AE平分∠BAC,∴∠EAB=∠EAC,DE垂直平分BC,∴BE=CE,延长AE到P,使EP=AE,连接BP,∵∠AEC=PEB,∴ΔAEC≌ΔPEB,∴∠EAC=∠P,AC=PB∴∠P=∠BAE,∴AB=PB,∴AB=AC.由∠BAE=∠CAE,AE=AE,∠ANE=∠AMC=90°得：ΔAEN≌ΔAEM,∴A...