# Second Bimedial Straight Line is Divisible Uniquely

## Theorem

In the words of Euclid:

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

## Proof

Let $AB$ be a second 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 medial rectangle.

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

WLOG let $AC > DB$.

$AD^2 + DB^2 < AC^2 + CB^2$
$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$

and:

$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$

Let $EF$ be a rational straight line.

Let $EK$ be a rectangle set out on $EF$ equal to the square on $AB$.

Let $EG = EK - AC^2 - CB^2$.

Then $HK = 2 \cdot AC \cdot CB$.

Similarly, let $EL = AD^2 + DB^2$.

Then $HL = 2 \cdot AD \cdot DB$.

Since the squares on $AC$ and $CB$ are by definition medial, it follows that $EG$ is medial.

But $EG$ is applied to the rational straight line $EF$.

$EH$ is a rational straight line which is incommensurable in length with $EF$.

Using the same reasoning:

$HN$ is a rational straight line which is incommensurable in length with $EF$.

We have that $AC$ and $CB$ are medial straight lines which are commensurable in square only.

Therefore $AC$ and $CB$ are incommensurable in length.

But:

$AC : CB = AC^2 : AC \cdot CB$
$AC^2$ is incommensurable with $AC \cdot CB$.

But we have that $AC$ and $CB$ are commensurable in square.

$AC^2 + CB^2$ is commensurable in square with $AC^2$.
$2 \cdot AC \cdot CB$ is commensurable with $AC \cdot CB$.
$AC^2 + CB^2$ is incommensurable with $2 \cdot AC \cdot CB$.

We have:

$EG = AC^2 + CB^2$

and:

$HK = 2 \cdot AC \cdot CB$

Therefore $EG$ is incommensurable with $HK$.

So by:

Proposition $1$ of Book $\text{VI}$: Areas of Triangles and Parallelograms Proportional to Base

and:

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

it follows that:

$EH$ is incommensurable in length with $HN$.

We also have that $EH$ and $HN$ are rational.

Therefore $EH$ and $HN$ are rational straight lines which are commensurable in square only.

Two rational straight lines commensurable in square only, when added together, form an irrational straight line which is a binomial.

Therefore $EN$ is a binomial straight line which is divided at $H$.

In the same way $EM$ and $MN$ can be shown to be rational straight lines commensurable in square only.

Thus $EN$ is a binomial straight line which is divided into terms in two places: $H$ and $M$.

We have that:

$AC^2 + CB^2 > AD^2 + DB^2$

Also:

$AD^2 + DB^2 > 2 \cdot AD \cdot DB$

Thus $EG = AC^2 + CB^2 > 2 \cdot AD \cdot DB = MK$

Thus $EH > MN$.

$EN$ can be divided into its terms in only one way.

This contradicts the conclusion that has been drawn: that $H$ and $M$ divide $EH$ into terms in two different ways.

So the supposition that $AB$ can be divided at $D$ such that $AD$ and $DB$ have the same properties as $AB$ and $CB$ must be false.

$\blacksquare$

## Historical Note

This proof is Proposition $44$ of Book $\text{X}$ of Euclid's The Elements.