AD,EF,BC相交于O点,且AO=OD,BO=OC,EO=OF.求证：△AEB≌△DFC
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先证明：△OAE≌△ODF,因为：二边及夹角（对等角）相等,得：AE=DF.
同理证得：△OBE≌△OCF,△OAB≌△OCD,得：EB=CF,AB=CD.
因为：AE=DF,EB=CF,AB=CD 三边相等.
所以：△AEB≌△DFC
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======以下答案可供参考======
供参考答案1:
AO=OD，BO=OC, ∠AOB=∠COD ==>△AOB≌△COD==>CD=ABBO=OC, EO=OF,∠BOE=∠FOC ==>△BOE≌△FOC==>FC=BEAO=OD, EO=OF,∠AOE=∠FOD ==>△AOE≌△FOD==>AE=FD三边相等==>△AEB≌△DFC