If the internal bisectors of the angles of the triangle ABC make angles $\alpha,\beta,\gamma$ with sides $a,b,c$ respectively then show that $a\sin 2\alpha+b\sin 2\beta+c\sin 2\gamma=0$
I tried to solve it.Let internal bisector $AI$ make angle $\beta$ with side $b$ and angle $\gamma$ with side $c$,let internal bisector $BI$ make angle $\alpha$ with side $a$,angle $\gamma$ with side $c$,internal bisector $CI$ make angle $\alpha$ with side $a$,angle $\beta$ with side $b$.
$\angle A=\beta+\gamma,\angle B=\alpha+\gamma,\angle C=\alpha+\beta$
$a\sin 2\alpha+b\sin 2\beta+c\sin 2\gamma=2R(\sin A\sin 2\alpha+\sin B\sin 2\beta+\sin C\sin2\gamma)=2R(\sin(\beta+\gamma)\sin 2\alpha+\sin(\alpha+\gamma)\sin 2\beta+\sin(\alpha+\beta)\sin2\gamma)$
I tried to further simplify it using $2\alpha+2\beta+2\gamma=\pi$ but could not achieve final result.Is my approach right?Please help me reach the final answer.