First, let the incenter of $AFB$ be $I_1$ and let the incenter of $CHD$ be $I_2$. Then let $FA=a$ and $FB=b$. Then by the properties of the incenter, horizontal distance from $I_1$ to $BC$ is $\frac{1-a+b}{2}$ and the horizontal distance from $I_2$ to $AD$ is the same due to symmetry Therefore, the horizontal distance from $I_1$ to $I_2$ is just $a-b$. But then notice that $AE=b$, and so $EF=a-b$. Therefore, the horizontal distance from $I_1$ to $I_2$ is just the diameter of each of the circles, or twice the radius By symmetry, this implies that the incircle of $AFB$ is tangent to the midline of $AD$ and $BC$. Now let $M$ be the midpoint of $AB$ and let the midline intersect $AF$ at $N$ Now because $MNFB$ has an inscribed circle and a pair of opposite right angles, then $NF=NM$ and $BF=BM=\frac{1}{2}$ From here, it is easy to see that $AF=\frac{\sqrt{3}}{2}$, so then $EF=\frac{\sqrt{3}-1}{2}$, and so the radius of the circle is just half that, and so it's $$\boxed{\frac{\sqrt{3}-1}{4}}$$