\documentclass[lang=en,11pt,blue,fancy]{elegantbook-uci}
% customized by Zhiqin Lu on 12/19/2020

%\usepackage{pgfplots}
\usepackage{tikz}
\usetikzlibrary{calc}
%\usetikzlibrary{shapes, positioning}
\usepackage{tkz-euclide}
%\usetikzlibrary{shapes.geometric}

\usepackage{amsmath,mathrsfs,amssymb,amscd,mathtools}
%\usepackage[framemethod=TikZ]{mdframed}
 



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\newcommand{\pa}{\partial}
\newcommand{\x}{{\bf x}}
\newcommand{\p}{{\bf p}}
\newcommand{\q}{{\bf q}}

\newcommand{\uu}{{\bf u}}
\newcommand{\vv}{{\bf v}}
\newcommand{\ww}{{\bf w}}
\newcommand{\xx}{{\bf x}}
\newcommand{\yy}{{\bf y}}


\newcommand{\ee}{{\bf e}}
\newcommand{\ff}{{\bf f}}
\newcommand{\tri}{\triangle}



\renewcommand{\proofname}{Proof.}
\newcommand{\elinkname}{External Link.}

% Marking the right angle

\def\dotMarkRightAngle[size=#1](#2,#3,#4){%
 \draw [red,thick]($(#3)!#1!(#2)$) -- 
       ($($(#3)!#1!(#2)$)!#1!90:(#2)$) --
       ($(#3)!#1!(#4)$);
}


\title{The Geometry of Triangles}
\subtitle{UCI Math}

\author{Zhiqin Lu}
\institute{University of California, Irvine}
\date{February 14, 2021}
\version{0.01}
\bioinfo{Bio}{Zhiqin Lu is a professor at UCI}

\extrainfo{\emph{Partially supported by a  grant.}}


\logo{2.png}
\cover{cover.jpg}








\begin{document}
\maketitle

\frontmatter
\tableofcontents

\mainmatter

\chapter{Pascal and Brianchon Theorems}
\begin{pro}{\yellow{Pascal Theorem}}{}
The hexagon $ABCDEF$ is inscribed to the circle. Assume that $AB$, $CD$ intersects at $X$; $BC$, $EF$ intersects at $Y$; and
$CD$, $FA$ intersects at $Z$. Then $X,Y,Z$ are collinear.
\begin{center}
\begin{tikzpicture}
\coordinate(o) at (0,0);
\draw[thick] (o) circle (1.5cm);
\coordinate(a) at (10:1.5cm);
\coordinate(b) at (45:1.5cm);
\coordinate(c) at (80:1.5cm);
\coordinate(d) at (110:1.5cm);
\coordinate(e) at (225:1.5cm);
\coordinate(f) at (310:1.5cm);

\tkzInterLL(a,b)(d,e)\tkzGetPoint{x}
\tkzInterLL(b,c)(e,f)\tkzGetPoint{y}
\tkzInterLL(c,d)(f,a)\tkzGetPoint{z}
\tkzDrawPolygon[black,thick](a,b,c,d,e,f)

\draw[thick,blue](b)--(x)node[above,black]{$X$};
\draw[thick,blue](d)node[left,black,xshift=1mm,yshift=2mm]{$D$}--(x);
\draw[thick,blue](b)node[below,xshift=-1mm,black]{$B$}--(y)node[right,black]{$Y$};
\draw[thick,blue](f)--(y);
\draw[thick,blue](c)node[below,black]{$C$}--(z)node[above,black,xshift=1mm,yshift=-.7mm]{$Z$};
\draw[thick,blue](a)node[right,black]{$A$}--(z);
\draw[red,thick](x)--(y)--(z);

\draw(e)node[below,black,xshift=-1mm]{$E$};
\draw(f)node[below,black]{$F$};


