, 2011

We work in the opposite direction. Suppose that $H'$ is the point where $DF$ intersects the line through $C$ parallel to $AB$. We need to show that $H' = H$. For this purpose it suffices to prove that $E$, $G$, $H'$ are collinear, which reduces to showing that if $G' \neq E$ is the common point of $EH'$ and the incircle, then $G' = G$.

Note that $H'$ and $B$ lie on the same side of $AC$. Hence $CH' \parallel AB$ gives $\angle ACH' = 180^\circ - \angle BAC$. Also, some homothety with center $D$ maps the segment $BF$ to the segment $CH'$. Thus the equality $BD = BF$ implies that $CH' = CD = CE$, i.e. the triangle $ECH'$ is isosceles and
$$\angle H'EC = \frac{1}{2}(180^\circ - \angle ECH') = \frac{1}{2}\angle BAC.$$
But $G'$ and $H'$ lie on the same side of $AC$, so $\angle G'EC = \angle H'EC$ and consequently
$$\angle G'FE = \angle G'EC = \angle H'EC = \frac{1}{2}\angle BAC$$
so that
$$\angle G'FA = \angle G'FE + \angle EFA = \frac{1}{2}\angle BAC + \frac{1}{2}(180^\circ - \angle FAE) = 90^\circ.$$
Hence $FG'$ is a diameter of the incircle and the desired equality $G' = G$ follows.