七下数学三角形证明题如图,在△ABE中,AB=AE,AD=AC,∠BAC=∠EAD,BD与CE交于点O,那么BD=EC吗

七下数学三角形证明题如图,在△ABE中,AB=AE,AD=AC,∠BAC=∠EAD,BD与CE交于点O,那么BD=EC吗 如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O.求证：∠CAB=∠EAD=∠BOC 如图 AB=AC AD=AE BD=CE,BD与CE相交于点O.求证∠BOC=∠EAD 如图,已知AB=AC,AD=AE,∠EAD=∠CAB,BD与CD相较于点O,求证说明：BD=CE． 如图,在△ABC和△ADE中,∠BAC=∠DAE=90°,AB=AC,AD=AE,CE与BD相交于点M,BD交AC于点N 如图,AB=AC,AD=AE,BD=CE,BD与CE相较与点o,说明∠cab=∠ead=∠boc 如图,在Rt△ABC和Rt△ADE中AB=AC,AD=AE,CE与BD相交于点M,BD交AC于点N.（1）证明BD=CE 如图,AB=AC,AD=AE,BD=CE,BD与CE相交于点O,求证：角CAB=角EAD=角BOC 初三相似三角形题如图,在△ABC中,AB=AC,∠BAC=90°,BD是中线,AE⊥BD,交BC于点E,求证BE=2EC 如图，AB=AC，AD=AE，BD、CE交于O，求证：AO平分∠BAC． 如图,在△ABC中,AB=AC,AD⊥BD,AE⊥CE,且AD=AE,BD和CE交于点O.说明∠EAB=∠DAC △ABC的三个顶点都在圆O上,AD平分∠BAC,交BC于点D,交圆O与点E,如何证明AB*AC=AD*AE,即△ABE相