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$\Delta ABC$ 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 $\angle IMF$ enter image description here

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 ?

RopuToran
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1 Answers1

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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, $\angle IMD = \angle LID$.

enter image description here

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.

Mick
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  • I assume you meant $\angle LIH=\angle IHK$. However, there does seem to be circular logic somewhere. To know that $B$ and $HK$ are parallel, you need to know that $I$, $D$, and $H$ are collinear. And you don't seem to be using how the points are constructed,. – Batominovski Aug 01 '18 at 16:02
  • @Batominovski No, it is $\angle LIH = \angle IKH$. Perhaps an added picture can make things more clear. I, D, and H are collinear because of (1); plus $\angle IDM = \angle IHK = 90^0$; and the two angles are in corresponding positions. – Mick Aug 01 '18 at 18:52
  • What do you mean "Let L be a point somewhere near F" ? – RopuToran Aug 02 '18 at 17:46
  • @RopuToran I just want to draw a line that is tangent to the circle IDM touching it at I. That is the blue line. I need to give that tangent line a name and therefore LI. L is also used (in the later stage) for naming some angles. For example, $\angle LID$. – Mick Aug 02 '18 at 18:04