Straight Line Commensurable with Medial Straight Line is Medial

Proof
Let $A$ be a medial straight line of length $\rho \sqrt [4] k$ where:
 * $\rho$ is a rational number
 * $k$ is a rational number whose square root is irrational.

Let $B$ be a straight line which is commensurable in length with $A$.

Thus $B$ has length $\lambda \rho \sqrt [4] k$ where $\lambda$ is a rational number.

Let $CD$ be a rational straight line.

Let a rectangle $CE$ be set up on $CD$ equal to the square on $A$.

Let the breadth of $CE$ be $DE$.

Then from Square on Medial Straight Line:
 * $ED$ is a rational straight line which is commensurable in length with CD.

Let a rectangle $CF$ be set up on $CD$ equal to the square on $B$.

Let the breadth of $CF$ be $DF$.

As $A$ is commensurable in length with $B$, the square on $A$ is also commensurable with the square on $B$.

But as $CE = A^2$ and $CF = B^2$, $CE$ is commensurable with $CF$.

From Areas of Triangles and Parallelograms Proportional to Base:
 * $EC : CF = ED : DF$

From Commensurability of Elements of Proportional Magnitudes:
 * $ED$ is commensurable with $DF$.

But $ED$ is a rational straight line which is commensurable in length with CD.

By, $DF$ is a rational straight line.

By Commensurable Magnitudes are Incommensurable to Same Magnitude:
 * $DF$ is incommensurable in length with $DC$.

Therefore $CD$ and $DF$ are rational and commensurable in square only.