如图,四边形ABCD是平行四边形,对角线AC、BD交于点O,过点O画直线EF分别交AD、BC于点E、F。求证:OE=OF
1个回答

解题思路:

A

BC

D

是平行四边形,

A

D

BC

OA

=

O

C

OAE

=

OC

B

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.