由正弦定理得 a sinA= b sinB= c sinC=2R, ∴a=2RsinA,b=2RsinB,c=2RsinC, 故有asin(B-C)+bsin(C-A)+csin(A-B) =2R[sinAsin(B-C)+sinBsin(C-A)+sinCsin(A-B)] =2R[sinA(sinBcosC-cosBsinC)+sinB(sinCcosA-cosCsinA)+sinC(sinAcosB-cosAsinB)] =2R(sinAsinBcosC-sinAcosBsinC+sinBsinCcosA-sinBcosCsinA+sinCsinAcosB-sinCcosAsinB)=0, ∴等式成立.