Let $ID\perp MN$, where $D\in MN$. Let also E the midpoint of arc AC. Then I,D,N,B are concyclic, $\angle INB=\angle IDB$. Therefore we just have to show that $IM//BD\Leftrightarrow\angle EIM=\angle IBD=\angle INM\Leftrightarrow IE^2=EM\cdot EN\Leftrightarrow AE^2=EM\cdot EN$. This is true since $\triangle AEM\sim\triangle NEA$.

Congrats! Very nice solution! The official solution uses $AP=IP$ (easy angle calculation) to obtain $\triangle IMP\sim \triangle INP$ via $\triangle AEM\sim\triangle NEA$. And it immediately follows the conclusion

Actually, I think that the official solution is even nicer. Just to point out, in Myth's post, $P = E$.

Myth wrote: 9.4, 10.3 Let $I$ be an incenter of $ABC$ ($AB<BC$), $M$ is a midpoint of $AC$, $N$ is a midpoint of circumcircle's arc $ABC$. Prove that $\angle IMA=\angle INB$. Short remark about language: "$M$ is a midpoint of $AC$" should be "$M$ is the midpoint of $AC$" instead (the segment AC has only one midpoint, so the definite article should be used). The same error appears in your next sentence. Just wanted to point this out (no offence - just a little bagatelle mistake you do from time to time). Cool problem from Badzyan, by the way. Here is a slightly different solution: Let $I_c$ and $I_a$ be the excenters of triangle ABC opposite to the vertices C and A. Then, it is well-known that the midpoint N of the arc ABC on the circumcircle of triangle ABC is the midpoint of the segment $I_cI_a$. Now, since the internal and the external angle bisector of an angle are perpendicular to each other, we have $AI_c\perp AI_a$ and $CI_c\perp CI_a$, so that $\measuredangle I_cAI_a=90^{\circ}$ and $\measuredangle I_cCI_a=90^{\circ}$; thus, the points A and C lie on the circle with diameter $I_cI_a$. Consequently, $\measuredangle I_aAC=\measuredangle I_aI_cC$. In other words, $\measuredangle IAC=\measuredangle II_cI_a$. Similarly, $\measuredangle ICA=\measuredangle II_aI_c$. Hence, the triangles IAC and $II_cI_a$ are similar. Now, in similar triangles, corresponding points form equal angles. Since the points M and N are the midpoints of the respective sides AC and $I_cI_a$ of the similar triangles IAC and $II_cI_a$ (and thus, corresponding points in these triangles), we thus have $\measuredangle IMA=\measuredangle INI_c$. In other words, < IMA = < INB. And we're done. darij

Valiowk wrote: Actually, I think that the official solution is even nicer. Just to point out, in Myth's post, $P = E$. Of course the official solution is nicer and if you have noticed, the point D in my solution is useless actually, because $\angle IMA=\angle INB\Leftrightarrow \angle EIM=\angle INM$ can also be shown by angle chasing, and after all, it is just the same as the official solution.

Sorry for my typo. Indeed, $P=E$ ($P$ comes from the official solution).

mecrazywong wrote: Of course the official solution is nicer and if you have noticed, the point D in my solution is useless actually, because $\angle IMA=\angle INB\Leftrightarrow \angle EIM=\angle INM$ can also be shown by angle chasing, and after all, it is just the same as the official solution. Yes, I realised that your solution is about the same as the official solution. It's just that yesterday, when I was reading your solution, I somehow didn't really like the part with the cyclic quadrilateral and parallel lines , because like you say, it can be done even more easily by angle chasing. What I meant was that if Myth feels that yours is a "very nice solution", then the official solution should be "nicer". Okay, I'm just being picky.