JBMO-SL 2020 G1

Let $\triangle A B C$ be an acute triangle. The line through $A$ perpendicular to $B C$ intersects $B C$ at $D$. Let $E$ be the midpoint of $A D$ and $\omega$ the the circle with center $E$ and radius equal to $A E$. The line $B E$ intersects $\omega$ at a point $X$ such that $X$ and $B$ are not on the same side of $A D$ and the line $C E$ intersects $\omega$ at a point $Y$ such that $C$ and $Y$ are not on the same side of $A D$. If both of the intersection points of the circumcircles of $\triangle B D X$ and $\triangle C D Y$ lie on the line $A D$, prove that $A B=A C$.