\documentclass[%
version=last,%
paper=a5,
fontsize=10pt,%
%headings=small,%
bibliography=totoc,%
%abstract=true,%
twoside,%
%titlepage=true,%
%open=any,%
cleardoublepage=empty,%
%parskip=half-,%
%draft=true,%
%DIV=classic,%
DIV=11,%
headinclude=true,%
%titlepage=false,%
pagesize]%
{scrartcl}
%{scrbook}
%{scrreprt}

%\renewcommand{\thechapter}{\Roman{chapter}}

%\usepackage[notref,notcite]{showkeys}

%\raggedright
\usepackage{hfoldsty,url,verbatim}
\usepackage{multicol}
\setlength{\columnseprule}{0.5pt}
\usepackage{moredefs,lips}
\usepackage{pstricks,pst-plot,pst-3d,pst-node}
\psset{plotpoints=200}

\usepackage[polutonikogreek,turkish]{babel}

\usepackage{gfsporson}
\newcommand{\gr}[1]{\foreignlanguage{polutonikogreek}{\textporson{#1}}} %  
                                                     %Greek text

\usepackage%[12hr]
{datetime}  % I formerly used code adapted from the
%\usdate                          % LaTeX Companion (2d ed), p. 871  

\usepackage[headsepline]{scrpage2}
\pagestyle{scrheadings}
\clearscrheadings
\ohead{\pagemark}
%\ihead{\headmark}
%\rehead{\leftmark}
\rehead{Analitik Geometri \"Ozeti}
\lohead{\today, \currenttime}
%\cehead{Analytic Geometry}

\usepackage[neverdecrease]{paralist}

\usepackage{relsize}

\renewenvironment{quote}{\begin{list}{}
{\relscale{.90}\setlength{\leftmargin}{0.05\textwidth}
\setlength{\rightmargin}{\leftmargin}}
\item\relax}
{\end{list}}

%%%%%%%%%%%%%%%%% Turkish numeration %%%%%%%%%%%%%%%%%

\usepackage{ifthen}
\usepackage{calc} % part of tools
\newcounter{rfp}\newcounter{ones}\newcounter{tens}
\usepackage{refcount} % part of oberdiek bundle

\newcommand{\sayfanumaraya}[1]{%
\setcounterpageref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
   {\value{ones}=6}%
   {ya}%
   {\ifthenelse%
       {\value{ones}=9}%
       {a}%
       {\ifthenelse%
           {\value{ones}=2\or\value{ones}=7}%
           {ye}%
           {\ifthenelse%
               {\value{ones}=0}%
               {\ifthenelse%
                   {\value{tens}=2\or\value{tens}=5}
                   {ye}
                   {e}}
               {e}}}}}
\newcommand{\sayfaya}[1]{sayfa \sayfanumaraya{#1}}

\newcommand{\sayfanumarada}[1]{%
%\setcounter{rfp}{\number\numexpr\getpagerefnumber{#1}}%
\setcounterpageref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
   {\value{ones}=6\or\value{ones}=9}%
   {da}%
   {\ifthenelse%
       {\value{ones}=3\or\value{ones}=4\or\value{ones}=5}%
       {te}%
       {\ifthenelse%
           {\value{ones}=0}%
           {\ifthenelse%
               {\value{tens}=7}%
               {te}
               {\ifthenelse%
                   {\value{tens}=4\or\value{tens}=6}%
                   {ta}
                   {\ifthenelse%
                       {\value{tens}=1\or\value{tens}=3\or\value{tens}=9}%
                       {da}%
                       {de}}}}%
           {de}}}}
\newcommand{\sayfada}[1]{sayfa \sayfanumarada{#1}}
\newcommand{\Sayfada}[1]{Sayfa \sayfanumarada{#1}}

\newcommand{\numaraya}[1]{%
\setcounterref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
   {\value{ones}=6}%
   {ya}%
   {\ifthenelse%
       {\value{ones}=9}%
       {a}%
       {\ifthenelse%
           {\value{ones}=2\or\value{ones}=7}%
           {ye}%
           {\ifthenelse%
               {\value{ones}=0}%
               {\ifthenelse%
                   {\value{tens}=2\or\value{tens}=5}
                   {ye}
                   {e}}
               {e}}}}}


\newcommand{\numarada}[1]{%
%\setcounter{rfp}{\number\numexpr\getpagerefnumber{#1}}%
\setcounterref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
   {\value{ones}=6\or\value{ones}=9}%
   {da}%
   {\ifthenelse%
       {\value{ones}=3\or\value{ones}=4\or\value{ones}=5}%
       {te}%
       {\ifthenelse%
           {\value{ones}=0}%
           {\ifthenelse%
               {\value{tens}=7}%
               {te}
               {\ifthenelse%
                   {\value{tens}=4\or\value{tens}=6}%
                   {ta}
                   {\ifthenelse%
                       {\value{tens}=1\or\value{tens}=3\or\value{tens}=9}%
                       {da}%
                       {de}}}}%
           {de}}}}

\newcommand{\numarayi}[1]{%
\setcounterref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
  {\value{ones}=1\or\value{ones}=5\or\value{ones}=8}%
  {i}%
  {\ifthenelse%
     {\value{ones}=2\or\value{ones}=7}%
     {yi}%
     {\ifthenelse%
        {\value{ones}=3\or\value{ones}=4}%
        {\"u}%
        {\ifthenelse%
           {\value{ones}=6}%
           {y\i}%
           {\ifthenelse%
              {\value{ones}=9}%
              {u}%
              {\ifthenelse%
                 {\value{tens}=7\or\value{tens}=8}%
                 {i}%
                 {\ifthenelse%
                    {\value{tens}=2\or\value{tens}=5}%
                    {yi}%
                    {\ifthenelse%
                       {\value{tens}=1\or\value{tens}=3}%
                       {u}%
                       {\ifthenelse%
                          {\value{tens}=4\or\value{tens}=6\or\value{tens}=9}%
                          {\i}
                          {\"u}}}}}}}}}}

\newcommand{\numaranin}[1]{%
\setcounterref{rfp}{#1}%
\setcounter{ones}{\value{rfp}-(\value{rfp}/10)*10}%
\setcounter{tens}{\value{rfp}/10-(\value{rfp}/100)*10}%
\arabic{rfp}'%
\ifthenelse%
  {\value{ones}=1\or\value{ones}=5\or\value{ones}=8}%
  {in}%
  {\ifthenelse%
     {\value{ones}=2\or\value{ones}=7}%
     {nin}%
     {\ifthenelse%
        {\value{ones}=3\or\value{ones}=4}%
        {\"un}%
        {\ifthenelse%
           {\value{ones}=6}%
           {n\i n}%
           {\ifthenelse%
              {\value{ones}=9}%
              {un}%
              {\ifthenelse%
                 {\value{tens}=7\or\value{tens}=8}%
                 {in}%
                 {\ifthenelse%
                    {\value{tens}=2\or\value{tens}=5}%
                    {nin}%
                    {\ifthenelse%
                       {\value{tens}=1\or\value{tens}=3}%
                       {un}%
                       {\ifthenelse%
                          {\value{tens}=4\or\value{tens}=6\or\value{tens}=9}%
                          {\i n}
                          {\"u}}}}}}}}}}

\newcommand{\Teorem}[1]{Teorem \ref{#1}}
\newcommand{\Teoreme}[1]{Teorem \numaraya{#1}}
\newcommand{\Teoremde}[1]{Teorem \numarada{#1}}
\newcommand{\Teoremi}[1]{Teorem \numarayi{#1}}
\newcommand{\Teoremin}[1]{Teorem \numaranin{#1}}

\usepackage{chngcntr}
\counterwithout{figure}{section}
\newcommand{\Sekil}[1]{\c Sekil \ref{#1}}
\newcommand{\Sekli}[1]{\c Sekil \numarayi{#1}}
\newcommand{\Seklin}[1]{\c Sekil \numaranin{#1}}
\newcommand{\Sekle}[1]{\c Sekil \numaraya{#1}}
\newcommand{\Sekilde}[1]{\c Sekil \numarada{#1}}

%%%%%%% END Turkish numeration %%%%%%%%%%%%%%%%%%%

\usepackage{amsmath,amssymb,amsthm,upgreek,mathrsfs}
\allowdisplaybreaks
\newcommand{\N}{\mathbb N}
\newcommand{\Q}{\mathbb Q}
\newcommand{\Qp}{\Q^+}
%\newcommand{\comp}{\mathbin{\&}}
\newcommand{\comp}[4]{(#1:#2)\mathbin{\&}(#3:#4)}
\newcommand{\ydp}[1]{\overrightarrow{#1}} % "yönlü doğru parçası"
\newcommand{\ise}{\;\mathrel{\text{ise}}\;}
\newcommand{\abs}[1]{\lvert#1\rvert}
\renewcommand{\emptyset}{\varnothing}
\renewcommand{\leq}{\leqslant}
\renewcommand{\geq}{\geqslant}

\renewcommand{\theequation}{\fnsymbol{equation}}
\newtheorem{theorem}{Teorem}
%\newtheorem{problem}{Problem}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Tan\i m}
%\newtheorem{exercise}{Al\i\c st\i rma}
%\renewcommand{\theexercise}{\Roman{exercise}}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{S\"oz}

\newcommand{\eup}[2][i]{\textbf{[\textsc{#1}.#2]}} % "Euclid's Proposition"

\begin{document}
\title{Analitik Geometri \"Ozeti}
\author{David Pierce}
\date{\today, \currenttime}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{http://mat.msgsu.edu.tr/~dpierce/}}
\maketitle
%\begin{multicols}2
\tableofcontents
%  \end{multicols}
\listoffigures

\section{Denklik ba\u g\i nt\i lar\i}

\begin{definition}
\textbf{Do\u gal say\i lar,} $1$, $2$, $3$, .\lips Bunlar 
$\N$ k\"umesini olu\c sturur:
  \begin{equation*}
  \N=\{1,2,3,\dots\}.
  \end{equation*}
(Bu ifadede $=$ i\c sareti \emph{ayn\i l\i\u g\i} g\"osterir,
yani $\N$ ve $\{1,2,3,\dots\}$ ayn\i\ k\"umedir.)
\end{definition}

\begin{remark}
\.Ilkokuldan bildi\u gimiz gibi 
iki do\u gal say\i\ toplanabilir ve \c carp\i labilir,
ve do\u gal say\i lar s\i ralan\i r.
\end{remark}

\begin{definition}
S\i ral\i\ ikililer,
\begin{equation*}
  (a,b)=(x,y)\iff a=x\And b=y
\end{equation*}
\"ozelli\u gini sa\u glar.
T\"um $A$ ve $B$ k\"umeleri i\c cin
\begin{equation*}
A\times B=\{(x,y)\colon x\in A\And y\in B\}.
\end{equation*}
\end{definition}



\begin{definition}
Bir $A$ k\"umesinin yans\i mal\i, simetrik, ve ge\c ci\c sli 
iki-konumlu $R$ ba\u g\i nt\i s\i,
$A$ k\"umesinin \textbf{denklik ba\u g\i nt\i s\i d\i r.}
$A$ k\"umesinin $b$ eleman\i n\i n
\textbf{$R$ ba\u g\i nt\i s\i na g\"ore denklik s\i n\i f\i}
veya \textbf{$R$-s\i n\i f\i,}
\begin{equation*}
  \{x\in A\colon x\mathrel Rb\}
\end{equation*}
k\"umesidir.%%%%%
\footnote{Bu k\"umeye ``denklik s\i n\i f\i'' demek, bir gelenektir.
K\"umeler kuram\i nda her k\"ume bir s\i n\i ft\i r,
ama her s\i n\i f k\"ume de\u gildir.
\"Orne\u gin $\{x\colon x\notin x\}$ s\i n\i f\i, k\"ume olamaz.}
%%%%%%%%%%%%%%%%
Bu denklik s\i n\i f\i\ i\c cin
\begin{equation*}
  [b]
\end{equation*}
k\i saltmas\i\ kullan\i labilir
(ama \Teoremde{thm:eq-cl}n sonra kullan\i lmayacak).
\end{definition}

