Side of Rational Plus Medial Area is Divisible Uniquely

Proof
Let $AB$ be a side of a rational plus a medial area.

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 rational rectangle.

From :
 * $2 \cdot AC \cdot CB$ is a rational rectangle.

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

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

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

and so:
 * $\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right) = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$

Since $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational:


 * $\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right)$ is rational.

But $\left({AC^2 + CB^2}\right)$ and $\left({AD^2 + DB^2}\right)$ are medial.

From this cannot be the case.

So there can be no such $D$.