\end{tikzpicture}
\end{center}

\end{pro}
\vspace{.2in}
\begin{remark}
The above is called Pascal Theorem, which was discovered by the French mathematician \emph{Blaise Pascal}. when he was 16 years old. The theorem can be generalized to the case to the case of conic section (see \href{https://en.wikipedia.org/wiki/Conic_section}{Wikipedia here}). When the conic section is degenerated to two lines, it is also called the \emph{\red{Pappus Hexagon Theorem}} below -- some people believed that Euclid knew this theorem before Pappus.
\end{remark}
\vspace{.2in}
\begin{theorem*}{(Pappus Theorem)}{}
In the following picture, the Hexagon $BC'AB'CA'$ is inscribed on the two black lines. Assume that $BC',B'C$ intersect at $X$; $CA', A'C$ intersect at $Y$, and $AB', B'A$ intersect at $Z$. Then $X,Y,Z$ are collinear.
\begin{center}
\begin{tikzpicture}
\draw[very thick](-1,0)--(8,0);
\coordinate (a) at (0,0);
\coordinate (b) at (3,0);
\coordinate (c) at (7,0);
\draw[very thick](-1,2.5)--(8,4);
\coordinate (a') at (1,17/6);
\coordinate (b') at (4,10/3);
\coordinate (c') at (6,11/3);

\draw[thick,blue](a)node[below,black]{$A$}--(b')node[above,black]{$B'$};
\draw[thick,blue](a)--(c');
\draw[thick,blue](b)node[below,black]{$B$}--(a')node[above,black]{$A'$};
\draw[thick,blue](b)--(c')node[above,black]{$C'$};
\draw[thick,blue](c)node[below,black]{$C$}--(a');
\draw[thick,blue](c)--(b');

\tkzInterLL(b,c')(b',c)\tkzGetPoint{x}
\tkzInterLL(a,c')(a',c)\tkzGetPoint{y}
\tkzInterLL(b,a')(b',a)\tkzGetPoint{z}

\draw[thick,red] (x)node[right,black,xshift=0mm]{$X$}--(y)node[below,black,xshift=0mm]{$Y$}--(z)node[left,black,xshift=0mm]{$Z$};

\end{tikzpicture}
\end{center}
\end{theorem*}

\newpage
\begin{proof}
We use analytic geometry to prove the Pascal Theorem. 

We assume the circle is the unit circle. 
Let the equations for 
\[
AB,BC,CD,DE,EF,FA\] be $\ell_1,\ell_2,\cdots,\ell_6$. These functions $\ell_j$ are linear functions. As a result, we consider two cubic polynomials $\ell_1\ell_3\ell_5$ and $\ell_2\ell_4\ell_6$. Obviously, these two polynomials pass the nine points $A,B,C,D,E,F,X,Y,Z$. 

We choose a general point $P$ in the circle. Choose a number $\lambda$ such that 
\[
(\ell_1\ell_3\ell_5+\lambda \ell_2\ell_4\ell_6)(P)=0.
\]

Here is a fundamental question: in general, if a cubic curve doesn't vanishing identically on the unit circle, then what is the maximum number of intersections? The answer is $6$, and we shall prove it. 


We can use complex numbers to write any cubic polynomials as 
\[
f(z)=Az^3+Bz^2\bar z+Cz\bar z^2+D\bar z^3+Ez^2+Fz\bar z+G\bar z^2+Hz+I\bar z+J=0.
\]
Let $z=e^{i\theta}$ be a point on the unit disk, then $\bar z=1/z$. If we multiply the above equation by $z^3$ on both sides, we get a degree $6$ polynomial of single variables. In general, such a $6$-degree polynomial has at most $6$ roots. Since $f(z)$ vanishes on severn points $A,B,C,D,E,F,P$ on the unit circle, it must be vanishing identically on the circle. As a result, we can factorize it as 
\[
f(z)=(|z|^2-1)\ell(z),
\]
where, by the degree consideration, $\ell(z)$ must be linear. Since $\ell$ passes $X,Y,Z$, we conclude that $X,Y,Z$ are collinear. 



