[BLACK-Dragon]

Given an acute-angled triangle $ABC$, let $H$ be the orthocenter. $A$ cirlcle passing through the points $B,C$ and a cirlcle with a diameter $AH$ intersect at two distinct points $X,Y$ . Let $D$ be the foot of the perpendicular drawn from $A$ to line $BC$, and let $K$ be the foot of the perpendicular drawn from $D$ to line $ XY $ . Show that $ \angle{BKD}=\angle{CKD} $