Side of Sum of Medial Areas is Irrational

Proof

 * Euclid-X-41.png

From :
 * Let $AB$ and $BC$ be two straight lines incommensurable in square such that:
 * $AB^2 + BC^2$ is medial
 * $AB \cdot BC$ is medial
 * $AB \cdot BC$ is incommensurable with the $AB^2 + BC^2$.

Let $DE$ be a rational straight line.

Using :
 * Let $DF$ be a rectangle set out on $DE$ equal to $AB^2 + BC^2$
 * Let $GH$ be a rectangle set out on $DE$ equal to $2 \cdot AB \cdot BC$.

From :
 * The rectangle $DH$ equals $AC^2$.

Since $AB^2 + BC^2$ is medial, $DF$ is medial.

We have that $DF$ is applied to the rational straight line $DE$.

Therefore from :
 * $DG$ is rational and incommensurable in length with $DE$.

For the same reason $GK$ is rational and incommensurable in length with $GF$, that is, $DE$.

Since $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$:
 * $DF$ is incommensurable with $GH$.

So by:

and:
 * $DG$ is also incommensurable in length with $GK$.
 * $DG$ is also incommensurable in length with $GK$.

Also, $DG$ and $GK$ are rational.

Therefore $DG$ and $GK$ are rational straight lines which are commensurable in square only.

Therefore $DK$ is irrational.

From :
 * $DK$ is binomial.

But $DE$ is rational.

Therefore from :
 * $DH$ is irrational.

Also, the side of the square equal to $DH$ is irrational.

But $AC$ is the side of the square equal to $DH$.

Therefore $AC$ is irrational.

Such a straight line is called the side of the sum of two medial areas.