ABC is a given triangle in which AB = AC .

Question

ABC is a given triangle in which $AB = AC $. The sides AB and AC are produced to P and Q respectively such that $BP.CQ = A B ^ 2 $Prove that the line PQ always passes through a fixed point.

Is there something that the questions hints here or are we supposed to put the triangle in the coordinate system and manually solve it.

Answer 1

For the solution involving coordinate geometry we can proceed as

Let $ A = (0,0)$ ,$ B = (a,b)$,$ C = (-a, b)$

let P and Q lie on extended lines AB and AC respectively

Now Given $ BP.CQ =AB^2 $

Therefore $ (BP/AB)(CQ/AC)=1$ since AB = AC. Let $ (BP/AB)= r and (CQ/AC)= 1/r$

By external section formula $P = [a(r+1) , b(r+1)]$ and $Q = [-a(r+1)/r , b(r+1)/r]$

Then line PQ is ${x-a(r+1)}{b(r-1)} = {y-b(r+1)}{a(r+1)}$ i.e. $r(bx - ay + 2ab) - (bx + ay - 2ab) = 0$ Therefore PQ pases through $(0,2b) $

Answer 2

Through $B$ draw a line parallel to $AC$, through $C$ draw a line parallel to $AB$, these two lines intersect at $X$. We prove that $X$ is the point you want.

In fact, for any $P$ on the extension of $AB$, link $PX$ and extend it to intersect (extension of) $AC$ at $Q'$, we just need to show $BP\cdot CQ'=AB^2$; this would imply $Q'=Q$ and we see $PQ$ passes through $X$.

To achieve this, we just observe $BQ : AB= PX : XQ'$ and $ CQ': AC= XQ': PX$; multiply these two we see $BQ\cdot CQ'/AB\cdot AC = 1$, that is $BQ\cdot CQ'=AC^2$.