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

## Theorem

In the words of Euclid:

A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.

## Proof

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 Proposition $12$ of Book $\text{VI}$: Construction of Fourth Proportional Straight Line, let it be contrived that:

$AB : CD = AE : CF$
$EB : FD = AB : CD$

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

$AE$ is commensurable in length with $CF$

and:

$EB$ is commensurable in length with $FD$.

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

$CF$ and $FD$ are medial.
$AE : CF = EB : FD$
$AE : EB = CF : FD$

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

$CF$ 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:

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

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 $AB$.

$\blacksquare$

## Historical Note

This proof is Proposition $67$ of Book $\text{X}$ of Euclid's The Elements.