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Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. 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 \[
\angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.

Let $M$ and $N$ be the reflection of $C$ wrt $B$ and $D$.Then $MSHC$ and $CHTN$ are cyclic.Now $S$ and $T$ are the midpoint of arc $MHC$ and $NHC$ of $(MHC)$ and $(NHC)$.This implies $HS$ and $HT$ are the external angle bisector of $\angle{MHC}$ and $\angle{NHC}$.After some easy angle chasing we can show that $\angle{MAS}=\angle{HAT}$ and $\angle{MHS}=\angle{AHT}$.So $M$ is the isogonal conjugate of $T$ wrt $\triangle SAH$.So $\angle{AST}=180-\angle{MSH}=\angle{MCH}$.So $CH\perp ST$.Again $\angle{AHT}=\angle{MHS}=\angle{SHP}$.So the circumcenter of $\triangle TSH$ lies on $AH$.Again $AH\perp BD$.So $BD$ touches circumcircle of $\triangle TSH$.

In my solution the reflections of $C$ wrt $B,D$ are $C_1,C_2$ respectively.

I also proved that $C_1$ is the isogonal conjugate of $T$ wrt $\triangle ASH$ in the same way Asif did. Thus it is easy to see that $\angle TSH=\angle C_1SB$. Now, $\angle AHT=\angle C_1HS=\angle C_1CS=\angle BC_1S=90^{\circ}-\angle C_1SB=90^{\circ}-\angle TSH$.
Again, $\angle AHT=90^{\circ}-\angle THD$. So $\angle THD=\angle TSH$ which means $BD$ is tangent to the circumcircle of $\triangle TSH$.