The bisector of the angle $\hat{A}$ passes through the midpoint of the arc $BC$ which does not contain the point $A$. Denote this point by $S$. $MS$ is the perpendicular bisector of $[BC]$, so $MS \parallel ID$ (both lines are perpendicular on $BC$).

Also, $\angle ASC = \angle ABC$ and $\angle SAC = \angle BAE$ proves that the triangles $SAC$ and $BAE$ are similar, so $\frac{AB}{BE} = \frac{AS}{SC}$.

Using the angle bisector theorem, we get $\frac{AB}{BE} = \frac{AI}{IE}$.

It is known that $SI = SC$ (one might compute the angles of triangle $SCI$). We have $\frac{AS}{SC} = \frac{AS}{SI} = \frac{AM}{MF}$.

We proved $\frac{AI}{IE} = \frac{AM}{MF}$.

Using Menelaus's theorem in the triangle $AEF$ and the transversal $I-X-M$ (where $\{X\} = IM \cap EF$), we have $\frac{AI}{IE} \cdot \frac{EX}{XF} \cdot \frac{MF}{MA} = 1$.

Using the equality we proved before, we find $\frac{EX}{XF} = 1$, which means that $X$ is the midpoint of $[EF]$.