2012 OIM Problems/Problem 5

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Problem

Let $ABC$ be a triangle and let $P$ and $Q$ be the points of intersection of the parallel to $BC$ by $A$ with the exterior bisectors of the angles $\angle B$ and $\angle C$, respectively. The perpendicular to $BP$ at $P$ and the perpendicular to $CQ$ at $Q$ intersect at $R$. If $I$ is the incenter of $ABC$, show that $AI = AR$.

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

Solution

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See also

OIM Problems and Solutions