ISL 2023 G2

Let $A B C$ be a triangle with $A C>B C$. Let $\omega$ be the circumcircle of triangle $A B C$ and let $r$ be the radius of $\omega$. Point $P$ lies on segment $A C$ such that $B C=C P$ and point $S$ is the foot of the perpendicular from $P$ to line $A B$. Let ray $B P$ intersect $\omega$ again at $D$ and let $Q$ lie on line $S P$ such that $P Q=r$ and $S, P, Q$ lie on the line in that order. Finally, let the line perpendicular to $C Q$ from $A$ intersect the line perpendicular to $D Q$ from $B$ at $E$. Prove that $E$ lies on $\omega$.