如图,∠DAC=∠CAE,AD=AE,D为BC上一点,且BD=DC,AC^2=AE^2+CE^2.求证:AB^2=AE^2+CE^2

如图,已知：∠DAC=∠CAE,CE⊥AB于E,AE=1/2（AD+AB）.求证：∠B+∠D=180° 如图,∠DAC=∠CAE,AD=AE,D为BC上一点,且BD=DC,AC^2=AE^2+CE^2.求证:AB^2=AE^2+CE^2 如图,AB/AD=BC/DE=AC/AE,求证：∠BAD=∠CAE 如图,B,D分别在AC,CE上,AD是∠CAE的平分线,BD‖AE,AB=BC.求证:AC=AE 如图AB=AD ∠BAD=∠CAE AC=AE 求证AB=AD 已知如图△ABC中,AB=BC,AD为中线,E为BC延长线上一点,且CE=CB,求证∠DAC=∠CAE 已知:如图,AB=AE,AC=AD,BC=DE,C、D在边BE上.求证:∠CAE=∠DAB 如图,D在BC上,E在AB上,AB=AC,∠B=∠CAE,BD=AE,求证:AD=CE 已知：如图,点D、E在△ABC的边BC上,AD=AE,∠BAD=∠CAE.求证：AB=AC 已知：如图,点D、E在△ABC的边BC上,AD=AE,∠BAD=∠CAE.求证：AB=AC 已知：如图,点D、E在△ABC的边BC上,AD=AE,∠BAD=∠CAE.求证：AB=AC【用三线合一】 如图,d.e是△abc的边bc上两点,且ab=ac,ae=ad,试说明∠BAD=∠CAE 如图,AB=AD,AC=AE,∠DAB=∠CAE.∠C与∠ 为什么? 已知：如图AB=AD,AC=AE,∠BAD=∠CAE ,求证：OA平分∠BOE 1、如图1111,AB=AD,∠BAD=∠CAE,AC=AE,求证：DE=BC 如图,AB=AD,AC=AE,∠BAD=∠CAE,连接BC、DE,求证：bc=de 如图,已知AB=AC,AD=AE,∠BAD=∠CAE.求证：BE=CD. 已知：如图,AB=AD,AC=AE,∠BAD=∠CAE,求证：BC=DE