Using Ratios for proof making If
$$a(y+z)=x,b(z+x)=y,c(x+y)=z$$
prove that 
$$\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ac)}=\frac{z^2}{c(1-ba)}$$
I haven't been able to get to the form in the proof, any suggestions on how to do it are welcome.

If $y+z=0,$ we have $x=0,$ which leads to $y=z=0.$ Assuming that the fractions in the final relation are well-defined, we have the desired equality.
We arrive to the same conclusion if $z+x=0$ or $x+y=0.$
Assume that $(x+y)(y+z)(z+x)\neq 0.$
Rewrite $$a(y+z)=x,\;b(z+x)=y,\;c(x+y)=z$$ as $$a=\frac{x}{y+z},\;b=\frac{y}{z+x},\;c=\frac{z}{x+y}.$$
Then $$\begin{aligned}a(1-bc)=&\frac{x}{y+z}\left(1-\frac{yz}{(z+x)(x+y)}\right)\\=&\frac{x^2(x+y+z)}{(y+z)(z+x)(x+y)}\end{aligned}$$
We have analogous expressions for $b(1-ca)$ and $c(1-ab).$
Finally, we obtain $$\frac{x^2}{a(1-bc)}=\frac{(y+z)(z+x)(x+y)}{x+y+z}=\frac{y^2}{b(1-ac)}=\frac{z^2}{c(1-ba)}.$$
