如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o
如图△abe中ab=ae,ad=ac,角bad=角eac,bc,ed交与点o
如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC,DE交于点O.试说明:BC=ED
如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC、DE交于点O.∠ABC=∠AED;
如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC、DE交于点O.
求解一道图形题7.如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC,BC,DE交与点O.求证:(1)△AB
如图,在△ABE中,AB=AE,AD=AC,∠BAD=∠EAC、BC、DE交于点O.求证(△ABC≌△AED)
如图,在△ABC中,AB=AC,D是BC中点,AE平分∠BAD交BC于点E,点O是AB上一点,⊙O过A、E两点,交AD于
如图所示,AB=AE,AC=AD,∠BAD=∠EAC,试说明ED=BC
如图,在△ABC中,AD是角平分线,点E在AB上,且AE=AC,EF‖BC,分别交AC、AD于点F、G,CE交AD于点O
如图已知ab=ad,ac=ae∠bad=∠eac是说明bc=de
如图2,△ABC中,AB>AC,AD平分△的外角∠EAC交BC的延长线于点D,在AB的反向延长线上截取AE=AC,
如图BD为圆O直径,AB=AC AD交BC于点E,AE=2,ED=4