三角形ABC中,a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=

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三角形ABC中,a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=
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三角形ABC中,a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=
三角形ABC中,a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=

三角形ABC中,a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=
正弦定理
a/sinA=b/sinB=c/sinC
令1/x=a/sinA=b/sinB=c/sinC
所以sinA=ax
sinB=bx
sinC=cx
所以原式=a(bx-cx)+b(cx-ax)+c(ax-bx)
=abx-acx+bcx-abx+acx-bxc
=0

a(sinB-sinC)+b(sinC-sinA)+c(sinA-sinB)=0
由正弦定理得
a:b:c=sinA:sinB:sinC
a(sinB-sincC)+b(sinC-sinA)+c(sinA-sinB)
=c[(a/c)(sinB-sincC)+(b/c)(sinC-sinA)+(sinA-sinB)]
=c[(sinA/sinC)(sinB-sincC)+(sinB/sinC)(sinC-sinA)+(sinA-sinB)]
=c[(sinAsinB/sinC)-sinA+sinB-(sinBsinA/sinC)+(sinA-sinB)]
=0