∠ABC =∠CBG-∠ABG = 90-∠ABG =∠ Abe-∠ ABG = ∠ EBG;
Because, in △ABC and △EBG, ∠ ACB = 90 = ∠ EGB, ∠ABC = ∠EBG, AB = BE.
So, △ ABC △ EBG
Available: BC = BG
Because, BF⊥DG, GE⊥DG,
So, BF∑GE,
Because BD = BC = BG,
So FD = FE, that is, point F is the midpoint of ED.