如右图,点P是△ABC的外角∠ACE的角平分线和内角∠ABC的角平分线的交点,若∠A=40,则∠P=多少

Question

如右图,点P是△ABC的外角∠ACE的角平分线和内角∠ABC的角平分线的交点,若∠A=40,则∠P=多少

Answer 1

∵CP平分∠ACE,
∴∠ECP=1/2∠ECA=1/2(∠ABC+∠A)=1/2∠ABC+1/2∠A,
∵BP平分∠ABC,∴∠CBP=1/2∠ABC,
又∠EAP=∠P+∠CBP
∴∠P=∠EAP-∠CBP
=1/2∠ABC+1/2∠A-1/2∠ABC
=1/2∠A
=20°.
欢迎追问.