ISL 1998 G8

Let $ABC$ be a triangle such that $\angle A=90^{\circ }$ and $\angle B<\angle C$ . The tangent at $A$ to the circumcircle $\omega$ of triangle $ABC$ meets the line $BC$ at $D$ . Let $E$ be the reflection of $A$ in the line $BC$ , let $X$ be the foot of the perpendicular from $A$ to $BE$ , and let $Y$ be the midpoint of the segment $AX$ . Let the line $BY$ intersect the circle $\omega$ again at $Z$ .

Prove that the line $BD$ is tangent to the circumcircle of triangle $ADZ$ .

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