That which produces Medial Whole with Medial Area is Irrational

Proof

 * Euclid-X-75.png

Let $AB$ be a straight line.

Let a straight line $BC$ such that:
 * $BC$ is incommensurable in square with $AB$
 * $AB^2 + BC^2$ is medial
 * twice the rectangle contained by $AB$ and $BC$ is medial
 * $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$

be cut off from $AB$.

Let $DI$ be a rational straight line.

Using :
 * Let $DE$ be a parallelogram set out on $DI$ equal to $AB^2 + BC^2$.

Let its breadth be $DG$.

Similarly:
 * Let $DH$ be a parallelogram set out on $DI$ equal to $2 \cdot AB \cdot BC$.

From :
 * $FE = AC^2$

We have that $AB^2 + BC^2$ is medial and equal to $DE$.

Therefore $DE$ is medial.

We have that $DE$ has been applied to the rational straight line $DI$ producing $DG$ as breadth.

So from :
 * $DG$ is rational and incommensurable in length with $DI$.

We have that $2 \cdot AB \cdot BC$ is medial.

But $2 \cdot AB \cdot BC = DH$.

Therefore $DH$ is medial.

We have that $AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$.

Therefore $DE$ is incommensurable with $DH$.

But from :
 * $DE : DH = DG : DF$

Therefore by :
 * $DG$ is incommensurable with $DF$.

But both $GD$ and $DF$ are rational.

Therefore $GD$ and $DF$ are rational straight lines which are commensurable in square only.

Therefore, by definition, $FG$ is an apotome.

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

Hence its "side" is irrational.

But $AC$ is the "side" of $FE$.

Therefore $AC$ is irrational.

Such a straight line is known as that which produces with a medial area a medial whole.