbf{Ar\c simet \"ozelli\u gini} varsay\i r.
Bu \"ozelli\u ge g\"ore
iki uzunlu\u gun
daha k\"u\c c\"u\u g\"un\"un bir kat\i,
daha b\"uy\"u\u g\"unden daha b\"uy\"ukt\"ur.

Pozitif ger\c cel say\i lar\i n da Ar\c simet \"ozelli\u gi vard\i r.
Ayr\i ca iki pozitif ger\c cel say\i n\i n b\"ol\"um\"u vard\i r,
ve bu b\"ol\"um, pozitif
ger\c cel bir say\i d\i r.
\"Oklid'de iki uzunlu\u gun oran\i\ vard\i r,
ve bu oran, pozitif ger\c cel bir say\i\ olarak anla\c s\i labilir.

  Asl\i nda analitik geometri yapmak i\c cin,
  Ar\c simet \"ozelli\u gine ihtiyac\i m\i z yoktur.
  Ayr\i ca oranlar negatif olabilir.
  Orant\i lar\i n \emph{tan\i m\i} olarak
  Thales Teoremi'ni varsayabiliriz.
  Bu durumda orant\i l\i l\i k ba\u g\i nt\i s\i n\i n
  ge\c ci\c sli oldu\u gunu g\"ostermek zorunday\i z.
  Bunun i\c cin, \emph{Pappus Teoremi'ni}
  ve \emph{Desargues Teoremi'ni} kullanaca\u g\i z.

  \section{Analik geometride baz\i\ tarihler}

  \begin{description}
  \item[M.\"O. 585'te]
    Thales'in \"onceden s\"oyledi\u gi g\"une\c s tutulmas\i\
    \cite[\textsc i.74]{Herodotus-Loeb}, \cite{Heath-Aristarchus}.
      \item[M.S. 4.\ y\"uzy\i l\i nda]
    Pappus Teoremi yay\i nlan\i r
    \cite{Jones-whole,Pappus-Pierce-2}.
  \item[1637'de] Descartes'\i n
    \emph{La G\'eometrie}'si yay\i nlan\i r.
  \item[1640'ta] Pascal,
    Pappus Teoremi'nin bir genelle\c stirmesini verir
    \cite{Pascal-Conics,Pascal-Coniques}.
  \item[1648'de] Desargues Teoremi yay\i nlan\i r
    \cite{Desargues-Thm}.
  \item[1899'da] Hilbert,
    Pappus ve Desargues Teoremi'nin
  analitik geometri yapmak i\c cin yetti\u gini g\"osterir \cite{MR0116216}.
  \item[1905'te]
  Hessenberg,
  Pappus Teoremi'nden
  Desargues Teore\-mi'ni
  kan\i tlar
  \cite{MR1511339}.
  \c Simdi
  Hilbert, Hessenberg'in kan\i t\i n\i\ kullanabilir
  \cite{Hilbert-10}.
  \.Ikisi,
  %Hilbert ve Hessenberg,
  Pappus Teoremi'ne
  \emph{Pascal Teoremi'ni} der.
\item[1957'de]
  Artin,
  Pappus ve Desargues Teoremi'nin
  analitik geometri i\c cin
  yetti\u gini
  Hilbert'inkinden (ve bizimkinden)
  ba\c ska bir \c sekilde
  g\"osterir
  \cite{Artin}.
  \end{description}


\section{Desargues Teoremi: Bildirme}

\"U\c c tane $AD$, $BE$, ve $CF$ do\u grusu
\Sekilde{fig:Des}ki gibi
\begin{figure}
              \psset{PointSymbol=none}
  \subfloat[]{\label{fig:Des-1}
    \begin{pspicture}(5,4)
      \pstGeonode[PosAngle={180,90,-90,90}]
        (0.5,1.2)A(2,3.4)B(1,0.6)C(4.8,3.4){E}
      \pstTranslation[PosAngle={135,-90}] B{E}{A,C}[D,F]
      \ncline A{D}\ncline B{E}\ncline C{F}
      \psset{linewidth=2pt}
      \ncline AB\ncline{D}{E}
      \ncline AC\ncline{D}{F}
      \psset{linestyle=dashed}
      \ncline BC\ncline{E}{F}
  \end{pspicture}}
  \hfill
  \subfloat[]{\label{fig:Des-2}
    \begin{pspicture}(1,0)(6,4)
      \pstGeonode[PosAngle={180,90,-90,0}]
        (0.5,1.2)A(2,3.4)B(1,0.6)C(5.5,1.5)O
      \pstHomO[HomCoef=0.55,PosAngle={135,90,-90}]O{A,B,C}[D,E,F]
      \ncline AO\ncline BO\ncline CO
      \psset{linewidth=2.4pt}
      \ncline AB\ncline{D}{E}
      \ncline AC\ncline{D}{F}
      \psset{linestyle=dashed}
      \ncline BC\ncline{E}{F}
    \end{pspicture}}
  \caption{Desargues Teoremi}
  \label{fig:Des}
\end{figure}
\begin{compactenum}[(a)]
\item
  ya birbirine paralel olsun,
\item
  ya da $O$ noktas\i nda kesi\c ssin.
\end{compactenum}
Ayr\i ca
\begin{align}\label{eqn:Des-hyp}
  AB&\parallel DE,&AC&\parallel DF
\end{align}
olsun.
\textbf{Desargues Teoremi'ne}
g\"ore
\begin{equation*}
  BC\parallel EF.
\end{equation*}
Bu teoremin birinci, ``parallel'' durumunu elde etmek i\c cin,
\"Oklid'in \emph{\"O\u geler}'inin
birinci kitab\i ndan
\"Onermeler 34 ve 30, birinci ortak kavram,
ve \"Onerme 33 yeter.

\section{Pappus Teoremi: Bildirme ve G\"osterme}

Desargues Teoremi'nin ikinci durumunu
g\"ostermek i\c cin,
\c Sekil \ref{fig:Pappus} veya \numarada{fig:Pappus-duz}ki gibi
\begin{figure}
  \centering
    \begin{pspicture}(10,6)
            \psset{PointSymbol=none}
            %     \psgrid
    \pstGeonode[PosAngle={0,0,90,-90}]
    (9.5,0.2){\TA}(9.5,5.8){\TC}(0.2,5){\TD}(0.2,2.5){\TF}
            \pstHomO[HomCoef=0.48,PosAngle=-90]{\TF}{\TA}[{\TE}]
    \pstTranslation[PointSymbol=none,PointName=none] {\TE}{\TC}{\TF}[x]
    \pstInterLL[PosAngle=90] {\TF}x{\TD}{\TC}{\TB}
            \ncline{\TD}{\TC}\ncline{\TF}{\TA}
            \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}

    \caption{Pappus Teoremi}\label{fig:Pappus}
\end{figure}
$A$ noktas\i\ $EF$'de,
$D$ noktas\i\ $BC$'de olsun,
ve tekrar \esitlikte{eqn:Des-hyp}ki gibi
$AB\parallel DE$ ve $AC\parallel DF$ olsun.
\textbf{Pappus Teoremi'ne}
g\"ore
\begin{equation}\label{eqn:Pap-conc}
  BF\parallel CE.
\end{equation}
Bunu kan\i tlamak i\c cin
\Sekilde{fig:Pappus-duz}ki gibi
\begin{figure}
  \centering
  \begin{pspicture}(10,6)
            \psset{PointSymbol=none}
            %     \psgrid
    \pstGeonode[PosAngle={180,180,-90,90}]
    (0.5,5.8){\TC}(0.5,0.2){\TE}(9.8,1){\TF}(9.8,3.5){\TB}
            \pstHomO[HomCoef=0.48,PosAngle=90]{\TB}{\TC}[{\TD}]
    \pstTranslation[PointSymbol=none,PointName=none] {\TD}{\TE}{\TB}[x]
    \pstInterLL[PosAngle=-90] {\TB}x{\TF}{\TE}{\TA}
       { \psset{linewidth=0pt,hatchangle=0}
         \pspolygon[fillstyle=solid,fillcolor=lightgray]
         ({\TA})({\TF})({\TC})({\TD})
        \pspolygon[fillstyle=vlines]({\TA})({\TE})({\TB})({\TD}) }
%            \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}

