Q13 Matemática  (Caucasus Mathematical Olympiad 2019)

Dado um triângulo $ABC$ com $BC=a$, $CA=b$, $AB=c$, $\angle BAC=\alpha$, $\angle CBA=\beta$, $\angle ACB=\gamma$. Prove que $$a\mathrm{sen}(\beta -\gamma )+b\mathrm{sen}(\gamma -\alpha )+c\mathrm{sen}(\alpha -\beta )=0.$$