Let $E$ be the tangency point of $A$-excircle with segment $BC$ and $R=2I-D$. Line tangent to incircle passing through $R$ is parallel to $BC$, so homothethy centered at $A$ transforming incircle on excircle moves $R$ to $E$, hence $R\in \overline{AE}$. Since $BD=\frac{AB+BC-AC}{2}=CE$ we know that $M$ is a midpoint of segment $DE$. Hence $MI\parallel AE$ so $MI$ bisects segment $AD$ because there is only one line passing through $M$ parallel to $AE$.QED