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 $AF=AG$.
Note that $w$ is an Apollonius circle, that means for any point $X$ on $w$, $\frac{BX}{XC}$ is constant. Thus $FD$ is the angle bisector of $\angle BFA$. So $D$ is the incenter of $\triangle ABF$. By angle chasing, $\angle FAG=\angle FBA$ and $\angle AGF=180-\angle FDA=90-\angle DBA$, so $AF=AG$.
also, note that if $O$ is the center of $w$ (midpoint of $BC$) then since $(EDBC)=-1$, we have $CA.CF=CD.CE=CB.CO$ so $BOFA$ is cyclic. the rest is a simple angle chasing.