Proof: By Euclid
(related to Proposition: Prop. 10.064: Square on Side of Rational plus Medial Area applied to Rational Straight Line)
- Let the same construction be made as in the (propositions) before this.
- Therefore, since $AB$ is the square root of a rational plus a medial (area) , 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 rational [Prop. 10.40].
- Therefore, since the sum of the (squares) on $AC$ and $CB$ is medial, $DL$ is thus medial.
- Hence, $DM$ is rational and incommensurable in length with $DE$ [Prop. 10.22].
- Again, since twice the (rectangle contained) by $ACB$ - that is to say, $MF$ - is rational, $MG$ (is) thus rational and commensurable (in length) with $DE$ [Prop. 10.20].
- $DM$ (is) thus incommensurable (in length) with $MG$ [Prop. 10.13].
- Thus, $DM$ and $MG$ are 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 fifth (binomial straight line).
- For, similarly (to the previous propositions), it can be shown that the (rectangle contained) by $DKM$ is equal to the (square) on $MN$, and $DK$ (is) incommensurable in length with $KM$.
- Thus, 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 $DM$ and $MG$ are [rational] (straight lines which are) commensurable in square only, and the lesser $MG$ is commensurable in length with $DE$.
- Thus, $DG$ is a fifth binomial (straight line) [Def. 10.9] .
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
