# Side of Area Contained by Rational Straight Line and First Apotome

## Theorem

In the words of Euclid:

If an area be contained by a rational straight line and a first apotome, the "side" of the area is an apotome.

## Proof

Let the area $AB$ be contained by the rational straight line $AC$ and the first apotome $AD$.

It is to be proved that the "side" of $AB$ is an apotome.

Let $DG$ be the annex of the first apotome $AD$.

Then, by definition:

$AG$ and $GD$ are rational straight lines which are commensurable in square only
the whole $AG$ is commensurable with the rational straight line $AC$
the square on $AG$ is greater than the square on $GD$ by the square on a straight line which is commensurable in length with $AG$.

Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $DG$ and deficient by a square figure.

that parallelogram divides $AG$ into commensurable parts.

Let $DG$ be bisected at $E$.

Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure.

Therefore $AF$ is commensurable with $FG$.

Through $E$, $F$ and $G$ let $EH$, $FI$ and $GK$ be drawn parallel to $AC$.

We have that $AF$ is commensurable with $FG$.

$AG$ is commensurable with each of $AF$ and $FG$.

But $AG$ is commensurable with $AC$.

each of the straight lines $AF$ and $FG$ is commensurable in length with $AC$.

We have that $AC$ is rational.

Therefore each of the straight lines $AF$ and $FG$ is rational.

each of the rectangles $AI$ and $FK$ is rational.

We have that $DE$ is commensurable in length with $EG$.

$DG$ is also commensurable in length with each of the straight lines $DE$ and $EG$.

But $DG$ is rational and incommensurable in length with $AC$.

each of the straight lines $DE$ and $ED$ is rational and incommensurable in length with $AC$.

Therefore from Proposition $21$ of Book $\text{X}$: Medial is Irrational:

each of the rectangles $DH$ and $EK$ is medial.

Let the square $LM$ be constructed equal to $AI$.

Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$.

the squares $LM$ and $NO$ are about the same diameter.

Let $PR$ be the diameter of $LM$ and $NO$.

We have that the rectangle contained by $AF$ and $FG$ equals the square on $EG$.

$AF : EG = EG : FG$

But we also have:

$AF : EG = AI : EK$
$EG : FG = EK : KF$
$AI : EK = EK : FK$

Therefore $EK$ is a mean proportional between $AI$ and $FK$.

$MN$ is a mean proportional between $LM$ and $NO$.

We have that:

$AI$ equals the square $LM$

and:

$KF$ equals the square $NO$.

Therefore:

$MN = EK$

But:

$EK = DH$

and:

$MN = LO$

Therefore $DK$ equals the gnomon $UVW$ and $NO$.

But:

$AK$ equals the squares $LM$ and $NO$.

Therefore the remainder $AB$ equals $ST$.

But $ST$ is the square on $LN$.

Therefore the square on $LN$ equals $AB$.

Therefore $LN$ is the "side" of $AB$.

Now it is to be shown that $LN$ is an apotome.

We have that each of the rectangles $AI$ and $FK$ is rational.

Also:

$AI = LM$

and:

$FK = NO$

Therefore each of the squares $LM$ and $NO$, that is the squares on $LP$ and $PN$, is rational.

Therefore $LP$ and $PN$ are both rational.

We have that $DH$ is medial and equals $LO$.

Therefore $LO$ is medial.

But $NO$ is rational.

Therefore $LO$ is incommensurable with $NO$.

$LO : NO = LP : PN$
$LP$ is incommensurable in length with $PN$.

We have that both $LP$ and $PN$ are rational.

Thus $LP$ and $PN$ are rational straight lines which are commensurable in square only.

Therefore by definition $LN$ is an apotome.

But $LN$ is the "side" of the area $AB$.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $91$ of Book $\text{X}$ of Euclid's The Elements.
It is the converse of Proposition $97$: Square on Apotome applied to Rational Straight Line.