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:△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,△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,△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△DBC中,已知AB=DB,AC=DC,则下列说法中错误的是()A △ABC全等于△DBC B ∠A=
如图,在△ABC和△DCB中,AB=DC AC=DB.AC与DB交于点M.(1)求证:△ABC≌
如图,△ABC≌△DBC,AB=CD ,AC=DB,BC=BC求证∠A=∠D
如图,在△ABC和△DCB中,AB=DC,AC=DB,AC与DB相交于点M (2)
如图,在△ABC和△DCB中,AB=DC,AC=DB,AC、DB相交于点M.
如图,D为△ABC内一点,且DB=DC,AB=AC,AD的延长线交BC于E点,求证:AE⊥BC.
如图,D为△ABC内一点,且DB=DC,AB=AC,AD的延长线交BC于E点,.求证AE⊥BC