Question 97
In an isosceles triangle $ABC$ with $BC=CA$, let $D$ be an interior point of the side $AB$ such that $AD\lt DB$. Let $P$ and $Q$ be interior points of the sides $BC$ and $CA$, respectively, such that $\angle DPB=\angle AQD=90^\circ$. The perpendicular bisector of $PQ$ meets $CQ$ at $E$, and the circumcircles of triangles $ABC$ and $QPC$ meet again at $F\ne C$.
Prove that if $P,E,F$ are collinear, then $\angle ACB=90^\circ$.
Prove that if $P,E,F$ are collinear, then $\angle ACB=90^\circ$.