(2003•海南)如图所示,△ABC≌△AEF,AB=AE,∠B=∠E,有以下结论:①AC=AE;②∠FAB=∠EAB;
已知:如图,AE=AC,AB=AD,∠EAB=∠CAD 已知:如图,AE=AC,AB=AD,∠EAB=∠CAD
已知如图,AE=AC,∠E=∠C,∠EAB=∠CAD,求证:AB=AD
如图,已知AB=AC,BC=DC,AE平分∠FAB,问:AE与AD是否垂直?为什么?
已知:如图,AE=AC,AB=AD,∠EAB=∠CAD
如图,AD,BD分别平分∠EAB和∠ABC,AE垂直EC于E,BC垂直EC于C.求证:AB=AE+BC
如图所示,在△ABC中,∠BAC=2∠B,AB=2AC,AE平分
如图所示,在直角三角形△ABC中 ∠ACB=90° 点D、E在边AB上,且有AE=AC BC=BD
已知,在△ABC中,∠CAB=2∠B,AE平分∠CAB交BC于E,AB=2AC 求证:AE=2CE
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
如图,AB=AC,AD=AE,∠EAB=∠DAC,问:△ABD与△ACE是否全等?∠D与∠E有什么关系?为什么?
如图,已知在△ABC和△AEF中,AB=AC,AE=AF,∠CAB=∠EAF,BE交FC于O点.
在△ABC中,AB=AC,AE‖BC,试说明AE平分∠DAC