Question 124
In triangle $ABC$ denote by $O$ and $H$ be the circumcenter and the orthocenter. The point $P$ is the reflection of $A$ with respect to $OH$. Assume that $P$ is not on the same side of $BC$ as $A$. Points $E$ and $F$ lie on sides $AB$ and $AC$, respectively, such that $BE = PC$ and $CF = PB$. Let $K$ be the intersection point of $AP$ and $OH$. Prove that $\widehat{EKF} = 90^\circ$.