# 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.*

(*The Elements*: Book $\text{X}$: Proposition $91$)

## 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.

From Proposition $17$ of Book $\text{X} $: Condition for Commensurability of Roots of Quadratic Equation:

- 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 $FG$.

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

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

But $AG$ is commensurable with $AC$.

Therefore from Proposition $12$ of Book $\text{X} $: Commensurability is Transitive Relation:

- 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.

Therefore from Proposition $19$ of Book $\text{X} $: Product of Rationally Expressible Numbers is Rational:

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

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

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

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

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

Therefore from Proposition $13$ of Book $\text{X} $: Commensurable Magnitudes are Incommensurable with Same Magnitude:

- 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 $FK$.

Therefore from Proposition $26$ of Book $\text{VI} $: Parallelogram Similar and in Same Angle has 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 $EG$.

Therefore from Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:

- $AF : EG = EG : FG$

But we also have:

- $AF : EG = AI : EK$

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

- $EG : FG = EK : KF$

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

- $AI : EK = EK : FK$

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 $NO$.

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

- $LO : NO = LP : PN$

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

- $LP$ 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.

## Sources

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