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.