$D$ is the incenter of $\triangle ABC$. $DE \perp BC$ ($E \in BC$). $AE \cap \bigcirc(A, B, C) = F$ ($F \not\equiv A$). $G$ is the midpoint of the larger arc of $BC$. $GF \cap \bigcirc(B, C, D) = {H}$ ($GH < GF$). Prove that the excenter of $A$ in $\triangle ABC$, the midpoint of $BC$ and $H$ are collinear.
Let the midpoint of $BC$ and the excenter of $A$ in $\triangle ABC$ be respectively $I$ and $K$.
What I am trying to prove is that $HI \parallel AE$ and $HK \parallel AF$. (Perhaps $EFIH$ and $AFKH$ are parallelograms.) But I don't exactly know how.
Extend $DE$ to meet the circumcircle of $BDC$ again at $I$. By Power of a Point, $DE\times IE=BE\times EC=AE\times EF$, so $A,D,I,F$ are concyclic. Therefore, $\angle DAF=\angle EIF$.
Now, extend IF to meet the circumcircle of $ABC$ at $G'$. Construct a line through $G'$ parallel to $DE$ (i.e. perpendicular to $BC$. Let this line meet $BC$ at $M$ and circumcircle of $ABC$ at $N$. Then we have $\angle FG'N=\angle FID=\angle ADF$ Therefore $AD$ meets $G'N$ at $N$. Since $AD$ meets circumcircle $ABC$ at the midpoint of arc $BC$, this midpoint is $N$ and therefore $G'=G$ and $M$ is the midpoint of $BC$. In essence, we have shown that $G,F,I$ are collinear.
Now, it is time to clean up. Let the $A$-excenter be $P$. Then it is very well known that $A,D,P$ are collinear and $DP$ is the diameter of circle $BDC$. This means that $\angle PID =90^{\circ}$ so $PI$ is parallel to $BC$. Therefore arc $BI$ equals to arc $CP$ in circle $BDC$. Therefore $\angle BHI=\angle CHP$. It suffices to show that $\angle BHI=\angle CHM$.
But it is equally well-known that $GB,GC$ are tangents to the circumcircle of $BDHC$. Since $G,H,I$ collinear, $GH$ is actually the $H$-symmedian of $\triangle BHC$. Therefore $\angle BHI=\angle CHM$ and we are done.