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
$$
\angle SHC - \angle BSC = 90^\circ \text{,} \quad\quad\
\angle THC - \angle DTC = 90^\circ \text{.}
$$
Prove that the circumcircle of triangle SHT is tangent to the line $BD$.
\begin{flushright}\emph{(Iran)}\end{flushright}
Školjka 
be a convex quadrilateral with 
. Point 
is the foot of the perpendicular from 
to 
. The points 
and 
are chosen on the sides 
and 
, respectively, in such a way that 
and 
Prove that the circumcircle of triangle SHT is tangent to the line