          \caption{Pappus Teoremi'nin d\"uzenlemesi}
          \label{fig:Pappus-duz}
          
        \end{figure}
  ${\TA}{\TD}$, ${\TB}{\TE}$, ve
${\TC}{\TF}$
\c cizilsin.
\"Oklid'in \"Onerme 37'sine g\"ore
  ${\TA}{\TB}\parallel {\TD}{\TE}$ oldu\u gundan
\begin{equation*}
  {\TA}{\TB}{\TD}={\TA}{\TB}{\TE}.
\end{equation*}
\.Ikinci ortak kavrama g\"ore
\begin{equation}\label{eqn:FBE}
  {\TF}{\TB}{\TD}{\TA}={\TA}{\TF}{\TB}+{\TA}{\TB}{\TD}
  ={\TA}{\TF}{\TB}+{\TA}{\TB}{\TE}
  ={\TF}{\TB}{\TE}.
\end{equation}
Benzer \c sekilde
${\TF}{\TD}\parallel {\TA}{\TC}$ oldu\u gundan
\begin{gather}\notag
  {\TF}{\TD}{\TA}={\TF}{\TD}{\TC},\\\label{eqn:FBC}
  {\TF}{\TB}{\TD}{\TA}
  ={\TF}{\TB}{\TD}+{\TF}{\TD}{\TA}={\TF}{\TB}{\TD}+{\TF}{\TD}{\TC}={\TF}{\TB}{\TC}.
\end{gather}
\c Simdi \eqref{eqn:FBE} ve \esitlikte{eqn:FBC}n
\begin{equation*}
  {\TF}{\TB}{\TE}={\TF}{\TB}{\TC}.
\end{equation*}
\"Onerme 39'a g\"ore
\esitlikte{eqn:Pap-conc}ki gibi ${\TB}{\TF}\parallel {\TC}{\TE}$.
%Pappus'un Teoremi kan\i tlanm\i\c st\i r.

Pappus'un Teoremi'nde
iki do\u gru ve bir alt\i gen vard\i r, ve
\begin{compactitem}
  \item
    alt\i gen\i n kenarlar\i ndan her biri
    do\u grular\i n birinden di\u gerine
    ge\c cer,
    \item
iki durumda alt\i genin kar\c s\i t kenarlar\i\
birbirine paraleldir.
\end{compactitem}
Sonu\c c olarak \"u\c c\"unc\"u durumda da
kar\c s\i t kenerlar paralellerdir.
\c Sekiller \ref{fig:Pappus} ve \numarada{fig:Pappus-duz}
alt\i gen $ABFDEC$ olur.

\section{Desargues Teoremi: G\"osterme}

\c Simdi
%Desargues Teoremi'nin ikinci durumunu kan\i tlamak i\c cin,
$AD$, $BE$, ve $CF$'nin ortak noktas\i\ $O$ olsun;
ayr\i ca $AB\parallel DE$ ve $AC\parallel DF$ olsun.
\.Iki durum vard\i r.
\begin{enumerate}[(a)]
\item
  E\u ger
    \Sekilde{fig:Hessen}ki gibi
\begin{figure}
  \centering
    \psset{PointSymbol=none}
\begin{pspicture}(10,7)
%    \psgrid[subgriddiv=1]
    \pstGeonode[PosAngle={180,180,-90,90}]
    (0.5,6.8)L(0.5,0.2)C(9.8,1)O(9.8,5.5)N
    \pstHomO[HomCoef=0.55,PosAngle=180]CL[A]
    \pstHomO[HomCoef=0.45,PosAngle=90]NL[B]
    \pstTranslation[PointName=none] ABN[x]
    \pstInterLL[PosAngle=-150]NxOA{D}
    \pstInterLL[PosAngle=75]NxOB{E}
    \pstTranslation[PointName=none] AC{D}[x]
    \pstInterLL[PosAngle=-90]{D}xOC{F}
    \pstInterLL[PosAngle=-90] L{D}OCM
    {\psset{linewidth=0pt,hatchangle=0}
    \pspolygon[fillstyle=solid,fillcolor=lightgray]
    (O)(N)(M)(L)(C)(B)
%    \pspolygon[fillstyle=solid,fillcolor=gray]
    \pspolygon[fillstyle=hlines]
    (O)(N)(M)(D)(F)(E)
    \pspolygon[fillstyle=vlines](O)(N)(D)(L)(A)(B)}
%    \uput*[90]({D}){$D$}
    \ncline OA\ncline OC
%    \ncline OB
%    \ncline ML
    \ncline LN
    \ncline NO
    \psset{linewidth=3pt}
    \ncline LC\ncline{D}{F}
    \ncline AB\ncline{D}{N}
    \psset{linestyle=dashed}
    \ncline BC\ncline{E}{F}    
    \ncline NM
\end{pspicture}
  
  \caption{Desargues i\c cin Hessenberg'in kan\i t\i}
  \label{fig:Hessen}
\end{figure}
$BACO$ bir paralelkenar de\u gilse,
o zaman
  $OB\nparallel AC$ varsay\i labilir.
  Bu durumda
$LM\parallel OB$ ve $D\in LM$ olsun.
  Pappus Teoremi ile
\begin{itemize}
\item
  $ONDLAB$ alt\i geninde $ON\parallel AL$,
\item
  $ONMLCB$ alt\i geninde $BC\parallel MN$,
\item
  $ONMDFE$ alt\i geninde $EF\parallel MN$.
\end{itemize}
Paralellik ge\c ci\c sli oldu\u gundan
$BC\parallel EF$.
\item
  E\u ger \Sekilde{fig:Hessen-2} gibi
\begin{figure}
  \centering
    \psset{PointSymbol=none}
\begin{pspicture}(10,5)
    \pstGeonode[PosAngle={-90,90,90}](0.2,0.6)C(1,4.4)A(9.8,4.4)B
    \pstTranslation ABC[O]
    \pstHomO[HomCoef=0.7,PosAngle={90,0,-90}]O{A,B,C}[D,G,F]
    \ncline OA\ncline OB\ncline OC
    \ncline AB
    \ncline AC\ncline DF
    \ncline BC\ncline{G}F
    \psset{linestyle=dotted}
    \ncline D{G}
  \end{pspicture}
  
  \caption{Hessenberg'in kan\i t\i n\i n 2.\ durumu}
  \label{fig:Hessen-2}
\end{figure}
 $BACO$ bir paralelkenar ise,
  o zaman $ABCO$ bir paralelkenar de\u gildir.
  Bu durumda
  $OB$'de bir $G$ i\c cin $FG\parallel CB$.
  \.Ilk durumdan $DG\parallel AB$,
  dolay\i s\i yla $G$ ve $E$ noktas\i\ ayn\i d\i r.
  \qedhere
\end{enumerate}
Desargues Teoremi'nin ikinci durumu kan\i tlanm\i\c st\i r.

