ISL 2014 G5

Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and $$\angle S H C-\angle B S C=90^{\circ}, \angle T H C-\angle D T C=90^{\circ} .$$ Prove that the circumcircle of triangle $S H T$ is tangent to the line $B D$. (Iran)