Proof: By Euclid
(related to Proposition: Prop. 10.087: Construction of Third Apotome)
- Let the rational (straight line) $A$ be laid down.
- And let the three numbers, $E$, $BC$, and $CD$, not having to one another the ratio which (some) square number (has) to (some) square number, be laid down.
- And let $CB$ have to $BD$ the ratio which (some) square number (has) to (some) square number.
- And let it have been contrived that as $E$ (is) to $BC$, so the square on $A$ (is) to the square on $FG$, and as $BC$ (is) to $CD$, so the square on $FG$ (is) to the (square) on $GH$ [Prop. 10.6 corr.] .
- Therefore, since as $E$ is to $BC$, so the square on $A$ (is) to the square on $FG$, the square on $A$ is thus commensurable with the square on $FG$ [Prop. 10.6].
- And the square on $A$ (is) rational.
- Thus, the (square) on $FG$ (is) also rational.
- Thus, $FG$ is a rational (straight line).
- And since $E$ does not have to $BC$ the ratio which (some) square number (has) to (some) square number, the square on $A$ thus does not have to the [square] on $FG$ the ratio which (some) square number (has) to (some) square number either.
- Thus, $A$ is incommensurable in length with $FG$ [Prop. 10.9].
- Again, since as $BC$ is to $CD$, so the square on $FG$ is to the (square) on $GH$, the square on $FG$ is thus commensurable with the (square) on on $GH$ [Prop. 10.6].
- And the (square) on $FG$ (is) rational.
- Thus, the (square) on $GH$ (is) also rational.
- Thus, $GH$ is a rational (straight line).
- And since $BC$ does not have to $CD$ the ratio which (some) square number (has) to (some) square number, the (square) on $FG$ thus does not have to the (square) on $GH$ the ratio which (some) square number (has) to (some) square number either.
- Thus, $FG$ is incommensurable in length with $GH$ [Prop. 10.9].
- And both are rational (straight lines).
- $FG$ and $GH$ are thus rational (straight lines which are) commensurable in square only.
- Thus, $FH$ is an apotome [Prop. 10.73].
-
So, I say that (it is) also a third (apotome) .
-
For since as $E$ is to $BC$, so the square on $A$ (is) to the (square) on $FG$, and as $BC$ (is) to $CD$, so the (square) on $FG$ (is) to the (square) on $HG$, thus, via equality, as $E$ is to $CD$, so the (square) on $A$ (is) to the (square) on $HG$ [Prop. 5.22].
- And $E$ does not have to $CD$ the ratio which (some) square number (has) to (some) square number.
- Thus, the (square) on $A$ does not have to the (square) on $GH$ the ratio which (some) square number (has) to (some) square number either.
- $A$ (is) thus incommensurable in length with $GH$ [Prop. 10.9].
- Thus, neither of $FG$ and $GH$ is commensurable in length with the (previously) laid down rational (straight line) $A$.
- Therefore, let the (square) on $K$ be that (area) by which the (square) on $FG$ is greater than the (square) on $GH$ [Prop. 10.13 lem.] .
- Therefore, since as $BC$ is to $CD$, so the (square) on $FG$ (is) to the (square) on $GH$, thus, via convertion, as $BC$ is to $BD$, so the square on $FG$ (is) to the square on $K$ [Prop. 5.19 corr.] 2.
- And $BC$ has to $BD$ the ratio which (some) square number (has) to (some) square number.
- Thus, the (square) on $FG$ also has to the (square) on $K$ the ratio which (some) square number (has) to (some) square number.
- $FG$ is thus commensurable in length with $K$ [Prop. 10.9].
- And the square on $FG$ is (thus) greater than (the square on) $GH$ by the (square) on (some straight line) commensurable (in length) with ($FG$).
- And neither of $FG$ and $GH$ is commensurable in length with the (previously) laid down rational (straight line) $A$.
- Thus, $FH$ is a third apotome [Def. 10.13] .
- Thus, the third apotome $FH$ has been found.
- (Which is) 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"
