IMO 2014 problem 3


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21. rujna 2014.
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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 to 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 \text{.}
Prove that line BD is tangent to the circumcircle of triangle TSH.
Izvor: International Mathematical Olympiad 2014, day 1