\chapter{Vekt\"orler ve oranlar\i}

\section{Vekt\"orler Grubu}

\"Oklid'de do\u gru par\c calar\i n e\c sitli\u gi
bir denklik ba\u g\i nt\i s\i d\i r.
\"Ozellikle birinci ortak kavrama g\"ore
e\c sitlik ge\c ci\c slidir.
E\c sitli\u ge g\"ore bir do\u gru par\c cas\i n\i n denklik s\i n\i f\i,
%do\u gru par\c cas\i n\i n
onun
\textbf{uzunlu\u gu} olarak
anla\c s\i labilir.

E\u ger her do\u gru kendisine paralel ise,
o zaman \"Onerme 30'a g\"ore paralellik de
bir denklik ba\u g\i nt\i s\i d\i r.

Desargues Teoremi'nin
ilk durumu sayesinde
y\"onl\"u do\u gru par\c calar\i n\i n,
a\c sa\u g\i daki ko\c sullar\i\ sa\u glayan
\emph{e\c sitlik} ba\u g\i nt\i s\i\ vard\i r,
ve bu ba\u g\i nt\i\
bir denklik ba\u g\i nt\i s\i d\i r.
\begin{enumerate}
\item 
  Herhangi $ABDC$ paralelkenari i\c cin
  \begin{equation*}
    \vec AB=\vec CD.
  \end{equation*}
  \item
    Herhangi $\vec AB$ y\"onl\"u do\u gru par\c cas\i\ ve
    $C$ noktas\i\ i\c cin,
  bir ve tek bir $D$ noktas\i\ i\c cin,
  \begin{equation*}
    \vec AB=\vec CD.
  \end{equation*}
\end{enumerate}
E\c sitli\u ge g\"ore
y\"onl\"u bir do\u gru par\c cas\i n\i n
denklik s\i n\i f\i,
bir \textbf{vekt\"ord\"ur.}

Bir vekt\"or daha vard\i r.
$\vec AA$'n\i n y\"onl\"u do\u gru par\c cas\i\ olmad\i\u g\i\ halde
her durumda
\begin{equation*}
  \vec AA=\vec BB
\end{equation*}
olsun, ve her $\vec AA$'n\i n temsil etti\u gi vekt\"or,
\begin{equation*}
  \bm0
\end{equation*}
olsun.
Ayr\i ca
tan\i ma g\"ore
\begin{gather*}
  \vec AB+\vec BC=\vec AC,\\
  -\vec AB=  \vec BA
\end{gather*}
olsun.
Bu tan\i mlar, vekt\"orlere ge\c cer,
ve herhangi $\bm a$, $\bm b$, ve $\bm c$ vekt\"or\"u i\c cin
\begin{gather*}
  (\bm a+\bm b)+\bm c=\bm a+(\bm b+\bm c),\\
  \bm b+\bm a=\bm a+\bm b,\\
  \bm a+\bm0=\bm a,\\
  \bm a+(-\bm a)=\bm0.
\end{gather*}
B\"oylece vekt\"orler
$V$ k\"umesini olu\c sturdu\u gunda
$(V,+,-,\bm0)$ bir \textbf{abelyan gruptur.}

$\vec CD=\vec AB$ oldu\u gunda
\begin{equation*}
  A+\vec CD=B
\end{equation*}
olsun.  $\vec AB$ bir $\bm a$ vekt\"or\"un\"u
temsil etti\u ginde
\begin{equation*}
  A+\bm a=B
\end{equation*}
olsun.  O zaman her durumda
\begin{equation*}
  A+(\bm b+\bm c)=(A+\bm b)+\bm c.
\end{equation*}
Bu \c sekilde vekt\"orler grubu,
d\"uzlemi \textbf{etki eder.}
K\i saca
\begin{compactitem}
\item
  iki vekt\"or\"un toplam\i\ vard\i r;
\item
  bir nokta ve bir vekt\"or\"un toplam\i\ vard\i r;
\item
  iki noktan\i n toplam\i\ yoktur.
\end{compactitem}

\section{Oranlar Cismi}

\label{oran}\.Iki paralel y\"onl\"u do\u gru par\c cas\i n\i n
\textbf{oran\i} vard\i r,
ve ayr\i ca onlar\i n temsil etti\u gi
vekt\"orlerin ayn\i\ oran\i\ vard\i r.
Desargues Teoremi'nin
ikinci durumu sayesinde bir oran,
a\c sa\u g\i daki ko\c sullar sa\u glayan
bir denklik s\i n\i f\i\ olarak tan\i mlanabilir.
\begin{enumerate}
\item 
  Thales Teoremi do\u grudur.
\item
  Herhangi $\rat{\vec OA}{\vec OB}$ oran\i\
  ve $\vec OC$ y\"onl\"u do\u gru par\c cas\i\ i\c cin,
  bir ve tek bir $D$ noktas\i\ i\c cin,
  \begin{equation*}
    \rat{\vec OA}{\vec OB}\prop\rat{\vec OC}{\vec OD}.
  \end{equation*}
\end{enumerate}
Burada e\u ger $\vec OA$ bir $\bm a$ vekt\"or\"un\"u
ve $\vec OB$ bir $\bm b$ vekt\"or\"un\"u temsil ederse,
o zaman $\rat{\vec OA}{\vec OB}$ oran\i
\begin{equation*}
  \bm a:\bm b
\end{equation*}
olarak yaz\i labilir.
Ayr\i ca $\bm0:\bm a$ oran\i\ vard\i r.
Her durumda bu oran ayn\i d\i r ve
\begin{equation*}
  0
\end{equation*}
olarak yaz\i labilir.
\c Simdi tan\i ma g\"ore
\begin{gather*}
  \rat{\bm a}{\bm c}+\rat{\bm b}{\bm c}=\rat{(\bm a+\bm b)}{\bm c},\\
  \rat{-\bm a}{\bm b}=-(\rat{\bm a}{\bm b}),\\
  (\rat{\bm a}{\bm b})\cdot(\rat{\bm b}{\bm c})=\rat{\bm a}{\bm c},\\
   \bm b\neq\bm 0\lto\rat{\bm b}{\bm a}=(\rat{\bm a}{\bm b})^{-1},\\
  \rat{\bm a}{\bm a}=1
\end{gather*}
olsun.
Pappus Teoremi sayesinde
\begin{equation*}
  (\rat{\bm a}{\bm b})\cdot(\rat{\bm c}{\bm d})
  \prop
  (\rat{\bm c}{\bm d})\cdot(\rat{\bm a}{\bm b}),
\end{equation*}
\c c\"unk\"u \Sekilde{fig:comm}
\begin{figure}
  \centering
  \psset{PointSymbol=none}
  \begin{pspicture}(10,4)
    \pstGeonode[PosAngle={180,0,0}](0.5,1.5)O(9.5,0.2)E(9.5,3.8)F
    \pstHomO[HomCoef=0.7,PosAngle=-90]OE[B]
    \pstHomO[HomCoef=0.5,PosAngle=-90]OB[A]
    \pstHomO[HomCoef=0.5,PosAngle=90]OF[D]
    \pstHomO[HomCoef=0.7,PosAngle=90]OD[C]
    \ncline OE\ncline OF\ncline AD\ncline BF\ncline CB\ncline DE
    \psset{linestyle=dashed}
    \ncline AC\ncline EF
  \end{pspicture}
  \caption{Oranlar \c carpmas\i n\i n de\u gi\c smelili\u gi}
  \label{fig:comm}
\end{figure}
e\u ger
\begin{align*}
  AD&\parallel BF,&BC&\parallel ED
\end{align*}
ise, o zaman
Thales Teoremi'nden
\begin{align*}
  \rat{\vec OA}{\vec OB}&\prop\rat{\vec OD}{\vec OF},&
  \rat{\vec OB}{\vec OE}&\prop\rat{\vec OC}{\vec OD},
\end{align*}
ve ayr\i ca
Pappus Teoremi'nden
\begin{equation*}
  AC\parallel EF,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
  \rat{\vec OA}{\vec OE}\prop\rat{\vec OC}{\vec OF},
\end{equation*}
ve sonu\c c olarak
\begin{multline*}
  (\rat{\vec OA}{\vec OB})\cdot(\rat{\vec OC}{\vec OD})
  \prop  (\rat{\vec OA}{\vec OB})\cdot(\rat{\vec OB}{\vec OE})\\
  \prop\rat{\vec OA}{\vec OE}
  \prop\rat{\vec OC}{\vec OF}
  \prop  (\rat{\vec OC}{\vec OD})\cdot(\rat{\vec OD}{\vec OF})\\
  \prop  (\rat{\vec OC}{\vec OD})\cdot(\rat{\vec OA}{\vec OB}).
\end{multline*}
B\"oylece
oranlar
$K$ k\"umesini olu\c sturdu\u gunda
\begin{itemize}
\item
  $(K,+,-,0)$ bir abelyan gruptur,
\item
  $(K\setminus\{0\},{}\cdot{},{}^{-1},1)$ bir abelyan gruptur,
\item
  $K$'de
    herhangi $a$, $b$, ve $c$ oran\i\ i\c cin
    \begin{equation*}
      a\cdot(b+c)=a\cdot b+a\cdot c.
    \end{equation*}
\end{itemize}
Bundan dolay\i\
$(K,{}\cdot{},+,-,1,0)$ bir \textbf{cisimdir.}

