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
\[ \angle CHS - \angle CSB = 90 ^ \circ, \angle THC - \angle DTC = 90^ \circ . \]
Prove that line \(BD\) is tangent to the circumcirlce of triangle \(TSH\).