证明：取AB中点D,连接ED∵AE平分∠BAC,∠BAC＝2∠B∴∠BAE＝∠CAE＝∠BAC/2＝∠B∴AE＝BE∵D是AB的中点∴ED⊥AD（三线合一）,AD＝AB/2∴∠ADE＝90∵AB＝2AC∴AC＝AB/2∴AC＝AD∵AE＝AE∴△ACE≌△ADE（SAS）∴∠...