Question 111
Let $ABC$ be an acute triangle, $P$ be the midpoint of the side $BC$ and $K$ be the foot of the altitude from $A$. Let $D$ be a point on the segment $AP$ such that $\angle BDC = 90^\circ$. Let the second intersection point of the circumcircle of $ADK$ and line $BC$ be $E$. Let the second intersection point of the circumcircle of $ABC$ and line $AE$ be $F$. Prove that $\angle AFD = 90^\circ$.