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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 $AC$.

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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!

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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.

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  • $\begingroup$ 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? $\endgroup$ – Sean Lee Apr 14 '20 at 15:35
  • $\begingroup$ 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. $\endgroup$ – A. K Apr 14 '20 at 15:42
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    $\begingroup$ 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? $\endgroup$ – Sean Lee Apr 14 '20 at 16:00
  • $\begingroup$ @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). $\endgroup$ – Calvin Lin Apr 15 '20 at 5:52
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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 $AC$.

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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.

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