<s>[INST] Proving the following trigonometric proportion $$\frac{a\sin(B-C)}{b^2-c^2}=\frac{b\sin(C-A)}{c^2-a^2}=\frac{c\sin(A-B)}{a^2-b^2}$$
A, B, C are angles of a triangle and a, b, c are the sides of a triangle.
I tried using various things such as sine rule and then replacing the various rations in terms of sides of triangle, however it produced no result. Urgent help is needed!
 [/INST] By the Law of Sines, we have
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
where $d$ is the diameter of the triangle's circumcircle. As a result, the relation to prove becomes (after eliminating some $d$s)
$$\frac{\sin A \;\sin(B-C)}{\sin^2B-\sin^2C} = \frac{\sin B \;\sin(C-A)}{\sin^2C-\sin^2A} = \frac{\sin C \;\sin(A-B)}{\sin^2A-\sin^2B} \qquad (\star)$$
Consider $\sin A\;\sin(B-C)$, where we know $A = \pi - (B+C)$ so that $\sin A = \sin(B+C)$. By a Product-to-Sum Identity,
$$\begin{align}
\sin(B+C) \sin(B-C) &= \frac{1}{2}\left(\;\cos((B+C)-(B-C))- \cos((B+C)+(B-C)) \;\right) \\[4pt]
&= \frac{1}{2}\left( \cos 2C - \cos 2B \right) \\[4pt]
&= \frac{1}{2}\left( \; ( 1 - 2 \sin^2 C ) - ( 1 - 2 \sin^2 B )\; \right) \\[4pt]
&= \sin^2 B - \sin^2 C
\end{align}$$
Thus, the left-most expression of $(\star)$ reduces to $1$; by symmetry, the other sides do, as well, so that all three are in fact equal. (Note that the expressions in the original problem reduce to $1/d$.)
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