Circumradius Party!

In $\triangle ABC$ , $AD$ is the perpendicular from $A$ onto $BC$ , $BE$ is the median and $CF$ is the internal angle bisector of $\angle ACB$ such that $D$ is on $BC$ , $E$ on $AC$ and $F$ on $AB$ . Suppose that $AD$ , $BE$ and $CF$ are concurrent. If $FE = ED = 6$ unis, find

$\large{\dfrac{R_{AEF} + R_{FBD} + R_{EDC} + R_{DEF}}{R_{ABC}}}$

Details and assumptions : $R_{XYZ} =$ length of the circumradius of $\triangle XYZ$