∠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.