Question 15
Let $ABC$ be a triangle such that $\angle A = 90^{\circ}$ and $\angle B \lt \angle C$. The tangent at $A$ to its circumcircle $\omega$ meets the line $BC$ at $D$. Let $E$ be the reflection of $A$ across $BC$, $X$ the foot of the perpendicular from $A$ to $BE$, and $Y$ the midpoint of $AX$. Let the line $BY$ meet $\omega$ again at $Z$. Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$.