2006 OIM Problems/Problem 5

Problem

Given a circle $\Gamma$, consider a quadrilateral $ABCD$ with its four sides tangent to $\Gamma$, with $AD$ tangent to $\Gamma$ at $P$ and $CD$ tangent to $\Gamma$ at $Q$. Let $X$ and $Y$ be the points where $BD$ intersect $\Gamma$, and $M$ the midpoint of $XY$. Show that $\angle AMP = \angle CMQ$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

OIM Problems and Solutions