(1)证明:连接OD,CD,
∵AC是⊙O的直径,
∴∠CDA=90°=∠BDC,
∵OE ∥ AB,CO=AO,
∴BE=CE,
∴DE=CE,
∵在△ECO和△EDO中
DE=CE
EO=EO
OC=OD ,
∴△ECO≌△EDO,
∴∠EDO=∠ACB=90°,
即OD⊥DE,OD过圆心O,
∴ED为⊙O的切线.
(2)过O作OM⊥AB于M,过F作FN⊥AB于N,
则OM ∥ FN,∠OMN=90°,
∵OE ∥ AB,
∴四边形OMFN是矩形,
∴FN=OM,
∵DE=4,OC=3,由勾股定理得:OE=5,
∴AC=2OC=6,
∵OE ∥ AB,
∴△OEC ∽ △ABC,
∴
OC
AC =
OE
AB ,
∴
3
6 =
5
AB ,
∴AB=10,
在Rt△BCA中,由勾股定理得:BC=
1 0 2 - 6 2 =8,
sin∠BAC=
BC
AB =
OM
OA =
8
10 ,
即
OM
3 =
4
5 ,
OM=
12
5 =FN,
∵cos∠BAC=
AC
AB =
AM
OA =
3
5 ,
∴AM=
9
5
由垂径定理得:AD=2AM=
18
5 ,
即△ADF的面积是
1
2 AD×FN=
1
2 ×
18
5 ×
12
5 =
108
25 .
答:△ADF的面积是
108
25 .