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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.

commentEdited by Orl.

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