(1)△ACM,△CBN 是等边三角形∴AC=MC CN=BC ∠ACN = ∠BCM = 180-60=120
∴△ACM ≌ △MCB ∴AN=BM
(2)在BF上取一点E,使得BE＝NF,由（1）知∠CNF＝∠CBE,由CB＝CN得 CBE≌△CNF,∴CF＝CE,∠BCE＝∠NCF,∴∠ECF＝60°即△CEF为等边三角形 ∴CF＝EF,∴BF＝CF＋NF.