# Area contained by Apotome and Binomial Straight Line Commensurable with Terms of Apotome and in same Ratio

## Theorem

In the words of Euclid:

*If an area be contained by an apotome and the binomial straight line the terms of which are commensurable with the terms of the apotome and in the same ratio, the "side" of the area is rational.*

(*The Elements*: Book $\text{X}$: Proposition $114$)

### Porism

In the words of Euclid:

*And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines.*

(*The Elements*: Book $\text{X}$: Proposition $114$ : Porism)

## Proof

Let the rectangle $AB \cdot CD$ be contained by the apotome $AB$ and the binomial straight line $CD$.

Let $CE$ be the greater term of $CD$.

Let:

- $CE$ be commensurable in length with $AF$
- $ED$ be commensurable in length with $FB$
- $CE : ED = AF : FB$

Let the "side" of the rectangle $AB \cdot CD$ be $G$.

It is to be demonstrated that $G$ is rational.

Let $H$ be a rational straight line.

Let a rectangle equal to $H^2$ be applied to $CD$ which produces $KL$ as breadth.

By definition, $KL$ is an apotome.

Let the terms of $KL$ be $KM$ and $ML$.

Let $KM$ and $ML$ be commensurable with the terms $CE$ and $ED$ of the binomial straight line $CD$.

- $CE$ and $ED$ are commensurable with the terms $AF$ and $FB$ of the apotome $CD$.

Therefore:

- $AF : FB = KM : ML$

and so:

- $AF : KM = BF : LM$

and so by Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:

- $AB : KL = AF : KM$

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

- $AF$ is commensurable in length with $KM$.

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

- $AB$ is commensurable in length with $KL$.

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

- $AB : KL = CD \cdot AB : CD \cdot KL$

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

- $CD \cdot AB$ is commensurable with $CD \cdot KL$.

But $CD \cdot KL = H^2$.

Therefore $CD \cdot AB$ is commensurable with $H^2$.

But $G^2 = CD \cdot AB$.

Therefore $G^2$ is commensurable with $H^2$.

But $H^2$ is rational.

Therefore $G^2$ is also rational.

Therefore $G$ is rational straight line.

But $G^2$ is the "side" of the rectangle $AB \cdot CD$.

Hence the result.

$\blacksquare$

## Historical Note

This proof is Proposition $114$ 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