Very nice triangles.

Since angle BFC is equal to angle BEC and they are both right the quadrilateral BFEC can be inscribed in a circle with radius BC. Since D is the midpoint FD and ED are radii in the circle and are therefore of equal length. Looking at the angle sum in triangle FAC the angle FCA is 30 degrees. Since angle FCA is an interior angle in the circumscribed circle, and angle FDE is a center angle, FDE is is $2*30 = 60$ degrees. Thus it is an equilateral triangle.

Let O be the intersection of (FC) and (BE).

Notice that angle FOE is 120. Therefore angle FOB = EOC = 60.

And because both triangle FOB and EOC are right angled : $\frac {FO} {BO} = \frac {EO} {OC} = \cos 60 = \frac 1 2$

We can then observe that triangles FOE and BOC are similar. Therefore $\frac {FE} {BC} = \frac {FO} {BO} = \frac {EO} {OC}$ . Since we know this ratio is $\frac 1 2$ , this means $\frac {FE} {BC}= \frac 1 2$ and because D is the midpoint of BC, we can finally conclude that FE = BD (1)

BFC is a right triangle and D is the midpoint of BC. Therefore BD=DC=DF (2)

BEC is also a right triangle, and in a similar fashion we show that DE = DC = BD (3)

From results (1) (2) and (3) we conclude that FE = ED = FD. Thus EFD is equilateral.