If any triangle `A B C` , that: `(asin(B-C))/(b^2-c^2)=(bsin(C-A))/(c^2-a^2)=(csin(A-B))/(a^2-b^2)`

Explanation

According to sine rule,
`a/(sinA) = b/(sinB) = c/(sinC)= k(say)`
`therefore a=k sinA, b=k sinB, c=k sinC` Now, `(asin(B-C))/(b^(2)-c^(2)) = (ksinAsin(B-C))/(k^(2)sin^(2)B-k^(2)sin^(2)C)`
`=(sinAsin(B-C))/(ksin(B+C)sin(B-C))`
`=(sinA)/(ksin(180^(@)-A)) = (sinA)/(k sinA)=1/k`.......(i)
`(bsin(C-A))/(c^(2)-a^(2)) = (ksinBsin(C-A))/(k^(2)sin^(2)C-k^(2)sin^(2)A)`
`=(sinBsin(C-A))/(k(sin(C+A)sin(C-A))`
`(sinB)/(ksin(180^(@)-B))=(sinB)/(ksinB)`
`=1/k` ........(ii)
`(csin(A-B))/(a^(2)-b^(2)) = (ksinCsin(A-B))/(k^(2)sin^(2)A-k^(2)sin^(2)B)`
`=(sinCsin(A-B))/(k(sin(A+B)sin(A-B))`
`=(sinC)/(ksin(180^(@)-C)`
`=(sinC)/(k sinC)=1/k`...........(iii)
From (i), (ii) and (iii), we get
`(a sin(B-C))/(b^(2)-c^(2)) = (bsin(C-A))/(c^(2)-a^(2)) = (c sin(A-B))/(a^(2)-b^(2))`