\"Orne\u gin
\begin{compactitem}
\item
  kesirli say\i r $\Q$,
\item
  ger\c cel say\i lar $\R$,
\end{compactitem}
cismini olu\c sturur.
Tamsay\i lar bir cismi olu\c sturmaz,
ama her $p$ asal say\i s\i\ i\c cin,
$p$ mod\"ul\"une g\"ore tamsay\i lar
$\F_p$ cismi olu\c stur.
Bu cisimde
\begin{equation}\label{eqn:p}
  \underbrace{1+\dots+1}_p=0,
\end{equation}
k\i saca $p=0$, olur.
Bununla birlikte $\Q$ ve $\R$'de
her $p$ i\c cin \eqref{eqn:p} yanl\i\c st\i r.
\c Sekillerimizde oranlar cismi $\R$'dir.
\label{2neq0}\emph{Kullanaca\u g\i m\i z her oranlar cisminde
  $2\neq0$ olur.}
(Di\u ger durumda hi\c c do\u gru par\c cas\i n\i n
orta noktas\i\ yoktur;
\sayfada{thm:mp}ki \"Onerme \numaraya{thm:mp} bak\i n.)


\section{Menelaus ve Ceva}

\begin{proposition}[Menelaus Teoremi
    \protect{\cite[H69--70]{Ptolemy-Toomer}}]
  \Sekilde{fig:Men}
  \begin{figure}
    \centering
    \psset{PointSymbol=none}
    \begin{pspicture}(6,6)
      \pstGeonode[PosAngle={-90,-90,90}](0.2,0.6)D(5.8,0.6)C(4,5.4)A
      \pstHomO[HomCoef=0.6,PosAngle=-90]DC[B]
      \pstHomO[HomCoef=0.7,PosAngle=45]CA[E]
      \pstInterLL[PosAngle=135]DEABF
      \ncline DC\ncline CA\ncline AB\ncline DE
      \pstTranslation[PointName=none]CAB
      \pstInterLL[PosAngle=135]B{B'}DEG
      \ncline[linestyle=dashed]BG
    \end{pspicture}
    \caption{Menelaus Teoremi}
    \label{fig:Men}
  \end{figure}
  \begin{equation*}
    (\rat{\vec AF}{\vec FB})
    \cdot(\rat{\vec BD}{\vec DC})
    \cdot(\rat{\vec CE}{\vec EA})
    =-1.
  \end{equation*}
\end{proposition}

\begin{proof}
  Thales'ten
  \begin{align*}
    \rat{\vec AF}{\vec FB}
    &\prop\rat{\vec AE}{\vec GB}\\
    &\prop(\rat{\vec AE}{\vec EC})\cdot(\rat{\vec EC}{\vec GB})\\
    &\prop(\rat{\vec AE}{\vec EC})\cdot(\rat{\vec DC}{\vec DB}).
    \qedhere
  \end{align*}
\end{proof}

\begin{proposition}[Ceva Teoremi]
  \Sekilde{fig:Ceva}
  \begin{figure}
    \centering
    \psset{PointSymbol=none}
    \begin{pspicture}(6,6)
      \pstGeonode[PosAngle={-90,-90,90}](0.2,0.6)B(5.8,0.6)C(3.5,5.4)A
      \pstHomO[HomCoef=0.6,PosAngle=-90]BC[D]
      \pstHomO[HomCoef=0.3,PosAngle=45]CA[E]
      \pstInterLL[PosAngle=135]ADBEG
      \pstInterLL[PosAngle=135] CGABF
      \ncline BC\ncline CA\ncline AB
      \psset{linestyle=dashed}
      \ncline AD\ncline BE\ncline CF
    \end{pspicture}
    \caption{Ceva Teoremi}
    \label{fig:Ceva}
  \end{figure}
  \begin{equation*}
    (\rat{\vec AF}{\vec FB})
    \cdot(\rat{\vec BD}{\vec DC})
    \cdot(\rat{\vec CE}{\vec EA})
    =1.
  \end{equation*}
\end{proposition}

\begin{proof}
  Menelaus Teoremi'nden
  \begin{gather*}
    -(\rat{\vec AF}{\vec FB})
    \prop(\rat{\vec AC}{\vec CE})\cdot(\rat{\vec EG}{\vec GB})\\
    -(\rat{\vec BD}{\vec DC})
    \prop(\rat{\vec BG}{\vec GE})\cdot(\rat{\vec EA}{\vec AC}).\qedhere
  \end{gather*}
\end{proof}