\begin{theorem}\label{thm:eq-cl}
  $R$, $A$ k\"umesinin denklik ba\u g\i nt\i s\i\ olsun, 
ve $b\in A$, $c\in A$ olsun.
%\begin{equation*}
%  \text{Ya }\qquad[b]=[c]\qquad\text{ ya da }\qquad[b]\cap[c]=\emptyset.
%\end{equation*}
  Ya
\begin{equation*}
[b]=[c]
%  \{x\in A\colon x\mathrel Rb\}=  \{x\in A\colon x\mathrel Rc\}
\end{equation*}
ya da
\begin{equation*}
[b]\cap[c]=\emptyset.
%  \{x\in A\colon x\mathrel Rb\}\cap \{x\in A\colon x\mathrel Rc\}=\emptyset.
\end{equation*}
\end{theorem}

\begin{definition}
  $\N\times\N$ k\"umesinin $\approx$ ba\u g\i nt\i s\i,
  \begin{equation*}
    (k,\ell)\approx(x,y)\iff ky=\ell x
  \end{equation*}
tan\i m\i n\i\ sa\u glas\i n.
\end{definition}

\begin{theorem}
  $\N\times\N$ k\"umesinin $\approx$ ba\u g\i nt\i s\i,
denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{definition}
$\N\times\N$ k\"umesinin $(k,\ell)$ eleman\i n\i n $\approx$-s\i n\i f\i,
  \begin{equation*}
    \frac k{\ell}
  \end{equation*}
veya $k/\ell$ \textbf{pozitif kesirli say\i s\i d\i r.}
Pozitif kesirli say\i lar,
\begin{equation*}
  \Qp
\end{equation*}
k\"umesini olu\c sturur.
\end{definition}

\begin{theorem}
  A\c sa\u g\i daki e\c sitlikler,
$\Qp$ k\"umesinin toplama ve \c carpma i\c slemleri i\c cin iyi tan\i md\i r:
  \begin{align*}
    \frac k\ell+\frac mn&=\frac{kn+\ell m}{\ell n},&\frac k\ell\cdot\frac mn&=\frac{am}{\ell n}.
  \end{align*}
Yani
\begin{equation*}
  \frac k\ell=\frac{k'}{\ell'}\And\frac mn=\frac{m'}{n'}\implies
\frac{kn+\ell m}{\ell n}=\frac{k'n'+\ell'm'}{\ell'n'}\And
\frac{km}{\ell n}=\frac{k'm'}{\ell'n'}.
\end{equation*}
Ayr\i ca $\Qp$ a\c sa\u g\i daki tan\i ma g\"ore s\i ralan\i r:
\begin{equation*}
  \frac k\ell<\frac mn\iff kn<\ell m.
\end{equation*}
\end{theorem}

\section{Uzunluklar}

\begin{definition}
  \"Oklid'deki gibi bir \textbf{do\u grunun} u\c c noktalar\i\ vard\i r,
ve \c cak\i\c san do\u grular \textbf{e\c sittir.}
(\"Ozel olarak do\u grular i\c cin e\c sitlik ayn\i l\i k de\u gildir.)
\end{definition}

\begin{theorem}
  Do\u grular\i n e\c sitli\u gi, denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{definition}
  Do\u grunun e\c sitlik s\i n\i f\i, do\u grunun \textbf{uzunlu\u gudur.}
K\"u\c c\"uk $a$, $b$, $c$, \lips Latin harfleri uzunluk g\"osterecek.
E\u ger bir $AB$ do\u grusunun uzunlu\u gu $c$ ise
\begin{equation*}
  AB=c
\end{equation*}
ifadesini yazar\i z.%%%%%
\footnote{Bu uygulama Descartes'\i n 1637 \emph{Geometri} kitab\i ndan gelir.}
\end{definition}

\begin{theorem}
\.Iki uzunluk toplanabilir, 
ve bir kesirli say\i\ bir uzunlu\u gu \c co\u galtabilir.
Toplama de\u gi\c smeli ve birle\c smelidir,
ve \c co\u galtma toplama \"uzerine da\u g\i l\i r.
E\u ger $a<b$ ise
\begin{equation*}
  a+x=b
\end{equation*}
denklemi \c c\"oz\"ulebilir.
\end{theorem}

\begin{theorem}\label{thm:prop}
E\u ger $CD=BK$ ve $GH=FL$ ise,
ve \Sekilde{fig:ABCDEF}ki gibi
\begin{figure}[ht]
\mbox{}\hfill
\psset{unit=4mm}
  \begin{pspicture}(0,-0.83)(4,3)
    \pspolygon(0,0)(4,3)(4,0)
\uput[d](0,0){$A$}
\uput[d](4,0){$B$}
\uput[u](4,3){$K$}
%\uput[ur](2,1.5){$e$}
%\uput[l](0,1.5){$a$}
%\uput[d](2,0){$b$}
  \end{pspicture}
\hfill\psset{unit=5mm}
  \begin{pspicture}(0,-0.5)(4,3.5)
    \pspolygon(0,0)(4,3)(4,0)
\uput[d](0,0){$E$}
\uput[d](4,0){$F$}
\uput[u](4,3){$L$}
%\uput[ur](2,1.5){$f$}
%\uput[l](0,1.5){$c$}
%\uput[d](2,0){$d$}
  \end{pspicture}
\hfill\mbox{}
  \caption{Benzer dik \"u\c cgenler}\label{fig:ABCDEF}
  
\end{figure}
$ABK$ ve $EFL$ \"u\c cgenlerinde $\angle ABK$ ve $\angle EFL$ dik
ve $\angle BAK=\angle FEL$ ise,
o zaman
\begin{equation*}
  (AB,CD)\mathrel{\mathscr R}(EF,GH)
\end{equation*}
olsun.
Bu $\mathscr R$ ba\u g\i nt\i s\i,
bir denklik ba\u g\i nt\i s\i d\i r.
Ayr\i ca sadece do\u grular\i n uzunlu\u guna ba\u gl\i d\i r.
\end{theorem}

\begin{definition}
E\u ger \Teoremde{thm:prop}ki gibi 
$(AB,CD)\mathrel{\mathscr R}(EF,GH)$ ise
$AB$, $CD$, $EF$, ve $GH$ do\u grular\i\ \textbf{orant\i l\i d\i r,}
ve
\begin{equation*}
  AB:CD::EF:GH
\end{equation*}
\textbf{orant\i s\i n\i} yazar\i z; ayr\i ca
$AB=a$, $CD=b$, $EF=c$, ve $GH=d$ ise
\begin{equation*}
  a:b::c:d
\end{equation*}
ifadesini yazar\i z.
Buradaki $AB:CD$ ve $a:b$ ifadeleri,
$(AB,CD)$ ve $(a,b)$ 
s\i ral\i\ ikililerinin denklik s\i n\i f\i n\i\ g\"osterir;
bu s\i n\i f, bir \textbf{orand\i r.}
Bu durumda $::$ simgesi, oranlar\i n \emph{ayn\i l\i\u g\i n\i} g\"osterir.
\end{definition}

\begin{theorem}
  \Sekilde{fig:parallel}
  \begin{figure}[ht]
    \centering
\psset{unit=3mm}
    \begin{pspicture}(0,-1)(12,10)
      \pspolygon(0,0)(12,0)(12,9)
\psline(4,0)(12,6)
%\psline[linestyle=dotted](6,0)(12,6)
\uput[d](0,0){$A$}
\uput[d](12,0){$B$}
\uput[u](12,9){$C$}
\uput[d](4,0){$D$}
\uput[r](12,6){$E$}
%\uput{3pt}[100](6,0){$F$}
%\uput[r](12,3){$a$}
%\uput[d](8,0){$b$}
%\uput[u](9,0){$c$}
%\uput[r](12,7.5){$d$}
%\uput[d](2,0){$e$}
    \end{pspicture}
    \caption{Paralellik ve orant\i l\i l\i k}\label{fig:parallel}
    
  \end{figure}
$ABC$ a\c c\i s\i\ dik ise
  \begin{equation*}
    AB:BC::DB:BE\iff AC\parallel DE.
  \end{equation*}
\end{theorem}

\begin{theorem}
  $a:b::a:c\implies b=c$.
\end{theorem}

\begin{theorem}\label{thm:axxb}
  Her $a:x::x:b$ orant\i s\i\ \c c\"oz\"ulebilir.
\end{theorem}

\begin{theorem}
  $a:b::d:e\And b:c::e:f\implies a:c::d:f$.
\end{theorem}

\begin{definition}
$c:d::b:e$ ise
  \begin{equation*}
    \comp abcd::b:e,
  \end{equation*}
ve $b:e$ oran\i, $a:b$ ve $c:d$ oranlar\i n\i n \textbf{bile\c skesidir.}
\end{definition}

\section{Alanlar}

\begin{theorem}
Ayn\i\ geni\c sli\u gi ve y\"uksekli\u gi olan dikd\"ortgenler e\c sittir.
\Sekilde{fig:ab=cx}
\begin{figure}[ht]
  \centering
\psset{unit=2mm}
  \begin{pspicture}(0,-1.5)(21,16.5)
    \pspolygon(0,0)(21,0)(21,14)(0,14)
\psline(0,10)(21,10)
\psline(6,0)(6,14)
\psline[linestyle=dotted](0,14)(21,0)
\uput[dl](0,0){$A$}
\uput[d](6,0){$B$}
\uput[dl](6,10){$C$}
\uput[l](0,10){$D$}
\uput[r](21,10){$E$}
\uput[dr](21,0){$F$}
\uput[ur](21,14){$H$}
\uput[ul](0,14){$G$}
\uput[u](6,14){$K$}
  \end{pspicture}
  \caption{Dikd\"ortgenlerin e\c sitli\u gi}\label{fig:ab=cx}
  
