Question

如图,∠A=50°,∠ABC和∠ACB的外角平分线交于点P，求∠P的度数

解题要详细哦！

Answer 1

解：在BC延长线上取点D
∵∠A+∠ABC+∠ACB＝180, ∠A＝50
∴∠ABC+∠ACB＝180-∠A＝180-50＝130
∵CP平分∠ACD, ∠ACD＝∠ABC+∠A＝∠ABC+50
∴∠DCP＝∠ACD/2＝（∠ABC+50）/2＝∠ABC/2+25
∵BP平分∠ABC
∴∠CBP＝∠ABC/2
∵∠DCP＝∠CBP+∠P
∴∠CBP+∠P＝∠ABC/2+25
∴∠ABC/2+∠P＝∠ABC/2+25
∴∠P＝25

追问

加一下qq好么，我图发不出去

追答

解：
∵∠A+∠ABC+∠ACB＝180，∠A＝50
∴∠ABC+∠ACB＝180-∠A＝180-50＝130
∵∠DBC＝180-∠ABC，PB平分∠DBC
∴∠PBC＝∠DBC/2＝（180-∠ABC）/2
∵∠ECB＝180-∠ACB，PC平分∠ECB
∴∠PCB＝∠ECB/2＝（180-∠ACB）/2
∴∠PBC+∠PCB
＝（180-∠ABC）/2+（180-∠ACB）/2
＝（360-∠ABC-∠ACB）/2
＝[360-（∠ABC+∠ACB）]/2
＝（360-130）/2
＝115