Problem

Source: 2018 Iran MO 3rd Geometry 3

Tags: geometry, circumcircle



$H$ is the orthocenter of acude triangle $ABC$.Let $\omega$ be the circumcircle of $BHC$ with center $O'$.$\Omega$ is the nine-point circle of $ABC$.$X$ is an arbitrary point on arc $BHC$ of $\omega$ and $AX$ intersects $\Omega$ at $Y$.$P$ is a point on $\Omega$ such that $PX=PY$.Prove that $O'PX=90$.