Another beauty hidden in a simple triangle (2) 
Let the inscribed circle of triangle $ABC$ touch its sides $BC, CA, AB$ at $F, G, E$. Denote the center of this circle by $D$, and the midpoint of $BC$ by $M$. Prove that lines $AM$, $EG$ and $DF$ concur.


My idea was to consider the intersection of lines $EG$ and $FD$ first, then to draw a line through intersection point $H$ parallel to $BC$. That line will intersect triangle at some points $I$ and $J$ (not shown). If I could prove that $IH=JH$, that would automatically prove that $AM$ has to pass through point $H$ as well. However, this day was so hot here in my city that I could not think more about this problem. I can only hope that it's more pleasant elsewhere :)
 A: 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.

A: 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 $M$. 
A: 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.

A: $IH = JH$ because $M$ is the center point of BC and triangle $AIJ$ is similar with respect to the intersection of its parallel base with line $AM$.
However, you have to show that these two sides are the same due to symmetry about the line $FD$, so the extension of $FD$ is shown to meet the intersection at $H$.
The following diagram achieves this by adding a symmetrical construction about $FD$ with $X = Y$.

