If $ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b}$, Prove that $x+y+z$ $=0$ 

Question  If $$ \frac{x}{b-c} = \frac{y}{c-a} = \frac{z}{a-b} $$ Prove that $x+y+z$ $=0$


I've attmempted this question by cross multiplying so that
$$ x(c-a)(a-b) = y(b-c)(a-b) = (b-c)(c-a)z $$
but that did not work
I also tried splitting the equation so 
$ \frac{x}{b-c} = \frac{y}{c-a}$  and $\frac{y}{c-a} = \frac{z}{a-b}$
and tried forming systems of equation but that didn't lead me to the proof as well...
Could someone please help!

Using $k$-method, let $\displaystyle \frac{x}{b-c}=\frac{y}{c-a}=\frac{z}{a-b}=k$, then $\displaystyle \sum_{xyz} x=\sum_{abc} (b-c)k=0$.
