Construction of Medial Straight Lines Commensurable in Square Only containing Rational Rectangle whose Square Differences Commensurable with Greater

Proof
Let $A$ and $B$ be rational straight lines which are commensurable in square only.

Let $A^2 > B^2$ such that:
 * $A^2 = B^2 + \rho^2$

where $\rho$ is a straight line which is commensurable in length with $A$.

This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater.

Let $C$ be a straight line such that $C^2 = A B$.

From Medial is Irrational, $A B$ is medial.

Therefore $C^2$ is medial.

So by definition $C$ is medial.

Let $C D = B^2$.

By hypothesis $B^2$ is rational.

Therefore $C D$ is rational.

We have that:
 * $A : B = A^2 : A B$

But:
 * $C^2 = A B$

and:
 * $C D = B^2$

and so:
 * $A : B = C^2 : C D$

But as:
 * $C^2 : C D = C : D$

it follows that:
 * $A : B = C : D$

But $A$ is commensurable in square only with $B$.

Therefore by Commensurability of Elements of Proportional Magnitudes:
 * $C$ is commensurable in square only with $D$.

We have that $C$ is medial.

Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $D$ is medial.

We have that:
 * $A : B = C : D$

and:
 * $A^2 = B^2 + \rho^2$

where $\rho$ is a straight line which is commensurable in length with $A$.

Therefore Commensurability of Squares on Proportional Straight Lines:
 * $C^2 = D^2 + \sigma^2$

where $\sigma$ is a straight line which is commensurable in length with $C$.

Therefore two medial straight lines $C$ and $D$ have been found which are commensurable in square only and which contain a rational rectangle.

Further, $C^2$ is greater than $D^2$ by the square on a straight line commensurable in length with $C$.

Similarly it can be proved that the square on $C$ exceeds the square on $D$ by the square on a straight line incommensurable in length with $C$, when the square on $A$ is greater than the square on $B$ by the square on a straight line incommensurable in length with $A$.