G1.2 Let $x, y$ and $z$ be positive numbers. Given that $\frac{x+z}{2 z-x}=\frac{z+2 y}{2 x-z}=\frac{x}{y}$. Find the value of $\frac{x}{y}$.
It is equivalent to $\frac{x+z}{2 z-x}=\frac{x}{y} \ldots$
(1) and $\frac{z+2 y}{2 x-z}=\frac{x}{y} \ldots$

From (2), $y z+2 y^{2}=2 x^{2}-x z \Rightarrow(x+y) z=2\left(x^{2}-y^{2}\right)$
$\Rightarrow x+y=0$ (rejected, $x>0$ and $y>0$ ) or $z=2(x-y)$
From (1): $x y+y z=2 x z-x^{2} \Rightarrow(2 x-y) z=x^{2}+x y \ldots$
Sub. (4) into (5): $2(x-y)(2 x-y)=x^{2}+x y$
$2\left(2 x^{2}-3 x y+y^{2}\right)=x^{2}+x y$
$3 x^{2}-7 x y+2 y^{2}=0$
$(3 x-y)(x-2 y)=0 \Rightarrow \frac{x}{y}=\frac{1}{3}$ or 2
When $y=3 x$, sub. into (3): $z=2(x-3 x)=-4 x$ (rejected, $x>0$ and $z>0$ )
$$
\therefore \frac{x}{y}=2
$$

Method $2 \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k \Rightarrow a=b k, c=d k, e=f k \Rightarrow \frac{a+c+e}{b+d+f}=\frac{b k+d k+f k}{b+d+f}=k$
$$
k=\frac{x}{y}=\frac{a+c+e}{b+d+f}=\frac{(x+z)+(z+2 y)+x}{(2 z-x)+(2 x-z)+y}=\frac{2 x+2 y+2 z}{x+y+z}=2(\because x+y+z>0)
$$

Remark: The original question is: Given that $\frac{x+z}{2 z-x}=\frac{z+2 y}{2 x-z}=\frac{x}{y}$. Find the value of $\frac{x}{y}$.
The question has more than one solution.