\end{proof}
\newpage


\begin{theorem*}{(Brianchon Theorem)}{}
The Hexagon $ABCDEF$ is circumscribed on a circle. Then $AD, BE$, and $CF$ are concurrent.
\begin{center}
\begin{tikzpicture}
\coordinate(o) at (0,0);
\draw[thick] (o) circle (2cm);
\coordinate(a_1) at (30:2);
\coordinate(b_1) at (80:2);
\coordinate(c_1) at (120:2);
\coordinate(d_1) at (2000:2);
\coordinate(e_1) at (280:2);
\coordinate(f_1) at (-20:2);




\tkzDefTangent[at=a_1](o)\tkzGetPoint{a_2}
\tkzDefTangent[at=b_1](o)\tkzGetPoint{b_2}
\tkzDefTangent[at=c_1](o)\tkzGetPoint{c_2}
\tkzDefTangent[at=d_1](o)\tkzGetPoint{d_2}
\tkzDefTangent[at=e_1](o)\tkzGetPoint{e_2}
\tkzDefTangent[at=f_1](o)\tkzGetPoint{f_2}

\tkzInterLL(a_1,a_2)(b_1,b_2)\tkzGetPoint{a}
\tkzInterLL(b_1,b_2)(c_1,c_2)\tkzGetPoint{b}
\tkzInterLL(c_1,c_2)(d_1,d_2)\tkzGetPoint{c}
\tkzInterLL(d_1,d_2)(e_1,e_2)\tkzGetPoint{d}
\tkzInterLL(e_1,e_2)(f_1,f_2)\tkzGetPoint{e}
\tkzInterLL(f_1,f_2)(a_1,a_2)\tkzGetPoint{f}




\draw[thick](a)node[above]{$A$}--(b)node[above]{$B$}--(c)node[left]{$C$}--(d)node[below]{$D$}--(e)node[below]{$E$}--(f)node[right]{$F$}--cycle;

\draw[green,thick](a)--(d);\draw[green,thick](b)--(e);\draw[green,thick](c)--(f);
\draw[blue,fill] (o)node[left,black]{$O$} circle(1pt);

\end{tikzpicture}
\end{center}

\end{theorem*}

\begin{remark} Pascal Theorem and Brainchon Theorem are two famous ``dual'' theorems. We can prove it using Pascal Theorem. 
\end{remark}

\vspace{.2in}


\begin{definition*}{(Pole and Polar)}{}
Let $O$ be the unit circle. The the pair $(A, PQ)$ is called the pair of pole and polar, where $A$ is the pole, and $PQ$ is the polar. 

Let $(x_0,y_0)$ be the coordinates of $A$. Then the equation of $PQ$ is
\[
x_0x+y_0y-1=0.
\]


\begin{center}
\begin{tikzpicture}[scale=1.4]
\coordinate (o) at (0,0);

\coordinate (R) at (0:1cm);


\draw[thick](o) circle(1cm);
\coordinate (a) at (-3,0);
\tkzDefTangent[from = a](o,R) \tkzGetPoints{p}{q}

\draw[thick](p)node[below]{$P$}--(q)node[above]{$Q$};
\draw[thick](q)-- (a)--(p);
\draw[thick] (a)--(o)node[right]{$O$};
\draw[fill](a)node[left]{$A$} circle(1pt);
\draw[fill](o)circle(1pt);

\end{tikzpicture}
\end{center}
\end{definition*}
\newpage

\begin{proof}
Let the coordinates of $A_i$ be $(x_i,y_i)$ for $1\leq i\leq 6$. Then the equations for $B_6B_1$ is
\[
\ell_1(x,y)=x_1x+y_1y-1.
\]
Similarly, the equations for $B_iB_{i+1}$ for $1\leq i\leq 5$ are
\[
\ell_i(x,y)=x_ix+y_iy-1.
\]
Using the Pascal Theorem, there is a number $\lambda$ such that
\[
\ell_1\ell_3\ell_5+\lambda\ell_2\ell_4\ell_6=C(x^2+y^2-1)(px+qy-1),
\]
where $C$ is a constant. We claim $(p,q)$ is on the lines $A_1A_4$, $A_2A_5$, and $A_3A_6$.

