三角形ABC中两个底角的外角的平分线相交于点P,求证∠P=90-1/2∠A

Question

三角形ABC中两个底角的外角的平分线
相交于点P,求证∠P=90-1/2∠A

Answer 1

证明：∠P=180-(∠PBC+∠PCB)
∠PBC+∠PCB=1/2(180-∠ABC+180-∠ACB)=180-1/2(∠ABC+∠ACB)
∠ABC+∠ACB=180-∠A
则∠PBC+∠PCB=90+1/2∠A
从而：∠P=180-(∠PBC+∠PCB)=90-1/2∠A