Side of Sum of Medial Areas is Irrational

Theorem

In the words of Euclid:

If two straight lines incommensurable in square which make the sum of the squares on them medial, and the rectangle contained by them medial and also incommensurable with the sum of the squares on them, be added together, the whole straight line is irrational; and let it be called the side of the sum of two medial areas.

Proof

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.

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 $2 \cdot AB \cdot BC$.
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 $DE$.

$DG$ 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:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

Proposition $11$ of Book $\text{X}$: Commensurability of Elements of Proportional Magnitudes:
$DG$ 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.

$DK$ is binomial.

But $DE$ is rational.

Therefore from Book $\text{X}$ Definition $4$: Rational Area:

$DH$ 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.

$\blacksquare$

Historical Note

This proof is Proposition $41$ of Book $\text{X}$ of Euclid's The Elements.