(related to Proposition: Prop. 10.021: Medial is Irrational)

- For let the rectangle $AC$ be contained by the rational straight lines $AB$ and $BC$ (which are) commensurable in square only.
- I say that $AC$ is irrational, and its square root is irrational - let it be called
**medial**.

- For let the square $AD$ have been described on $AB$.
- $AD$ is thus rational [Def. 10.4] .
- And since $AB$ is incommensurable in length with $BC$.
- For they were assumed to be commensurable in square only.
- And $AB$ (is) equal to $BD$.
- $DB$ is thus also incommensurable in length with $BC$.
- And as $DB$ is to $BC$, so $AD$ (is) to $AC$ [Prop. 6.1].
- Thus, $DA$ [is] [incommensurable]bookofproofs$1095 with $AC$ [Prop. 10.11].
- And $DA$ (is) rational.
- Thus, $AC$ is irrational [Def. 10.4] .
- Hence, its square root [that is to say, the square root of the square equal to it] is also irrational [Def. 10.4] - let it be called
**medial**. - (Which is) the very thing it was required to show.∎

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