China Southeastern Mathematical Olympiad

Join $AD$, and it is obvious that $\angle ADB=90^{\circ }$. Then join $AI$ and $DM$. Suppose $DM$ intersects $AC$ at $G$. Since $\angle ABI=\angle DBM$, we obtain $\frac{AB}{BD}=\frac{BI}{BM}$. Hence, $△ABI\sim △DBM$, and $$\angle DMB=\angle AIB=90^{\circ }+\frac{1}{2}\angle ACB.$$ Join $IG$, $IC$, $IM$, then $$\angle IMG=\angle DMB-90^{\circ }=\frac{1}{2}\angle ACB=\angle GCI.$$ Therefore, $I$, $M$, $C$, $G$ are concyclic, and $IG\perp AC$. Consequently, since $G$ and $N$ represent the same point, we conclude that $M$, $N$, $D$ are collinear.