Problem

Source: Mongolia TST 2011 Test 3 #2

Tags: geometry, circumcircle, incenter, angle bisector, geometry unsolved



Given a triangle $ABC$, the internal and external bisectors of angle $A$ intersect $BC$ at points $D$ and $E$ respectively. Let $F$ be the point (different from $A$) where line $AC$ intersects the circle $w$ with diameter $DE$. Finally, draw the tangent at $A$ to the circumcircle of triangle $ABF$, and let it hit $w$ at $A$ and $G$. Prove that $AF=AG$.