在三角形abc中,角a等于α,三角形abc的平分线或外角平分线交于p,且角p=β,试探求下列各图中α与β的关请帮忙证明角β=90°+二分之一α要过程【∵,D:\My Documents\My Pictures\无标题.bmp
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![在三角形abc中,角a等于α,三角形abc的平分线或外角平分线交于p,且角p=β,试探求下列各图中α与β的关请帮忙证明角β=90°+二分之一α要过程【∵,D:\My Documents\My Pictures\无标题.bmp](/uploads/image/z/7229845-37-5.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2abc%E4%B8%AD%2C%E8%A7%92a%E7%AD%89%E4%BA%8E%CE%B1%2C%E4%B8%89%E8%A7%92%E5%BD%A2abc%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E6%88%96%E5%A4%96%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8Ep%2C%E4%B8%94%E8%A7%92p%3D%CE%B2%2C%E8%AF%95%E6%8E%A2%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%9B%BE%E4%B8%AD%CE%B1%E4%B8%8E%CE%B2%E7%9A%84%E5%85%B3%E8%AF%B7%E5%B8%AE%E5%BF%99%E8%AF%81%E6%98%8E%E8%A7%92%CE%B2%3D90%C2%B0%2B%E4%BA%8C%E5%88%86%E4%B9%8B%E4%B8%80%CE%B1%E8%A6%81%E8%BF%87%E7%A8%8B%E3%80%90%E2%88%B5%2CD%3A%5CMy+Documents%5CMy+Pictures%5C%E6%97%A0%E6%A0%87%E9%A2%98.bmp)
在三角形abc中,角a等于α,三角形abc的平分线或外角平分线交于p,且角p=β,试探求下列各图中α与β的关请帮忙证明角β=90°+二分之一α要过程【∵,D:\My Documents\My Pictures\无标题.bmp
在三角形abc中,角a等于α,三角形abc的平分线或外角平分线交于p,且角p=β,试探求下列各图中α与β的关
请帮忙证明角β=90°+二分之一α要过程【∵,
D:\My Documents\My Pictures\无标题.bmp
在三角形abc中,角a等于α,三角形abc的平分线或外角平分线交于p,且角p=β,试探求下列各图中α与β的关请帮忙证明角β=90°+二分之一α要过程【∵,D:\My Documents\My Pictures\无标题.bmp
【如果是∠B、∠C的外角平分线交于P】
∵PB是∠B的外角平分线
∴∠PBC=1/2(∠A+∠C)
∵PC是∠C的外角平分线
∴∠PCB=1/2(∠A+∠B)
∴∠P=180°-∠PBC-∠PCB
= 180°-1/2(∠A+∠C)-1/2(∠A+∠B)
= 180°-1/2(∠A+∠B+∠C)-1/2∠A
= 180°-1/2*180°-1/2∠A
= 90°-1/2∠A
∠A=α,∠P=β
即:β = 90°-1/2α
【没图,如果是∠B的平分线与∠C的外角平分线交于P,则β=1/2α】
证明:
令∠C的外角为∠ACD
则∠ACD=∠A+∠B
∵PB是∠B的平分线
∴∠PBC =∠PBA = 1/2∠B
∵PC是∠C的外角平分线
∴∠PCA=1/2∠ACD=1/2(∠A+∠B)
∴∠P = 180°-∠PBC-∠PCB
= 180°-1/2∠B-(∠C+∠PCA)
= 180°-1/2∠B-{∠C+1/2(∠A+∠B)}
= 180°-1/2∠B-∠C-1/2∠A-1/2∠B
= 180°-∠B-∠C-1/2∠A
= (180°-∠B-∠C)-1/2∠A
= ∠A-1/2∠A
= 1/2∠A
∠A=α,∠P=β
即:β = 1/2α
a=2b