In order to prove this, let $P=(p_1,q_1)$ be the intersection of $B_1B_6$ and $B_3B_4$. Then we have
\[
x_1p_1+y_1q_1-1=0,\qquad x_4p_1+y_4q_1-1=0.
\]
Moreover, we have
\[
pp_1+qq_1-1=0.
\]
Thus the three points $A_1,A_4$ and $(p,q)$ are on the line
\[
p_1x+q_1y-1=0.
\]
This completes the proof. 
\begin{center}
\begin{tikzpicture}
\coordinate(o) at (0,0);
\draw[thick] (o) circle (2cm);
\coordinate(a_1) at (30:2);
\coordinate(b_1) at (80:2);
\coordinate(c_1) at (120:2);
\coordinate(d_1) at (2000:2);
\coordinate(e_1) at (280:2);
\coordinate(f_1) at (-20:2);




\tkzDefTangent[at=a_1](o)\tkzGetPoint{a_2}
\tkzDefTangent[at=b_1](o)\tkzGetPoint{b_2}
\tkzDefTangent[at=c_1](o)\tkzGetPoint{c_2}
\tkzDefTangent[at=d_1](o)\tkzGetPoint{d_2}
\tkzDefTangent[at=e_1](o)\tkzGetPoint{e_2}
\tkzDefTangent[at=f_1](o)\tkzGetPoint{f_2}

\tkzInterLL(a_1,a_2)(b_1,b_2)\tkzGetPoint{a}
\tkzInterLL(b_1,b_2)(c_1,c_2)\tkzGetPoint{b}
\tkzInterLL(c_1,c_2)(d_1,d_2)\tkzGetPoint{c}
\tkzInterLL(d_1,d_2)(e_1,e_2)\tkzGetPoint{d}
\tkzInterLL(e_1,e_2)(f_1,f_2)\tkzGetPoint{e}
\tkzInterLL(f_1,f_2)(a_1,a_2)\tkzGetPoint{f}




\draw[thick](a)node[above]{$A_1$}--(b)node[above]{$A_2$}--(c)node[left]{$A_3$}--(d)node[below]{$A_4$}--(e)node[below]{$A_5$}--(f)node[right]{$A_6$}--cycle;

\draw[green,thick](a)--(d);\draw[green,thick](b)--(e);\draw[green,thick](c)--(f);
\draw[blue,fill] (o)node[left,black]{$O$} circle(1pt);

\draw(b_1)node[above,red]{$B_1$};
\draw(c_1)node[above,red,xshift=-2mm,yshift=-1mm]{$B_2$};
\draw(d_1)node[left,red]{$B_3$};
\draw(e_1)node[below,red]{$B_4$};
\draw(f_1)node[right,red]{$B_5$};
\draw(a_1)node[right,red]{$B_6$};

\draw[red](b_1)--(c_1)--(d_1)--(e_1)--(f_1)--(a_1)--cycle;
\end{tikzpicture}
\end{center}


\end{proof}

\nocite{*}

 
\bibliography{geometry-bib}

\end{document}



\chapter{Ceva and Menelaus Theorems}
\begin{pro}{\yellow{Menelaus Theorem}}{}
In the following $\tri ABC$. $D,E,F$ are points on $BC$, $CA$, and $AB$, respectively. Assume that $D,E,F$ are collinear. Then
\[
\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot \frac{CE}{EA}=1
\]
\begin{center}
\begin{tikzpicture}[scale=1.2]
\tkzDefPoint (0,0){b}
\tkzDefPoint (4,0){c}




\tkzInterCC[R](b,3.2cm)(c, 3.9cm)
\tkzGetPoints{a}{a'}

\tkzDrawPolygon[black,  thick](a,b,c)




\tkzDefMidPoint(a,c)\tkzGetPoint{e}









\draw(a)node[above]{$A$};



\coordinate(f) at ($(a)!.7!(b)$);
\tkzInterLL(e,f)(b,c)\tkzGetPoint{d}
\draw[thick](b)node[below]{$B$}--(d)node[below]{$D$}--(e)node[right,yshift=1mm]{$E$};;

