# 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!

• @SeanLee AK is correct. For another way, observe that $\angle B_1 A I = \frac{ \alpha + \beta } { 2 } = \angle IAB_1$ so $B_1 A = B_1 I$. This is a "well known" property of $B_1$. Note that AK is not saying "$IA = IB_1 = IC$" (in which case I agree with you that it's true only if $I$ is the circumcenter). – Calvin Lin Apr 15 '20 at 5:52
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 $$AC$$.
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.