Straight Line Commensurable with Major Straight Line is Major

Proof

 * Euclid-X-66.png

Let $AB$ be a major.

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

It is to be shown that $CD$ is also a major.

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

Let $AE$ be the greater term.

By definition, $AE$ and $EB$ are straight lines such that:
 * $AE$ and $EB$ are incommensurable in square
 * $AE^2 + EB^2$ is rational
 * $AE \cdot EB$ 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 :
 * $AB : BE = CD : DF$

Therefore by :
 * $AB^2 : BE^2 = CD^2 : DF^2$

Using a similar line of reasoning to the above:
 * $AB^2 : AE^2 = CD^2 : CF^2$

and:
 * $AB^2 : AE^2 + EB^2 = CD^2 : CF^2 + FD^2$

Therefore by :
 * $AB^2 : CD^2 = AE^2 + EB^2 : CF^2 + FD^2$

But $AB^2$ is commensurable with $CD^2$.

Therefore $AE^2 + EB^2$ is commensurable with $CF^2 + FD^2$.

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

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

Similarly, $2 \cdot AE \cdot EB$ is commensurable with $2 \cdot CF \cdot FD$.

By :
 * $2 \cdot AE \cdot EB$ is medial.

Therefore $2 \cdot CF \cdot FD$ is medial.

Thus it has been demonstrated that $CF$ and $FD$ are straight lines such that:
 * $CF$ and $FD$ are incommensurable in square
 * $CF^2 + FD2$ is rational
 * $CF \cdot FD$ is medial.

Thus by definition $CD$ is a major.