Question 17
Let $ABCD$ be a convex quadrilateral with $AB \gt AD$ and $\angle B = \angle D = 90^\circ$. Let $P$ be the point on side $AB$ such that $AP = AD$. Lines $PD$ and $BC$ intersect at $Q$. The perpendicular line to $AC$ through $Q$ intersects $AB$ at $R$. Let $S$ be the foot of the perpendicular from $D$ to $AC$. Prove that $\angle PSQ = \angle RCP$.