如图所示,在△ABC中,∠A=α,△ABC的内角平分线或外角平分线交于点P,且∠P=β,试探求下列

2025-05-17 14:33:33
推荐回答(2个)
回答(1):

1、α=2β
∵BP平分∠ABC,那么∠3=∠4=1/2∠ABC
CP平分∠ACB外角。那么∠1=∠2=1/2(180°-∠ACB)=90°-1/2∠ACB
∴∠2=∠P+∠4=β+1/2∠ABC
∴90°-1/2∠ACB=β+1/2∠ABC
90°-1/2(∠ACB+∠ABC)=β
90°-1/2(180°-∠A)=β
1/2∠A=β
∴α=2β
2、∵BP平分∠ABC外角,CP平分∠ACB外角
∴∠CBP=1/2(180°-∠ABC)=90°-1/2∠ABC
∠BCP=1/2(180°-∠ACB)=90°-1/2∠ACB
∴∠P=180°-(∠CBP+∠BCP)
=180°-(90°-1/2∠ABC+90°-1/2∠ACB)
=1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=90°-1/2∠A
即β=90°-1/2α

回答(2):

角b=角a/2

角a/2+角b=90度