Problem

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Tags: geometry, trigonometry, circumcircle, ratio, projective geometry, geometry proposed



Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE = CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF = FG$.