Question 136
In acute triangle $ABC$, $H$ is the orthocenter. Let $\omega$ be the circumcircle of triangle $BHC$, and $O'$ be its center. $\Gamma$ is the circle that passes through the midpoints of the edges of triangle $ABC$. Point $X$ is an arbitrary point lying on arc $BHC$ of $\omega$. Line $AX$ intersects $\Gamma$ for the second time at point $Y$, where $X$ lies between $A$ and $Y$. $P$ is a point on $\Gamma$ such that $PX = PY$. Prove that $\angle O'PX = 90^\circ$.