An Elegant Collinearity Problem Involving the Incircle

A problem from a vietnamese Facebook group.

Problem. Let $ABC$ be a triangle, and let its incircle, with center $I$, be tangent to $BC$ at $D$. Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively. Prove that $M$, $I$, and $N$ are collinear. 

Proof. Let $BC=a$, $CA=b$, $AB=c$, let $s$ be the semiperimeter, $\Delta=[ABC]$, and let $r$ be the inradius. Assume $b>c$; the case $b<c$ is symmetric, while $b=c$ is immediate. Since $BD=s-b$ and $N$ is the midpoint of $BC$,
$$DN=\frac a2-(s-b)=\frac{b-c}{2}.$$
As $ID\perp BC$ and $ID=r$,
$$\tan\angle DNI=\frac{r}{DN}=\frac{2r}{b-c}.$$
Let $M'$ be the foot of the perpendicular from $M$ to $BC$. Since $M$ is the midpoint of $AD$,
$$[CDM]=\frac12[CDA]=\frac14\,b(s-c)\sin C.$$
On the other hand,
$$[CDM]=\frac12(s-c)MM',$$
hence
$$MM'=\frac{b\sin C}{2}.$$
By the cosine law in $\triangle ACD$,
$$AD^2=b^2+(s-c)^2-2b(s-c)\cos C.$$
Since $MD=AD/2$,
$$DM'^2=DM^2-MM'^2=\frac14\bigl(b\cos C-(s-c)\bigr)^2.$$
Using
$$\sin^2\frac C2=\frac{(s-a)(s-b)}{ab},\qquad\cos^2\frac C2=\frac{s(s-c)}{ab},$$
we get
$$\cos C=\cos^2\frac C2-\sin^2\frac C2=\frac{s(s-c)-(s-a)(s-b)}{ab},$$
so
$$b\cos C-(s-c)=\frac{(s-a)(b-c)}{a}.$$
Therefore
$$DM'=\frac{(s-a)(b-c)}{2a}.$$
Since $M',D,N$ occur in this order,
$$NM'=ND+DM'=\frac{b-c}{2}+\frac{(s-a)(b-c)}{2a}=\frac{s(b-c)}{2a}.$$
Thus
$$\tan\angle DNM=\frac{MM'}{NM'}=\frac{ab\sin C}{s(b-c)}=\frac{2\Delta}{s(b-c)}=\frac{2r}{b-c}.$$
Hence
$$\tan\angle DNM=\tan\angle DNI,$$
and therefore $M,I,N$ are collinear.