\draw(c)node[below]{$C$};
\draw(f)node[above,xshift=-.5mm]{$F$};


\end{tikzpicture}
\end{center}

\end{pro}

\vspace{.2in}

\begin{remark}
The inverse theorem is also true, that is, if the above equation is valid, then $D,E,F$ are collinear. 
\end{remark}
\begin{elink}
\href{https://en.wikipedia.org/wiki/Menelaus\%27s_theorem}{Here is the Menelaus Theorem on Wikipedia.}
\end{elink}

\vspace{.2in}



\renewcommand{\proofname}{Proof:}

\begin{proof} We draw $BK$ parallel to $CA$. The idea is to represent all three fractions to the quotients on the line $DE$. 
First, we have
\[
\frac{AF}{FB}=\frac{FE}{KF},\qquad\frac{BD}{DC}=\frac{DK}{DE}
\]
We then have
\[
\frac{CE}{EA}=\frac{CE}{BK}\cdot\frac{BK}{EA}=\frac{DE}{DK}\cdot\frac{KF}{FE}.
\]
\begin{center}
\begin{tikzpicture}[scale=1.2]
\tkzDefPoint (0,0){b}
\tkzDefPoint (4,0){c}




\tkzInterCC[R](b,3.2cm)(c, 3.9cm)
\tkzGetPoints{a}{a'}

\tkzDrawPolygon[black,  thick](a,b,c)




\tkzDefMidPoint(a,c)\tkzGetPoint{e}









\draw(a)node[above]{$A$};



\coordinate(f) at ($(a)!.7!(b)$);
\tkzInterLL(e,f)(b,c)\tkzGetPoint{d}
\draw[thick](b)node[below]{$B$}--(d)node[below]{$D$}--(e)node[right,yshift=1mm]{$E$};;

\draw(c)node[below]{$C$};
\draw(f)node[above,xshift=-.5mm]{$F$};


\tkzDefParallelogram(b,a,c)
\tkzGetPoint{p}
\tkzInterLL(p,b)(d,e)
\tkzGetPoint{k}
\draw(k) node[above]{$K$};
\draw[red,dashed,thick](k)--(b);
\end{tikzpicture}
\end{center}
Putting the above all together, we prove the theorem. 
\end{proof}
\vspace{.2in}



\renewcommand{\aa}{{\bf a}}
\newcommand{\bb}{{\bf b}}
\newcommand{\cc}{{\bf c}}
\newcommand{\dd}{{\bf d}}
%\newcommand{\ee}{{\bf e}}
%\newcommand{\ff}{{\bf f}}


\begin{proof}
Algebraic proof. Let $A,B,C$ be represented by vectors $\aa,\bb,\cc$, respectively. Assume that 
\[
\frac{AF}{FB}=\lambda,\qquad\frac{CE}{EA}=\mu,\qquad\frac{AF}{FB}=\nu.
\]
Let $D,E,F$ be represented by vectors $\dd,\ee,\ff$. Then we have
\[
\ee=\frac{1}{1+\mu}\cc+\frac{\mu}{1+\mu}\aa,\qquad\ff=\frac{1}{1+\nu}\aa+\frac{\nu}{1+\nu}\bb.
\]
The vector $\dd$ is a little special. Since $D$ is outside the line segment $BC$, we have
\[
\dd=\frac{1}{1-\lambda}\bb+\frac{-\lambda}{1-\lambda}\cc
\]
Since $D,E,F$ are collinear, we can write one vector, say $\dd$, as a linear combination of $\ee,\ff$, that is, we can have
\[
\dd=c_1\ee+c_2\ff,
\]
which is
\[
\frac{1}{1-\lambda}\bb+\frac{-\lambda}{1-\lambda}\cc=\frac{c_1}{1+\mu}\cc+\frac{c_1\mu}{1+\mu}\aa+\frac{c_2}{1+\nu}\aa+\frac{c_2\nu}{1+\nu}\bb.
\]
From the above equation, we get
\[
\frac{1}{1-\lambda}=\frac{c_2\nu}{1+\nu},\quad \frac{-\lambda}{1-\lambda}=\frac{c_1}{1+\mu},\quad \frac{c_1\mu}{1+\mu}+\frac{c_2}{1+\nu}=0.
\]
From the above first two equations, we have
\[
c_1=\frac{-\lambda(1+\mu)}{1-\lambda},\quad c_2=\frac{1+\nu}{(1-\lambda)\nu}.
\]
Substituting these into the third equation and simplifying, we get $\lambda\mu\nu=1$.
  \end{proof}
  

