If $x(b-c)+y(c-a)+z(a-b)=0$ then show that .. I am stuck with the following problem that says:

If $x(b-c)+y(c-a)+z(a-b)=0$ then show that 
  $$\frac{bz-cy}{b-c}=\frac{cx-az}{c-a}=\frac{ay-bx}{a-b}$$ where $a \neq b \neq c.$

Can someone point me in the right direction? Thanks in advance for your time .

HINT:
$$-(a-b)z=x(b-c)+y(c-a)$$
$$\frac{bz-cy}{b-c}=\dfrac{-b\{x(b-c)+y(c-a)\}-c(a-b)y}{(b-c)(a-b)}=?$$
