Straight Line Commensurable with Bimedial Straight Line is Bimedial and of Same Order

Proof

 * Euclid-X-66.png

Let $AB$ be bimedial.

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

It is to be shown that $CD$ is bimedial, and that the order of $CD$ is the same as the order of $AB$.

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

Let $AE$ be the greater medial.

By definition, $AE$ and $EB$ are medial straight lines which are commensurable in square only.

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

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

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

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

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

But by hypothesis $AE$ and $EB$ are medial.

Therefore by :
 * $CF$ and $FD$ are medial.

From :
 * $AE : CF = EB : FD$

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

But by hypothesis $AE$ and $EB$ are commensurable in square only.

Therefore by :
 * $CF$ and $FD$ are commensurable in square only.

But $CF$ and $FD$ are medial.

Therefore, by definition, $CD$ is bimedial.

It remains to be demonstrated that $CD$ is of the same order as $AB$.

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

Therefore:
 * $AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$

Therefore by :
 * $AE^2 : CF^2 = AE \cdot EB : CF \cdot FD$

But:
 * $AE^2$ is commensurable with $CF^2$.

Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.

Suppose $AB$ is a first bimedial.

Then $AE \cdot EB$ is rational.

It follows that $CF \cdot FD$ is rational.

Thus by definition $CD$ is a first bimedial.

Suppose otherwise that $AB$ is a second bimedial.

Then $AE \cdot EB$ is medial.

It follows that $CF \cdot FD$ is medial.

Thus by definition $CD$ is a second bimedial.

Thus $CD$ is of the same order as $AB$.