Question

已知,如图O是△ABC的内角∠ABC和外角∠ACE的平分线的交点.

Answer 1

∵∠A+∠ABC+∠ACB＝180

∴∠ABC+∠ACB＝180-∠A

∵∠ACE＝180-∠ACB,CO平分∠ACE

∴∠OCE＝∠ACE/2＝（180-∠ACB）/2＝90-∠ACB/2

∵BO平分∠ABC

∴∠OBC＝∠ABC/2

∵∠OCE是△OBC的外角

∴∠OCE＝∠BOC+∠OBC＝∠BOC+∠ABC/2

∴∠BOC+∠ABC/2＝90-∠ACB/2

∴∠BOC＝90-（∠ABC+∠ACB）/2＝90-（180-∠A）/2＝∠A/2

∴当∠A＝46时,∠BOC＝∠A/2＝23°

∠A=n°时,∠BOC＝n°/2