证明：在BP的延长线上取点D,使PC＝PD,连接CD∵等边△ABC∴AC＝BC,∠BAC＝∠ACB＝60∵∠BAC+∠BPC+∠ABP+∠ACP＝360,∠ABP+∠ACP＝180∴∠APC＝360-180-60＝120∴∠CPD＝180-∠BPC＝60∵PC＝PD∴等边△PCD∴PC＝DC,∠PCD＝60∴∠ACB＝∠PCD∵∠ACP＝∠ACB+∠BCP,∠BCD＝∠PCD+∠BCP∴∠ACP＝∠BCD∴△ACP≌△BCD （SAS）∴BD＝PA∵PB+PD＝BD∴PB+PC＝BD∴PB+PC＝PA