Let $ABC$ be a triangle. The internal angular bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ meet the circumcircle of the triangle $ABC$ at the points $A_1$, $B_1$, and $C_1$, respectively. Suppose that $B_1C_1$ meets $AB$ at $M$, and $A_1B_1$ meets $BC$ at $N$. Prove that $MN$ is parallel to $AC$.
So far, I have managed to prove that $M,I,N$ are collinear, where $I$ is the incentre of the circle and $AA_1$ is perpendicular to $B_1C_1$. I have also attempted to prove the result using radical axis / Brianchon's theorem but to no avail. Can anyone help me with this? Any help is greatly appreciated!