How can I prove that $MN$ is parallel to $AC$?

Question

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!

Answer 1

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.

Answer 2

Let $\alpha:=\dfrac12\,\angle BAC$, $\beta:=\dfrac12\,\angle CBA$, and $\gamma:=\dfrac12\,\angle ACB$. It follows that $$\angle B_1C_1C=\angle B_1BC=\beta=\angle B_1BA=\angle B_1A_1A\,.$$ This means $$\angle MC_1I=\angle B_1C_1C=\beta=\angle B_1A_1A=\angle MBI\,.$$ Therefore, $IMC_1B$ is a cyclic quadrilateral. Thus, $$\angle MIC_1=\angle MBC_1=\angle ABC_1=\angle ACC_1=\gamma\,.$$ This means $MI\parallel AC$.

Similarly, $$\angle NA_1I=\angle B_1A_1A=\beta=\angle B_1BC=\angle NBI\,.$$ Thus, $INA_1B$ is also a cyclic quadrilateral. That is, $$\angle NIA_1=\angle NBA_1=\angle CBA_1=\angle CAA_1=\alpha\,.$$ This means $NI\parallel AC$. Thus, $MI$ and $NI$ are both parallel lines to $AC$ that pass through $I$. Ergo, they are the same line. This shows that $MN$ passes through $I$ and is parallel to $AC$.

Answer 3

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.