$\triangle ABC$ 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 $\angle RM_1M$.