Question 44
Let $ABC$ be a triangle with circumcircle $\omega$, circumcenter $O$, and orthocenter $H$. Let $K$ be the midpoint of $AH$. The perpendicular to $OK$ at $K$ intersects $AB$ and $AC$ at $P$ and $Q$, respectively. The lines $BK$ and $CK$ intersect $\omega$ again at $X$ and $Y$, respectively. Prove that the second intersection of the circumcircles of triangles $KPY$ and $KQX$ lies on $\omega$.