如图 ,.已知AF=AB,AE=AC,∠BAF=∠CAE

1.如图,已知AB=AE,AF=AC,BF=CE,∠AFC=55°,∠E=40°,求∠BAF的度数 急!大神求解啊啊啊!AF＝AB,AE＝AC,∠BAF＝∠CAE,请说明∠1＝∠BAF的理由 已知：如图,AB=AD,AC=AE,∠BAD=∠CAE,求证：BC=DE 已知:如图,AB=AD,AC=AE,∠CAE=∠BAD.求证△EAD≌△CAB 已知:如图,AB/AD=BC/DE=AC/AE,求证:∠BAD=∠CAE 如图,已知AB分之AE = BC分之ED = AC分之AD 证明∠BAD=∠CAE 如图,AB/AD=BC/DE=AC/AE,求证：∠BAD=∠CAE 如图,等腰三角形ABC中,AB=AC,点D是AC上一动点,点E在BD的延长线上,且AB=AE,AF平分∠CAE交DE于点 如图,AB=AE,BC=ED,∠B=∠E,∠BAF=∠EAF,求证AF垂直CD 如图,AB=AE,BC=ED,∠B=∠E,∠BAF=∠EAF,求证AF⊥CD 已知:如图,AB=AC,AE=AF,AE⊥EC,AF⊥BF,则∠1=∠2吗 已知:如图,AB=AC,AE=AF,AE⊥EC,AF⊥BF,则∠1=∠2吗?为什么?