# 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.*

(*The Elements*: Book $\text{X}$: Proposition $67$)

## 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:

- $AB : CD = AE : CF$

Therefore by Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

- $EB : FD = AB : CD$

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

Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

- $AE$ is commensurable in length with $CF$

and:

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

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

Therefore by Proposition $23$ of Book $\text{X} $: Straight Line Commensurable with Medial Straight Line is Medial:

- $CF$ and $FD$ are medial.

From Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

- $AE : CF = EB : FD$

Therefore by Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

- $AE : EB = CF : FD$

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

Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:

- $CF$ 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 Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:

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

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*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions