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