Question 97
Given a triangle $ABC$, the internal bisector of $\hat{A}$ intersects $BC$ at $D$ and the external bisector of $\hat{A}$ intersects $BC$ at $E$. Let $C_1$ be the circumcircle of $ADE$. $AC$ intersects $C_1$ at $F$. Let $C_2$ be the circumcircle of $ABF$ and the tangent of $C_2$ at $A$ intersects $C_1$ at $G$. Prove that $AF = AG$.