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IMO 2014 Problem 3


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

\displaystyle \angle CHS -\angle CSB = 90^\circ,\ \angle THC-\angle DTC = 90^\circ.

Prove that the line {BD} is tangent to the circumcircle of triangle {TSH}.

IMO 2014 Problem 3 (Day 1)

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Categories: Geometry, IMO, Problem Solving Tags: ,
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