如图(1),△ABC和△ADE中,已知AB=AC,AD=AE,且∠CAB=∠EAD

如图(1),△ABC和△ADE中,已知AB=AC,AD=AE,且∠CAB=∠EAD 如图 △ABC和△ADE中,已知AB=AC,AD=AE且∠CAB=∠EAD,请说明BE=CD成立的理由 如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD.求证CE=BD 如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD=90° 如图,等腰三角形ABC和等腰三角形ADE中,AB=AC,AD=AE,∠CAB=∠EAD,试说明：△ACE≌ΔABD 如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD=90°.求证:1：求证CE=BD；CE⊥BD. 在△ABC和△ADE中,AB=AC,AD=AE,∠CAB=∠EAD.试说明CE=BD. 如图,在△ABC和△ADE中,AC=AB,AE=AD,∠CAB=∠EAD=90°.求证CE=BD；CE⊥BD. 如图在三角形ABC和三角形ADE中,已知AB=AC,AD=AE且角BAC=角EAD,点d在bc上 已知:如图,AB=AD,AC=AE,∠CAE=∠BAD.求证△EAD≌△CAB 已知如图等腰三角形ABC与DEF AB=AC AD=AE 且∠CAB=∠EAD 求证CE=BD 三角形ABC和三角形ADE中,AB=AC,AD=AE.且角CAB=角EAD,求证CE=BD