证⊿ACM≌⊿ANB,∠ACM＝∠ANB.∠MPN＝∠PCN＋∠PNC＝∠PCA＋∠ACN＋∠PNC＝∠PNA＋∠PNC＋∠ACN＝∠ACN＋∠ANC＝120..若在NB上截取NQ＝CP,连结AQ,利用“S.A.S”可证得⊿ACP≌⊿ANQ,所以∠CAP＝∠NAQ,∠QAC＋∠NAQ＝6...