Solving the system $a^2-c^2=x^2-z^2$, $ab=xy$, $ac=xz$, $bc=yz$ I've stumbled upon these equations, and am struggling to find a manual way to solve this in $\Re$:

$$\begin{align}
a^2-c^2&=x^2-z^2 \\
ab&=xy \\
ac&=xz \\
bc&=yz\end{align}$$

I've used Wolfram Alpha to compute this and I found that this is only possible if:
$$a=x, \quad y=b, \quad z=c$$
Is there an easy manual way to compute this?
I've tried moving around variables with no luck.

Let uppercase letters denote the logarithm of the corresponding lowercase parameter, in absolute value. The last three equations read
$$\begin{cases}X+Y&=A+B,\\Y+Z&=B+C,\\Z+X&=C+A,\end{cases}$$
which is an easy system, giving the single solution $X=A,Y=B,Z=C$. Hence $(x,y,z)=(\pm a,\pm b,\pm c)$.
Now to discuss the signs, it suffices to remark that $x=a,y=b,z=c$ is a solution regardless the signs of $a,b,c$, and if we try to change the sign of one unknown, we need to change the other two. Finally
$$(x,y,z)=(a,b,c)\lor (x,y,z)=-(a,b,c).$$
In all cases, the first equation is implicitly fulfilled.