\end{figure}
$ABCD$ ve $CEHK$ dikd\"ortgenleri e\c sittir ancak ve ancak $GC$ ve $CF$
bir do\u grudad\i r.
\end{theorem}

\begin{definition}
  Dikd\"ortgenin \textbf{alan\i,} onun e\c sitlik s\i n\i f\i d\i r.
Geni\c sli\u gi $a$ ve y\"uksekli\u gi $b$ olan dikd\"ortgenin alan\i
\begin{equation*}
  a\cdot b
\end{equation*}
veya $ab$ ile g\"osterilir.  Ayr\i ca $a\cdot a$ alan\i
\begin{equation*}
  a^2
\end{equation*}
ile g\"osterilir.
\end{definition}

\begin{theorem}
  Uzunluklar\i n \c carpmas\i\ de\u gi\c smelidir
ve toplama \"uzerine da\u g\i l\i r.
Ayr\i ca
\begin{equation*}
  ab=ac\implies b=c.
\end{equation*}
\end{theorem}

\begin{theorem}
  $a:b::c:d\iff ad=bc \iff a:c::b:d$.
\end{theorem}

\begin{theorem}
 $a:b::c:d\implies a:b::a\pm c:b\pm d$.
\end{theorem}

\begin{theorem}\label{thm:well}
  $ab=de\And ac=df\implies b:c::e:f$.
\end{theorem}

\begin{theorem}\label{thm:sol}
Her $ab=cx$ denklemi \c c\"oz\"ulebilir.
\end{theorem}

\begin{definition}
\Teorem{thm:well} ve \numaraya{thm:sol} g\"ore
alanlar\i n \textbf{oran\i}
\begin{equation*}
  ab:ac::b:c
\end{equation*}
ile tan\i mlanabilir.
\end{definition}

\begin{theorem}
$ab:cd::ab:ef\implies cd=ef$.
\end{theorem}

\begin{theorem}
  $ab:cd::ef:gh\implies ab:ef::cd:gh$.
\end{theorem}

\begin{theorem}
  $ab:cd::ef:gh\implies ab:cd::ab+ef:cd+gh$.
\end{theorem}

\begin{theorem}
$a:b::c:d\iff a^2:b^2::c^2:d^2$.  
\end{theorem}

\begin{definition}
  A\c c\i lar\i\ s\i ras\i yla e\c sit olan \"u\c cgenler \textbf{benzerdir.}
\end{definition}

\begin{theorem}
  Benzer \"u\c cgenlerin kenarlar\i\ orant\i l\i d\i r,
yani $ABC$ ve $DEF$ benzer ise
\begin{equation*}
  AB:BC::DE:EF.
\end{equation*}
\end{theorem}

\begin{theorem}
  \Sekilde{fig:parallel2}
  \begin{figure}[ht]
    \centering
\psset{unit=3mm}
    \begin{pspicture}(0,-1)(18,10)
%\psgrid
      \pspolygon(0,0)(12,0)(18,9)
\psline(4,0)(16,6)
%\psline[linestyle=dotted](6,0)(12,6)
\uput[d](0,0){$A$}
\uput[d](12,0){$B$}
\uput[u](18,9){$C$}
\uput[d](4,0){$D$}
\uput[dr](16,6){$E$}
    \end{pspicture}
    \caption{Paralellik ve orant\i l\i l\i k}\label{fig:parallel2}
    
  \end{figure}
$ABC$ herhangi \"u\c cgen olsun.
  \begin{equation*}
    AB:BC::DB:BE\iff AC\parallel DE.
  \end{equation*}
\end{theorem}

\begin{theorem}
  $\comp abcd::ac:bd$.
\end{theorem}

\section{Koni kesitleri ve paraboller}

\begin{definition}\label{def:axis}
Bir daire
ve ayn\i\ d\"uzlemde olmayan bir nokta,
bir \textbf{koniyi} 
(\gr{k~wnos} ``\c cam kozala\u g\i'')
belirtir.
Daire, koninin \textbf{taban\i d\i r,}
ve nokta, koninin \textbf{tepe noktas\i d\i r.}
Koninin \textbf{y\"uzeyi,}
tepe noktas\i ndan taban\i n s\i n\i r\i na giden do\u grular taraf\i ndan
olu\c sturulur.
Koninin tepe noktas\i ndan taban\i n merkezine giden do\u gru,
koninin \textbf{eksenidir} (\gr{>'axwn} ``dingil'').
Her durumda,
ekseni i\c ceren her d\"uzlem,
koniyi bir \"u\c cgende keser.
Bu \"u\c cgene \textbf{eksen \"u\c cgeni} denebilir.
\end{definition}

\begin{remark}
Koninin ekseni, koninin taban\i na dik olmayabilir.
Koninin eksen \"u\c cgeninin taban\i,
koninin taban\i n\i n bir \c cap\i d\i r.
\end{remark}

\begin{theorem}\label{thm:cone}
Bir koninin bir eksen \"u\c cgeni,
 \Sekilde{fig:ax-base}ki gibi
\begin{figure}[ht]
\psset{unit=8mm}
\mbox{}\hfill
\begin{pspicture}(0,-0.5)(4,4)
%\psgrid
\pspolygon(0,0)(4,0)(1,4)
\psset{linestyle=dashed}
\psline(1,0)(1.75,3)
\psline(1,0)(2,2.67)
\psline(1,0)(1.5,3.33)
\uput[l](1,4){$A$}
\uput[l](0,0){$B$}
\uput[r](4,0){$C$}
\uput[d](1,0){$F$}
\uput[ur](1.75,3){$G$}
\uput[ur](1.5,3.33){$G$}
\uput[ur](2,2.67){$G$}
\psdots(1,0)
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\hfill\mbox{}
\caption{Koninin eksen \"u\c cgeni ve taban\i}\label{fig:ax-base}
\end{figure}
taban\i\ $BC$ olan $ABC$ \"u\c cgeni olsun.
Koninin taban\i n\i n $DE$ kiri\c si \c cizilsin,
ve bu kiri\c s, $BC$ \c cap\i na dik olsun.
O zaman kiri\c s, 
\c cap taraf\i ndan bir $F$ noktas\i nda ikiye b\"ol\"un\"ur, ve
\begin{equation}\label{eqn:DF^2}
  DF^2=BF\cdot FC.
\end{equation}
\end{theorem}

\begin{definition}\label{def:conic}
\Teorem{thm:cone} durumunda
$DE$ kiri\c sini i\c ceren bir d\"uzlem,
eksen \"u\c cgeninin $AC$ kenar\i n\i\
bir $G$ noktas\i nda kessin.
O zaman bu d\"uzlem,
koninin y\"uzeyini
\Sekilde{fig:curve}ki gibi
\begin{figure}[ht]
\centering
\psset{unit=8mm}
\begin{pspicture}(-2,-0.5)(2,3.5)
%\psgrid
\psTilt{104.04}{
\psplot{-1.732}{1.732}{3 x x mul sub}
\psline(0,0)(0,3)
\psline(-1.414,1)(1.414,1)
}
\psline(-1.732,0)(1.732,0)
\uput[d](-1.732,0){$D$}
\uput[d](1.732,0){$E$}
\uput[d](0,0){$F$}
\uput[u](-0.75,3){$G$}
\uput[l](-1.6,1){$K$}
\uput[r](1.2,1){$L$}
\uput[ur](-0.2,1){$M$}
\end{pspicture}
\caption{Bir koni kesiti}\label{fig:curve}
\end{figure}
bir $DGE$ e\u grisinde keser.
Bu e\u griye \textbf{koni kesiti} denir.
$DE$ do\u grusu, e\u grinin bir kiri\c sidir.
\end{definition}

\begin{theorem}\label{thm:conic}
$KL$ do\u grusu,
yukar\i daki koni kesitinin ba\c ska bir kiri\c si olsun,
ve bu kiri\c s,
$DE$ kiri\c sine paralel olsun.
$KL$ kiri\c si ve $FG$ do\u grusu bir $M$ noktas\i nda kesi\c sir.
Ayr\i ca koninin taban\i na paralel olan
ve $KL$ kiri\c sini i\c ceren bir d\"uzlem vard\i r.
Bu d\"uzlem,
\begin{compactitem}
\item 
$ABC$ \"u\c cgenini
$BC$ taban\i na paralel olan bir $NP$ do\u grusunda keser, ve
\item
koninin kendisini, \c cap\i\ $NP$ olan bir dairede keser.
\end{compactitem}
\Sekle{fig:2bases}
\begin{figure}[ht]
\psset{unit=8mm}
\begin{pspicture}(-0.5,-0.5)(4,4)
%\psgrid
\pspolygon(0,0)(4,0)(1,4)
\psline(0.25,1)(3.25,1)
\psset{linestyle=dashed}
\psline(1,0)(2.00,2.67)
\psline(1,0)(1.75,3.00)
\psline(1,0)(1.50,3.33)
\uput[l](1,4){$A$}
\uput[l](0,0){$B$}
\uput[r](4,0){$C$}
\uput[d](1,0){$F$}
\uput[ur](1.75,3){$G$}
\uput[ur](1.5,3.33){$G$}
\uput[ur](2,2.67){$G$}
\uput[l](0.25,1){$N$}
\uput[r](3.25,1){$P$}
\uput[dr](1.3,1){$M$}
\psdots(1,0)
\end{pspicture}
\hfill
\begin{pspicture}(0,-1)(3.5,3.5)
\pscircle(1.5,2){1.5}
\psline(0,2)(3,2)
\psset{linestyle=dashed}
\psline(0.9,0.636)(0.9,3.364)
\psline(1.0,0.586)(1.0,3.414)
\psline(1.1,0.556)(1.1,3.444)
\uput[l](0,2){$N$}
\uput[r](3,2){$P$}
\uput[dl](1,0.586){$K$}
\uput[ul](1,3.414){$L$}
\uput[ur](1,2){$M$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\caption{Koninin eksen \"u\c cgeni ve tabanlar\i}\label{fig:2bases}
\end{figure}
bak\i n.
Koni kesitinin $LK$ kiri\c si,
bu yeni dairenin kiri\c sidir,
ve dairenin $NP$ \c cap\i na diktir,
dolay\i s\i yla $KM=ML$.
Bu \c sekilde $GF$ \i\c s\i n\i,
$DGE$ koni kesitinin
$DE$ kiri\c sine paralel olan her kiri\c si ikiye b\"oler.
\end{theorem}

\begin{definition}
Tan\i m \ref{def:conic}
ve \Teoremde{thm:conic} 
$G$ noktas\i, koni kesitinin \textbf{k\"o\c sesidir,}
ve $GF$ \i\c s\i n\i\
koni kesitinin bir \textbf{\c cap\i d\i r,}
\c c\"unk\"u $DE$ kiri\c sine paralel olan kiri\c sleri ikiye b\"oler.
E\u ger \c cap,
ikiye b\"old\"u\u g\"u 
ve birbirine paralel olan
kiri\c slere dik ise,
ona \textbf{eksen} denir.
Ama her durumda
$DE$ kiri\c sinin $DF$ (veya $EF$) yar\i s\i na
\textbf{ordinat} denir,
ve \c cap\i n $GF$ par\c cas\i na,
$DF$ ordinat\i na kar\c s\i l\i k gelen \textbf{absis} denir.
\end{definition}

\begin{remark}
O zaman $KM$ ve $LM$ do\u grular\i\ da
ordinatt\i r,
ve onlara kar\c s\i l\i k gelen absis, $GM$ do\u grusudur.
\end{remark}

\begin{theorem}\label{thm:para}
\Sekilde{fig:2bases}ki durumda $FG\parallel BA$ olsun.
O zaman 
\begin{equation*}
GM:GF::ML^2:FE^2.
\end{equation*}
Sonu\c c olarak
bir $\ell$ uzunlu\u gu i\c cin,
koni kesitinin herhangi ordinat\i n\i n uzunlu\u gu $y$
ve ordinata kar\c s\i l\i k gelen absisin uzunlu\u gu $x$ ise
\begin{equation*}
  y^2=\ell x.
\end{equation*}
Ayr\i ca
\begin{equation*}
  \ell:GA::CB^2::CA\cdot CB.
\end{equation*}
\end{theorem}

\begin{definition}
  \Teoremde{thm:para}ki koni kesiti
\textbf{parabold\"ur}
(\gr{parabol'h} ``uygulama, yerle\c stirme''),
ve $\ell$, parabol\"un \textbf{parametresidir}
ve parabol\"un \textbf{dikey kenar\i n\i n} uzunlu\u gudur.%%%%%
\footnote{Dikey kenar\i n Latincesi, \emph{latus rectum.}}
\end{definition}

\begin{theorem}[Menaechmus]
  Parametreleri $a$ ve $b$ olan paraboller ile
  \begin{equation*}
    a:x::x:y::y:b
  \end{equation*}
orant\i lar\i\
\Sekilde{fig:Men}ki gibi \c c\"oz\"ulebilir.
\begin{figure}[ht]
\psset{unit=2cm}
  \centering
  \begin{pspicture}(0,-0.25)(3,2)
    \psplot02{x x mul 2 div}
\psplot03{x sqrt}
\psline(0,2)(0,0)(3,0)
\psline(0,1.2599)(1.5874,1.2599)(1.5874,0)
\uput[dr](1.5874,1.2599){$A$}
\uput[d](1.5874,0){$B$}
\uput[dl](0,0){$C$}
\uput[l](0,1.2599){$D$}
  \end{pspicture}
  \caption{\.Iki orta orant\i l\i}\label{fig:Men}
  
