\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

Prove Menelanou's theorem: \lq \lq If a line cuts the sides of a
triangle $ABC$ in the points $C_1, A_1, B_1$ then $\ds
\frac{AC_1}{C_1B} \cdot \frac{BA_1}{A_1C} \cdot \frac{CB_1}{B_1A}
= -1 $".

Show, conversely, that if the above product of the ratios is -1
then $C_1, A_1, B_1$ are colinear.

\vspace{.25in}

{\bf Answer}

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\begin{picture}(4,2)
\put(0.5,0){\line(1,0){6}}

\put(0.5,0){\line(1,1){2}}

\put(2.5,2){\line(1,-1){2}}

\put(1.5,1.5){\line(3,-1){5}}

\put(0.2,0){\makebox(0,0){$A$}}

\put(2.5,2.3){\makebox(0,0){$B$}}

\put(4.6,0.2){\makebox(0,0){$C$}}

\put(4,1){\makebox(0,0){$A_1$}}

\put(6,.3){\makebox(0,0){$B_1$}}

\put(1.9,1.75){\makebox(0,0){$C_1$}}

\end{picture}

${}$

Let $A,B,C,$ have position vectors ${\bf a, \, b, \, c,}$

Then $\ds {\bf c}_1 = \frac{\mu_1{\bf a} + \lambda_1{\bf
b}}{\lambda_1 + \mu_1} \hspace{.2in} {\bf a}_1 = \frac{\mu_2{\bf
b} + \lambda_2{\bf c}}{\lambda_2 + \mu_2} \hspace{.2in}{\bf b}_1 =
\frac{\mu_3{\bf c} + \lambda_3{\bf a}}{\lambda_3 + \mu_3}$

${}$

$C_1 A_1 B_1$ are collinear if and only if for some $t\epsilon{\bf
R}$,  ${\bf b}_1 = t{\bf c}_1 + (1-t){\bf a}_1$

Thus $$ \frac{\mu_3{\bf c} + \lambda_3{\bf a}}{\lambda_3 + \mu_3}
= t  \frac{\mu_1{\bf a} + \lambda_1{\bf b}}{\lambda_1 + \mu_1} +
(1-t) \frac{\mu_2{\bf b} + \lambda_2{\bf c}}{\lambda_2 + \mu_2}$$

i.e. iff

$\left. \begin{array}{rcl} \ds \frac{\mu_3}{\lambda_3 + \mu_3} & =
& (1-t) \ds \frac{\lambda_2}{\lambda_2 + \mu_2} \\ 0 & = & t \ds
\frac{\lambda_1}{\lambda_1 + \mu_1} + (1-t)\frac{\mu_2}{\lambda_2
+ \mu_2} \\ \ds \frac{\lambda_3}{\lambda_3 + \mu_3} & = & t \ds
\frac{\mu_1}{\lambda_1+\mu_1} \end{array} \right\}$ Comparing of
${\bf a, b, c}$ independently.

From the third and first equations, substituting in the second we
have,

\begin{eqnarray*} 0 & = & \frac{\lambda_3}{\lambda_3 + \mu_3} \cdot
\frac{\lambda_1 + \mu_1}{\mu_1}\cdot
\frac{\lambda_1}{\lambda_1+\mu_1} + \frac{\mu_3}{\lambda_3
+\mu_3}\cdot\frac{\lambda_2+\mu_2}{\lambda_2}\cdot
\frac{\mu_2}{\lambda_2 + \mu_2} \\ & = &
\frac{\lambda_3\lambda_1}{\mu_1} + \frac{\mu_3 \mu_2}{\lambda_2}
\\
\\ -1 & = & \frac{\lambda_1 \lambda_2 \lambda_3 }{\mu_1 \mu_2
\mu_3}
\end{eqnarray*}

\end{document}