\chapter{Koordinatlar}

\section{Noktalar}

E\u ger $A$ noktas\i\ $OB$ do\u grusunda ise,
\begin{equation*}
  (\rat{\vec OA}{\vec OB})\cdot\vec OB=\vec OA
\end{equation*}
olsun.
O zaman herhangi $a$ ve $b$ oran\i\
ve herhangi $\bm c$ ve $\bm d$ vekt\"or\"u i\c cin
\begin{gather*}
  a\cdot(\bm c+\bm d)=a\cdot\bm c+a\cdot\bm d,\\
  (a+b)\cdot\bm c=a\cdot\bm c+b\cdot\bm c,\\
  a\cdot(b\cdot\bm c)=(a\cdot b)\cdot\bm c,\\
  1\cdot\bm c=\bm c.
\end{gather*}
B\"oylece vekt\"orler grubu,
oranlar cismi alt\i nda bir \textbf{vekt\"or uzay\i d\i r.}
Ayr\i ca $OAB$ bir \"u\c cgen oldu\u gunda
herhangi $P$ noktas\i\ i\c cin,
girdileri $K$'de olan
bir ve tek bir $(a,b)$ s\i ral\i\ ikilisi
\begin{equation*}
  \vec OP=x\cdot\vec OA+y\cdot\vec OB
\end{equation*}
denklemini sa\u glar.
Asl\i nda e\u ger
\Sekilde{fig:coord}ki gibi
\begin{figure}
  \centering
  \psset{PointSymbol=none}
  \begin{pspicture}(7,5)
%    \psgrid
    \pstGeonode[PosAngle={-90,-90,180,90}](0.2,0.6)O(4,0.6)A(1,4.8)B(6.8,3)P
    %\psdot(A)
    \pstTranslation[PointName=none]AOP
    \pstInterLL[PosAngle=180] OBP{P'}D
    \pstTranslation[PointName=none]BOP
    \pstInterLL[PosAngle=-90] OAP{P'}C
        \pspolygon[fillstyle=solid,fillcolor=lightgray]
    (O)(A)(B)
    \pspolygon[fillstyle=hlines]
    (O)(C)(P)(D)
%    \ncline OA\ncline OB\ncline AB
%    \psset{linestyle=dashed}
%    \ncline PC\ncline PD\ncline CA
  \end{pspicture}
  \caption{Kartezyan koordinatlar}
  \label{fig:coord}
\end{figure}
$P$ ne $OA$'da ne $OB$'de ise,
o zaman $OA$'da olan bir $C$ i\c cin
ve $OB$'de olan bir $D$ i\c cin
$OCPD$ bir paralelkenard\i r,
ve bu durumda
\begin{equation*}
  \vec OP=\vec OC+\vec OD,
\end{equation*}
dolay\i s\i yla
\begin{align*}
  a&=\rat{\vec OC}{\vec OA},&
  b&=\rat{\vec OD}{\vec OB}.
\end{align*}

Yukar\i da $(O,A,B)$ s\i ral\i\ \"u\c cl\"us\"u,
bir \textbf{kartezyan koordinatlar sistemini} kurur.
E\u ger
\begin{align*}
  A&=O+\vec a,&B&=O+\bm b
\end{align*}
ise, o zaman ayn\i\ sistem
\begin{equation*}
  (O,\vec OA,\vec OB)\quad\text{ veya }\quad
  (O,\bm a,\bm b)
\end{equation*}
olarak yaz\i labilir.
Bu sistemde $(a,b)$'nin girdileri,
$P$'nin
\textbf{kartezyan koordinatlar\i d\i r,}
ve
\begin{equation*}
  P=(a,b)
\end{equation*}
e\c sitli\u gi yaz\i labilir.

\begin{proposition}
  Bir kartezyan sisteme g\"ore
  \begin{align*}
    A&=(a_1,a_2),&
    B&=(b_1,b_2),&
    C&=(c_1,c_2),&
    D&=(d_1,d_2)
  \end{align*}
  olsun.  O zaman
  \begin{equation*}
    \vec AB=\vec CD\liff(b_1-a_1,b_2-a_2)=(d_1-c_1,d_2-c_2).
  \end{equation*}
\end{proposition}

\begin{proposition}\label{thm:mp}
  Bir kartezyan sisteme g\"ore
  \begin{align*}
    A&=(a_1,a_2),&
    B&=(b_1,b_2)
  \end{align*}
  olsun.
  E\u ger oranlar cisminde $2\neq0$ ise,
  O zaman $AB$ do\u gru par\c cas\i n\i n
  orta noktas\i\ vard\i r,
  ve bu noktan\i n koordinatlar\i,
  \begin{equation*}
    \left(\frac{a_1+b_1}2,\frac{a_2+b_2}2\right).
  \end{equation*}
\end{proposition}

\section{Do\u grular}

Bir $P$ noktas\i n\i n
bir $CD$ do\u grusunda olmas\i\ i\c cin,
bir oran\i n
\begin{equation*}
  \vec OP=\vec OC+t\cdot\vec CD
\end{equation*}
denklemini sa\u glamas\i,
gerek ve yeter bir ko\c suldur.
$(O,A,B)$ sistemine g\"ore
\begin{align*}
  C&=(c_1,c_2),&D&=(d_1,d_2)
\end{align*}
olsun, yani
\begin{align*}
  \vec OC&=c_1\cdot\vec OA+c_2\cdot\vec OB,&
    \vec OD&=d_1\cdot\vec OA+d_2\cdot\vec OB
\end{align*}
olsun.
O zaman
\begin{align*}
  &\phantom{{}={}}\vec OC+t\cdot\vec CD\\
  &=\vec OC+t\cdot(\vec CO+\vec OD)\\
  &=(1-t)\cdot\vec OC+t\cdot\vec OD\\
  &=(1-t)\cdot(c_1\cdot\vec OA+c_2\cdot\vec OB)
  +t\cdot(d_1\cdot\vec OA+d_2\cdot\vec OB)\\
  &=\bigl((1-t)c_1+td_1\bigr)\cdot\vec OA
  +\bigl((1-t)c_2+td_2\bigr)\cdot\vec OB\\
  &=\bigl(c_1+t(d_1-c_1)\bigr)\cdot\vec OA
  +\bigl(c_2+t(d_2-c_2)\bigr)\cdot\vec OB.
\end{align*}
B\"oylece
\begin{align*}
  &\phantom{{}\liff{}}\vec OC+t\cdot\vec CD=x\cdot\vec OA+y\cdot\vec OB\\
  &\liff x=c_1+t(d_1-c_1)\land y=c_2+t(d_2-c_2)\\
  &\liff x-c_1=t(d_1-c_1)\land y-c_2=t(d_2-c_2)\\
  &\liff
  \begin{cases}
    (d_2-c_2)(x-c_1)=t(d_1-c_1)(d_2-c_2)\land{}\\
    (d_1-c_1)(y-c_2)=t(d_1-c_1)(d_2-c_2).
  \end{cases}
\end{align*}
Sonu\c c olarak
bir noktan\i n $CD$'de olmas\i\ i\c cin
noktan\i n kartezyan koordinatlar\i n\i n
\begin{equation}\label{eqn:d_2-c_2}
  (d_2-c_2)(x-c_1)=(d_1-c_1)(y-c_2)
\end{equation}
denklemini sa\u glamas\i,
gerek ve yeter bir ko\c suldur.
Ayn\i\ denklem
\begin{equation*}
  (d_2-c_2)x+(c_1-d_1)y+c_2d_1-c_1d_2=0
\end{equation*}
bi\c ciminde yaz\i labilir.

Tam tersine $a$ ve $b$'den en az birinin $0$'dan farkl\i\ oldu\u gu
herhangi
\begin{equation}\label{eqn:abc}
  ax+by+c=0
\end{equation}
denkleminin bir $(c_1,c_2)$ \c c\"oz\"um\"u vard\i r.
Bu durumda
denklem
\begin{equation*}
a(x-c_1)=-b(y-c_2)
\end{equation*}
bi\c ciminde yaz\i labilir,
dolay\i s\i yla herhangi $t$ i\c cin
\begin{equation*}
  (-bt+c_1,at+c_2)
\end{equation*}
noktas\i\ \esitligin{eqn:abc} bir \c c\"oz\"um\"ud\"ur.
E\u ger $(d_1,d_2)$, $(c_1,c_2)$'den farkl\i\ bir \c c\"oz\"um ise,
o zaman \eqref{eqn:abc},
\eqref{eqn:d_2-c_2}
olarak yaz\i labilir.

\begin{proposition}
  \eqref{eqn:abc} ve $dx+ey+f=0$
  taraf\i ndan tan\i mlanm\i\c s do\u grular\i n paralel olmas\i\ i\c cin
  \begin{equation*}
    ae=bd,
  \end{equation*}
  gerek ve yeter bir ko\c suldur.
\end{proposition}

E\u ger bir $K$ cismi verilirse,
o zaman
$K^2$ kartezyan \c carp\i m\i n\i n do\u grular\i,
$a$ ve $b$'den en az birinin $0$'dan farkl\i\ olmak \"uzere
\eqref{eqn:abc} gibi bir denklem taraf\i ndan tan\i mlas\i n.
O zaman Pappus ve Desargues Teoremleri do\u grudur,
ve \sayfada{oran}ki gibi elde edilen oranlar,
$K$ cismini olu\c sturur.

