Let $\{D\} = AI \cap BC$. As $AB < AC$, $D$ lies between $B$ and $M$ and $\angle ACB < \angle ABC$.
We have $\angle IDB = \angle DAC + \angle ACB < \angle DAB + \angle ABD = \angle ADC$, therefore angle $IDB$ is acute.
Let $F$ and $E$ be the projections of $I$ onto $AB$ and $BC$, respectively. It follows that $E \in (BD) \subset BM$. Triangles $IBF$ and $IBE$ are congruent and so are triangles $IFA$ and $IEM$, therefore $BA = BM = \frac{BC}{2}$ and triangles $IBA$ and $IBM$ are congruent.

We have: $\angle MID = \angle IDB - \angle IMB = \angle DAC + \angle ACD - \angle IAB = \angle ACD$.
It follows that $\angle AIM = 180^\circ - \angle ACB$ (1).
Let $H$ be the projection of $B$ onto the line $AC$. It follows that $BH \le AB = \frac{BC}{2}$, which shows that $\angle ACB \le 30^\circ$ (2).
From (1) and (2) we obtain that $\angle AIM \ge 180^\circ - 30^\circ = 150^\circ$.