The incircle of $\triangle ABC$ has centre $I$ and touches $BC$ at $D$. Prove that $M,I,N$ are collinear

The first highlighted portion uses the angle bisector theorem twice, using the fact that $I$ lies on the bisectors of the angles at $A$ and $C$.

The second highlighted section just applies known formulas for $BD$, $BN$ and $BL$ in terms of $a, b, c$, where $BL$ is obtained from the angle bisector theorem again and the formula for $BN$ is a simple consequence of the fact that the tangent points of the incircle divide the triangle sides into three pairs of equal lengths.

For an alternative proof using simple geometry, consider the excircle of $ABC$ opposite $A$ and the point $E$ where it touches $a$, as well as the point $F$ on the incircle opposite to $D$. Clearly scaling $M'IN$ around $D$ by factor $2$ yields $AFE$, and $AFE$ are collinear as seen by the scaling around $A$ mapping the incircle to the excircle.