Question 22
In convex quadrilateral $ABCD$, $AB \perp AD$ and $AD = DC$. Let point $E$ lie inside segment $BC$, and point $F$ lie on the extension of $DE$ such that $\angle ABF = \angle DEC \gt 90^\circ$. Let $O$ be the circumcenter of triangle $CDE$. Let $P$ be a point on the extension of $FO$ such that $FP = FB$. Let segment $BP$ intersect $AC$ at point $Q$.
Prove: $\angle AQB = \angle DPF$.
Prove: $\angle AQB = \angle DPF$.