Question 159
Let $ABC$ be an acute triangle, and let $D$ be the foot of the altitude from $A$. The circle with centre $A$ passing through $D$ intersects the circumcircle of triangle $ABC$ in $X$ and $Y$, in such a way that the order of the points on this circumcircle is: $A$, $X$, $B$, $C$, $Y$. Show that $\angle BXD = \angle CYD$.