You make an auxiliary line from D, make it parallel to BC, and intersect with AF at one point, suppose G.

So DG//BC

It is easy to prove that triangle BEF and triangle DEG are congruent.

(Angle BEF= Angle EBF= Angle EDG BE=ED)

So DG=BF

Obviously DG is the center line of triangular AFC.

So FC=2DG=2BF.

So BF/FC= 1/2.