Question

In triangle ABC, prove the following:

$\frac{a^2\sin \left(B-C\right)}{\sin A}+\frac{b^2\sin \left(C-A\right)}{\sin B}+\frac{c^2\sin \left(A-B\right)}{\sin C}=0$

Answer 1

Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$

Then,

Consider the LHS of the equation $\frac{a^2\sin \left(B-C\right)}{\sin A}+\frac{b^2\sin \left(C-A\right)}{\sin B}+\frac{c^2\sin \left(A-B\right)}{\sin C}=0$.

$$\begin{gathered} \mathrm{LHS}=\frac{a^2\sin \left(B-C\right)}{\sin A}+\frac{b^2\sin \left(C-A\right)}{\sin B}+\frac{c^2\sin \left(A-B\right)}{\sin C} \\ =\frac{k^2{\sin }^{2}A\sin \left(B-C\right)}{\sin A}+\frac{k^2{\sin }^{2}B\sin \left(C-A\right)}{\sin B}+\frac{k^2{\sin }^{2}C\sin \left(A-B\right)}{\sin C} \\ =k^2\sin A\sin \left(B-C\right)+k^2\sin B\sin \left(C-A\right)+k^2\sin C\sin \left(A-B\right) \\ =k^2\left[\sin A\left(\sin B\cos C-\sin C\cos B\right)+\sin B\left(\sin C\cos A-\sin A\cos C\right)+\sin C\left(\sin A\cos B-\sin B\cos A\right)\right] \\ =k^2\left(\sin A\sin B\cos C-\sin A\sin C\cos B+\sin B\sin C\cos A-\sin A\sin B\cos C+\sin A\sin C\cos B-\sin C\sin B\cos A\right) \\ =0=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$