∫ 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