\end{figure}
Parametresi $b$ olan parabol\"un bir ordinat\i\ $AB$ ve
ona kar\c s\i l\i k gelen absis $CB$ ise, ve
parametresi $a$ olan parabol\"un bir ordinat\i\ $AD$ ve
ona kar\c s\i l\i k gelen absis $CD$ ise,
ve her parabol\"un ordinatlar\i\
di\u ger parabol\"un \c cap\i na paralel ise,
o zaman $CB$ ve $CD$ do\u grular\i n\i n uzunluklar\i\
yukar\i daki orant\i lar\i\ \c c\"ozer.
\end{theorem}

\section{Hacimler}

\begin{definition}
  Dik paralely\"uz\"un \textbf{hacmi,}
onun e\c sitlik s\i n\i f\i d\i r.
Geni\c sli\u gi $a$, y\"uksekli\u gi $b$,
ve derinli\u gi $c$ olan dik paralely\"uz\"un hacmi
\begin{equation*}
  a\cdot b\cdot c
\end{equation*}
veya $abc$ ile g\"osterilir.
\end{definition}

\begin{theorem}
  $abc=bac=bca$ ve $ab(c+d)=abc+abd$.
\end{theorem}

\begin{theorem}
  $abc=ade\implies bc=de$.
\end{theorem}

\begin{theorem}
  $ab:cd::e:f\iff abf=cde$.
\end{theorem}

\section{Hiperboller}

\begin{theorem}\label{thm:hyp}
\Sekilde{fig:ax-base} koni kesitinin $GF$ \c cap\i\
$G$ noktas\i n\i n \"otesine uzat\i l\i rsa,
\Sekilde{fig:hyper}ki gibi
\begin{figure}[ht]
\begin{pspicture}(-1,-2)(4,4)
\psline(1,0)(2,4)(0,0)(4,0)(1,2)
\psline(0,0)(-0.5,-1)(5.5,-1)(4,0)
\psline(1,0)(0.744,-1)
\psline[linestyle=dashed](0.744,-1)(2.798,-1.515)(3.476,1.200)(3.076,1.302)
\psline[linestyle=dashed](2.647,-0.412)(2.798,-1.515)
\uput[l](-0.5,-1){$N$}
\uput[d](0.744,-1){$M$}
\uput[ur](5.5,-1){$P$}
\uput[d](2.798,-1.515){$U$}
%\uput[r](3.476,1.200){$V$}
\uput[l](1,2){$A$}
\uput[l](0,0){$B$}
\uput[ur](4,0){$C$}
%\uput{3pt}[120](1,0){$F$}
\uput[ur](1,0){$F$}
\uput[ur](1.438,1.714){$G$}
\uput[l](2,4){$X$}
\psset{linestyle=dashed}
\psline(1,0)(2.647,-0.412)(2,4)
\psline(1.438,1.714)(3.076,1.302)(2.647,-0.412)
\psline(2.370,1.479)(1.941,-0.235)
\psline(2.932,3.765)(3.647,3.588)(3.076,1.302)
\uput[ur](2.370,1.479){$H$}
\uput[dl](2.647,-0.412){$R$}
\uput[ur](3.076,1.302){$S$}
\uput[dl](1.941,-0.235){$T$}
\psline(2.370,1.479)(2.932,3.765)(2,4)
\uput[u](2.932,3.765){$Y$}
\uput[ur](3.647,3.588){$Z$}
\psset{linestyle=dotted}
\psline(0.5,0)(1,2)
\uput[d](0.5,0){$J$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-2)(4.5,4)
\pscircle(2,2){2}
\psline(0,2)(4,2)
\psline(1,0.268)(1,3.732)
\psline[linestyle=dashed](1,0.268)(2.732,0.268)(2.732,2)
\uput[l](0,2){$B$}
\uput[r](4,2){$C$}
\uput[dl](1,0.268){$D$}
\uput[ul](1,3.732){$E$}
\uput[ur](1,2){$F$}
\end{pspicture}
\caption{Konide hiperbol}\label{fig:hyper}
\end{figure}
$BA$ do\u grusunun uzat\i lmas\i n\i\ bir $X$ noktas\i nda kessin.
$FR$ do\u grusu, $GF$ \c cap\i na dik olsun ve
\begin{equation}\label{eqn:FRFG}
  FR\cdot FG=DF^2
\end{equation}
e\c sitli\u gini sa\u glas\i n.  
$MU\parallel FR$ olsun, ve
(gerekirse uzat\i lm\i\c s)
$XR$ ve $MU$, $U$ noktas\i nda kesi\c ssin.  O zaman
\begin{equation*}
  GM\cdot MU=KM^2.
\end{equation*}
($KM$, \Sekilde{fig:2bases}ki gibidir.)
 $AJ\parallel XF$ olsun;
o zaman
\begin{equation*}
  GH:GX::BJ\cdot JC:AJ^2.
\end{equation*}
$GH$ do\u grusunun uzunlu\u gu $\ell$ olsun,
ve $GX$ do\u grusunun uzunlu\u gu $2a$ olsun.
Koni kesitinin herhangi bir ordinat\i n\i n uzunlu\u gu $y$
ve bu ordinata kar\c s\i l\i k gelen absisin uzunlu\u gu $x$ ise
\begin{equation*}
  2ay^2=2a\ell x+\ell x^2.
\end{equation*}
\end{theorem}

\begin{remark}
\Sekle{fig:hyper-graph}
\begin{figure}[ht]
  \centering
\psset{unit=8mm}
  \begin{pspicture}(-2,-4.4)(6,3.333)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]04{t t t 2 add mul 6 div sqrt add
                                 t t 2 add mul 2 div sqrt}
\parametricplot[linewidth=2pt]05{t t t 2 add mul 6 div sqrt sub
                                 0 t t 2 add mul 2 div sqrt sub}
\psline(1.419,-2.739)(4.581,2.739)
\psline(-2,0)(6,0)
\psdots(0,0)(0,1.333)(-2,0)
\psline(0,0)(0,1.333)
\uput[d](-1,0){$2a$}
\uput[l](0,0.667){$\ell$}
\uput[d](1.5,0){$x$}
\uput[-30](3.791,1.369){$y$}
\uput[-30](2.209,-1.369){$y$}
\psset{linestyle=dotted}
\psline(3,3.333)(-2,3.333)(-2,0)
\psline(-2,1.333)(3,1.333)
\psline(-2,0)(3,3.333)(3,0)
\psline(0,1.333)(0,3.333)
\psline(3,0)(5.739,-1.581)(4.157,-4.320)(1.419,-2.739)
  \end{pspicture}
  \caption{$2ay^2=2a\ell x+\ell x^2$ hiperbol\"u}\label{fig:hyper-graph}
  
\end{figure}
bak\i n; buradaki $\ell$-i\c saretli do\u gru,
koni kesitinin d\"uzlemine dik olarak d\"u\c s\"un\"ulebilir.
\end{remark}

\begin{definition}
\Teoremde{thm:hyp}
alan\i\ $y^2$ olan kare,
alan\i\ $\ell x$ olan dikd\"ortgenini a\c st\i\u g\i ndan,
koni kesitine
\textbf{hiperbol} (\gr{<uperbol'h} ``a\c sma'') denir;
$GH$ do\u grusu,
hiperbol\"un \textbf{dikey kenar\i d\i r;}
dikey kenar\i n $\ell$ uzunlu\u gu, hiperbol\"un
\textbf{parametresidir;}
$GX$ do\u grusu, hiperbol\"un \textbf{yanlamas\i na kenar\i d\i r;}%%%%%
\footnote{\gr{pl'agia pleur'a}; Latincesi \emph{latus transversum.}}
%%%%%%%%%%%%%
yanlamas\i na kenar\i n orta noktas\i,
hiperbol\"un \textbf{merkezidir.}
\end{definition}


\begin{remark}
\Sekilde{fig:hyper}
$FS$ ve $FH$ dikd\"ortgenlerinin fark\i\ $TS$ dikd\"ortgendir,
ve bu dikd\"ortgen $GY$ dikd\"ortgenine benzerdir.%%%%%
\footnote{Bu $GY$ dikd\"ortgeni, hiperbol\"un \textbf{\c seklidir}
(\gr{e>~idos}).}
%%%%%%%%%%%%%%%%%
\end{remark}

\section{\.I\c saretli uzunluklar ve elipsler}

\begin{definition}
Bir y\"on ile donat\i lm\i\c s bir do\u gru,
bir \textbf{y\"onl\"u do\u grudur.}
E\u ger $AB$, $A$'dan $B$'ye y\"on ile donat\i l\i rsa,
olu\c san y\"onl\"u do\u gru
\begin{equation*}
  \ydp{AB}
\end{equation*}
bi\c ciminde yaz\i labilir.
$\ydp{AA}$,
\textbf{yoz} veya \textbf{dejenere} y\"onl\"u do\u grudur ve
$A$ noktas\i\ olarak anla\c s\i labilir.
\begin{comment}
  

$A$, $B$, ve $C$ bir do\u gruda ise
\begin{align*}
-\ydp{AB}&=\ydp{BA},&
  \ydp{AB}+\ydp{BC}&=\ydp{AC}.
\end{align*}


