Question 100
Let $ABC$ be an isosceles triangle with $BC=CA$. Let $D$ lie in the interior of side $AB$ and satisfy $AD\lt DB$. Choose $P,Q$ in the interiors of sides $BC,CA$, respectively, with $\angle DPB=\angle DQA=90^\circ$. The perpendicular bisector of $PQ$ meets the line $CQ$ at $E$, and the circumcircles of $ABC$ and $CPQ$ meet again at $F\ne C$. Given that $P,E,F$ are collinear, prove that $\angle ACB=90^\circ$.
The English source specifies the segment $CQ$ instead of the line; the intersection lies on that segment automatically, as verified in the solutions.
The English source specifies the segment $CQ$ instead of the line; the intersection lies on that segment automatically, as verified in the solutions.