First Bimedial Straight Line is Divisible Uniquely

Theorem

In the words of Euclid:

A first bimedial straight line is divided at one point only.

(The Elements: Book $\text{X}$: Proposition $43$)

Proof

Let $AB$ be a first bimedial straight line.

Let $AB$ be divided at $C$ to create $AC$ and $CB$ such that:

$AC$ and $CB$ are medial straight lines

$AC$ and $CB$ are commensurable in square only

$AC$ and $CB$ contain a rational rectangle.

Let $AB$ be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$.

From Proposition $4$ of Book $\text{II} $: Square of Sum:

$AB^2 = \paren {AC + CB}^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

and:

$AB^2 = \paren {AD + DB}^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$

and so:

$\paren {AC^2 + CB^2} - \paren {AD^2 + DB^2} = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$

By definition, $AC \cdot CB$ and $AD \cdot DB$ are both rational areas.

So by Proposition $4$ of Book $\text{II} $: Product of Rationally Expressible Numbers is Rational:

$2 AC \cdot CB$ and $2 AD \cdot DB$ are both rational areas.

But $AB$

Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area.

As:

$AC$ and $CB$ are medial straight lines

and

$AD$ and $DB$ are medial straight lines

it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all medial areas.

But from Proposition $26$ of Book $\text{X} $: Medial Area not greater than Medial Area by Rational Area this cannot be the case.

From this contradiction it follows that $D$ cannot divide $AB$ into medial straight lines commensurable in square only and containing a rational rectangle.

$\blacksquare$

Historical Note

This proof is Proposition $43$ 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