△ABC中,证明( b-c )sinA+( c-a )sinB+( a-b )sinC = 0

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△ABC中,证明( b-c )sinA+( c-a )sinB+( a-b )sinC = 0

利用正弦定理:
a / sinA = b / sinB = c / sinC = 2R(R为三角形外接圆的半径)
所以:
sinA = a / 2R
sinB = b / 2R
sinC = c / 2R
代入,得:
( b-c )sinA+( c-a )sinB+( a-b )sinC
= (b - c)*a / 2R + (c - a)*b / 2R + (a - b)*c / 2R
= (ab - ac + bc - ab + ac - bc) / 2R
= 0