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 1. 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.

Proof 2. This problem is essentially a degenerate case of Newton’s theorem for tangential quadrilaterals: in every tangential quadrilateral, the center of the inscribed circle lies on the line joining the midpoints of the two diagonals. As one of the vertices of the quadrilateral approaches the point of tangency $D$ on $BC$, the quadrilateral degenerates into triangle $ABC$, and the Newton line becomes precisely the line containing the midpoint $M$ of $AD$, the incenter $I$, and the midpoint $N$ of $BC$.