作BH⊥PC于H,在射线HC上截取HG=HP,连结BG（如图）；∴BG=BP∴∠G＝∠P在正六边形中BC=BA,∠ABC=120°,弧BC=60°∴弧BC所对的圆周角∠P＝30°∴∠PBG=180°－∠P－∠G＝180°－2∠P＝120°＝∠ABC∴∠CBG＝∠ABP∴△B...