International Mathematical Olympiad 2014 Problem 3

Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90°$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and $$\angle CHS - \angle CSB = 90°, \quad \angle THC - \angle DTC = 90°.$$

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