\section{Kareli denklemler}

Bir kartezyan sistemde,
koordinatlar\i\
\begin{equation*}%\label{eqn:el}
  x^2+y^2=1
\end{equation*}
denklemini sa\u glayan noktalar
standart \textbf{elipsi,}
\begin{equation*}%\label{eqn:hyp}
  x^2-y^2=1
\end{equation*}
denklemini sa\u glayan noktalar
standart \textbf{hiperbol\"u,}
\begin{equation}\label{eqn:par}
  x=y^2
\end{equation}
denklemini sa\u glayan noktalar
standart \textbf{parabol\"u}
olu\c sturur.
Verilen e\u grilerin her biri,
bir \textbf{koniktir.}
Elips ve hiperbolden her biri,
\textbf{merkezlidir,}
ve denklemi
  \begin{equation*}%\label{eqn:cc}
    x^2\pm y^2=1
  \end{equation*}
bi\c ciminde yaz\i labilir;
asl\i nda e\u grinin merkezi $O$'dur.

Koniklerin adlar\i\
daha sonra a\c c\i klanacaktir.

Pergeli Apollonius'un tan\i m\i na g\"ore,
e\u ger bir $(O,A,B)$ sisteminde
bir e\u grinin i\c cerdi\u gi her
\begin{equation*}
  O+x\cdot\vec OA+y\cdot\vec OB
\end{equation*}
noktas\i\ i\c cin e\u gri
$O+x\cdot\vec OA-y\cdot\vec OB$ noktas\i n\i\ da
i\c cerirse,
o zaman $OA$
e\u grinin bir \textbf{diametresidir.}
\"Orne\u gin $OA$, standart parabol\"un bir diametresidir,
Ayn\i\ \c sekilde hem $OA$ hem de $OB$,
standart merkezli koniklerin diametresidir.

\begin{proposition}[Apollonius \protect{\cite[\textsc i.50]{MR1660991}}]
  Herhangi $(O,A,B)$ sisteminde
  standart merkezli bir coni\u gin merkezinden ge\c cen her do\u gru
  bir diametredir.
  Asl\i nda e\u ger konik,
  koordinatlar\i\ $(a,b)$ olan bir $C$ noktas\i n\i\ i\c cerirse,
  o zaman $D$'nin
  koordinatlar\i n\i n $(b,\mp a)$ oldu\u gu
  $(O,C,D)$ sisteminin
  standart merkezli bir coni\u gi ayn\i d\i r.
\end{proposition}

\begin{proof}
  \Sekle{fig:cc} bak\i n.
  \begin{figure}
  \centering
  \psset{PointSymbol=none}
  \subfloat[Elips]{%
    \begin{pspicture}*(-5,-3.3)(5,3.7)
%      \psgrid%[subgriddiv=1]
      \newcommand{\points}{47}
      \pstGeonode[PosAngle={135,0,90}](0,0)O(4.4,1)A(0.5,3.1)B
      \ncline OA\ncline OB
      \multido{\ix=1+1}{\points}{
        \pstHomO[PointName=none,
          HomCoef=\ix\space\points\space 1 add div 360 mul cos]OA[X_\ix]
        \pstHomO[PointName=none,
          HomCoef=\ix\space\points\space 1 add div 360 mul sin]OB[Y_\ix]
\pstTranslation[PointName=none]O{X_\ix}{Y_\ix}[Z_\ix]}
\pstGenericCurve[GenCurvFirst=A,
GenCurvLast=A]{Z_}1{\points}
      \psset{PointName=none,linestyle=dotted}
      \pstTranslation AOO[A']\ncline O{A'}
      \pstTranslation BOO[B']\ncline O{B'}
      \psset{PointName=none}
      \pstHomO[HomCoef=0.6]O{A,B'}[a,b]
      \pstHomO[HomCoef=0.8]O{B,A}[c,d]
      \psset{PointName=default}
      \pstTranslation[PosAngle=45] Oac[C]
      \pstTranslation[PosAngle=-45] Obd[D]
      \psset{linestyle=dashed}
      \ncline OC\ncline OD
    \end{pspicture}}

      \subfloat[Hiperbol]{%
    \begin{pspicture}*(-5,-3.4)(5,3.4)
%      \psgrid%[subgriddiv=1]
      \newcommand{\points}{9}
      \pstGeonode[PosAngle={45,0,90}](0,0)O(2.5,-0.5)A(-0.2,1.5)B
      \ncline OA\ncline OB
      \multido{\ix=1+1}{\points}{
        \pstHomO[PointName=none,
          HomCoef=0.5 \ix\space 1 sub \points\space 1 sub div sub 4 mul
        ]OB[Y_\ix]
        \pstHomO[PointName=none,
          HomCoef=0.5 \ix\space 1 sub \points\space 1 sub div sub 4 mul
          dup mul 1 add sqrt
        ]OA[X_\ix]
\pstTranslation[PointName=none]O{X_\ix}{Y_\ix}[Z_\ix]
        \pstHomO[PointName=none,
          HomCoef=0.5 \ix\space 1 sub \points\space 1 sub div sub 4 mul
          dup mul 1 add sqrt neg
        ]OA[X_\ix]
\pstTranslation[PointName=none]O{X_\ix}{Y_\ix}[W_\ix]}
\pstGenericCurve{Z_}1{\points}
\pstGenericCurve{W_}1{\points}
      \psset{PointName=none,linestyle=dotted}
      \pstTranslation AOO\ncline O{O'}
      \pstTranslation BOO\ncline O{O'}
      \psset{PointName=none}
      \pstHomO[HomCoef=-1.25]O{A,B}[a,b]
      \pstHomO[HomCoef=0.75]O{B,A}[c,d]
      \psset{PointName=default}
      \pstTranslation[PosAngle=-150] Oac[C]
      \pstTranslation[PosAngle=-90] Obd[D]
      \psset{linestyle=dashed}
      \ncline OC\ncline OD
    \end{pspicture}}
\caption{Merkezli konikler}
  \label{fig:cc}
\end{figure}
Varsay\i ma g\"ore
  $a^2\pm b^2=1$ ve
  \begin{gather*}
    \vec OC=a\cdot\vec OA+b\cdot\vec OB,\\
    \vec OD=b\cdot\vec OA\mp a\cdot\vec OB.
  \end{gather*}
  \c Simdi
  \begin{equation*}
    \vec OP=s\cdot\vec OC+t\cdot\vec OD
  \end{equation*}
  olsun.  O zaman
  \begin{align*}
    \vec OP
    &=s\cdot(a\cdot\vec OA+b\cdot\vec OB)
    +t\cdot(b\cdot\vec OA\mp a\cdot\vec OB)\\
    &=(sa+tb)\cdot\vec OA+(sb\mp ta)\cdot\vec OB.
  \end{align*}
  Ayr\i ca
  \begin{align*}
    &\phantom{{}={}}(sa+tb)^2\pm(sb\mp ta)^2\\
    &=s^2\cdot(a^2\pm b^2)+t^2(b^2\pm a^2)\\
    &=s^2\pm t^2.\qedhere
  \end{align*}
\end{proof}

