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82问答网 > 如图，已知四边形ABCD中，AB=DC，AD=BC，点E在BC上，点F在AD上，AF=CE

如图，已知四边形ABCD中，AB=DC，AD=BC，点E在BC上，点F在AD上，AF=CE

EF的对角线BD交于O，求证：BD与EF互相平分

2025-03-14 20:43:56

推荐回答（2个）

回答1：

连接BF、DE
∵AD=BC，AB=DC
∴ABCD是平行四边形
∴AD∥BC
∵AE=CE
∴AD-AF=BC-CF
那么DF=BE
∵DF∥BE
∴BFDE是平行四边形
∴BD与EF互相平分

回答2：

因为AD=BC,AB=CD
所以，四边形ABCD为平行四边形
所以，AD//BC,AD=BC
因为，AF=EC
所以，BE=DF
又因为，角OBE=角FDO,角BOE=角FOD
所以，角BEO=角OFD
所以三角形BEO和三角形FDO全等
所以，BO=DO 所以O是BD中点

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