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Proof: By Euclid

(related to Proposition: Prop. 10.062: Square on Second Bimedial Straight Line applied to Rational Straight Line)

• Let $AB$ be a second bimedial (straight line) having been divided into its (component) medial (straight lines) at $C$, such that $AC$ is the greater segment.
• And let $DE$ be some rational (straight line).
• And let the parallelogram $DF$, equal to the (square) on $AB$, have been applied to $DE$, producing $DG$ as breadth.
• I say that $DG$ is a third binomial (straight line).

• Let the same construction be made as that shown previously.
• And since $AB$ is a second bimedial (straight line), having been divided at $C$, $AC$ and $CB$ are thus medial (straight lines) commensurable in square only, and containing a medial (area) [Prop. 10.38].
• Hence, the sum of the (squares) on $AC$ and $CB$ is also medial [Prop. 10.15], [Prop. 10.23 corr.] .
• And it is equal to $DL$.
• Thus, $DL$ (is) also medial.
• And it is applied to the rational (straight line) $DE$.
• $MD$ is thus also rational, and incommensurable in length with $DE$ [Prop. 10.22].
• So, for the same (reasons), $MG$ is also rational, and incommensurable in length with $ML$ - that is to say, with $DE$.
• Thus, $DM$ and $MG$ are each rational, and incommensurable in length with $DE$.
• And since $AC$ is incommensurable in length with $CB$, and as $AC$ (is) to $CB$, so the (square) on $AC$ (is) to the (rectangle contained) by $ACB$ [Prop. 10.21 lem.] , the (square) on $AC$ (is) also incommensurable with the (rectangle contained) by $ACB$ [Prop. 10.11].
• And hence the sum of the (squares) on $AC$ and $CB$ is incommensurable with twice the (rectangle contained) by $ACB$ - that is to say, $DL$ with $MF$ [Prop. 10.12], [Prop. 10.13].
• Hence, $DM$ is also incommensurable (in length) with $MG$ [Prop. 6.1], [Prop. 10.11].
• And they are rational.
• $DG$ is thus a binomial (straight line) [Prop. 10.36].
• So, we must show that (it is) also a third (binomial straight line).
• So, similarly to the previous (propositions), we can conclude that $DM$ is greater than $MG$, and $DK$ (is) commensurable (in length) with $KM$.
• And the (rectangle contained) by $DKM$ is equal to the (square) on $MN$.
• Thus, the square on $DM$ is greater than (the square on) $MG$ by the (square) on (some straight line) commensurable (in length) with ($DM$) [Prop. 10.17].
• And neither of $DM$ and $MG$ is commensurable in length with $DE$.
• Thus, $DG$ is a third binomial (straight line) [Def. 10.7] .
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"