\begin{proposition}[Apollonius \protect{\cite[\textsc i.49]{MR1660991}}]
    Bir $(O,\bm a,\bm b)$ sisteminde
  \begin{enumerate}[1)]
  \item
    standart parabol,
    $0$ olmayan her $t$ oran\i\ i\c cin
    \begin{equation*}
      (O,t^2\cdot\bm a,t\cdot\bm b)
    \end{equation*}
    sisteminin
    standart parabol\"u ile ayn\i d\i r;
  \item
    standart parabol,
    \begin{equation*}
      (O+\bm a+\bm b,\bm a,-2\bm a-\bm b)
    \end{equation*}
         sisteminin
    standart parabol\"u ile ayn\i d\i r;
  \item
  standart parabol\"un diametrisine paralel olan
  her do\u gru bir diametredir.
  \end{enumerate}
\end{proposition}

\begin{proof}
  \begin{asparaenum}
    \item
  \Sekle{fig:par} bak\i n\lips
\begin{figure}
  \centering
  \psset{PointSymbol=none}
%  \subfloat[]{
    \begin{pspicture}*(-3.2,-3)(6.8,3)
%      \psgrid[subgriddiv=1]
      \newcommand{\points}{9}
      \pstGeonode[PosAngle={-135,0,180}](0,0)O(2.5,-0.5)A(-1,2.8)B
      \ncline OA\ncline OB
      \multido{\ix=1+1}{\points}{
        \pstHomO[PointName=none,
          HomCoef=0.5 \ix\space 1 sub \points\space 1 sub div sub 4 mul
        ]OB[Y_\ix]
        \pstHomO[PointName=none,
          HomCoef=0.5 \ix\space 1 sub \points\space 1 sub div sub 4 mul
          dup mul
        ]OA[X_\ix]
\pstTranslation[PointName=none]O{X_\ix}{Y_\ix}[Z_\ix]}
      \pstGenericCurve{Z_}1{\points}
      \pstHomO[HomCoef=0.64,PosAngle=90]OA[C]
      \pstHomO[HomCoef=-0.8,PosAngle=180]OB[D]
      \psset{linestyle=dashed}
      \ncline OD
      \psset{PointName=none,linestyle=dotted}
      \pstTranslation OAB[U]
      \pstTranslation OCD[V]
      \ncline BU\ncline UA\ncline DV\ncline VC
    \end{pspicture}
    %}
    \caption{Parabol}
    \label{fig:par}
\end{figure}
\item
  \Sekle{fig:parab} bak\i n.
\begin{figure}
  \centering
  \psset{PointSymbol=none}
%  \subfloat[]{
        \begin{pspicture}(-3.2,-3)(6.8,4)
%      \psgrid[subgriddiv=1]
      \newcommand{\points}{9}
      \pstGeonode[PosAngle={-135,0,90}](0,0)O(2.5,-0.5)A(-1,2.8)B
      \ncline OA\ncline OB
      \multido{\ix=1+1}{\points}{
        \pstHomO[PointName=none,
          HomCoef=0.67 \ix\space 1 sub \points\space 1 sub div sub 3 mul
        ]OB[Y_\ix]
        \pstHomO[PointName=none,
          HomCoef=0.67 \ix\space 1 sub \points\space 1 sub div sub 3 mul
          dup mul
        ]OA[X_\ix]
\pstTranslation[PointName=none]O{X_\ix}{Y_\ix}[Z_\ix]}
\pstGenericCurve{Z_}1{\points}
      \psset{PointName=none,linestyle=dotted}
      \pstTranslation BOO\ncline O{O'}
      \psset{PointName=default}
      \pstTranslation[PosAngle=90] OAB[U]
      \pstTranslation OAU[C]
      \pstTranslation[PosAngle=180] AOO[D]
      \pstTranslation[PointName=none] DUU
      \ncline DO\ncline OC\ncline BU\ncline U{U'}\ncline UA
      \psset{linestyle=dashed}
      \ncline UD\ncline UC
        \end{pspicture}
        %}

    \caption{Parabol}
  \label{fig:parab}
\end{figure}
Orada
\begin{align*}
  O+\bm a&=A,&
  O+\bm b&=B,
\end{align*}
ve ayr\i ca
\begin{align*}
  O+\bm a+\bm b&=U,&
  U+\bm a&=C,&
  U-2\bm a-\bm b&=D.
\end{align*}
  E\u ger
  \begin{equation*}
    \vec UP=s\cdot\vec UC+t\cdot\vec UD
  \end{equation*}
  ise, o zaman
  \begin{align*}
    \vec UP
    &=s\cdot\vec OA-t\cdot(2\cdot\vec OA+\vec OB)\\
    &=(s-2t)\cdot\vec OA-t\cdot\vec OB,
  \end{align*}
  dolay\i s\i yla 
  \begin{align*}
    \vec OP
    &=\vec UP+\vec OU\\
    &=(s-2t)\cdot\vec OA-t\cdot\vec OB+\vec OA+\vec OB\\
    &=(s-2t+1)\cdot\vec OA-(t-1)\vec OB.
  \end{align*}
  Ayr\i ca
  \begin{align*}
    &\phantom{{}={}}
    s-2t+1-(t-1)^2\\
    &=s-2t+1-t^2+2t-1\\
    &=2-t^2.
  \end{align*}
\item
  \lips\qedhere
  \end{asparaenum}
\end{proof}

E\u ger $a$, $b$, ve $c$ oran\i ndan
en az biri $0$ de\u gilse,
\begin{equation*}
  ax^2+bxy+cy^2+dx+ey+f=0
\end{equation*}
denkleminin bir koni\u gi tan\i mlad\i\u g\i n\i\
g\"osterece\u giz.

\begin{proposition}
  Katsay\i lar\i\ oran olan
  herhangi   $f(x,y)$ polinomu i\c cin
  bir $(O,\bm a,\bm b)$ sisteminde
  \begin{equation*}
    f(x,y)=0
  \end{equation*}
  taraf\i ndan tan\i mlanm\i\c s e\u gri,
  \begin{enumerate}[1)]
  \item
    $(O-c\cdot\bm a,\bm a,\bm b)$ sisteminde
    $f(x+c,y)=0$,
  \item
    $(O-c\cdot\bm b,\bm a,\bm b)$ sisteminde
    $f(x,y+c)=0$,
  \item
    $(O,c\inv\cdot\bm a,\bm b)$ sisteminde
    $f(cx,y)=0$,
  \item
    $(O,\bm a,c\inv\cdot\bm b)$ sisteminde
    $f(x,cy)=0$
  \end{enumerate}
  taraf\i ndan tan\i mlanm\i\c s e\u gri ile ayn\i d\i r.
\end{proposition}

\begin{proof}
  \begin{align*}
    O+x\cdot\bm a+y\cdot\bm b
    &=(O-c\cdot\bm a)+(x+c)\cdot\bm a+y\cdot\bm b\\
    &=(O-c\cdot\bm b)+x\cdot\bm a+(y+c)\cdot\bm b\\
    &=O+cx\cdot(c\inv\cdot\bm a)+y\cdot\bm b\\
    &=O+x\cdot\bm a+cy\cdot(c\inv\cdot\bm b).\qedhere
  \end{align*}
  \begin{comment}
    
    \begin{gather*}
      O+c\cdot a+(x\cdot\bm a+y\cdot\bm b)
      =O+(x+c)\cdot\bm a+y\cdot\bm b,\\
      O+c\cdot b+(x\cdot\bm a+y\cdot\bm b)
      =O+x\cdot\bm a+(y+c)\cdot\bm b,\\
      O+x\cdot(c\cdot\bm a)+y\cdot\bm b
      =O+cx\cdot\bm a+y\cdot\bm b,\\
      O+x\cdot\bm a+y\cdot(c\cdot\bm b)
      =O+x\cdot\bm a+cy\cdot\bm b.\qedhere
    \end{gather*}
  \end{comment}
\end{proof}

