Method 1: Since $DE\parallel CB$, so $\angle DEB=\angle CBD=\angle DBE$, which means that $DE=BE=\frac {AB}{2}$. Since $FE=\frac {CB}{2}$, thus $DF=DE-EF=\frac {AB-CB}{2}$. Since $FN=CF-CN=\frac {AC-(AC+CB-AB)}{2}=\frac {AB-CB}{2}=DF$, thus we have $\angle DNF=\angle MDN=\angle CMN=\angle CNM$, which implies that $D,N,M$ are collinear. Method 2: It's proven above that $DE=BE=AE$, thus $AD\perp DB$. Let the tangency of the incircle on $AB$ be J, thus $MJ\perp DB$ $\Rightarrow$ $MJ\parallel AD$. Let $G$ be the intersection of $AD$ and $CB$, thus $D$ is the midpoint of $AG$, which means that $M(J,D;A,G)=-1=M(J,N;A,G)$ and that implies $D,N,M$ are collinear.

Another solution: Let $DM\cap AC=T$. We will prove that $N\equiv T$ It is obvious that $\angle ADB=90^{\circ}$.We know $\angle ABD=\angle IBM$.It implies that $\triangle ABD \sim \triangle IBM \Rightarrow \frac{AB}{BD}=\frac{IB}{BM}\Rightarrow \triangle ABI\sim \triangle DBM\Rightarrow \angle DMB=\angle AIB=90^{\circ}+\frac{\angle ACB}{2}\Rightarrow \angle IMT=\frac{\angle ACB}{2}=\angle TCI$ $I,M,C,T$ are concyclic. So $T\equiv N$

Mine is different: Sine $EF \parallel BC$ and $\angle ABI =\angle CBI$ $\Rightarrow$ $\angle ABD = \angle EDB = \angle DBC$ so $ED=EB=EA$ $\Rightarrow$ $AD \perp DB$. Now we will show that if $BI$ intersects $MN$ at $P$ then $AP \perp BP$. Let $\angle A=2 \alpha$ , $\angle B=2 \beta$ and $\angle C= 2 \theta$. So $\alpha+ \beta+\theta=90$. Since $CM=CN$ then $\angle CMN= \angle CNM= \alpha+\beta$ so $\angle BPM= \beta = \angle IAN$ $\Rightarrow$ points $A, N, P, I$ are concyclic. Since $IN \perp AN$ $\Rightarrow$ $AP \perp IP$. So $D=P$.

Dear Mathlinkers, you can see http://perso.orange.fr/jl.ayme vol. 4 An unlikely concurrence Sincerely Jean-Louis