Ok, sorry it took so long, but I still don't have the answer. Here's my work (not that anyone's going to read through the whole thing:
Let $x=AD$, then $AF=DF=x$ also, since the triangle is equilateral. This means the radius of the circle is $r=\frac{x\sqrt{3}}{3}$.
Now, since $\angle BFD=120^\circ$, we can use the cosine law on \(\displaystyle \triangle BDF\)and solve for $FB$:
$$BD^2=DF^2+FB^2-2\times DF\times FB\times \cos \angle BFD$$
$$16 = x^2 + FB^2 -2x\times FB \times\frac{-1}{2}$$
$$FB^2+xBF-16=0$$
$$FB=\frac{-x+\sqrt{64-3x^2}}{2}$$
Similarly, we can use the cosine law on \(\displaystyle \triangle CDF\) to find
\(\displaystyle CD=\frac{-x+\sqrt{16-3x^2}}{2}\)
Because they're tangents, we know that $CD$ and $FB$ are equal to $CE$ and $BE$ respectively, hence,
$$BC=CD+FB=-x+\frac{\sqrt{16-3x^2}+\sqrt{64-3x^2}}{2}$$
--------------------------------------------------------------------------
A well known formula for the radius of an incircle is $r=\frac{A}{s}$, where $r$ is the radius, $A$ is the area of the triangle, and $s$ is the semiperimeter.
The area of the triangle is
$$\frac{\sin 60^\circ}{2}\times AC\times AB = \frac{\sqrt{3}}{4}\bigg(\frac{x+\sqrt{16-3x^2}}{2}\bigg)\bigg(\frac{x+\sqrt{64-3x^2}}{2}\bigg)$$
and the semiperimeter is $AD+BC$, which is
$$\frac{\sqrt{64-3x^2}+\sqrt{16-3x^2}}{2}$$
Hence we get the following equation for $x$:
$$\frac{\sqrt{3}}{4}\bigg(\frac{x+\sqrt{16-3x^2}}{2}\bigg)\bigg(\frac{x+\sqrt{64-3x^2}}{2}\bigg)=\frac{x\sqrt{3}}{3}\bigg(\frac{\sqrt{ 64-3x^2}+\sqrt{16-3x^2}}{2}\bigg)$$
which, after simplication of course, does not reveal any quadratic roots.