Straight Line Commensurable with Side of Sum of two Medial Areas

Proof

 * Euclid-X-66.png

Let $AB$ be the side of the sum of two medial areas.

Let $CD$ be commensurable in length with $AB$.

It is to be shown that $CD$ is also the side of the sum of two medial areas.

Let $AB$ be divided into its terms by $E$.

Let $AE > EB$.

By definition, $AE$ and $EB$ are straight lines such that:
 * $AE$ and $EB$ are incommensurable in square
 * $AE^2 + EB^2$ is medial
 * $AE \cdot EB$ is medial.

Using, let it be contrived that:
 * $AB : CD = AE : CF$

Therefore by :
 * $EB : FD = AB : CD$

Therefore by :
 * $AE : CF = EB : FD$

But $AB$ is commensurable in length with $CD$.

Therefore from :
 * $AE$ is commensurable in length with $CF$

and:
 * $EB$ is commensurable in length with $FD$.

We have that:
 * $AE : CF = EB : FD$

Therefore by :
 * $AE : EB = CF : FD$

and by by :
 * $AB : BE = CD : DF$

Using a similar line of reasoning to :
 * $CF$ and $FD$ are incommensurable in square
 * $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$
 * $AE \cdot EB$ is commensurable with $CF \cdot FD$.

We have that $AE^2 + EB^2$ is medial.

Therefore $CF^2 + FD^2$ is medial.

Similarly we have that $AE \cdot EB$ is medial.

Therefore $CF \cdot FD$ is medial.

Hence:
 * $CF$ and $FD$ are incommensurable in square
 * $CF^2 + FD^2$ is medial
 * $CF \cdot FD$ is medial.

Thus by definition $CD$ is the side of the sum of two medial areas.