∠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.