# 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.

### 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.

## 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:

$AF : KM = BF : LM$
$AB : KL = AF : KM$
$AF$ is commensurable in length with $KM$.
$AB$ is commensurable in length with $KL$.
$AB : KL = CD \cdot AB : CD \cdot KL$
$CD \cdot AB$ 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.