在三角形ABC中AB=AC,角A=120度,AB的垂直平分线交BC于M,交AB于E,AC的垂直平分线交BC于N,交AC于F,求证：BM=MN=NC.过点A做AH⊥BC于H.不妨设 BC = 6 ,则有：∠BAH = ∠CAH = 60°,BH = CH = 3 ；AB = BH / sin∠BAH = 2...