2014 IMO Problems/Problem 3

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 $\triangle{TSH}.$

Solution

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

2014 IMO (Problems) • Resources
Preceded by Problem 2	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 4
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2014 IMO Problems/Problem 3

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