\end{comment}
E\u ger $ABDC$ ve $DCEF$ paralelkenar ise,
o zaman
\begin{align*}
  \ydp{AB}&=\ydp{CD},&  \ydp{AB}&=\ydp{EF},
\end{align*}
\"Ozel olarak $\ydp{AA}=\ydp{BB}$.
\end{definition}

\begin{theorem}
Do\u grular\i n paralelli\u gi
ve y\"onl\"u do\u grular\i n e\c sitli\u gi,
denklik ba\u g\i nt\i s\i d\i r.
\end{theorem}

\begin{definition}
  Y\"onl\"u do\u grunun e\c sitlik s\i n\i f\i, \textbf{vekt\"ord\"ur.}
\end{definition}

\begin{definition}
  Her paralellik s\i n\i f\i\ i\c cin bir y\"on \textbf{pozitif,}
di\u ger y\"on \textbf{negatif} olsun.
O zaman her (yoz olmayan) y\"onl\"u do\u gru ya pozitif ya negatiftir.
Bir y\"onl\"u do\u grunun pozitifli\u gi veya negatifli\u gi,
y\"onl\"u do\u grunun \textbf{i\c saretidir.}
\end{definition}

\begin{theorem}
$A$, $B$, ve $C$ bir do\u gruda olsun.
O zaman $\ydp{AB}$ ve $\ydp{BC}$ 
y\"onl\"u do\u grular\i n\i n i\c saretleri ayn\i d\i r 
ancak ve ancak $AB<AC$ ve $BC<AC$.  
\end{theorem}

\begin{theorem}\label{thm:sign}
A\c sa\u g\i daki ko\c sulu sa\u glayan $\mathscr R$ ba\u g\i nt\i s\i\
bir denklik ba\u g\i nt\i s\i d\i r:
$\ydp{AB}\mathrel{\mathscr R}\ydp{CD}$ ancak ve ancak
 $\ydp{AB}$ ve $\ydp{CD}$ 
y\"onl\"u do\u grular\i n\i n i\c saretleri ayn\i\ ve $AB=CD$.
\end{theorem}

\begin{definition}
  \Teoremde{thm:sign}ki denklik ba\u g\i nt\i s\i na g\"ore
bir y\"onl\"u do\u grunun denklik s\i n\i f\i,
\textbf{y\"onl\"u do\u grunun uzunlu\u gudur.}
\end{definition}

\begin{remark}
Bir do\u grunun uzunlu\u gu, yeni tan\i m\i\ alabilir:
$\ydp{AB}$ ve $\ydp{BA}$ y\"onl\"u do\u grular\i n\i n uzunluklar\i n\i n
hangisi pozitif ise, 
$AB$ do\u grusunun uzunlu\u gu olarak al\i nabilir.
Bu tan\i m\i\ ba\c slang\i\c ctan kullanabildik.
\end{remark}

\begin{definition}
K\"u\c c\"uk $a$, $b$, $c$, \lips Latin harfleri,
y\"onl\"u do\u grunun 
uzunlu\u gunu (yani i\c saretli uzunlu\u gunu) g\"osterecek.
Yoz y\"onl\"u do\u grunun uzunlu\u gu,
\begin{equation*}
  0
\end{equation*}
olsun, ve $\ydp{AB}=c$ ise
\begin{equation*}
  -c=\ydp{BA}
\end{equation*}
olsun.
\end{definition}

\begin{theorem}
\.Iki i\c saretli uzunluk toplanabilir, ve tan\i ma g\"ore
$A$, $B$, ve $C$ bir do\u gruda ve
\begin{align*}
  \ydp{AB}&=d,&\ydp{BC}&=e,&\ydp{AC}&=f
\end{align*}
ise
\begin{equation*}
  d+e=f.
\end{equation*}
Bu durumda toplama de\u gi\c smeli ve birle\c smelidir;
ayr\i ca
\begin{gather*}
a+0=a,\\
a+(-a)=0.
\end{gather*}
\end{theorem}

\begin{definition}
  $\begin{gathered}[t]
  a-b=a+(-b),\\
-a\cdot b=-(ab)=a\cdot(-b),\\
-a\cdot bc=a\cdot(-b)\cdot c=ab\cdot(-c)=-(abc).
  \end{gathered}$
\end{definition}

\c Simdi hiperbol\"un $2ay^2=2a\ell x+\ell x^2$ denkleminde 
$x$ ve $y$ negatif olabilir.
Ayr\i ca $a$ negatif olabilir, 
ama bu durumda tan\i mlanan e\u gri hiperbol de\u gildir:

\begin{definition}
$\ell>0$ ve $a>0$ ise
  \begin{equation*}
2ay^2=2a\ell x-\ell x^2
  \end{equation*}
  denklemi,
\textbf{dikey kenar\i n\i n} uzunlu\u gu $\ell$ olan,
\textbf{yanlamas\i na kenar\i n\i n} uzunlu\u gu $2a$ olan 
\textbf{elipsi} (\gr{>'elleiyis} ``eksiklik'') tan\i mlar
(ama ordinatlar\i n \c capa a\c c\i s\i n\i\ se\c cilmeli).
\Sekle{fig:ell-graph}
\begin{figure}[ht]
  \centering
\psset{unit=3cm}
  \begin{pspicture}(0,-1.051)(2,1.5)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]02{t t 2 t sub mul 6 div sqrt add
                                   t 2 t sub mul 2 div sqrt}
\parametricplot[linewidth=2pt]02{t t 2 t sub mul 6 div sqrt sub
                                 0 t 2 t sub mul 2 div sqrt sub}
\psline(0.9484,-0.6667)(1.7182,0.6667)
\psline(0,0)(2,0)
\psdots(0,0)(0,1.333)(2,0)%(1,0)
\psline(0,0)(0,1.333)
\pcline[offset=8pt](0,1.3333)(2,1.3333)
\ncput*{$2a$}
%\uput[d](1,0){$2a$}
\pcline[offset=8pt](0,0)(0,1.333)
\ncput*{$\ell$}
%\uput[l](0,0.667){$\ell$}
\uput[d](0.6667,0){$x$}
\uput[150](1.141,-0.333){$y$}
\psset{linestyle=dotted}
\psline(0,1.333)(2,0)
\psline(1.3333,0)(1.3333,1.3333)
\psline(2,0.4444)(0,0.4444)
\psline(0,1.3333)(2,1.3333)(2,0)
\psline(1.3333,0)(2,-0.385)(1.615,-1.051)(0.9484,-0.6667)
  \end{pspicture}
  \caption{$2ay^2=2a\ell x-\ell x^2$ elipsi}\label{fig:ell-graph}
  
\end{figure}
bak\i n.
Hiperboldeki gibi elipsin \textbf{merkezi,}
yanlamas\i na kenar\i n\i n orta noktas\i d\i r.
Hiperbol ve elips, \textbf{merkezli koni kesitidir.}
\end{definition}

\begin{theorem}
\Teoremde{thm:hyp}
%\Sekilde{fig:ax-base} 
koni kesitinin $GF$ \c cap\i\
$F$ noktas\i n\i n \"otesine uzat\i l\i rsa
ve $AB$ do\u grusunun uzat\i lmas\i n\i\ keserse,
hiperbol\"un yerine
elips \c c\i kar.
\end{theorem}

\begin{remark}
\c Simdi her koni kesiti ya parabol ya hiperbol ya da elipstir.
Pergeli Apollonius bu adlar\i\ vermi\c stir.
Parabol olmayan her koni kesiti merkezlidir.
\end{remark}

\section{Eksenler}

\begin{definition}
D\"uzlemde iki do\u gru bir $O$ noktas\i nda kesi\c ssin.
Do\u grular\i n birine \textbf{$x$ ekseni,}
di\u gerine \textbf{$y$ ekseni} densin,
ve $O$ noktas\i na \textbf{ba\c slang\i\c c noktas\i} densin.
\end{definition}

\begin{theorem}\label{thm:coord}
  D\"uzlemde her $A$ noktas\i\ i\c cin
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
$ABOC$ paralelkenard\i r.
Tam tersine $b$ ve $c$ i\c saretli uzunluk olmak \"uzere, 
herhangi bir $(b,c)$ s\i ral\i\ ikilisi i\c cin,
$x$ ekseninde bir ve tek bir $B$ i\c cin,
$y$ ekseninde bir ve tek bir $C$ i\c cin,
d\"uzlemde bir ve tek bir $A$ i\c cin
\begin{align*}
  \ydp{OB}&=b,&\ydp{OC}&=c,
\end{align*}
ve $ABOC$ paralelkenard\i r.
\end{theorem}

\begin{definition}
  \Teoremde{thm:coord} $b$, $A$ noktas\i n\i n \textbf{$x$ koordinat\i d\i r,}
ve $c$, $A$ noktas\i n\i n \textbf{$y$ koordinat\i d\i r.}
\Sekilde{fig:axes}ki gibi
\begin{figure}[ht]
  \centering
  \begin{pspicture}(-4,-0.866)(1,2.598)
    \psline{->}(-4,0)(1.5,0)
\psline{->}(-0.5,-0.866)(1.5,2.598)
\psline[linestyle=dotted](1,1.732)(-2,1.732)(-3,0)
\uput[120](-2,1.732){$A$}
\uput[120](1,1.732){$C$}
\uput[-30](1,1.732){$c$}
\uput[120](-3,0){$B$}
\uput[d](-3,0){$b$}
\uput[-60](0,0){$O$}
\uput[ur](1.5,0){$x$}
\uput[r](1.5,2.598){$y$}
  \end{pspicture}
  \caption{Koordinatlar}\label{fig:axes}
  
\end{figure}
$B$ noktas\i na $b$ yaz\i labilir, ve
$C$ noktas\i na $c$ yaz\i labilir.
\end{definition}

\begin{remark}
\Sekilde{fig:2branches}ki
\begin{figure}[ht]
  \centering
\psset{unit=7.5mm}
  \begin{pspicture}(-8,-4)(6,4)
% transverse side 2, upright side 4/3, angle 60 degrees
\psset{plotpoints=200}
\parametricplot[linewidth=2pt]04{t t t 2 add mul 6 div sqrt add
                                   t t 2 add mul 2 div sqrt}
\parametricplot[linewidth=2pt]05{t t t 2 add mul 6 div sqrt sub
                                 0 t t 2 add mul 2 div sqrt sub}
\parametricplot[linewidth=2pt]04{-2 t t t 2 add mul 6 div sqrt add sub
                                  0 t t 2 add mul 2 div sqrt sub}
\parametricplot[linewidth=2pt]05{-2 t t t 2 add mul 6 div sqrt sub sub
                                 t t 2 add mul 2 div sqrt}
\psline{->}(-8,0)(6,0)
\uput[d](6,0){$x$}
\psline{->}(-2.309,-4)(2.309,4)
\uput[ul](2.309,4){$y$}
\psdots(0,0)(-2,0)%(0,1.333)
%\psline(0,0)(0,1.333)  % latus rectum
\uput[dr](-2,0){$-2a$}
\uput[120](0,0){$O$}
%\uput[l](0,0.667){$\ell$}
\psdots(3,0)(-5,0)(1.581,2.739)(-1.581,-2.739)
\uput[dr](3,0){$b$}
\uput[dr](-5,0){$-2a-b$}
\uput[120](1.581,2.739){$c$}
\uput[dr](-1.581,-2.739){$-c$}
\psset{linestyle=dotted}
\pspolygon(1.419,-2.739)(4.581,2.739)(-3.419,2.739)(-6.581,-2.739)
  \end{pspicture}
  \caption{\.Ikinci dal\i\ ile $2ay^2=2a\ell x+\ell x^2$ hiperbol\"u}\label{fig:2branches}
  
\end{figure}
koordinatlar\i\ $(b,c)$ olan nokta hiperboldeyse,
koordinatlar\i\ $(b,-c)$, $(-2a-b,c)$, ve $(-2a-b,-c)$ olan noktalar\i\ da
hiperboldedir.
\end{remark}

\begin{theorem}
  Denklemi $2ay^2=2a\ell x+\ell x^2$ olan hiperbol\"u verilsin,
ama yeni $st$ eksenleri se\c cilsin.
E\u ger
\begin{compactitem}
\item
  $s$ ekseni, $x$ eksenidir, ve
\item
$t$ ekseni, hiperbol\"un merkezinden ge\c cer ve $y$ eksenine paralel ise,
\end{compactitem}
o zaman yeni $st$ eksenlerine g\"ore hiperbol\"un denklemi,
\begin{equation*}
  2at^2=\ell s^2-\ell a^2.
\end{equation*}
\end{theorem}

\begin{theorem}\label{thm:hyp-cen}
\.I\c saretli uzunluklar\i n oran\i
  \begin{equation*}
    a:b::c:d\iff ad=bc
  \end{equation*}
kural\i na g\"ore tan\i mlanabilir.
Oranlar\i n toplam\i
\begin{equation*}
  (a:c)+(b:c)::(a+b):c
\end{equation*}
kural\i na g\"ore tan\i mlanabilir.
\end{theorem}

\begin{remark}
  \c Simdi oranlar\i\ say\i lar gibi kullanabiliriz.
\end{remark}

\begin{definition}
  $a:a$ oran\i
  \begin{equation*}
    1
  \end{equation*}
olarak yaz\i ls\i n, ve $a:b$ oran\i
\begin{equation*}
  \frac ab
\end{equation*}
veya $a/b$ bi\c ciminde yaz\i ls\i n.
O zaman hiperbol\"un $2ay^2=2a\ell x+\ell x^2$ denklemi
\begin{equation*}
  y^2=\ell x+\frac{\ell}{2a}x^2
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu denkleme \textbf{Apollonius denklemi} diyelim.
\Teoreme{thm:hyp-cen} g\"ore, 
farkl\i\ eksenlere g\"ore,
hiperbol\"un $2ay^2=\ell x^2-\ell a^2$ denklemi de vard\i r;
bu denklem
\begin{equation*}
  \frac{x^2}{a^2}-\frac{y^2}{\ell a/2}=1
\end{equation*}
bi\c ciminde yaz\i labilir.
Bu denkleme \textbf{merkez denklemi} diyelim.
\end{definition}