  Let
  \[
  \dd'=(1-\lambda')\dd,\qquad   \ee'=(1+\mu')\ee,\qquad   \ff'=(1+\nu')\ff,
  \]
  where $\lambda'=\lambda,\mu=-\mu',\nu=-\nu'$. Then
  \[
  \dd'=\bb-\lambda'\cc,\qquad   \ee'=\cc-\mu'\aa,\qquad  \ff'=\aa-\nu'\bb,\qquad
  \]

\begin{proposition*}{}{}
We have
 \[
  \dd'=-\lambda'\ee'-\lambda'\mu'\ff'.
  \]
\end{proposition*}
\begin{proof}
  Since $\lambda'\mu'\nu'=1$, the we have
  \[
  \dd'=-\lambda'\ee'-\lambda'\mu'\ff'.
  \]
  \end{proof}

Using the above theorem, we shall revisit the Ceva Theorem.
\begin{theorem*}{Ceva Theorem}{}
In the following picture, the lines $AD,BE, CF$ are concurrent if and only if
\[
\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1.
\]
\begin{center}
\begin{tikzpicture}[scale=1]
\draw[black,thick] (0,0)node[below]{$A$}--(6,0)node[below]{$B$}--(4,4)node[above]{$C$}--cycle;
\draw[black,thick] (4,4)--(3.592,0)node[below]{$F$};
\draw[black, thick] (2.54,2.54)node[left,yshift=2mm,xshift=1mm]{$E$}--(6,0);
\draw[black, thick] (4.924,2.154)node[right,yshift=2mm,xshift=-1mm]{$D$}--(0,0);
\node at (3.5,1.3) [black]{$G$};
 \end{tikzpicture}

  \end{center}
  \end{theorem*}
  
  \begin{proof}
  Let $\ell_1, \ell_2,\ell_3$ be the equations of the lines $BC, CA, AB$\footnote{$\ell_j$ are liner functions, for example, if $\ell$ is an equation of a line, then $\ell(\xx)=ax+by-c$, where $\xx=(x,y)$.}.
  As before, we have
  \[
  \dd=\frac{1}{1+\lambda}\bb+\frac{\lambda}{1+\lambda}\cc,\quad  \ee=\frac{1}{1+\mu}\cc+\frac{\mu}{1+\mu}\aa,\quad\ff=\frac{1}{1+\nu}\aa+\frac{\nu}{1+\nu}\bb.
  \]
  We claim that 
  \[
  \ell_2-\frac{\ell_2(\bb)}{\lambda\ell_3(\cc)}\ell_3
  \]
  is the equation of $AD$.
Since
\[
\frac{\ell_2(\bb)}{\lambda\ell_3(\cc)}\cdot\frac{\ell_3(\cc)}{\mu\ell_1(\aa)}\cdot\frac{\ell_1(\aa)}{\nu\ell_2(\bb)}=1,
\]
we know 
  \[
    \ell_2-\frac{\ell_2(\bb)}{\lambda\ell_3(\cc)}\ell_3
    \]
    can be written as a linear combination of 
    \[
     \ell_3-\frac{\ell_3(\cc)}{\mu\ell_1(\aa)}\ell_1,\qquad  \ell_1-\frac{\ell_1(\aa)}{\nu\ell_2(\bb)}\ell_2.
    \]
    Thus $AD, BE, CF$ are concurrent. 
  \end{proof}

