https://artofproblemsolving.com/community/q1h1808444p12035727

Here is nice solution. We can start with this and using sine law $\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}=2R$ $2Ra \sin(\beta-\gamma) =2Ra \sin(\beta)\cos(\gamma)-2Ra \sin(\gamma)\cos(\beta)=ab \cos(\gamma)-ac \cos(\beta)$ $2Rb \sin(\gamma-\alpha)=2Rb \sin(\gamma)\cos(\alpha)-2Rb \sin(\alpha)\cos(\gamma)=bc \cos(\alpha)-ab \cos(\gamma)$ $2Rc \sin(\alpha-\beta)=2Rc \sin(\alpha)\cos(\beta)-2Rc \sin(\beta)\cos(\alpha)=ac \cos(\beta)-bc \cos(\alpha)$ Now, we can add up these equations to get: $2Ra \sin(\beta-\gamma)+2Rb \sin(\gamma-\alpha)+2Rc \sin(\alpha-\beta)=ab \cos(\gamma)-ac \cos(\beta)+bc \cos(\alpha)-ab \cos(\gamma)+ac \cos(\beta)-bc \cos(\alpha)=0$ $2R(a\sin(\beta-\gamma)+b\sin(\gamma-\alpha)+c\sin(\alpha-\beta))=0$, and finally: $a\sin(\beta-\gamma)+b\sin(\gamma-\alpha)+c\sin(\alpha-\beta)=0$ $Q.E.D.$

$a=2R \sin(\alpha)$$,b=2R \sin(\beta)$ and $c=2R \sin(\gamma)$ We need to prove that : $$ \sin(\alpha) \sin(\beta-\gamma) + \sin(\beta) \sin(\gamma-\alpha) + \sin(\gamma) \sin(\alpha-\beta) = 0.$$On expansion of LHS we get three pairs of identical terms which cancel each other.