(related to Proposition: Prop. 10.059: Root of Area contained by Rational Straight Line and Sixth Binomial)

- For let the area $ABCD$ be contained by the rational (straight line) $AB$ and the sixth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, such that $AE$ is the greater term.
- So, I say that the square root of $AC$ is the square root of (the sum of) two medial (areas).

- For let the same construction be made as that shown previously.
- So, (it is) clear that $MO$ is the square root of $AC$, and that $MN$ is incommensurable in square with $NO$.
- And since $EA$ is incommensurable in length with $AB$ [Def. 10.10] , $EA$ and $AB$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $AK$ - that is to say, the sum of the (squares) on $MN$ and $NO$ - is medial [Prop. 10.21].
- Again, since $ED$ is incommensurable in length with $AB$ [Def. 10.10] , $FE$ is thus also incommensurable (in length) with $EK$ [Prop. 10.13].
- Thus, $FE$ and $EK$ are rational (straight lines which are) commensurable in square only.
- Thus, $EL$ - that is to say, $MR$ - that is to say, the (rectangle contained) by $MNO$ - is medial [Prop. 10.21].
- And since $AE$ is incommensurable (in length) with $EF$, $AK$ is also incommensurable with $EL$ [Prop. 6.1], [Prop. 10.11].
- But, $AK$ is the sum of the (squares) on $MN$ and $NO$, and $EL$ is the (rectangle contained) by $MNO$.
- Thus, the sum of the (squares) on $MNO$ is incommensurable with the (rectangle contained) by $MNO$.
- And each of them is medial.
- And $MN$ and $NO$ are incommensurable in square.
- Thus, $MO$ is the square root of (the sum of) two medial (areas) [Prop. 10.41].
- And (it is) the square root of $AC$.
- (Which is) the very thing it was required to show.∎

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