由正弦定理得
===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.