How can I prove that $MN$ is parallel to $AC$? 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!
Showing MN is parallel to AC can be done by proving that B1C1 is the perpendicular bisector of AI. It is well-known that B1A=B1I=B1C, and C1 satisfy a similar relation. By this property we conclude that B1C1 is the perpendicular bisector of AI. And now that you have proved M,I,N are collinear, it suffices to notice that MI=MA (M is on the perpendicular bisector of AI), and so MIA=MAI=IAC, which implies that MI is parallel to AC. And we're done.
In the standard notation by the law of sines we obtain:
$$\frac{BM}{BB_1}=\frac{\sin\frac{\gamma}{2}}{\sin\frac{\beta+\gamma}{2}}$$ or
$$BM=\frac{BB_1\sin\frac{\gamma}{2}}{\cos\frac{\alpha}{2}}.$$
By the same way:$$BN=\frac{BB_1\sin\frac{\alpha}{2}}{\cos\frac{\gamma}{2}}.$$
Id est, it's enough to prove that
$$\frac{BM}{c}=\frac{BN}{a}$$ or 
$$\frac{\sin\frac{\gamma}{2}}{\cos\frac{\alpha}{2}\sin\gamma}=\frac{\sin\frac{\alpha}{2}}{\cos\frac{\gamma}{2}\sin\alpha},$$ which is obvious.