(related to Proposition: Prop. 10.047: Side of Sum of Two Medial Areas is Divisible Uniquely)

- Let $AB$ be [the square root of (the sum of) two medial (areas)] which has been divided at $C$, such that $AC$ and $CB$ are incommensurable in square, making the sum of the (squares) on $AC$ and $CB$ medial, and the (rectangle contained) by $AC$ and $CB$ medial, and, moreover, incommensurable with the sum of the (squares) on ($AC$ and $CB$) [Prop. 10.41].
- I say that $AB$ cannot be divided at another point fulfilling the prescribed (conditions).

- For, if possible, let it have been divided at $D$, such that $AC$ is again manifestly not the same as $DB$, but $AC$ (is), by hypothesis, greater.
- And let the rational (straight line) $EF$ be laid down.
- And let $EG$, equal to (the sum of) the (squares) on $AC$ and $CB$, and $HK$, equal to twice the (rectangle contained) by $AC$ and $CB$, have been applied to $EF$.
- Thus, the whole of $EK$ is equal to the square on $AB$ [Prop. 2.4].
- So, again, let $EL$, equal to (the sum of) the (squares) on $AD$ and $DB$, have been applied to $EF$.
- Thus, the remainder - twice the (rectangle contained) by $AD$ and $DB$ - is equal to the remainder, $MK$.
- And since the sum of the (squares) on $AC$ and $CB$ was assumed (to be) medial, $EG$ is also medial.
- And it is applied to the rational (straight line) $EF$.
- $HE$ is thus rational, and incommensurable in length with $EF$ [Prop. 10.22].
- So, for the same (reasons), $HN$ is also rational, and incommensurable in length with $EF$.
- And since the sum of the (squares) on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $AC$ and $CB$, $EG$ is thus also incommensurable with $GN$.
- Hence, $EH$ is also incommensurable with $HN$ [Prop. 6.1], [Prop. 10.11].
- And they are (both) rational (straight lines).
- Thus, $EH$ and $HN$ are rational (straight lines which are) commensurable in square only.
- Thus, $EN$ is a binomial (straight line) which has been divided (into its component terms) at $H$ [Prop. 10.36].
- So, similarly, we can show that it has also been (so) divided at $M$.
- And $EH$ is not the same as $MN$.
- Thus, a binomial (straight line) has been divided (into its component terms) at different points.
- The very thing is absurd [Prop. 10.42].
- Thus, the square root of (the sum of) two medial (areas) cannot be divided (into its component terms) at different points.
- Thus, it can be (so) divided at one [point] only.∎

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