IMO 2014 Problem 3

Convex quadrilateral $A B C D$ has $\angle A B C = \angle C D A = 90^{\circ}$ . Point $H$ is the foot of the perpendicular from $A$ to $\overline{{B D}}$ . Points $S$ and $T$ lie on sides $A B$ and $A D$ , respectively, such that $H$ lies inside triangle $S C T$ and

$$\angle C H S - \angle C S B = 90^{\circ}, \angle T H C - \angle D T C = 90^{\circ}.$$

Prove that line $B D$ is tangent to the circumcircle of triangle $T S H$