\section{Dik eksenler}


\begin{theorem}
  Parabolde \c capa paralel olan her do\u gru, yeni bir \c capt\i r.
\Sekilde{fig:parab-new}ki gibi
\begin{figure}[ht]

\centering
\psset{unit=4cm}
\begin{pspicture}(-0.5625,-0.1)(1.5625,1.35)
\psline(-0.5625,0)(1.5625,0) % old x axis
%\psline(0,0)(-1.25,1.25) % old y axis
\psline(0.8125,0.75)(-0.1875,0.75)%(-0.75,0.75) % new x axis
\psline(-0.5625,0)(-0.1875,0.75)%(0.1825,1.5) % new y axis
\parametricplot{0}{1.25}{t t mul t sub
                       t} % parabola x = y^2 - y 
\psline(-0.1825,0.75)(0.5625,0) % old ordinate of new origin
\psline(0.0625,0.75)(0.3125,1.25) % new ordinate of point
\psline(0.3125,1.25)(1.5625,0) % old ordinate of point
\uput[d](0,0){$A$}
\uput[ul](-0.1875,0.75){$B$}
\uput[u](0.3125,1.25){$C$}
\uput[d](0.5625,0){$D$}
\uput[d](1.5625,0){$E$}
\uput[d](-0.5625,0){$F$}
\uput[d](0.0625,0.75){$H$}
\uput[ur](0.8125,0.75){$G$}
\uput[u](-0.28125,0){$a$}
\uput[u](0.28125,0){$a$}
\uput[u](1.0625,0){$x-a$}
\uput[ul](-0.375,0.375){$c$}
\uput[ur](0.1875,0.375){$b$}
\uput[ur](1.1875,0.375){$b$}
\uput[u](-0.0625,0.75){$s$}
\uput[u](0.4375,0.75){$x-a-s$}
\uput[dr](0.1875,1){$t$}
\uput[ur](0.5625,1){$y-b$}
  \end{pspicture}
  
  \caption{Parabol\"un yeni \c cap\i}\label{fig:parab-new}
  
\end{figure}
\begin{compactenum}[1)]
  \item
$ABC$ e\u grisi, \c cap\i\ $FE$ ve k\"o\c sesi $A$ olan parabol,
\item
$BD$ ve $CE$ ordinat,
\item
$FA=AD$, ve 
\item
$BG\parallel FE$,
$CH\parallel BF$ 
\end{compactenum}
olsun.
A\c sa\u g\i daki i\c saretli uzunluklar\i\ tan\i mlans\i n:
\begin{align*}
  &
  \begin{gathered}
    \ydp{AD}=a,\\
\ydp{DB}=b,
  \end{gathered}&
&\begin{gathered}
 \ydp{AE}=x,\\
\ydp{EC}=y,   
  \end{gathered}&
&\begin{gathered}
 \ydp{BH}=s,\\
\ydp{HC}=t,   
  \end{gathered}&
\ydp{FB}&=c.
\end{align*}
Parabol\"un dikey kenar\i n\i n uzunlu\u gu $\ell$ ise
\begin{equation*}
  \frac m{\ell}=\frac{c^2}{b^2}
\end{equation*}
olsun.  O zaman
\begin{equation*}
  y^2=\ell x
\end{equation*}
oldu\u gundan
\begin{equation*}
  t^2=ms.
\end{equation*}
\end{theorem}

\begin{theorem}
  Parabol\"un bir (ve tek bir) \c cap\i\ i\c cin
ordinatlar \c capa diktir
(yani Tan\i m \numarada{def:axis}ki gibi parabol\"un ekseni
ve tek bir ekseni vard\i r).
\end{theorem}

\begin{theorem}
  Hiperbol\"un merkezinden ge\c cen ve hiperbol\"u kesen her do\u gru, 
hiperbol\"un yeni bir \c cap\i d\i r.
  \Sekilde{fig:hyper-new}ki gibi
  \begin{figure}[ht]
    \centering
    \psset{unit=3cm}
    \begin{pspicture}(0,-0.167)(2.6,2.567)
\psline(0,0)(2.6,0)     % old diameter, slope 0
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex, slope -2
\psline(1.4,2.4)(2.6,0)     % old ordinate of point
\psline(1.4,0.933)(1.866,0)
\psline(1.4,0.933)(2.133,0.933)
\parametricplot{0}{2.4}{t t mul 1 add sqrt t 2 div sub t}
\psline(0,0)(1.4,0.933)     % new diameter, slope 2/3
\psline(1.4,2.4)(1.4,0.933) % new ordinate of point
\psline(0.866,0.577)(0.866,0)
\uput[d](1,0){$A$}
\uput[ul](0.866,0.577){$B$}
\uput[u](1.4,2.4){$C$}
\uput[d](0,0){$D$}
\uput[d](1.155,0){$E$}
\uput[d](2.6,0){$F$}
\uput[d](0.866,0){$G$}
\uput[ul](1.4,0.933){$H$}
\uput[ur](2.133,0.933){$K$}
\uput[d](1.866,0){$L$}
    \end{pspicture}
    \caption{Hiperbol\"un yeni \c cap\i}\label{fig:hyper-new}
    
  \end{figure}
  \begin{compactenum}[1)]
    \item
  $ABC$ e\u grisi, merkezi $D$ olan ve \c cap\i\ $DF$ olan hiperbol,
\item
$BE$ ve $CF$ ordinat,
\item
$DG:DA::DA:DE$,
\item
$CH\parallel BG$, $HK\parallel DA$, $HL\parallel BE$
  \end{compactenum}
olsun.
A\c sa\u g\i daki i\c saretli uzunluklar tan\i mlans\i n:
\begin{align*}
&\begin{gathered}
\ydp{DF}=x,\\
\ydp{FC}=y,
 \end{gathered}&
&\begin{gathered}
\ydp{DH}=s,\\
\ydp{HC}=t,
 \end{gathered}&
&\begin{gathered}
\ydp{DE}=c,\\
\ydp{EB}=d,
 \end{gathered}&
&\begin{gathered}
\ydp{DA}=a,\\
\ydp{GE}=e,
 \end{gathered}&
&\begin{gathered}
\ydp{DB}=f,\\
\ydp{GB}=g.
 \end{gathered}
\end{align*}
(\Sekle{fig:hyper-detail} bak\i n.)
\begin{figure}[ht]
    \psset{unit=3cm,labelsep=2pt}
%\mbox{}\hfill
    \begin{pspicture}(0,-0.167)(1.866,2.567)
\psline(0,0)(1.866,0)     % old diameter
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex
\psline(1.4,0.933)(1.866,0)
\psline(0,0)(1.4,0.933)     % new diameter
\uput[207](1.01,0.288){$d$}
\uput[-56](0.433,0.288){$f$}
\uput[124](0.7,0.467){$s$}
\uput[27](1.633,0.467){$\displaystyle\frac dfs$}
\uput[u](0.577,0){$c$}
\uput[d](0.933,0){$\displaystyle\frac cfs$}
    \end{pspicture}
\hfill
    \begin{pspicture}(0.866,-0.167)(2.133,2.567)
\psline(0.866,0)(1.866,0)     % old diameter
\psline(0.866,0.577)(1.155,0) % old ordinate of new vertex
\psline(1.4,2.4)(2.133,0.933)     % old ordinate of point
\psline(1.4,0.933)(1.866,0)
\psline(1.4,0.933)(2.133,0.933)
\psline(0.866,0.577)(1.4,0.933)     % new diameter
\psline(1.4,2.4)(1.4,0.933) % new ordinate of point
\psline(0.866,0.577)(0.866,0)
\uput[27](1.01,0.288){$d$}
\uput[l](0.866,0.288){$g$}
\uput[l](1.4,1.667){$t$}
\uput[d](0.990,0){$e$}
\uput[d](1.767,0.933){$\displaystyle\frac egt$}
\uput[27](1.767,1.667){$\displaystyle\frac dgt$}
    \end{pspicture}
%\hfill\mbox{}
    \caption{Hiperbol\"un benzer \"u\c cgenleri}\label{fig:hyper-detail}
    
  \end{figure}
Hiperbol\"un dikey kenar\i n\i n uzunlu\u gu $\ell$
ve $2b^2=\ell a$ ise
\begin{equation*}
  \frac{x^2}{a^2}-\frac{y^2}{b^2}=1
\end{equation*}
oldu\u gundan
\begin{equation*}
\frac{s^2}{f^2}-\frac{t^2}{g^2c/e}=1.
\end{equation*}
\end{theorem}

\begin{theorem}
  Hiperbol\"un bir (ve tek bir) \c cap\i\ i\c cin
ordinatlar \c capa diktir,
yani hiperbol\"un bir (ve tek bir) ekseni vard\i r.
\end{theorem}

\section{Uzakl\i k}

\begin{remark}
  Dik \"u\c cgenle $x^2=a^2+b^2$ denkleminin \c c\"oz\"um\"u 
bulunabilir.
\end{remark}

\begin{definition}
  $x^2=a^2+b^2$ denkleminin (pozitif) \c c\"oz\"um\"u
  \begin{equation*}
    \sqrt{a^2+b^2}.
  \end{equation*}
\end{definition}

\begin{definition}
  Eksenler verilirse,
``koordinatlar\i\ $(a,b)$ olan nokta'' ifadesinin yerine
``$(a,b)$ noktas\i'' diyebiliriz.
\end{definition}

\begin{theorem}
  Eksenler dik ise $(a,b)$ noktas\i n\i n $(c,d)$ noktas\i ndan uzakl\i\u g\i
  \begin{equation*}
    \sqrt{(a-c)^2+(b-d)^2}.
  \end{equation*}
\end{theorem}

\begin{definition}\label{def:slope}
  Eksenler dik ve $a\neq c$ ise ucu $(a,b)$ ve $(c,d)$ olan do\u grunun
\textbf{e\u gimi}
\begin{equation*}
  \frac{b-d}{a-c}.
\end{equation*}
\end{definition}

\begin{remark}
Tan\i m \numarada{def:slope} 
eksenlerin dik olmas\i\ gerekmez ama normaldir.
\end{remark}

\begin{theorem}
  Paralel do\u grular\i n e\u gimleri ayn\i d\i r.
Dik eksene g\"ore,
$a\neq c$ ise $(a,b)$ ve $(c,d)$ 
noktalar\i ndan ge\c cen u\c csuz do\u grunun noktalar\i,
\begin{equation*}
  y=\frac{d-b}{c-a}\cdot(x-a)+b
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
E\u gimi $e/f$ olan ve $(a,b)$ noktas\i ndan ge\c cen u\c csuz do\u grunun
noktalar\i,
\begin{equation*}
  y=\frac ef\cdot(x-a)+b
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
$y$ eksenine paralel olan ve $(a,b)$ noktas\i ndan ge\c cen u\c csuz do\u grunun
noktalar\i,
\begin{equation*}
  x=a
\end{equation*}
denklemini sa\u glayan noktalar\i d\i r.
\end{theorem}

