Figure shows a rectangle $ABCD$ and an isosceles triangle $\triangle DEC$.
$AD=BC=z$;$AB=DC=y$;$DE=CE=x$
One solution is as follows.
We know that the pythagoras theorem holds for a right triangle $\triangle EBC$.
In the right triangle, $\triangle EBC$.
$z^2+(\frac{y}{2})^{2}=x^2$ ----(|)
We also know that, AREA of $\triangle DBC$ = AREA of $\triangle DEC$.
$\frac{1}{2}(y)(z)=\frac{1}{2}(x)(x)$ ----(||)
We need to show, $y^2=x^2+x^2$
From, (|) and (||) we have, $z=\frac{x^2}{y}$, $x^2=\frac{y^2}{4}+z^2$
Now, $x^2=\frac{y^2}{4}+\frac{x^4}{y^2}$
$(4x^2)y^2=y^4+4x^4$
$y^4-(4x^2)y^2+4x^4=0$
$y^4-((2x^2)y^2+(2x^2)y^2)+4x^4=0$
$y^4-(2x^2)y^2-(2x^2)y^2+4x^4=0$
$y^2(y^2-2x^2)-(2x^2)(y^2-2x^2)=0$
$(y^2-2x^2)^2=0$
$y^2-2x^2=0$
$y^2=2x^2$
$y^2=x^2+x^2$ ----(|||)