Side of Sum of Two Medial Areas is Divisible Uniquely

Theorem

In the words of Euclid:

The side of the sum of two medial areas is divided at one point only.

Proof

Let $AB$ be a side of a sum of two medial areas.

Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:

$AC$ and $CB$ are incommensurable in square
$AC^2 + CB^2$ is medial
$AC$ and $CB$ contain a medial rectangle
$AC \cdot CB$ is incommensurable with $AC^2 + CB^2$.

Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.

Let $AC > BD$.

$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

and:

$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$

Let $EF$ be a rational straight line.

Let $EG$ be a rectangle set out on $EF$ equal to $AC^2 + CB^2$

and

Let $HK$ be a rectangle set out on $EF$ equal to $2 \cdot AC \cdot CB$.

Therefore $EK$ is a rectangle equal to $AB^2$.

Let $EL$ be a rectangle set out on $EF$ equal to $AD^2 + DB^2$

The remainder $MK$ is a rectangle set out on $EF$ equal to $2 \cdot AD \cdot DB$.

By hypothesis, $AC^2 + CB^2$ is medial.

We have that $AC^2 + CB^2$ is applied to a rational straight line $EF$.

$HE$ is rational and incommensurable in length with $EF$.

For the same reason:

$HN$ is rational and incommensurable in length with $EF$.

Since $AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$:

$EG$ is incommensurable with $GN$.

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

it follows that:

$EH$ is incommensurable with $HN$.

Also, $EH$ and $HN$ are rational.

Therefore $EH$ and $HN$ are rational straight lines commensurable in square only.

Therefore from Proposition $36$ of Book $\text{X}$: Binomial is Irrational:

$EN$ is a binomial straight line divided at $H$.

Using the same argument we can deduce:

$EN$ is a binomial straight line divided at $M$.

So there can be no such $D$.

$\blacksquare$

Historical Note

This proof is Proposition $47$ of Book $\text{X}$ of Euclid's The Elements.