Question

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°，求∠P的度数．

Answer 1

∵∠ACD是△ABC的外角
∴∠ACD=∠A+∠ABC=70°+∠ABC
∵CP是∠ACD的平分线
∴∠DCP=

1

2

∠ACD=

1

2

（70°+∠ABC）=35°+

1

2

∠ABC
∵BP是∠ABC的平分线
∴∠CBP=

1

2

∠ABC
∵∠DCP是△BCP的外角
∴∠DCP=∠CBP+∠P
35°+

1

2

∠ABC=

1

2

∠ABC+∠P
∴∠P=35°．