Does it follow that a+b+c=x+y+z=m+n+p If
$$abc=xyz=mnp$$
$a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=m^4+n^4+p^4-2m^2n^2-2n^2p^2-2p^2m^2$
$$\frac{x}{m}=\frac{n}{b}=\frac{c}{z}$$
Then $a+b+c=x+y+z=m+n+p$?
$a,b,c$ positive and are lenghts of triangles,
$x,y,z$ same, $m,n,p$ same.
I edited it.

Hint: Note that
$$
a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2=(a + b + c)(a + b - c)(a - b + c)(a - b - c).
$$
