a. Triangle $AFE$ is isosceles with $AE = AF$, and $AI \perp FE$, hence $\angle AEF = 90^\circ - \angle A/2$. In the same way from the isosceles triangle $CDE$ we get $\angle DEC = 90^\circ - \angle C/2$. As a consequence $\overline{MED} = 180^\circ - \overline{AEF} - \overline{DEC} = 180^\circ - (180^\circ - (\hat{A} + \hat{C})/2) = 45^\circ$. (*)

As $\triangle MDC \equiv \triangle MEC$, we obtain $MD = ME$ (**). By () and (*), the triangle $\triangle MED$ is right angled and isosceles. As a consequence $DN \perp EF$ and, because $AI \perp EF$, we obtain $DN \parallel AI$. As $AN \parallel ID$ we conclude that the quadrilateral $ANDI$ is a parallelogram, so $AI = ND$.

b. We have $\overline{EFD} = 180^\circ - \overline{AFE} - \overline{BFD} = (\hat{A} + \hat{B})/2 = 90^\circ - \hat{C}/2 = \overline{DIC}$ so $\triangle FMD \sim \triangle IDC$. We conclude $\frac{FM}{ID} = \frac{MD}{DC}$ which implies
$$FM = \frac{ID \cdot MD}{DC} = \frac{EI \cdot EM}{EC}.$$