在三角形ABC中，a2=b(b+c)是A=2B的什么条件？(
\u4e09\u89d2\u5f62ABC\u4e2d,\u5df2\u77e5a2-c2=2b,\u4e14acosC=3ccosA,\u6c42b\u6839\u636e\u6b63\u5f26\u5b9a\u7406\uff0ca,c\u4e0esinA\uff0csinC\u6210\u6b63\u6bd4\uff0ca/sinA=c/sincsinAcosC=3cosAsinC\u6240\u4ee5a*cosC=3cosA*c\u6839\u636e\u4f59\u5f26\u5b9a\u7406\uff0ccosC=\uff08a^2+b^2-c^2\uff09/2ab\uff0ccosA=\uff08c^2+b^2-a^2\uff09/2bc\uff0c\u6240\u4ee5a*cosC=\uff08a^2+b^2-c^2\uff09/2b\uff0c3cosA*c=3\uff08c^2+b^2-a^2\uff09/2b\uff0c\u56e0\u4e3aa*cosC=3cosA*c\uff0c\u6240\u4ee5a^2+b^2-c^2=3\uff08c^2+b^2-a^2\uff09\uff0c\u56e0\u4e3aa^2-c^2=2b,\u6240\u4ee5b^2+2b=3\uff08b^2-2b\uff09\uff0c\u5373b^2-4b=0\uff0c\u56e0\u4e3ab\u4e3a\u4e09\u89d2\u5f62ABC\u8fb9\u957f\uff0c\u6240\u4ee5b\u4e0d\u7b49\u4e8e0\u6240\u4ee5b=4
\u5e94\u9009A\uff0c\u5145\u8981\u6761\u4ef6\u3002
1\u3001\u5145\u5206\u6027\uff0c\u8bbe\u5df2\u77e5a^2=b(b+c)
\u5ef6\u957fCA\u81f3E\uff0c\u4f7fAE=AB\uff0c\u8fde\u7ed3BE\uff0cEC=b+c,<ACB=<ECB,a/(b+c)=b/a,\u25b3BCA\u223d\u25b3ECB
<E=<ABC,<BAC=<E+<ABE,\u4e09\u89d2\u5f62EBA\u4e3a\u7b49\u8170\u4e09\u89d2\u5f62\uff0c<E=<EBA,BAC=2<ABC,\u8fd9\u662f\u5145\u5206\u6027\u3002
2\u3001\u5fc5\u8981\u6027
\u8bbe\u5df2\u77e5<A=2<B
\u540c\u6837\uff0c\u5ef6\u957fCA\u81f3E\uff0c\u4f7fAE=AB\uff0c\u8fde\u7ed3BE\uff0c
<BAC=<E+<ABE\uff0cEA=BA\uff0c\u4e09\u89d2\u5f62EBA\u4e3a\u7b49\u8170\u4e09\u89d2\u5f62\uff0c<E=<ABC\uff0c<BAC=2<E=2<ABC,<ABC=<E,<ACB=<BCE,
\u25b3BCA\u223d\u25b3ECB
BC/EC=AC/BC,BC^2=EC*AC,EC=AB+AC
\u2234a^2=b(b+c)
\u8bc1\u6bd5\u3002
A,\u597d\u50cf\u662f\u8003\u7eb2\u4e0a\u7684
\u5145\u5206\u6027
a^2=b(b+c)
a^2=b^2+bc
b^2+c^2-a^2=-bc+c^2
cosA=-1/2+0.5(c/b)
cosA=-1/2+0.5(sinC/sinB)
2sinBcosA=-sinB+sin(A+B)
sinB=sinAcosB-sinBcosA
sinB=sin(A-B)
\u6240\u4ee5B=A-B\u6216\u8005B+A-B=\u03c0
B+A-B=\u03c0\u4e0d\u53ef\u80fd
\u6240\u4ee5A=2B
\u5fc5\u8981\u6027
\u7531A=2B
\u2234sinA=sin2B
sinBcosC+sinCcosB=2sinBcosB
bcosC+ccosB=2bcosB
\u8fd9\u91cc\u662f\u5c04\u5f71\u516c\u5f0fbcosC+ccosB=a
a=2bcosB
a=2b[(a^2+c^2-b^2)/2ac]
a^2c=ba^2+bc^2-b^3
a^2(c-b)=bc^2-b^3
a^2=b(b+c)
\u8fd8\u4e0d\u61c2\u95ee\u6211\u54c8
具体证明过程如下：
1.充分性
因为 A=2B
所以 sinC=sin(A+B)=sin3B
所以(sinB+sinC)/sinA=[1-(sinB)^2+3(cosB)^2)]/2cosB=2cosB
此处用到了正弦三倍角公式：sin3B=-(sinB)^3+3sinB(cosB)^2
因为 sinA/sinB=2sinBcosB/sinB=2cosB=(sinB+sinC)/sinA
所以 a/b=(b+c)/a
所以 a^2=b*(b+c)
2.必要性
因为 a^2=b(b+c),s (sinA)^2=(sinB)^2+sinBsin(A+B)
所以 (sinA+sinB)(sinA-sinB)=sinBsin(A+B)
所以 4sin[(A+B)/2]*cos[(A-B)/2]*cos[(A+B)/2]*sin[(A-B)/2]=sinBsin(A+B)
此处用到了和差化积的公式：
sinA+sinB=2sin[(A+B)/2]*cos[(A-B)/2]
sinA-sinB=2cos[(A+B)/2]*sin[(A-B)/2]
所以 sin(A+B)sin(A-B)=sinBsin(A+B)
所以 sin(A-B)=sinB
所以 A=2B
证明完毕
希望能够帮到你。
充分非必要条件
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