Proof: By Euclid
(related to Proposition: Prop. 10.049: Construction of Second Binomial Straight Line)
- For since, inversely, as the number BA is to AC, so the (square) on GF (is) to the (square) on FE [Prop. 5.7 corr.] 1, and BA (is) greater than AC, the (square) on GF (is) thus [also] greater than the (square) on FE [Prop. 5.14].
- Let (the sum of) the (squares) on EF and H be equal to the (square) on GF.
- Thus, via convertion, as AB is to BC, so the (square) on FG (is) to the (square) on H [Prop. 5.19 corr.] 2.
- But, AB has to BC the ratio which (some) square number (has) to (some) square number.
- Thus, the (square) on FG also has to the (square) on H the ratio which (some) square number (has) to (some) square number.
- Thus, FG is commensurable in length with H [Prop. 10.9].
- Hence, the square on FG is greater than (the square on) FE by the (square) on (some straight line) commensurable in length with (FG).
- And FG and FE are rational (straight lines which are) commensurable in square only.
- And the lesser term EF is commensurable in length with the rational (straight line) D (previously) laid down.
- Thus, EG is a second binomial (straight line) [Def. 10.6] .
- (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"