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

From sine law, we have,

`a/sinA = b/sinB = c/sinC = k`

`=> a = ksinA, b = ksin B, c = ksinC`

Now, `(asin(B-C))/(b^2-c^2) = (ksinAsin(B-C))/(k^2(sin^2B-sin^2C))`

`= sin(pi-(B+C)sin(B-C))/(k(sin^2B-sin^2C))`

`= (sin(B+C)sin(B-C))/(k(sin^2B-sin^2C))`

We know, `sin(B+C)sin(B-C) = sin^2B-sin^2C`

So, it becomes,

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