How do I show that the pythagoras theorem holds for the specific case of an "isosceles right triangle"?

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, $x^2=\frac{y^2}{4}+\frac{x^4}{y^2}$

$(4x^2)y^2=y^4+4x^4$

$y^4-(4x^2)y^2+4x^4=0$

$y^4-((2x^2)y^2+(2x^2)y^2)+4x^4=0$

$y^4-(2x^2)y^2-(2x^2)y^2+4x^4=0$

$y^2(y^2-2x^2)-(2x^2)(y^2-2x^2)=0$

$(y^2-2x^2)^2=0$

$y^2-2x^2=0$

$y^2=2x^2$

$y^2=x^2+x^2$ ----(|||)

Step 1: Assume the Pythagorean theorem for right triangles.

Step 2: Realize that an isosceles right triangle is a right triangle.

Step 3: Conclude that the Pythagorean theorem for isosceles right triangles is a special case (and therefore free from) the Pythagorean theorem for right triangles.

In addition, I can't help but notice the similarity between this and your previous questions https://math.stackexchange.com/q/182332/9754 and https://math.stackexchange.com/q/184003/9754, even though those are both several months old.