∠∠DAE =∠EAF。

∴eg=ed=ec；

And EF=EF, then rt ⊿ EGF ≌ rt ⊿ ECF (HL), ∠ Feg = ∠ FEC;

The same can be proved: Rt⊿ADE≌Rt⊿AGE,∠AEG=∠AED.

∴∠AED+∠FEC=90 degrees, so ∠ DAE = ∠ FEC; ∠C=∠D=90 degrees.

∴⊿ECF∽⊿ADE,FC/ED=EC/AD, that is, FC/(BC/2)/BC = 1/2, FC = (1/4) BC.