asin(B − C) + bsin(C − A) + csin(A − B) = 0

asin(B − C) + bsin(C − A) + csin(A − B) = 0

asin(B − C) = a[sin(B) × cos(C) − cos(B) × sin(C)]            ...(1)

now, sin(A)/a = sin(B)/b = sin(C)/c = k (say)

Therefore, sinA = ak, sinB = bk, sinC = ck

Substituting values of sinB and sinC in (1) and using cosine formula:

asin(B − C) = a[bk((a² + b² − c²)/(2ab)) − ck((c² + a² − b²)/(2ac))]

= (k/2)(a² + b² − c² − c² + a² + b²)

= k(b² − c²)

Similarly, bsin(C − A) = k(c² − a²)

and csin(A − B) = k(a² − b²)

Hence, LHS = k(b² − c² + c² − a² + a² − b²) = 0 = RHS