\begin{remark}
\c Su anda Descartes'\i n ortaya koydu\u gu uyla\c s\i m uygundur:
\end{remark}

\begin{definition}
Bir \textbf{birim} uzunlu\u gu se\c cilirse,
\begin{equation*}
  1
\end{equation*}
olarak yaz\i labilir.
E\u ger $a\cdot b=c\cdot 1$ ise,
o zaman $ab$ alan\i\ $c$ olarak anla\c s\i labilir.
Bu \c sekilde alan, hacim, oran---her \c sey bir uzunluk olur.
\"Ozel olarak e\u gim, bir harf ile yaz\i labilir.
\end{definition}

\begin{theorem}
  Dik eksenlere ve birim uzunlu\u guna g\"ore
$y$ eksenine paralel olmayan do\u grunun denkli\u gi
\begin{equation*}
  y=mx+b
\end{equation*}
bi\c ciminde yaz\i labilir, ve bunun gibi her denklem,
e\u gimi $m$ olan ve $(0,b)$ noktas\i ndan ge\c cen do\u gruyu tan\i mlar.
Benzer \c sekilde $a\neq0$ veya $b\neq0$ ise
(yani $a^2+b^2\neq0$ ise)
\begin{equation*}
  ax+by+c=0
\end{equation*}
denklemi bir do\u gru tan\i mlar,
ve her do\u grunun denklemi bu \c sekilde yaz\i labilir.
\end{theorem}

\begin{definition}
  \Sekilde{fig:cos} $\angle BAC$ dik ise
%$ABC$ \"u\c cgeninde $B$ a\c c\i s\i\ dik ve
%  \begin{align*}
%    AB&=c,&AC&=b,&\angle BAC&=\alpha
%  \end{align*}
%ise
\begin{equation*}
  \cos\alpha=\frac bc.
\end{equation*}%
\begin{figure}[ht]
\psset{unit=8mm}
\centering
\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(2,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](2,2){$C$}
\uput[r](2,1){$a$}
\uput[ul](1,1){$b$}
\uput[22.5](0,0){$\alpha$}
\end{pspicture}
    \caption{Kosin\"us tan\i m\i}\label{fig:cos}
    
  \end{figure}
Burada $\alpha$, $\angle BAC$ a\c c\i s\i n\i n e\c sitlik s\i n\i f\i\ olarak
anla\c s\i labilir,
ve $\cos\alpha$, a\c c\i n\i n \textbf{kosin\"us\"ud\"ur.}
Dik a\c c\i n\i n \"ol\c c\"us\"u
\begin{equation*}
  \frac{\uppi}2.
\end{equation*}
O zaman
\begin{equation*}
  \cos\frac{\uppi}2=0,
\end{equation*}
ve $\beta$ geni\c s a\c c\i\ ise
\begin{equation*}
  \cos\beta=-\cos(\uppi-\beta).
\end{equation*}
\end{definition}

\begin{theorem}[Kosin\"us Teoremi]
  \Sekilde{fig:tri}
  \begin{equation*}
    a^2=b^2+c^2-2bc\cos\alpha.
  \end{equation*}
  \begin{figure}[ht]
\psset{unit=8mm}
\mbox{}\hfill
\begin{pspicture}(-1,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(-1,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](-1,2){$C$}
\uput[ur](0.5,1){$a$}
\uput[dl](-0.5,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill
\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(0,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](0,2){$C$}
\uput[ur](1,1){$a$}
\uput[l](0,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill
\begin{comment}
  

\begin{pspicture}(0,-0.5)(2,2.5)
  \pspolygon(0,0)(2,0)(1,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](1,2){$C$}
\uput[ur](1.5,1){$a$}
\uput[ul](0.5,1){$b$}
\uput[ur](0,0){$\alpha$}
\end{pspicture}
\hfill


\end{comment}
\begin{pspicture}(0,-0.5)(3,2.5)
  \pspolygon(0,0)(2,0)(3,2)
\uput[d](0,0){$A$}
\uput[d](1,0){$c$}
\uput[d](2,0){$B$}
\uput[u](3,2){$C$}
\uput[dr](2.5,1){$a$}
\uput[ul](1.5,1){$b$}
\uput{9pt}[18](0,0){$\alpha$}
\end{pspicture}
\hfill\mbox{}
    \caption{Kosin\"us Teoremi}\label{fig:tri}
    
  \end{figure}
\end{theorem}

\begin{definition}
  Dik eksenlere g\"ore
  \begin{gather*}
    (a,b)\cdot(c,d)=ac+bd,\\
\lVert(a,b)\rVert=\sqrt{(a,b)\cdot(a,b)}=\sqrt{a^2+b^2}.
  \end{gather*}
$(a,b)$ noktas\i, $(0,0)$ ba\c slang\i\c c noktas\i ndan
$(a,b)$ noktas\i na giden y\"onl\"u do\u gru olarak anla\c s\i labilir.
\end{definition}

\begin{theorem}
  Dik eksenlere g\"ore $(a,b)$ ve $(c,d)$ aras\i ndaki a\c c\i\ $\theta$ ise
  \begin{equation*}
    (a,b)\cdot(c,d)=\lVert(a,b)\rVert\cdot\lVert(c,d)\rVert\cdot\cos\theta.
  \end{equation*}
\end{theorem}

\begin{theorem}[Cauchy--Schwartz E\c sitsizli\u gi]
  \begin{equation*}
(ac+bd)^2\leq(a^2+b^2)\cdot(c^2+d^2).
  \end{equation*}
\end{theorem}

\begin{definition}[mutlak de\u ger]
  $\abs a=
  \begin{cases}
    a,&\text{ e\u ger $a\geq0$ ise},\\
-a,&\text{ e\u ger $a<0$ ise}.
  \end{cases}$
\end{definition}

\begin{theorem}
  Dik eksenlere ve birim uzunlu\u guna g\"ore
$(s,t)$ noktas\i n\i n $ax+by+c=0$ do\u grusuna uzakl\i\u g\i
  \begin{equation*}
    \frac{\abs{as+bt+c}}{\sqrt{a^2+b^2}}.
  \end{equation*}
\end{theorem}

\section{Dik eksenlere g\"ore koni kesitleri}

\begin{definition}
  $a\neq0$ ise
  \begin{equation*}
    \frac a{\infty}=0.
  \end{equation*}
\end{definition}

\begin{remark}
\c Simdi $\ell>0$ ve $a\neq0$ ise,
dik eksenlere g\"ore,
\begin{equation*}
  y^2=\ell x-\frac{\ell}{2a}x^2
\end{equation*}
Apollonius denklemi,
ekseni $x$ ekseni olan
ve k\"o\c sesi ba\c slang\i\c c noktas\i\ olan
\begin{itemize}
\item 
$a<0$ durumunda hiperbol\"u,
\item
$a=\infty$ durumunda parabol\"u,
\item
$a>0$ durumunda elipsi
\end{itemize}
tan\i mlar.
\Sekle{fig:all} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=1.5cm,plotpoints=200,labelsep=3pt}
  \begin{pspicture}(-3,-2.74)(3,2.74)
\psline{->}(-3,0)(3,0)
\psline{->}(0,-2.74)(0,2.74)
\psset{linewidth=1.6pt}
    \psplot03{x x x mul 2 div add sqrt}
    \psplot03{x x x mul 2 div add sqrt neg}
    \psplot{-3}{-2}{x x x mul 2 div add sqrt}
    \psplot{-3}{-2}{x x x mul 2 div add sqrt neg}
    \psplot0{2.3}{x x x mul 1 div add sqrt}
    \psplot0{2.3}{x x x mul 1 div add sqrt neg}
    \psplot{-3}{-1}{x x x mul 1 div add sqrt}
    \psplot{-3}{-1}{x x x mul 1 div add sqrt neg}
    \psplot03{x                   sqrt}
    \psplot03{x                   sqrt neg}
    \psplot02{x x x mul 2 div sub sqrt}
    \psplot02{x x x mul 2 div sub sqrt neg}
    \psplot01{x x x mul 1 div sub sqrt}
    \psplot01{x x x mul 1 div sub sqrt neg}
\uput[dr](1,0){$a=1/2$}
\uput[ur](2,0){$a=1$}%(1.75,0.4677)
\uput[dr](2,1.414){$a=\infty$}
\uput[r](2,2){$a=-1$}
\uput[ul](2,2.4495){$a=-1/2$}
  \end{pspicture}
  \caption{$y^2=x-x^2/2a$ koni kesitleri}\label{fig:all}
  
\end{figure}
\end{remark}

\begin{remark}
  $\ell>0$ ve $a\neq0$ (ve $a\neq\infty$) olsun, ve
  \begin{align*}
    b&>0,&2b^2=\ell a
  \end{align*}
olsun.
O zaman dikey kenar\i\ $\ell$ olan,
yanlamas\i na kenar\i\ $2\abs a$ olan merkezli koni kesitlerinin 
merkez denklemi
\begin{equation*}
  \frac{x^2}{a^2}\pm\frac{y^2}{b^2}=1.
\end{equation*}
\Sekle{fig:central} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=1.2cm,plotpoints=200}
  \begin{pspicture}(-4,-2)(4,2)
\psline{->}(-4,0)(4,0)
\psline{->}(0,-2)(0,2)
\psset{linestyle=dotted}
\psline(-4,-2)(4,2)
\psline(-4,2)(4,-2)
\pspolygon(-2,-1)(-2,1)(2,1)(2,-1)
\psset{linestyle=solid,linewidth=1.6pt}
    \psplot{-2}2{1 x x mul 4 div sub sqrt}
    \psplot{-2}2{1 x x mul 4 div sub sqrt neg}
    \psplot24{1 x x mul 4 div sub neg sqrt}
    \psplot24{1 x x mul 4 div sub neg sqrt neg}
    \psplot{-2}{-4}{1 x x mul 4 div sub neg sqrt}
    \psplot{-2}{-4}{1 x x mul 4 div sub neg sqrt neg}
\uput[ur](2,0){$a$}
\uput[dl](-2,0){$-a$}
\uput[ur](0,1){$b$}
\uput[ur](0,-1){$-b$}
  \end{pspicture}
  \caption{$x^2/4\pm y^2=1$ koni kesitleri}\label{fig:central}
  
