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!

\begin{align}
x+y+z&=\frac{b-c}{a-b}z+\frac{c-a}{a-b}z+z\\
&=z\left(\frac{b-c+c-a+a-b}{a-b}\right)\\
&=z\cdot 0\\
&=0
\end{align}
