ISL 2020 G1

Let $A B C$ be an isosceles triangle with $B C=C A$, and let $D$ be a point inside side $A B$ such that $A D<D B$. Let $P$ and $Q$ be two points inside sides $B C$ and $C A$, respectively, such that $\angle D P B=\angle D Q A=90^{\circ}$. Let the perpendicular bisector of $P Q$ meet line segment $C Q$ at $E$, and let the circumcircles of triangles $A B C$ and $C P Q$ meet again at point $F$, different from $C$. Suppose that $P, E, F$ are collinear. Prove that $\angle A C B=90^{\circ}$. (Luxembourg)