如图,已知△ABC中,∠B的平分线与∠C的外角平分线相交于点P,若∠A=70°,则∠P=________.
网友回答
35°
解析分析:由三角形外角性质得,∠ACD=∠A+∠ABC=70°+∠ABC;角平分线的定义,求得∠DCP=∠ACD=(70°+∠ABC)=35°+∠ABC,∠CBP=∠ABC;再由三角形外角性质得,∠DCP=∠CBP+∠P即35°+∠ABC=∠ABC+∠P,求得∠P=35°.
解答:∵∠ACD是△ABC的外角
∴∠ACD=∠A+∠ABC=70°+∠ABC
∵CP是∠ACD的平分线
∴∠DCP=∠ACD=(70°+∠ABC)=35°+∠ABC
∵BP是∠ABC的平分线
∴∠CBP=∠ABC
∵∠DCP是△BCP的外角
∴∠DCP=∠CBP+∠P
35°+∠ABC=∠ABC+∠P
∴∠P=35°.
故