Proving the ratios Now, I tried this by the usual determinant way of doing it. However, I somehow always end up with the result: $a=b=c$. And then, I get thoroughly baffled at what to do next and how to generate square terms out of the variables. So, how is it to be done really? What should be the approach?

If $$a(y+z)=x,~~b(z+x)=y,~~c(x+y)=z,$$
prove that $$\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ca)}=\frac{z^2}{c(1-ab)}$$

 A: My idea would be trying to calculate $a(1 - bc)$ instead of $x^2$.
Multiplying all three terms gives you
$$
abc(x + y)(y + z)(z + x) = xyz,
$$
then
$$
a(1 - bc)(x + y)(y + z)(z + x) = x(x + y)(z + x) - xyz = x^2(x + y + z).
$$
If $x + y + z \neq 0$, then
$$
\frac{x^2}{a(1 - bc)} = \frac{(x + y)(y + z)(z + x)}{x + y + z},
$$
and similar for the other two fractions, which proves the desired equation.
If $x + y + z = 0$, then
\begin{align*}
-ax = x \Rightarrow a &= -1 \text{ or } x = 0. \\
-by = y \Rightarrow b &= -1 \text{ or } y = 0. \\
-cx = x \Rightarrow c &= -1 \text{ or } z = 0.
\end{align*}
There are $8$ possible combinations here, and some of them introduce exceptions to your equation. You should be able to plug them in and find out the results easily.
A: By solving $b(z+x)=y$, $c(x+y)=z$ with respect to $y,z$, we get
$$z=\frac{cx(b+1)}{1-bc}$$
(multiply the first equation by $c$, add the second equation and find $z$).
By solving $a(y+z)=x$, $b(z+x)=y$ with respect to $x,y$, we get
$$x=\frac{az(b+1)}{1-ab}.$$
Hence
$$x\cdot\frac{cx(b+1)}{1-bc}=x\cdot z=\frac{az(b+1)}{1-ab}\cdot z$$
that is (if $b=-1$ then $x=z=0$ and the equalities hold trivially)
$$\frac{cx^2}{1-bc}=\frac{az^2}{1-ab}\Rightarrow \frac{x^2}{a(1-bc)}=\frac{z^2}{c(1-ab)}.$$
By symmetry you can obtain the other equality.
P.S. Here we are assuming that $ab\not=1$, $bc\not=1$, $ac\not=1$, $a\not=0$, $b\not=0$, $c\not=0$.
A: You can write the system of first three identities as
$$
\left( {\begin{array}{*{20}c}
   { - 1} & a & a  \\
   b & { - 1} & b  \\
   c & c & { - 1}  \\
 \end{array} } \right)\left( {\begin{array}{*{20}c}
   x  \\
   y  \\
   z  \\
 \end{array} } \right) = \left( {\begin{array}{*{20}c}
   0  \\
   0  \\
   0  \\
 \end{array} } \right)
$$
and for it to have solutions you shall impose that the determinant be null
$$
\left| {\left( {\begin{array}{*{20}c}
   { - 1} & a & a  \\
   b & { - 1} & b  \\
   c & c & { - 1}  \\
 \end{array} } \right)} \right| = 0\quad  \Rightarrow \quad ab + ac + bc + 2abc = 1
$$
so that, actually, the three parameters $a,b,c$ are not independent.
Solving for the null vector, with the condition above, you get (eliminating for instance $a$):
$$
\frac{x}
{{1 - bc}} = \frac{y}
{{b(1 + c)}} = \frac{z}
{{c(1 + b)}}
$$
To pass to the squares, and get a more symmetrical formulation, you can follow the answer given by @Robert Z.
