如图,CP平分角ACB,BP是三角形ABC的外角角ABE的平分线,若角A=70°,求角P的度数

Question

如图,CP平分角ACB,BP是三角形ABC的外角角ABE的平分线,若角A=70°,求角P的度数

Answer 1

∵CP平分角ACB
∴∠PCB=1/2∠ACB=1/2(180°-∠A-∠ABC)=90°-1/2∠A-1/2∠ABC
∵BP平分∠ABE
∴∠PBE=1/2∠ABE=1/2(180°-∠ABC)=90°-1/2∠ABC
∵∠PBE=∠PCB+∠P=90°-1/2∠A-1/2∠ABC+∠P
∴90°-1/2∠A-1/2∠ABC+∠P=90°-1/2∠ABC
∠P=1/2∠A
∠P=35°