Problem

Source: FKMO 2018 Day 1 Problem 2

Tags: geometry, circumcircle



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 $AK$.