Home > Geometry, IMO, Problem Solving > IMO 2014 Problem 3

## 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)

Advertisements
Categories: Geometry, IMO, Problem Solving Tags: ,
1. No comments yet.
1. No trackbacks yet.