Binomial Straight Line is Divisible into Terms Uniquely

Proof
Let $AB$ be a binomial straight line.

Let $AB$ be divided into its terms at $C$.

That is, let the terms of $AB$ be $AC$ and $CB$.

By definition of binomial, $AC$ and $CB$ are rational straight lines commensurable in square only.

Suppose there exists a point $D$ on $AB$ different from $C$, such that $AD$ and $DB$ are also rational straight lines which are commensurable in square only.

Let $AC \ne DB$, otherwise $AD = BC$ and $AB$ will be divided in the same way by $D$ as it is $C$.

Similarly, the points $C$ and $D$ are not equidistant from the point of bisection of $AB$.

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

As:
 * $AC$ and $CB$ are rational straight lines

and
 * $AD$ and $DB$ are rational straight lines

it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all rational areas.

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

Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area.

But by definition, $2 AC \cdot CB$ and $2 AD \cdot DB$ are both medial.

But from this cannot be the case.

From this contradiction it follows that $D$ cannot divide $AB$ into rational straight lines which are commensurable in square only.