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$\triangle ABC$ is equilateral with a circle $\omega$ inscribed in it. MN is a tangent of $\omega$ and it intersects $AC$ and $BC$ at points $M$ and $N$ respectively. $AM_1=MC$ and $BN_1=CN$. $D,E,F$ touch the circle. $O$ is the center of $\omega$ and $OH_1 = r$. Prove that $M_1N_1$ intersects the center of $\omega$.

I've tried adding some additional segments (as you can see on the 2nd image). I've created $AR$ such that $AR=CN \Rightarrow MN=M_1R$ and $BP$ such that $BP=MC \Rightarrow PN_1=MN$. $\triangle CMN = \triangle AM_1R = \triangle BN_1P$. And if I want to prove that $M_1R$ and $N_1P$ touch $\omega$, I could say that it's because of symmetry (tell me if I'm wrong). That's all I've tried so far. Solving this problem would be equivalent to finding out that $M_1N_1$ bisects both the angle $\angle PN_1N$ and $\angle RM_1M$.

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  • $\begingroup$ "Prove that $M_1N_1$ bisects the center of $\omega$." Do you mean, "Prove that $M_1N_1$ contains the center of $\omega$."? (The title of your question suggests so.) $\endgroup$ – Blue Dec 14 '13 at 17:56
  • $\begingroup$ Yes, does the sentence imply something else? It does cut the point. English isn't my native language, though. $\endgroup$ – user26486 Dec 14 '13 at 17:57
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    $\begingroup$ Your English is good; there's just this one bit of vocabulary confusion. Usually, "bisects" means "divides into two equal pieces". I think you're using "bisects" where you mean "intersects", whose basic meaning is "meets". So, earlier in your question, you want to write that "$MN$ intersects $AC$ and $BC$ at $M$ and $N$." And then later, "Prove that $M_1N_1$ intersects the center of $\omega$." $\endgroup$ – Blue Dec 14 '13 at 18:13
  • $\begingroup$ You are so close, I really don't like to reveal the answer to keep its joy for you :-). And something else, I don't see a clear symmetry to say that those triangles are equal by symmetry. $\endgroup$ – hhsaffar Dec 14 '13 at 18:46
  • $\begingroup$ Hint: Try to prove $\angle M_1ON_1=180$ $\endgroup$ – hhsaffar Dec 14 '13 at 18:53
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I draw $M_1R$ tangent to $\omega$.

$M_1E=EM, OE=OE,E=90 \to \triangle OEM_1=\triangle OEM \\ \to \angle RM_1E =\angle NME = 2\angle OM_1E \to \angle RM_1A=\angle NMC \to \triangle ARM_1 = \triangle MNC$

$\angle EOF=360-\angle C - \angle E -\angle F = 120$, so to prove $\angle M_1ON_1=180$ we need to prove that $\angle EOM_1 + \angle FON_1=60$

$\angle RM_1E= \angle NME,\angle MNF= \angle PN_1F$

$\angle EMN + \angle FNM= 360-\angle FEC-\angle EFC= 240$

$\angle EOM_1 + \angle FON_1=90 - \angle OM_1E+90-\angle FN_1O = 180-\frac{\angle RM_1E+ \angle FN_1P}{2}\\=180-\frac{\angle EMN + \angle FNM}{2}=180-120=60 $

And this is what we wanted to prove.

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  • $\begingroup$ Ohh, I've proved it. To solve the problem I can show that $\angle N_1OM=2\alpha$, where $\alpha = {1\over2}\angle M_1MN$. I also create a variable $\beta$ such that $2\beta=\angle MNN_1$. By knowing that angles of a quadrilateral add up to $360^{\circ}$, I can deduce that if $\angle N_1OM=2 \alpha$, then $$\alpha + 2\alpha +\beta +2\beta=360^{\circ}\Rightarrow 3\alpha + 3\beta = 360^{\circ}\Rightarrow \alpha + \beta=120^{\circ}$$ and that is if and only if $M_1, O$ and $N_1$ are in a straight line (because $\angle MON_1$ has to be equal to $2\alpha$ in order for the equality above to be true). $\endgroup$ – user26486 Dec 14 '13 at 20:12
  • $\begingroup$ Dear @mathh , I am in a rush so I couldn't check your proof, but I updated my answer with a more complete version of my solution. $\endgroup$ – hhsaffar Dec 14 '13 at 21:02
  • $\begingroup$ But we don't know yet that $\alpha+\beta=120^{\circ}$. We can find that out: $$\angle CM_1N_1+\angle CN_1M_1+\angle C=\alpha+\beta+60^{\circ}=180^{\circ} \Rightarrow \alpha+\beta=120^{\circ}$$ I've already said that this can only be true when $M1$, $O$ and $N1$ are in a straight line ⇒ they are indeed in a straight line. $\endgroup$ – user26486 Dec 14 '13 at 21:32

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