解连接AC,AD
∵AB=AE,BC=DE且∠ABC=∠AED
∴⊿ABC≌⊿AED
∴AC=AD
又∵CF=DF,AF=AF
∴⊿AFC≌⊿AFD
∴∠AFC=∠AFD又∵∠AFC+∠AFD=180°
∴AF⊥CD