(related to Proposition: Prop. 10.073: Apotome is Irrational)

- For let the rational (straight line) $BC$, which commensurable in square only with the whole, have been subtracted from the rational (straight line) $AB$.
- I say that the remainder $AC$ is that irrational (straight line) called an
**apotome**.

- For since $AB$ is incommensurable in length with $BC$, and as $AB$ is to $BC$, so the (square) on $AB$ (is) to the (rectangle contained) by $AB$ and $BC$ [Prop. 10.21 lem.] , the (square) on $AB$ is thus incommensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.11].
- But, the (sum of the) squares on $AB$ and $BC$ is commensurable with the (square) on on $AB$ [Prop. 10.15], and twice the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.6].
- And, inasmuch as the (sum of the squares) on $AB$ and $BC$ is equal to twice the (rectangle contained) by $AB$ and $BC$ plus the (square) on $CA$ [Prop. 2.7], the (sum of the squares) on $AB$ and $BC$ is thus also incommensurable with the remaining (square) on $AC$ [Prop. 10.13], [Prop. 10.16].
- And the (sum of the squares) on $AB$ and $BC$ is rational.
- $AC$ is thus an irrational (straight line) [Def. 10.4] .
- And let it be called an
**apotome**. - (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"