△ABC中,a、b、c是三内角∠A、∠B、∠C所对的边,求证:a^2=b(b+c)是∠A=2∠B的充要条件.
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![△ABC中,a、b、c是三内角∠A、∠B、∠C所对的边,求证:a^2=b(b+c)是∠A=2∠B的充要条件.](/uploads/image/z/2528678-38-8.jpg?t=%E2%96%B3ABC%E4%B8%AD%2Ca%E3%80%81b%E3%80%81c%E6%98%AF%E4%B8%89%E5%86%85%E8%A7%92%E2%88%A0A%E3%80%81%E2%88%A0B%E3%80%81%E2%88%A0C%E6%89%80%E5%AF%B9%E7%9A%84%E8%BE%B9%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%5E2%3Db%28b%2Bc%29%E6%98%AF%E2%88%A0A%3D2%E2%88%A0B%E7%9A%84%E5%85%85%E8%A6%81%E6%9D%A1%E4%BB%B6.)
△ABC中,a、b、c是三内角∠A、∠B、∠C所对的边,求证:a^2=b(b+c)是∠A=2∠B的充要条件.
△ABC中,a、b、c是三内角∠A、∠B、∠C所对的边,求证:a^2=b(b+c)是∠A=2∠B的充要条件.
△ABC中,a、b、c是三内角∠A、∠B、∠C所对的边,求证:a^2=b(b+c)是∠A=2∠B的充要条件.
证明:先证a²=b(b+c)是∠A=2∠B的充分条件.
由a²=b²+c²-2bccosA得cosA=(b²+c²-a²)/(2bc),又由a²=b(b+c)得
c=(a²-b²)/b,代入cosA=(b²+c²-a²)/(2bc)得
cosA=[b²+(a²-b²)²/b²-a²]/[2(a²-b²)]=(a²-2b²)/(2b²)
=a²/(2b²)-1=sin²A/(2sin²B)-1=(1-cos²A)/(1-cos2B)-1,即1+cosA=(1-cos²A)/(1-cos2B),1=(1-cosA)/(1-cos2B),
1-cosA=1-cos2B,cosA=cos2B,A=2B或A=2π-2B.若A=2π-2B,则A+2B=2π,A/2+B=π,A+B>π,故A=2π-2B不合题意,所以A=2B.
再证a²=b(b+c)是∠A=2∠B的必要条件.
若A=2B,则sinA=sin2B=2sinBcosB,cosB=sinA/(2sinB)=a/(2b),又cosB=(a²+c²-b²)/(2ac),所以a/(2b)=(a²+c²-b²)/(2ac),b(a²+c²-b²)/(a²c)=1,
b(a²+c²-b²)=a²c,a²b+bc²-b²b=a²c,a²b-a²c=b²b-bc²,a²(b-c)=b(b²-c²),
a²=b(b+c).