如图2,o是角ABC与外角角ACD的平分线BO和CO的交点,试分析角BOC与角A有怎样的关系?请说明理由.

证明：∵∠A+∠ABC+∠ACB＝180°∴∠ABC+∠ACB＝180°-∠A∵∠ACD＝180-∠ACB,CO平分∠ACD∴∠2＝∠ACD/2＝（180°-∠ACB）/2＝90°-∠ACB/2∵BO平分∠ABC∴∠1＝∠ABC/2∵∠2是△OBC的外角∴∠2＝∠BOC+∠1＝∠BOC+∠ABC/2∴∠BOC+∠ABC/2＝90°-∠ACB/2∴∠BOC＝90°-（∠ABC+∠ACB）/2＝90°-（180°-∠A）/2＝∠A/2∴∠A＝2∠BOC