Given: ABC is a triangle.
According to sine rule
asinA=bsinB=csinC=k ...(i)
We have to prove:
a2sin(B−C)sinA+b2sin(C−A)sinB+c2sin(A−B)sinC=0
LHS=a2sin(B−C)sinA+b2sin(C−A)sinB+c2sin(A−B)sinC
Putting the values from equation (i)
LHS=k2sin2Asin(B−C)sinA+k2sin2Bsin(C−A)sinB+k2sin2Csin(A−B)sinC
=k2sinAsin(B−C)+k2sinBsin(C−A)+k2sinCsin(A−B)
=k2[sinA(sinBcosC−sinCcosB)+sinB(sinCcosA−sinAcosC)+sinC(sinAcosB−sinBcosA)]
=k2(sinAsinBcosC−sinAsinCcosB+sinBsinCcosA−sinAsinBcosC+sinAsinCcosB−sinBsinCcosA)
= 0
= R.H.S
Hence, proved.