∠AFE =∠ Ade = 90 =∠ABG；;

AG=AG, so: △ abg △ afg.

② Let BG = X, CD = 3DE and DE = AB = 6, then DE=FE=2.

△ abg △ afg, then BG = FG = X, CG = BC-BG = 6-X.

CG 2+CE 2 = EG 2 (Pythagorean Theorem)

Then: (6-x) 2+4 2 = (2+x) 2, x=3. Then BG=3, GC=6-x=3.

Then BG=GC. (the method is stupid. )

③BG=GF=GC=3

Then ∠ GCF = ∠ GFC, ∠ GCF+∠ GFC+∠ FGC =180 = 2 ∠ FCG+∠ FGC.

∠AGB=∠AGF，∠AGB+∠AGF+∠FGC = 180 = 2∠AGB+∠FGC

Then ∠FCG=∠AGB

So ag∨cf

④GF/GE=GF/(GF+FE)=3/5

Then the area of △FGC is 3/5 of △EGC.

The area of △EGC is GC*CE/2=3*4/2=6.

Then the area of delta delta △△FGC is 3.6.