$\Delta ABC$ inscribed $(O)$, incenter $I$, $(I)$ touch $BC$ at $D$. $AD \cap (O)=${$A;E$}. Let $M$ be the midpoint of $BC$ and $N$ be the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $G$ ($G$ lies in $ABC$). $(AGE) \cap (BIC)=${$G;F$}. Prove $MB$ bisects $\angle IMF$
I have proved $\overline{G,D,F}$ by using power of a point. I need to prove $I,D,G,M$ concyclic (haven't proved yet), then the problem will be much easier ... Any ideas ?
Since $\angle IDM = 90^0$, IM is the diameter of (IDM). Let L be a point somewhere near F. LI can be drawn tangent to (IDM) at I. Then, $\angle IMD = \angle LID$.
Produce IM to K such that IM = MK. Then, M is the center of the circle passing I, and K cutting MF at H. Note that LI is also tangent to circle (IHK) at I. Then, $\angle LIH = \angle IKH$.
Combining the two results, we can say that (1) BDMC // HK and (2) I, D, H are collinear. By midpoint theorem, ID = DH. Required result follows.