已知△ABC ,∠ABC、∠ACB的平分线BE、CF交于点O(1)若∠ABC=80°,∠ACB=60° 求∠A、∠BOC

∵∠ABC＝80,∠ACB＝60
∴∠A＝180-∠ABC-∠ACB＝180-80-60＝40
∵BO平分∠ABC,CO平分∠ACB
∴∠ABO＝∠CBO＝∠ABC/2＝40,∠ACO＝∠BCO＝∠ACB/2＝30
∴∠BOC＝180-（∠CBO+∠BCO）＝180-（40+30）＝110°
∵∠ABC＝70,∠ACB＝50
∴∠A＝180-∠ABC-∠ACB＝180-70-50＝60
∵BO平分∠ABC,CO平分∠ACB
∴∠ABO＝∠CBO＝∠ABC/2＝35,∠ACO＝∠BCO＝∠ACB/2＝25
∴∠BOC＝180-（∠CBO+∠BCO）＝180-（35+25）＝120°
3、∠BOC＝90+∠A,对任意三角形都成立
证明：
∵BO平分∠ABC,CO平分∠ACB
∴∠ABO＝∠CBO＝∠ABC/2,∠ACO＝∠BCO＝∠ACB/2
∴∠BOC＝180-（∠CBO+∠BCO）
＝180-（∠ABC+∠ACB）/2
＝180-（180-∠A）/2
＝90+∠A/2

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

(1) 求∠A、∠BOC的度数∠A=180°-∠ABC-∠ACB=180°-80°-60°=40°,∠BOC=180°-0.5(∠ABC+∠ACB)=80°
2、套上一问过程 ∠A=180°-∠ABC-∠ACB=180°-120°=60°,∠BOC=180°-0.5(∠ABC+∠ACB)=120°∠BOC
3、2∠A=∠BOC 成立 因为∠A=180°-∠ABC-∠ACB，∠BOC=180°-0.5(∠ABC+∠ACB) 代换可得