Problem

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Tags: geometry, parallel, incircle, Kharkiv



The incircle $\omega$ of triangle $ABC$ touches its sides $BC, CA$ and $AB$ at points $D, E$ and $E$, respectively. Point $G$ lies on circle $\omega$ in such a way that $FG$ is a diameter. Lines $EG$ and $FD$ intersect at point $H$. Prove that $AB \parallel CH$.