Show that $\angle BFD = \angle CFD$ Given a $\triangle ABC$ whose incenter is $I$ and its incircle is $\gamma = \odot (I,ID)$ with $D \in BC$.
Also, define $\{E,D\} = \gamma \cap AD$ and $F$ as the midpoint of $ED$.
Show that $\angle BFD = \angle CFD$

my attempt: this looks a lot like Blanchet's theorem. So I figured that point $G$, the harmonic conjugate of $D$ wrt $BC$ ($\mathcal H (B,C;D,G)$) could be useful and indeed if we show that $G,F$ and $I$ are collinear, the problem will be solved.
A curious fact is that $G$ is in the polar of $A$ wrt to $\gamma$. This is due the Gergonne point which is on $AD$.
 A: Let $X$ and $Y$ be the tangency points of lines $AB$ and $AC$ to the incircle, respectively. Let $G$ be the pole of line $AD$ with respect to the incircle; note that $G$ is on $BC$ (as $BC$ is the polar of $A$). Because $IG\perp ED$ and $ED$ is a chord of the incircle, $I$, $F$, and $G$ are collinear, and so $\angle GFD=90^\circ$. Also, inverting about the incircle and scaling by a factor of $2$ maps $G$ to $E$, $B$ to $X$, and $C$ to $Y$, so
$$-1=(D,E;X,Y)=(D,G;B,C).$$
Thus, since $FD$ and $FG$ are perpendicular and $(D,G;B,C)=-1$, $FD$ and $FG$ must be the internal and external angle bisectors of $\angle BFC$, as desired.
A: You can prove it by showing that the tangent of incircle at $E$ passes through $G$. We use this fact:
Let $A,B$ be points and $\Gamma$ be a circle. If $A$ lies on the polar of $B$ wrt $\Gamma$, then $B$ lies on the polar of $A$ wrt $\Gamma$. (This is proved by power of a point theorem.)
Now $G$ lies on the polar of $A$ wrt the incircle, $A$ lies on the polar of $G$ wrt the incircle. Since $DG$ is tangent to the incircle, $D$ also lies on the polar of $G$, so the line $AD$ is the polar of $G$. As $E$ is on $AD$, $G$ is on the polar of $E$, which is the tangent at $E$.
