Another beauty hidden in a simple triangle (2)

You had the right idea. From $H$ draw a line parallel to $BC$ and call its intersections with $AB$ and $AC$ in order $I$ and $J$. Note that the quadrilaterals $IHDF$ and $HGJD$ are inscribed since $\angle DGJ=\angle DHJ$ and $\angle IHD=\angle IFD$ are all right angles. It follows that $\angle GDJ=\angle GHJ=\angle IHF=\angle IDF$. So the two triangles $\triangle DFI$ and $\triangle DGJ$ are congruent for having a leg and a non-right angle congruent. It follows that $DJ$ and $DI$ are congruent and so $HJ$ is congruent to $HI$ and so the line through $A$ and $H$ must go through $M$.

Proof

Let $AM$ intersect $EG$ at $H$. We are to prove $HF$ is the line of the diameter.

Draw a line $l$ passing through $A$ and parallel to $BC$. Let $EG$ intersect $l$ at $X$. Thus, we may readily see that $(AB,AC|AM,AX)=-1,$ which shows that the polar line of $H$ with respect to the incircle is $AX$. Hence, $DH \perp AX ||BC$, then $F,D,H$ are colinear. The proof is completed so far.

Another Proof

Denote the the antipodal point of $F$ as $J$. Draw a line $l$ passing through $J$, parallel to $BC$, and intersecting $AB, AC$ at $K, L$ respectively.

Notice that $BCLK$ is a tangential quadrilateral. By Newton's Theorem(Page 156-157) ,which is a degenerated form of Brianchon's theorem, we may obtain $BL,CK,FJ$ are concurrent, namely at $H$. But $KL||BC$, hence $AH$ bisects $BC$ and $KL$. We are done.