Proof: By Euclid
(related to Proposition: Prop. 10.065: Square on Side of Sum of two Medial Area applied to Rational Straight Line)
- For let the same construction be made as in the previous (propositions).
- And since $AB$ is the square root of (the sum of) two medial (areas), having been divided at $C$, $AC$ and $CB$ are thus incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, moreover, the sum of the squares on them incommensurable with the (rectangle contained) by them [Prop. 10.41].
- Hence, according to what has been previously demonstrated, $DL$ and $MF$ are each medial.
- And they are applied to the rational (straight line) $DE$.
- Thus, $DM$ and $MG$ are each rational, and incommensurable in length with $DE$ [Prop. 10.22].
- And since the sum of the (squares) on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $AC$ and $CB$, $DL$ is thus incommensurable with $MF$.
- Thus, $DM$ (is) also incommensurable (in length) with $MG$ [Prop. 6.1], [Prop. 10.11].
- $DM$ and $MG$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $DG$ is a binomial (straight line) [Prop. 10.36].
- So, I say that (it is) also a sixth (binomial straight line).
- So, similarly (to the previous propositions), we can again show that the (rectangle contained) by $DKM$ is equal to the (square) on $MN$, and that $DK$ is incommensurable in length with $KM$.
- And so, for the same (reasons), the square on $DM$ is greater than (the square on) $MG$ by the (square) on (some straight line) incommensurable in length with ($DM$) [Prop. 10.18].
- And neither of $DM$ and $MG$ is commensurable in length with the (previously) laid down rational (straight line) $DE$.
- Thus, $DG$ is a sixth binomial (straight line) [Def. 10.10] .
- (Which is) the very thing it was required to show.
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"