Question

如图,已知△ABC的一条内角平分线与一条外角平分线交于点P,求证：∠P=1/2∠A

Answer 1

证明：

∵∠ACD＝∠A+∠ABC,CP平分∠ACD

∴∠PCD＝∠ACD/2＝（∠A+∠ABC）/2

∵BP平分∠ABC

∴∠PBC＝∠ABC/2

∴∠PCD＝∠P+∠PBC＝∠P+∠ABC/2

∴∠P+∠ABC/2＝（∠A+∠ABC）/2

∴∠P＝∠A/2