Side of Sum of Two Medial Areas is Divisible Uniquely

Proof

 * Euclid-X-47.png

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 $AC > BD$.

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

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

Let $EF$ be a rational straight line.

Using :
 * 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 $AB^2$.

Using :
 * 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 $EF$.

So from :
 * $HE$ 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:

and:

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

Using the same argument we can deduce:
 * $EN$ is a binomial straight line divided at $M$.

But this contradicts.

So there can be no such $D$.