\end{figure}
\end{remark}

\begin{definition}
$x^2/a^2-y^2/b^2=0$ denklemi,  
$x^2/a^2-y^2/b^2=1$ hiperbol\"un \textbf{asimptotlar\i n\i} tan\i mlar.
Yani hiperbol\"un asimptotlar\i, $y=\pm(b/a)x$ do\u grular\i d\i r.
\end{definition}

\begin{theorem}
  $y^2=\ell x+(\ell/2a)x^2$ hiperbol\"un asimptotlar\i n\i n denklemi
  \begin{equation*}
    y=\pm\sqrt{\frac{\ell}{2a}}\cdot(x-a).
  \end{equation*}
\end{theorem}

\begin{definition}
Denklemi $y^2=\ell x$ olan parabol\"un \textbf{odak noktas\i}
\begin{equation*}
  \left(\frac{\ell}4,0\right)
\end{equation*}
ve \textbf{do\u grultman do\u grusu}
\begin{equation*}
  x+\frac{\ell}4=0.
\end{equation*}
\end{definition}

\begin{theorem}
  Denklemi $y^2=\ell x$ olan parabol\"un noktalar\i,
odak noktas\i na ve do\u grultmana uzakl\i\u g\i\ ayn\i\ olan noktalard\i r.
\end{theorem}

\begin{definition}
Denklemi $x^2/a^2-y^2/b^2=1$ olan hiperbol\"un
\textbf{odak noktalar\i}
\begin{equation*}
  (\pm\sqrt{a^2+b^2},0),
\end{equation*}
(s\i ras\i yla) \textbf{do\u grultman do\u grular\i}
\begin{equation*}
  x=\pm\frac{a^2}{\sqrt{a^2+b^2}},
\end{equation*}
ve \textbf{d\i\c smerkezlili\u gi}
\begin{equation*}
  \frac{\sqrt{a^2+b^2}}a.
\end{equation*}
\Sekle{fig:hyp-foc} bak\i n.
\begin{figure}[ht]
  \centering
\psset{unit=6mm,plotpoints=200}
  \begin{pspicture}(-9,-6)(9,6)
%\psset{linewidth=1pt}
\psline{->}(-9,0)(9,0)
\psline{->}(0,-6)(0,6)
\psset{linestyle=dotted}
%\psline(-4.5,-6)(4.5,6) % asymptote
%\psline(-4.5,6)(4.5,-6)
\psline(3,0)(0,5)
\pscircle(0,0)3
\psline(0,3)(1.8,0)
\pspolygon(-3,-4)(-3,4)(3,4)(3,-4)
\psdots(5,0)(-5,0)(3,0)(-3,0)(0,4)(0,-4)
\pscircle(0,0)5
\psset{linestyle=solid,linewidth=1.6pt}
\psline(1.8,-6)(1.8,6) % directrix
\psline(-1.8,-6)(-1.8,6)
    \psplot3{5.4}{x x mul 9 div 1 sub sqrt 4 mul}
    \psplot3{5.4}{x x mul 9 div 1 sub sqrt 4 mul neg}
    \psplot{-5.4}{-3}{x x mul 9 div 1 sub sqrt 4 mul}
    \psplot{-5.4}{-3}{x x mul 9 div 1 sub sqrt 4 mul neg}
\uput[ur](3,0){$a$}
\uput[ul](-3,0){$-a$}
\uput[dl](0,4){$b$}
\uput[ul](0,-4){$-b$}
\uput[dr](5,0){$\sqrt{a^2+b^2}$}
\uput[dl](-5,0){$-\sqrt{a^2+b^2}$}
  \end{pspicture}  
  \caption{Hiperbol\"un odaklar\i\ ve do\u grultmanlar\i}\label{fig:hyp-foc}
  
\end{figure}
Ayr\i ca $0<b<a$ ise
denklemi $x^2/a^2+y^2/b^2=1$ olan elipsin
\textbf{odak noktalar\i}
\begin{equation*}
  (\pm\sqrt{a^2-b^2},0),
\end{equation*}
(s\i ras\i yla) \textbf{do\u grultman do\u grular\i}
\begin{equation*}
  x=\pm\frac{a^2}{\sqrt{a^2-b^2}},
\end{equation*}
ve \textbf{d\i\c smerkezlili\u gi}
\begin{equation*}
  \frac{\sqrt{a^2-b^2}}a.
\end{equation*}
\Sekle{fig:ell-foc} bak\i n.
\begin{figure}[ht]
\psset{unit=1.15cm,plotpoints=200}
  \centering
  \begin{pspicture}(-4.5,-2)(4.5,2)
\psline{->}(-4.5,0)(4.5,0)
\psline{->}(0,-2)(0,2)
\psset{linestyle=dotted}
\pspolygon(-2,-1.732)(2,-1.732)(2,1.732)(-2,1.732)
\psarc(0,1.732)2{180}{360}
\psline(1,0)(1,1.732)
\psarc(0,1.732)1{180}{360}
\psline(0,0.732)(2,1.732)
\psline(0,-0.268)(4,1.732)(2,1.732)
\psdots(1,0)(-1,0)(0,1.732)(0,-1.732)(2,0)(-2,0)(4,0)(-4,0)
\psset{linewidth=1.6pt,linestyle=solid}
    \psplot{-2}2{1 x x mul 4 div sub 3 mul sqrt}
    \psplot{-2}2{1 x x mul 4 div sub 3 mul sqrt neg}
\psline(4,-2)(4,2)
\psline(-4,-2)(-4,2)
\uput[d](1,0){$\sqrt{b^2-a^2}$}
\uput[ur](2,0){$a$}
\uput[d](-1,0){$-\sqrt{b^2-a^2}$}
\uput[ul](-2,0){$-a$}
\uput[dl](0,1.732){$b$}
\uput[ul](0,-1.732){$-b$}
\uput[dl](4,0){$\frac a{\sqrt{a^2-b^2}}$}
\uput[dr](-4,0){$\frac{-a}{\sqrt{a^2-b^2}}$}
  \end{pspicture}
  \caption{Elipsin odaklar\i\ ve do\u grultmanlar\i}\label{fig:ell-foc}
\end{figure}
\end{definition}

\begin{remark}
\begin{figure}[ht]
\mbox{}\hfill
\psset{unit=2cm,plotpoints=200}
  \begin{pspicture}(-0,-0.25)(2,1)
\uput[d](0,0){$A$}
\uput[d](1,0){$B$}
\uput[d](1.4142,0){$C$}
\uput[d](0.7071,0){$D$}
\psline(0,0)(1.4142,0)
\psdots(1.4142,0)(0,0)
\psset{linewidth=1.6pt}
\psline(0.7071,0)(0.7071,1)
    \psplot1{1.4142}{1 x x mul sub neg sqrt}
  \end{pspicture}  
\hfill
  \begin{pspicture}(-2.3094,-0.5)(0,1)
\uput[d](0,0){$A$}
\uput[d](-2,0){$B$}
\uput[d](-1.7320,0){$C$}
\uput[d](-2.3094,0){$D$}
\psline(-2.3094,0)(0,0)
\psdots(-1.7320,0)(0,0)
\psset{linewidth=1.6pt}
    \psplot{-2}0{1 x x mul 4 div sub sqrt}
\psline(-2.3094,0)(-2.3094,1)
  \end{pspicture}
\hfill\mbox{}
  \caption{Odak ve do\u grultman}\label{fig:foc-dir}
  
\end{figure}
\Sekilde{fig:foc-dir}ki gibi
merkezli koni kesitinin merkezi $A$, ve
(merkezin ayn\i\ taraf\i nda olan) k\"o\c sesi $B$, ve oda\u g\i\ $C$ ise,
ve do\u grultman\i, koni kesitinin eksenini $D$ noktas\i nda keserse,
tan\i ma g\"ore koni kesitinin d\i\c smerkezlili\u gi $AC:AB$, ama
\begin{equation*}
  AC:AB::AB:AD,
\end{equation*}
dolay\i s\i yla
\begin{equation*}
  BC:BD::AC:AB,
\end{equation*}
ve sonu\c c olarak koni kesitinin d\i\c smerkezlili\u gi $BC:BD$.
\end{remark}


\begin{theorem}\label{thm:ecc}
  Merkezli koni kesitinin noktalar\i,
bir odak noktas\i na 
ve ona kar\c s\i l\i k gelen do\u grultman do\u grusuna 
uzakl\i klar\i n\i n oran\i n\i n d\i\c smerkezlilik oldu\u gu noktalard\i r.
Yani 
%koni kesitinin d\i\c smerkezlili\u gi $\epsilon$ ise,
\Sekilde{fig:ecc}
\begin{figure}[ht]
  \centering
\psset{unit=2.4cm,plotpoints=200}
  \begin{pspicture}(-2,-1.414)(2,1.414)
\psline(-1.4142,0)(1.4142,0)
\psset{linestyle=dotted}
\psline(-1.4142,0)(-1.5,1.1180)(0.7071,1.1180)
\psline(-1.5,1.1180)(1.4142,0)
\psdots(1.4142,0)(-1.4142,0)(0,0)
\psset{linestyle=solid,linewidth=1.6pt}
\psline(0.7071,-1.414)(0.7071,1.414)
\psline(-0.7071,-1.414)(-0.7071,1.414)
    \psplot1{1.732}{1 x x mul sub neg sqrt}
    \psplot1{1.732}{1 x x mul sub neg sqrt neg}
    \psplot{-1}{-1.732}{1 x x mul sub neg sqrt}
    \psplot{-1}{-1.732}{1 x x mul sub neg sqrt neg}
\uput[d](0,0){$A$}
\uput[dl](-1,0){$B$}
\uput[dr](1,0){$B'$}
\uput[l](-1.4142,0){$C$}
\uput[r](1.4142,0){$C'$}
\uput[dr](-0.7071,0){$D$}
\uput[dl](0.7071,0){$D'$}
\uput[dl](-1.5,1.1180){$E$}
\uput[ul](-0.7071,1.1180){$F$}
\uput[r](0.7071,1.1180){$F'$}
  \end{pspicture}  

\psset{unit=1.6cm}
  \begin{pspicture}(-2,-2)(2,2)
\psline(-2.8284,0)(2.8284,0)
\psset{linestyle=dotted}%](-2,-1.4142)(2,-1.4141)(2,1.4142)(-2,1.4142)
\psline(-1.4141,0)(1,1.2247)(1.4142,0)
\psline(-2.8284,1.2247)(2.8284,1.2247)
\psdots(1.4142,0)(-1.4142,0)(0,0)
\psset{linewidth=1.6pt,linestyle=solid}
    \psplot{-2}2{1 x x mul 4 div sub 2 mul sqrt}
    \psplot{-2}2{1 x x mul 4 div sub 2 mul sqrt neg}
\psline(2.8284,-1.4142)(2.8284,1.4142)
\psline(-2.8284,-1.4142)(-2.8284,1.4142)
\uput[d](0,0){$A$}
\uput[dl](-2,0){$B$}
\uput[dr](2,0){$B'$}
\uput[d](-1.4142,0){$C$}
\uput[d](1.4142,0){$C'$}
\uput[l](-2.8284,0){$D$}
\uput[r](2.8284,0){$D'$}
\uput[ur](1,1.2247){$E$}
\uput[l](-2.8284,1.2247){$F$}
\uput[r](2.8284,1.2247){$F'$}
  \end{pspicture}
  \caption{D\i\c smerkezlilik}\label{fig:ecc}
\end{figure}
($CB=C'B'$ ve $BD=B'D'$ oldu\u gundan)
a\c sa\u g\i daki ko\c sullar denktir:
\begin{itemize}
\item 
$E$ noktas\i\ koni kesitinde,
\item
$CE:EF::CB:BD$,
\item
$C'E:EF'::CB:BD$.
\end{itemize}
\end{theorem}

\begin{remark}
$BC:BD::AB:AD::BB':DD'$
oldu\u gundan
merkezli koni kesitinin $E$ noktalar\i\ i\c cin
(ve sadece bu noktalar i\c cin)
\begin{equation*}
  C'E\pm CE:EF'\pm EF::BB':DD'.
\end{equation*}
Elipste $EF'+EF=FF'=DD'$,
dolay\i s\i yla
\begin{equation*}
  CE+C'E=BB'.
\end{equation*}
Hiperbolde $E$, soldaki daldaysa $EF'-EF=DD'$, dolay\i s\i yla
\begin{equation*}
  C'E-CE=BB'.
\end{equation*}
\end{remark}


\end{document}