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 🙂
Best Answer
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 $M$.