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如图,△ABC中,AB=AC,AD⊥BC,CE⊥AB,AE=CE.求证: (1)△AEF≌△CEB; (2)AF=2CD.
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如图在△ABC中AB=AC作AD⊥AB交BC的延长线于点D.作AE∥BDCE⊥AC且AECE相交于点
如图在△ABC中CA=CBAD⊥BCBE⊥ACAB=5AD=4则AE=.
如图点DE在△ABC的边BC上AB=ACBD=CE.求证AD=AE.
如图∠1=∠2要证明△ABC≌△ADE还需补充的条件是
AB=AD,AC=AE
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已知点D.E分别在△ABC的边AB.AC上下列给出的条件中不能判定DE∥BC的是A.BDAB=CEA
如图4Rt△ABC中AB⊥ACAD⊥BCBE平分∠ABC交AD于点E.EF∥AC下列结论一定成立的是
AB=BF
AE=ED
AD=DC
∠ABE=∠DFE
如图△ABC中AB=ACAD⊥BCCE⊥ABAE=CE.求证1△AEF≌△CEB2AF=2CD.
如图点D.E.在△ABC的边BC上AD=AEBD=CE求证AB=AC.
已知如图在△ABC中AB=ACAD是BC边上的中线AE∥BCCE⊥AE垂足为E.1求证△ABD≌△C
已知点D.E.分别在△ABC的边AB.AC上下列给出的条件中不能判定DE∥B.C.的是
)BD︰AB = CE︰AC; (
)DE︰BC = AB︰AD; (
)AB︰AC = AD︰AE; (
)AD︰DB = AE︰EC.
如图△ABC中AB=ACAD⊥BCCE⊥ABAE=CE.求证1△AEF≌△CEB2AF=2CD.
如图△ABC中AB=ACAD⊥BCCE⊥ABAE=CE.求证1△AEF≌△CEB2AF=2CD.
如图点D.E.在△ABC的边BC上AB=ACAD=AE求证BD=CE
如图在△ABC中CA=CBAD⊥BCBE⊥ACAB=5AD=4则AE=.
如图在△ABC中AB=ACAD⊥BCBE⊥AC则下列结论不正确的是
BD=DC
CE=AE
∠BAD=∠CAD
∠CBE=∠DAC
如图在△ABC中AB=AC作AD⊥AB交BC的延长线于点D.作AE∥BDCE⊥AC且AECE相交于点
如图在△ABC中AB=AC作AD⊥AB交BC的延长线于点D.作AE∥BDCE⊥AC且AECE相交于点
如图在△ABC中AB=AC点D.E.在BC上连接AD.AE如果只添加一个条件使得∠DAB=∠EAC则
AD=DE
BD=CE
AD=AE
BE=CD
如图在△ABC中AB=ACAD.CE是△ABC的两条中线P.是AD上一个动点则下列线段的长度等于BP
BC
CE
AD
AC
如图△ABC中AB=ACAD⊥BCCE⊥ABAE=CE.求证1△AEF≌△CEB2AF=2CD.
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