Second Bimedial is Irrational

Proof

 * Euclid-X-38.png

Let $AB$ and $BC$ be medial straight lines which are commensurable in square only.

Let $AB$ and $BC$ contain a medial rectangle.

Let $DE$ be a rational straight line.

Using :
 * Let $DF$ be a parallelogram set out on $DE$ equal to the square on $AC$.

Let its breadth be $DG$.

From :
 * $AC^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$

Let $EH$ be a rectangle applied to $DE$ whose area equals $AB^2 + BC^2$.

The rectangle $HF$ which remains from $DF$ having had $EH$ removed is therefore equal to $2 \cdot AB \cdot BC$.

Since each of $AB$ and $BC$ are medial, both of $AB^2$ and $BC^2$ are also medial.

By hypothesis, $2 \cdot AB \cdot BC$ is also medial.

As:
 * $EH = AB^2 + BC^2$

and:
 * $FH = 2 \cdot AB \cdot BC$

each of the rectangles $EH$ and $FH$ is medial.

Each of $EH$ and $FH$ is applied to the rational straight line $DE$.

From :
 * Each of $DH$ and $HG$ is rational and incommensurable in length with $DE$.

We have that $AB$ is incommensurable in length with $BC$.

We also have that:
 * $AB : BC = AB^2 : AB \cdot BC$

So by :
 * $AB^2$ is incommensurable with $AB \cdot BC$.

But from :
 * $AB^2 + BC^2$ is commensurable with $AB^2$

and from :
 * $2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$.

Therefore from :
 * $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.

But:
 * $EH = AB^2 + BC^2$

and:
 * $HF = 2 \cdot AB \cdot BC$

Therefore $EH$ is incommensurable with $HF$.

So from:

and:

it follows that $DH$ is incommensurable in length with $HG$.

Therefore $DH$ and $HG$ are rational straight lines which are commensurable in square only.

By :
 * $DG$ is irrational.

But $DE$ is rational.

From :
 * The rectangle contained by an irrational and a rational straight line is irrational.

Therefore $DF$ is irrational.

From :
 * The side of a square equal to an irrational area is irrational.

But $AC$ is the side of a square equal to $DF$.

Therefore $AC$ is irrational.

Such a straight line is called second bimedial.