Question 57
Let $ABCD$ be a cyclic quadrilateral with circumcircle $\omega$ and circumradius $r$. The diagonals $AC$ and $BD$ intersect at $P$. Suppose that $AD = DP$. Let $S$ be the foot of the perpendicular from $P$ to the line $AB$. Point $Q$ lies on line $SP$ such that $PQ = r$ and $S, P, Q$ lie on the line in that order. Let the line perpendicular to $CQ$ from $A$ intersect the line perpendicular to $DQ$ from $B$ at $E$. Prove that $E$ lies on $\omega$.