Question 80
Let $ABC$ be an acute triangle and let $H$ be its orthocenter. Suppose that a circle going through the points $B$, $C$ and the circle having the line segment $AH$ as its diameter intersect at two distinct points $X$ and $Y$. Let $D$ be the foot of the perpendicular line drawn from $A$ to the line $BC$, and $K$ be the foot of the perpendicular line drawn from $D$ to the line $XY$. Prove that $\angle BKD = \angle CKD$ must hold.