Question 61
Let $ABCD$ be a convex quadrilateral with $\angle B = \angle D = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. The points $S$ and $T$ are chosen on the sides $AB$ and $AD$, respectively, in such a way that $H$ lies inside triangle $SCT$ and
$$
\angle SHC - \angle BSC = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}.
$$
Prove that the circumcircle of triangle $SHT$ is tangent to the line $BD$.
$$
\angle SHC - \angle BSC = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}.
$$
Prove that the circumcircle of triangle $SHT$ is tangent to the line $BD$.