Question

If a line $ax+by+c=0$ cuts the sides $BC,CA$ and $AB$ of triangle $ABC$ at points $P,Q$ and $R$ respectively, the prove that
$\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$

Answer 1

$P$ is $(\frac{\lambda x_3+x_2}{\lambda +1}.\frac{\lambda y_3+y_2}{\lambda +1})$ where $\lambda =\frac{BP}{PC}$
$P$ lies on line $ax+by+c=0$
$a\frac{(\lambda x_3+x_2)}{\lambda +1}+b\frac{(\lambda y_3+y_2)}{\lambda +1}+c=0$
$\therefore \lambda (a{c}_2+b{y}_3+c)+(a{x}_2+b{y}_2+c)=0$
$\therefore \lambda =-\frac{(a{x}_2+b{y}_2+c)}{(a{x}_3+b{y}_3+c)}=-\frac{l_2}{l_3}$ where $\lambda =\frac{BP}{PC}$
Similarly, $\mu =-\frac{l_3}{l_1}$ where $\mu =\frac{CQ}{QA}$
and $v=-\frac{l_1}{l_2}$ where $v=\frac{AR}{RB}$
$\therefore \lambda \mu v=\left(\frac{l_2}{l_3}\right)\times \left(\frac{l_3}{l_1}\right)\times \left(\frac{l_1}{l_2}\right)=-1$