We have $\frac{AI}{ID} = \frac{AB}{BD} = \frac{EC}{DE} = \frac{BE}{DE} = \frac{IE}{DE}$, which proves the first relation.

Let $G = EF \cap BC$. We know that $GD$ is the bisector of the angle $\angle EGI$, hence $\frac{GI}{GE} = \frac{ID}{DE} = \frac{AI}{IE}$. (1)

Triangles $FBE$ and $BGE$ are similar (AA), therefore $EB^2 = EF \cdot EG = EI^2$. This means that triangles $FIE$ and $IGE$ are also similar (SAS), which leads to $\frac{FI}{IE} = \frac{GI}{GE}$. (2)

From (1) and (2) we obtain $FI = AI$, and the conclusion.

Let $O$ be the circumcenter of triangle $ABC$. We show that triangles $AIO$ and $II'E$ are similar, which proves that the points $I, O, F, E$ are co-cyclic.

$OE$ and $II'$ are both perpendicular to $BC$, hence they are parallel. It follows that $\angle I'IE \equiv \angle OEI \equiv \angle OAI$.

We prove that $\frac{AI}{II'} = \frac{AO}{IE}$. From the angle bisector theorem we obtain $\frac{AI}{ID} = \frac{AB}{BD} = \frac{AC}{CD} = \frac{AB + AC}{BC} = \frac{b+c}{a}$, hence $\frac{AI}{AD} = \frac{b+c}{a+b+c}$. As $AD = l_a = \frac{2bc \cos \frac{A}{2}}{b+c}$, we obtain $AI = \frac{2bc \cos \frac{A}{2}}{a+b+c}$.

But $IE = EB$ and $\frac{EB}{BC} = \frac{\sin \frac{A}{2}}{\sin A} = \frac{1}{2 \cos \frac{A}{2}}$. Moreover, $EI = EB = \frac{a}{2 \cos \frac{A}{2}}$.

It follows that, $AI \cdot IE = \frac{abc}{a+b+c}$.

On the other hand, $AO \cdot II' = R \cdot 2r = \frac{2RS}{p} = \frac{abc}{a+b+c}$, hence $AO \cdot II' = AI \cdot IE$, which, together with $\angle I'IE \equiv \angle OAI$, proves the similarity of the triangles $AIO$ and $II'E$.

We deduce that $\angle EIO \equiv \angle II'F \equiv \angle OEF \equiv \angle OFE$, hence $IOEF$ is cyclic (the order of the points and the arguments depend slightly on the configuration).

It follows that $\angle IOF \equiv \angle IEI' \equiv \angle AOI$. We also have $\angle IAO \equiv \angle IEO \equiv \angle IFO$, which means that triangles $AIO$ and $IFO$ are congruent (SAA), therefore $IF = IA$.