Proving $\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$ 
Prove that
$$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$$

I tried making $\sin 80^\circ=\sin(50^\circ+30^\circ)$, but it didn't go well. I also tried using, maybe, trig periodicities, but I still can't get it.

$\frac{\sin 30^\circ}{\sin 50^\circ}= \frac{\sin 40^\circ}{\sin 80^\circ}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{\sin (2(\cdot40^\circ))}$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{\sin 40^\circ}{2 \cdot \sin 40^\circ \cdot \cos 40^\circ }$
$\frac{\frac{1}{2}}{\sin 50^\circ} = \frac{1}{2\cdot \cos 40^\circ }$
$\cos 40^\circ = \sin 50^\circ $
and that's true because $\sin x = \cos(\frac{\pi}{2}-x)$.
