如图,△ABC与△DCB的顶点A和D在BC的同侧,AB=DC,AC=BD,AC与BD相交于点O,求证：OA=OD

如图,△ABC与△DCB的顶点A和D在BC的同侧,AB=DC,AC=BD,AC与BD相交于点O,求证：OA=OD 如图,三角形ABC与DCB的顶点在A和D在BC的同侧,AB=DC,AC=BD,AC与BD相交于点O,求证：OA=OD 如图,在△ABC和△DCB中,AC与BD相交于点O,AB=DC,AC=BD,求证:△ABC≌△DCB. 如图,在△ABC和△DCB中,AC与BD相交于点O,AB=DC,AC=BD.求证：△ABC≌△DCB 要用最简便的方法 如图：△ABC和△DCB中,AC与BD相交于O,AB=DC,AC=BD. 如图：△ABC和△DBC的顶点A和D在BC的同旁，AB=DC，AC=DB，AC和DB相交于点O，求证：∠A=∠D． 如图所示,已知△ABC与收件箱DBC的顶点A和D在BC在同侧,AB＝DC,AC＝DB,AC和DB交于点O,那么OA与OD 如图,AC和BD相交于点O,且AB平行DC,OC=OD,求证：OA=OB 如图：AC和BD相交于点O,OA=OC,OB=OD,求证：DC//AB 如图,△ABC和△DCB的顶点A和点D在BC的同旁,AB=DC,AC=DB,AC交DB于点O,试说明:△AOB≌△DOC 如图在△ABC和△DCB中,AC与BD相交于点O.AB=DC.AC=DB.试说明：∠ABC=∠DCB 如图,△ABC中,AB=AC,AB+AC=BD+DC,AC、BD交于O；求证：OA>OD