Let the line passing through $C$ and perpendicular to the line $SC$ intersect the line $AB$ at $Q$ (see Figure 1). Then
$$\angle SQC = 90^{\circ} - \angle BSC = 180^{\circ} - \angle SHC,$$
which implies that the points $C$, $H$, $S$, and $Q$ lie on a common circle. Moreover, since $SQ$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $SHC$ lies on the line $AB$. Similarly, we prove that the circumcentre $L$ of triangle $CHT$ lies on the line $AD$.

Figure 1

In order to prove that the circumcircle of triangle $SHT$ is tangent to $BD$, it suffices to show that the perpendicular bisectors of $HS$ and $HT$ intersect on the line $AH$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $AKH$ and $ALH$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that
$$\begin{equation*} \frac{AK}{KH} = \frac{AL}{LH} . \tag{1} \end{equation*}$$
We present two proofs of this equality.

First proof. Let the lines $KL$ and $HC$ intersect at $M$ (see Figure 2). Since $KH = KC$ and $LH = LC$, the points $H$ and $C$ are symmetric to each other with respect to the line $KL$. Therefore $M$ is the midpoint of $HC$. Denote by $O$ the circumcentre of quadrilateral $ABCD$. Then $O$ is the midpoint of $AC$. Therefore we have $OM \parallel AH$ and hence $OM \perp BD$. This together with the equality $OB = OD$ implies that $OM$ is the perpendicular bisector of $BD$ and therefore $BM = DM$.

Since $CM \perp KL$, the points $B$, $C$, $M$, and $K$ lie on a common circle with diameter $KC$. Similarly, the points $L$, $C$, $M$, and $D$ lie on a circle with diameter $LC$. Thus, using the sine law, we obtain
$$\frac{AK}{AL} = \frac{\sin \angle ALK}{\sin \angle AKL} = \frac{DM}{CL} \cdot \frac{CK}{BM} = \frac{CK}{CL} = \frac{KH}{LH},$$
which finishes the proof of (1).

Figure 2

Second proof. If the points $A$, $H$, and $C$ are collinear, then $AK = AL$ and $KH = LH$, so the equality (1) follows. Assume therefore that the points $A$, $H$, and $C$ do not lie in a line and consider the circle $\omega$ passing through them (see Figure 3). Since the quadrilateral $ABCD$ is cyclic,
$$\angle BAC = \angle BDC = 90^{\circ} - \angle ADH = \angle HAD.$$
Let $N \neq A$ be the intersection point of the circle $\omega$ and the angle bisector of $\angle CAH$. Then $AN$ is also the angle bisector of $\angle BAD$. Since $H$ and $C$ are symmetric to each other with respect to the line $KL$ and $HN = NC$, it follows that both $N$ and the centre of $\omega$ lie on the line $KL$. This means that the circle $\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).

Figure 3