# 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.

• Hi there, i understood the part where MI = MA, but i do not understand why B1A = B1I = B1C. sorry if this changes anything, but I is the incentre of the triangle ABC. Could u kindly explain to me why B1A=B1I=B1C? Apr 14, 2020 at 15:35
• Since B1 is on the angle bisector, it's the midpoint of arc AC not containing A, hence B1C=B1A. To prove B1I=B1A observe that: B1AI = B1AA1 = B1AC + CAA1 = B1BC + CAA1 = B/2 + A/2 and B1IA = 180 - AIB = 180 - (90 + C/2) = 90 - C/2 = A/2 + B/2 hence B1I=B1A.
– ak4
Apr 14, 2020 at 15:42
• hi thanks for this, but this only works if I is the centre of the circle. However, as mentioned above, I is the Incentre of the triangle and not the centre of the circle! unless I'm incorrect about this. any advices? Apr 14, 2020 at 16:00
• @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). Apr 15, 2020 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.