Question

The equal sides $AB$ and $AC$ of an isosceles
triangle $ABC$ are produced to point $P$ and $Q$
respectively such that $BP.CQ=A{B}^2$. Prove that the line $PQ$ always passes through a fixed point.

Answer 1

let the base $BC$ be taken along axis of $x$ and its mid-point be chosen as origin so that the points $B$ and $C$ are $(-a,0)$ and $(a,0)$ respectively. The third vertex $A$ will lie on $y-$axis at $(0,b)$ say. By given condition
$\frac{BP}{AB}=\frac{AB}{CQ}=\lambda sayor\frac{BP}{AB}=\frac{AC}{CQ}=\lambda \therefore AB=AC$
$B$ divides $PA$ in the ratio $\lambda :1$ and $C$ divides $AQ$ in the ratio $\lambda :1$ as shown in the figure. Hence by ratio formula the co-ordinate of $P$ are
$[-a(\lambda +1),-\lambda b]$ and $Q$ is $[a\frac{\lambda +1}{\lambda },\frac{b}{\lambda }]$
Slop of $PQ$ by $=\frac{y_2-y_1}{x_2-x_1}$ is $\frac{b}{a}\frac{\lambda -1}{\lambda +1}$
Hence equation of $PQ$ is
$y+\lambda b=\frac{b}{a}\frac{\lambda -1}{\lambda +1}[x+a(\lambda +1)]$
Or $a(\lambda +1)y+ab\lambda (\lambda +1)=b(\lambda -1)x+ab({\lambda }^2-1)$
Cancel ${ab\lambda }^2$ and collect the terms of $\lambda$
$(ay+bx+ab)+\lambda (bx-ay-ab)=0$
Above is of the form $u+\lambda v=0$ which represents a family of straight lines passing through the intersection of $u=0$ i.e. the point the points $(0,-b)$ which is a fixed point.