IMO 2014 Problem 3
IMO 2014 Problem 3 is a plane geometry problem from the International Mathematical Olympiad 2014, Problem 3: line BD tangent to the circle through T, S and H.
Problem
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 $TSH$.
Source International Mathematical Olympiad 2014, Problem 3
The diagram
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