Question 191
Let $ABC$ be a triangle with $AC \gt BC$. Let $\omega$ be the circumcircle of triangle $ABC$ and let $r$ be the radius of $\omega$. Point $P$ lies on segment $AC$ such that $BC = CP$ and point $S$ is the foot of the perpendicular from $P$ to line $AB$. Let ray $BP$ intersect $\omega$ again at $D$ and let $Q$ lie on line $SP$ such that $PQ = r$ and $S, P, Q$ lie on the line in that order. Finally, let the line perpendicular to $CQ$ from $A$ intersect the line perpendicular to $DQ$ from $B$ at $E$.
Prove that $E$ lies on $\omega$.
Prove that $E$ lies on $\omega$.