# Prove that the excenter of $A$ in $\triangle ABC$, the midpoint of $BC$ and $H$ are collinear.

$$D$$ 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.