一、辅助线:1、过A点做射线AX使∠PAX = 10?腺鵄X = 30?2、过B点做射线BY使∠PBY= 20?蝈于点M,交AC于点N.二、证明:1、由原题得知:∠APB = 150?舷哖C = 110?喯嘝C = 100?2、∠BAP = ∠MAP =10?舷咮P = ∠MBP =20?希犂q点是△ABM内心,所以∠AMP = ∠BMP =60?虾煜嘝M =100?嚒螧PC,所以点M在PC上.3、由以上推出∠BMP = ∠PMA =∠AMN = ∠NMC=60?腺鵄M = ∠ACM =30?可以推出AN=CN且BN⊥AC; 4、所以AB=AC,△ABC是等腰三角形.