\vspace{.2in}
\begin{elink}
Menelaus Theorem and Ceva Theorem are examples of ``dual theorems'' in projective geometry. See \href{https://www.heldermann-verlag.de/jgg/jgg11/j11h1beni.pdf}{here} for a unified proof using cross ratios.  In particular, the idea of cross ratios is sometimes used in AIME.
\end{elink}

\begin{theorem*}{(Desargues Theorem)}{}
We consider the triangles  $\tri ABC$ and $\tri A'B'C'$. Assume that the lines $BC$, $B'C'$ intersect at $P$, $CA$ and $C'A'$ intersect at $Q$, and $AB$, $A'B'$ inersect at $R$. Then $P,Q,R$ are collinear if and only if $AA',BB'$ and $CC'$ are concurrent. 
\begin{center}
\begin{tikzpicture}[scale=.6]
\tikzstyle{point1}=[ball color=cyan, circle, draw=black, inner sep=1pt]
\tikzstyle{point2}=[ball color=green, circle, draw=black, inner sep=1pt]
\tikzstyle{point3}=[ball color=red, circle, draw=black, inner sep=1pt]
\node(v1) at (0,8) [ball color=blue, circle, draw=black, inner sep=1pt] {};
\node (v2) at (0,6) [point1] {};
\node (v3) at (2,5.5) [point1] {};
\node (v4) at (1.5,4) [point1] {};
\node (v5) at (0,0) [point2] {};
\node (v6) at (2.75*2,8-2.75*2.5) [point2] {};
\node (v7) at (1.5*1.5,8-1.5*4) [point2] {};
\draw (v1) node[above]{$\scriptstyle O$}-- (v2) -- (v5)node[below]{$\scriptstyle B'$};
\draw (v1) -- (v3) -- (v6);
\draw (v1)-- (v4) -- (v7);
\draw (v2)node[left]{$\scriptstyle B$}-- (v3) -- (v4)-- (v2);
\draw (v5) -- (v6)node[above,xshift=1mm]{$\scriptstyle C'$} -- (v7)node[below]{$\scriptstyle A'$} -- (v5);
\node (v8) at (intersection of v2--v3 and v5--v6) [point3] {};
\node (v9) at (intersection of v2--v4 and v5--v7) [point3] {};
\node (v10) at (intersection of v3--v4 and v6--v7) [point3] {};
\draw (v3)node[above,xshift=1mm]{$\scriptstyle C$} -- (v8)node[above,xshift=1mm]{$\scriptstyle P$} -- (v6);
\draw (v4) node[right]{$\scriptstyle A$}-- (v9) node[above,xshift=1mm]{$\scriptstyle R$}-- (v7);
\draw (v4) -- (v10) -- (v7);
\draw (v8) -- (v9)-- (v10)node[left]{$\scriptstyle Q$};
\end{tikzpicture}
\end{center}
\end{theorem*} The line $PQR$ is called  the \emph{\red{axis of perspectivity}}, and the point $O$ is called the \emph{\red{center of perspectivity}}. The theorem is known as \emph{\red{Perspective principle}} in painting. See 
\href{https://en.wikipedia.org/wiki/Desargues\%27s_theorem}{the Wikipedia for details.}

\vspace{.2in}

\begin{proof}
Here we provide a proof using the Menelaus Theorem only. On $\tri OB'C'$, since $B,C,P$ are collinear, we must have
\[
\frac{OB}{BB'}\cdot\frac{B'P}{PC'}\cdot\frac{C'C}{CO}=1.
\]
On $\tri OC'A'$, since $C,A,Q$ are collinear, we have
\[
\frac{C'C}{CO}\cdot\frac{OA}{AA'}\cdot\frac{A'Q}{QC'}=1.
\]
On $\tri OA'B'$, since $A,B,R$ are collinear, we have
\[
\frac{OB}{BB'}\cdot\frac{B'R}{RA'}\cdot\frac{A'A}{AO}=1.
\]
From the above 3 equations, we get
\[
\frac{A'Q}{Q'C}\cdot\frac{C'P}{PB'}\cdot\frac{B'R}{RA'}=1.
\]
Therefore, $P,Q,R$ are collinear. 
\end{proof}
\newpage







