If the equal sides AB and AC (each equal to a) of a right angled isosceles triangle ABC be produced to P and Q so that BP×CQ=AB2, then the line PQ always passes through the fixed point

We take A as the origin and AB and AC as x-axis and y-axis respectively.Let AP = h, AQ = k. Equation of the line PQ isxh+yk=1             ------(1)Given, BP×CQ=AB2⇒h-ak-a=a2⇒hk-ak-ah+a2=a2 or  ak + ha = hkor ah+ak=1           ------(2)From (2), it follows that line (1) i.e., PQ passes through the fixed point (a, a)