Medial Area not greater than Medial Area by Rational Area

Proof
Let $\rho \cdot \sqrt k \rho$ and $\rho \cdot \sqrt \lambda \rho$ be two medial areas which have been applied to the same rational number $\rho$.

Their difference is then:
 * $\left({\sqrt k - \sqrt \lambda}\right) \rho^2$

Let $x = \sqrt k \rho$ and $y = \sqrt \lambda \rho$.

Let:
 * $\rho \left({x - y}\right) = \rho z$

Suppose $\rho z$ is a rational area.

Then $z$ must be a rational straight line and:
 * $(1): \quad z \frown \rho$

where $\frown$ denotes commensurability in length.

Since $\rho x$ and $\rho y$ are medial areas, it follows that $x$ and $y$ are rational and:
 * $(2): \quad x \smile \rho$ and $y \smile \rho$

where $\smile$ denotes incommensurability in length.

Then from $(1)$ and $(2)$:
 * $y \smile z$

Then:

But:

But $y^2 + z^2$ is rational.

Therefore $x^2$ must be irrational.

Therefore $x$ must be irrational.

But from $(2)$ $x$ is rational.

The result follows from this contradiction.