(related to Proposition: Prop. 10.041: Side of Sum of Medial Areas is Irrational)

- For let the two straight lines, $AB$ and $BC$, incommensurable in square, (and) fulfilling the prescribed (conditions), be laid down together [Prop. 10.35].
- I say that $AC$ is irrational.

- Let the rational (straight line) $DE$ be laid out, and let (the rectangle) $DF$, equal to (the sum of) the (squares) on $AB$ and $BC$, and (the rectangle) $GH$, equal to twice the (rectangle contained) by $AB$ and $BC$, have been applied to $DE$.
- Thus, the whole of $DH$ is equal to the square on $AC$ [Prop. 2.4].
- And since the sum of the (squares) on $AB$ and $BC$ is medial, and is equal to $DF$, $DF$ is thus also medial.
- And it is applied to the rational (straight line) $DE$.
- Thus, $DG$ is rational, and incommensurable in length with $DE$ [Prop. 10.22].
- So, for the same (reasons), $GK$ is also rational, and incommensurable in length with $GF$ - that is to say, $DE$.
- And since (the sum of) the (squares) on $AB$ and $BC$ is incommensurable with twice the (rectangle contained) by $AB$ and $BC$, $DF$ is incommensurable with $GH$.
- Hence, $DG$ is also incommensurable (in length) with $GK$ [Prop. 6.1], [Prop. 10.11].
- And they are rational.
- Thus, $DG$ and $GK$ are rational (straight lines which are) commensurable in square only.
- Thus, $DK$ is irrational, and that (straight line which is) called binomial [Prop. 10.36].
- And $DE$ (is) rational.
- Thus, $DH$ is irrational, and its square root is irrational [Def. 10.4] .
- And $AC$ (is) the square root of $HD$.
- Thus, $AC$ is irrational - let it be called the square root of (the sum of) two medial (areas).
- (Which is) the very thing it was required to show.∎

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