\"Onermenin her par\c cas\i n\i n ba\c ska bir bi\c cimi vard\i r.
\"Orne\u gin
\begin{itemize}
\item
  $(O,\bm a,\bm b)$ ve $(O+c\cdot\bm a,\bm a,\bm b)$ sisteminde
  s\i ras\i yla
  \begin{align*}
    f(x+c,y)&=0,&f(x,y)&=0
  \end{align*}
  taraf\i ndan tan\i mlanm\i\c s e\u griler birbiriyle ayn\i d\i r;
\item
  $(O,\bm a,\bm b)$ ve $(O,+c\cdot\bm a,\bm b)$ sisteminde
  s\i ras\i yla
  \begin{align*}
    f(cx,y)&=0,&f(x,y)&=0
  \end{align*}
  taraf\i ndan tan\i mlanm\i\c s e\u griler birbiriyle ayn\i d\i r.  
\end{itemize}
\begin{comment}
  
  \begin{proposition}
    Bir kartezyan sistemde
    \begin{equation*}%\label{eqn:abcdef}
      ax^2+bxy+cy^2+dx+ey+f=0
    \end{equation*}
    taraf\i ndan tan\i mlanm\i\c s \c sekil
    ya en \c cok iki nokta i\c cerir,
    ya da b\"ut\"un d\"uzlemidir,
    ya da bir do\u grudur,
    ya da bir sistemin standart bir koni\u gidir.
  \end{proposition}

  \begin{proof}
    Verilen sistem $(O,\bm a,\bm b)$ olsun.
    Birka\c c tane durum vard\i r.
    \begin{asparaenum}
    \item
      \fbox{$a$, $b$, ve $c$'den her biri $0$} olsun.
      O zaman iddia do\u grudur.
    \item
      \fbox{$b\neq0$ ama $a=c=0$} olsun.
      \begin{multline*}
        b(x+y)(x-y)+d(x+y)+e(x-y)+f\\
        =bx^2-by^2+(d+e)x+(d-e)y+f
      \end{multline*}
      ve
      \begin{align*}
        (x+y)\cdot\bm a+(x-y)\cdot\bm b
        =
        x\cdot(\bm a+\bm b)+y\cdot(\bm a-\bm b)
      \end{align*}
      oldu\u gundan e\u grimiz
      $(O,\bm a+\bm b,\bm a-\bm b)$
      sisteminde
      \begin{equation*}
        bx^2-by^2+(d+e)x+(d-e)y+f=0
      \end{equation*}
      taraf\i ndan tan\i mlan\i r.
    \item
      \fbox{$a\neq0$} olsun.
      O zaman
      %    \fbox{$a=1$} varsay\i labilir.  Ayr\i ca
      \begin{multline*}
        ax^2+bxy+cy^2+dx+ey+f\\
        =a\left(x+\frac{by}{2a}\right)^2
        +\left(c-\frac{b^2}{4a}\right)y^2+dx+ey+f\\
        =a\left(x+\frac{by}{2a}\right)^2
        +\left(c-\frac{b^2}{4a}\right)y^2\\
        +d\left(x+\frac{by}{2a}\right)+\left(e-\frac{bd}{2a}\right)y+f.
      \end{multline*}
      Bu durumda
      \begin{equation*}
        \left(x+\frac{by}{2a}\right)\cdot\bm a
        +y\cdot\bm b
        =x\cdot\bm a+y\cdot\left(\frac{b}{2a}\cdot a+\bm b\right)
      \end{equation*}
      oldu\u gundan
      e\u grimiz
      $(U,\bm a,(b/2a)\cdot a+\bm b)$
      sisteminde
      \begin{equation*}
        ax^2+\left(c-\frac{b^2}{4a}\right)y^2
        +dx+\left(e-\frac{bd}{2a}\right)y+f=0
      \end{equation*}
      taraf\i ndan tan\i mlan\i r.
    \item
      \fbox{$c\neq0$} olsun.
      O zaman e\u grimiz $(O,\bm b,\bm a)$ sisteminde
      \begin{equation*}
        cx^2+bxy+ay^2+ex+dy+f=0    
      \end{equation*}
      taraf\i ndan tan\i mlan\i r.
    \end{asparaenum}
    \c Simdi $(O,\bm a,\bm b)$ sisteminde e\u grimizin
    \begin{equation*}
      x^2+cy^2+dx+ey+f=0
    \end{equation*}
    taraf\i ndan tan\i mland\i\u g\i n\i\ varsayabiliriz.
    \"U\c c kalan durum vard\i r.
    \begin{asparaenum}
    \item
      \fbox{$c=e=0$} olsun.  O zaman
      \begin{equation*}
        x^2+dx+f
        =\left(x+\frac d2\right)^2+f-\frac{d^2}4
      \end{equation*}
      oldu\u gundan
      e\u grimiz $U=O+(d/2)\cdot\bm a$ olmak \"uzere
      $(U,\bm a,\bm b)$ sisteminde
      \begin{equation*}
        x^2+f-\frac{d^2}4=0
      \end{equation*}
      taraf\i ndan tan\i mlan\i r \c c\"unk\"u
      \begin{equation*}
        \vec UP=x\cdot\bm a+y\cdot b
        \liff
        \begin{aligned}[t]
          \vec OP
          &=\vec OU+x\cdot\bm a+y\cdot b\\
          &=\left(x+\frac d2\right)\cdot\bm a+y\cdot\bm b.
        \end{aligned}
      \end{equation*}
      Bu durumda ``e\u grimiz''
      en \c cok iki nokta i\c cerir.
    \item
      \fbox{$c=0$ ama $e\neq0$} olsun.
      O zaman
      \begin{equation*}
        x^2+dx+ey+f
        =\left(x+\frac d2\right)^2+e\left(y+\frac fe-\frac{d^2}{4e}\right)
      \end{equation*}
      oldu\u gundan
      e\u grimiz
      \begin{equation*}
        U=O+\frac d2\cdot\bm a+\left(\frac fe-\frac{d^2}{4e}\right)\cdot\bm b
      \end{equation*}
      olmak \"uzere
      $(U,\bm a,\bm b)$ sisteminde
      \begin{equation*}
        x^2+ey=0
      \end{equation*}
      taraf\i ndan
      ve $(U,e\cdot\bm b,\bm a)$ sisteminde \eqref{eqn:par}
      taraf\i ndan
      tan\i mlan\i r.
    \item
      \fbox{$c\neq0$ and $e\neq0$} olsun.
      O zaman
      \begin{multline*}
        x^2+cy^2+dx+ey+f\\
        =
        \left(x+\frac d2\right)^2+c\left(y+\frac e{2c}\right)^2
        +f-\frac{d^2}4-\frac{e^2}{4c}
      \end{multline*}
      oldu\u gundan
      e\u grimiz
      \begin{equation*}
        U=O+\frac d2\cdot\bm a+\frac e{2c}\cdot\bm b
      \end{equation*}
      olmak \"uzere
      $(U,\bm a,\bm b)$ sisteminde
      \begin{equation*}
        x^2+cy^2=
        \frac{d^2}4+\frac{e^2}{4c}-f
      \end{equation*}
      taraf\i ndan
      tan\i mlan\i r\lips\qedhere
    \end{asparaenum}
  \end{proof}
\end{comment}
\chapter{Uzunluklar}


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