Question 183
Let $ABC$ be an acute triangle. The line through $A$ perpendicular to $BC$ intersects $BC$ at $D$. Let $E$ be the midpoint of $AD$ and $\omega$ the circle of center $E$ and radius $AE$. The line $BE$ intersects $\omega$ at $X$ such that $X$ and $B$ are not on the same side of $AD$, and the line $CE$ intersects $\omega$ at $Y$ such that $C$ and $Y$ are not on the same side of $AD$. If two intersection points of the circumcircles of triangles $BDX$ and $CDY$ lie on the line $AD$, prove that $AB = AC$.