Problem

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Tags: geometry, incenter, right triangle



In a triangle $ABC$, $\angle B=90^o$ and $\angle A=60^o$, $I$ is the point of intersection of its angle bisectors. A line passing through the point $I$ parallel to the line $AC$, intersects the sides $AB$ and $BC$ at the points $P$ and $T$ respectively. Prove that $3PI+IT=AC$ . (Anton Trygub)