Since $\angle CII_a = \angle BII_a = 90^\circ$ the excenter $I_a$ belongs to the circumcircle of the triangle $BIC$. Moreover in the triangle $CDA$, $CI, CI_a$ are the internal and external bisector, respectively, and therefore the points $A, D$ are the conjugate harmonics of the points $I, I_a$. Since $M$ is the midpoint of the segment $AD$, from the relation of Newton we have:
$$MA^2 = MI \cdot MI_a. \qquad (1)$$
Taking the power of $M$ with respect to the circle $(I, B, C)$ we have:
$$MI \cdot MI_a = MF \cdot MB. \qquad (2)$$
From (1) and (2) it follows that:
$$MA^2 = MF \cdot MB,$$
Which means that $MA$ is tangent of the circumcircle of the triangle $AFB$. Therefore
$$\angle MBA = \angle FAM. \qquad (3)$$
and finally
$$\begin{align*} \angle FAC + \angle FCA &= \angle FAM + \frac{\angle A}{2} + \angle FCI + \frac{\angle C}{2} = \\ & (3) \qquad \angle MBA + \frac{\angle A}{2} + \angle FBI + \frac{\angle C}{2} = \frac{\angle A}{2} + \frac{\angle B}{2} + \frac{\angle C}{2} = 90^\circ. \end{align*}$$
Hence: $AF \perp FC$.

Figure 2

We draw from $C$ the parallel line to the bisector $AD$ which meets the line $AB$ at point $N$ and the line $AM$ at point $Z$. Taking angle equalities we have:
$$\angle ACN = \angle DAC = \frac{\angle A}{2}, \quad \angle ANC = \angle BAD = \frac{\angle A}{2}$$
And so the triangle $ACN$ is isosceles. Since $M$ is the midpoint of $AD$ it follows that $Z$ is the midpoint of $CN$. Hence $AZ \perp CN \Rightarrow \angle AZC = 90^\circ$.
Therefore, in order to have $\angle AFC = 90^\circ$, it is enough to prove that the quadrilateral $CFAZ$ is cyclic. In fact, we have
$$\angle CFZ = 180^\circ - \angle BFC = 180^\circ - \angle BIC \\ = 180^\circ - \left(90^\circ + \frac{\angle A}{2}\right) = 90^\circ - \frac{\angle A}{2} = \angle CAZ.$$

Figure 3