∫ DE ⊥ AB, ∴ Deb = 90, so: ∠ EDB = 180-90-45 = 45.

∵ac∥bf,∴∠fbc= 180-∠ACB = 180-90 = 90

That is, δδBFD is an isosceles right triangle.

∴BF=BD, and d is the midpoint of BC.

∴CD=BD=EF

AC = BC，∠ACD =∠FBC = 90°，BF = CD

∴δcbf≌δacd(sas)

∴∠CAD=∠BCF

∠∠CAD+∠ADC = 90。

∴∠BCF+∠ADC=90

Therefore: ∠ CGD =180-(∠ BCF+∠ ADC) =180-90 = 90.

∴AD⊥CF