解题思路:
∵
A
BC
D
是平行四边形,
∴
A
D
∥
BC
,
OA
=
O
C
∴
∠
OAE
=
∠
OC
B
又
∵
∠
AOE
=
"
∠
C
OF
"
∴
△
AOE
≌
△
C
OF
∴
O
E
=
O
F
.
根据平行四边形的性质可得AD∥BC ,OA=OC,根据平行线的性质可得∠OAE=∠OCB,再结合对顶角∠AOE=∠COF即可证得△AOE≌△COF,从而得到结论.
解题思路:
∵
A
BC
D
是平行四边形,
∴
A
D
∥
BC
,
OA
=
O
C
∴
∠
OAE
=
∠
OC
B
又
∵
∠
AOE
=
"
∠
C
OF
"
∴
△
AOE
≌
△
C
OF
∴
O
E
=
O
F
.
根据平行四边形的性质可得AD∥BC ,OA=OC,根据平行线的性质可得∠OAE=∠OCB,再结合对顶角∠AOE=∠COF即可证得△AOE≌△COF,从而得到结论.