# Construction of Second Apotome of Medial is Unique

## Theorem

In the words of Euclid:

To a second apotome of a medial straight line only one medial straight line can be annexed which is commensurable with the whole in square only and which contains with the whole a medial rectangle.

## Proof Let $AB$ be the first apotome of a medial straight line.

Let $BC$ be added to $AB$ such that:

$AC$ and $CB$ are medial straight lines
$AC$ and $CB$ are commensurable in square only
$AC \cdot CB$ is a medial rectangle.

It is to be proved that no other medial straight line can be added to $AB$ which is commensurable in square only with the whole and which contains with the whole a medial rectangle.

Suppose $BD$ can be added to $AB$ so as to fulfil the conditions stated.

Then by definition of the first apotome of a medial straight line, $AD$ and $DB$ are such that:

$AD$ and $DB$ are medial straight lines
$AD$ and $DB$ are commensurable in square only
$AD \cdot DB$ is a medial rectangle.

Let $EF$ be a rational straight line.

Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth.

Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth.

Therefore from Proposition $7$ of Book $\text{II}$: Square of Difference:

the remainder $EL$ equals $AB^2$.

Thus $AB$ equals the "side" of $EL$.

Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth.

But $EL = AB^2$.

Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$.

We have that $AC$ and $CB$ are medial straight lines.

Therefore $AC^2$ and $CB^2$ are also medial.

Also $AC^2 + CB^2 = EG$.

So by:

Proposition $15$ of Book $\text{X}$: Commensurability of Sum of Commensurable Magnitudes

and:

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

it follows that:

$EG$ is medial.

But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth.

$EM$ is rational and incommensurable in length with $EF$.

We have that $AC \cdot CB$ is a medial rectangle.

$2 \cdot AC \cdot CB$ is a medial rectangle.

But $2 \cdot AC \cdot CB = HG$.

Therefore $HG$ is medial.

Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth.

$HM$ is rational and incommensurable in length with $EF$.

Also, $AC$ and $CB$ 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 $AC^2 + CB^2$ is commensurable with $AC^2$.

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

We have that:

$EG = AC^2 + CB^2$

and:

$GH = 2 \cdot AC \cdot CB$

Therefore:

$EG$ is incommensurable with $HG$.
$EG : HG = EM : HM$
$EM$ is incommensurable in length with $MH$.

But both $EM$ and $MH$ are rational straight lines.

Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only.

Therefore $EH$ is an apotome, and $HM$ an annex to it.

Similarly it can be shown that $HN$ is also an annex to it.

Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole.

From Proposition $79$ of Book $\text{X}$: Construction of Apotome is Unique, this is impossible.

The result follows.

$\blacksquare$

## Historical Note

This proof is Proposition $81$ of Book $\text{X}$ of Euclid's The Elements.