FKMO 2018 Problem 2

Triangle $ABC$ satisfies $\angle ABC < \angle BCA < \angle CAB < 90^{\circ}$ . $O$ is the circumcenter of triangle $ABC$ , and $K$ is the reflection of $O$ in $BC$ . $D,E$ is the foot of perpendicular line from $K$ to line $AB$ , $AC$ , respectively. Line $DE$ meets $BC$ at $P$ , and a circle with diameter $AK$ meets the circumcircle of triangle $ABC$ at $Q(\neq A)$ . If $PQ$ cuts the perpendicular bisector of $BC$ at $S$ , then prove that $S$ lies on the circle with diameter $AK$ .