Proof: By Euclid

Let $A$ be a rational (straight line), and $BC$ a binomial (straight line), of which let $DC$ be the greater term.
And let the (rectangle contained) by $BC$ and $EF$ be equal to the (square) on $A$.
I say that $EF$ is an apotome whose terms are commensurable (in length) with $CD$ and $DB$, and in the same ratio, and, moreover, that $EF$ will have the same order as $BC$.

fig112e

For, again, let the (rectangle contained) by $BD$ and $G$ be equal to the (square) on $A$.
Therefore, since the (rectangle contained) by $BC$ and $EF$ is equal to the (rectangle contained) by $BD$ and $G$, thus as $CB$ is to $BD$, so $G$ (is) to $EF$ [Prop. 6.16].
And $CB$ (is) greater than $BD$.
Thus, $G$ is also greater than $EF$ [Prop. 5.16], [Prop. 5.14].
Let $EH$ be equal to $G$.
Thus, as $CB$ is to $BD$, so $HE$ (is) to $EF$.
Thus, via separation, as $CD$ is to $BD$, so $HF$ (is) to $FE$ [Prop. 5.17].
Let it have been contrived that as $HF$ (is) to $FE$, so $FK$ (is) to $KE$.
And, thus, the whole $HK$ is to the whole $KF$, as $FK$ (is) to $KE$.
For as one of the leading (proportional magnitudes is) to one of the following, so all of the leading (magnitudes) are to all of the following [Prop. 5.12].
And as $FK$ (is) to $KE$, so $CD$ is to $DB$ [Prop. 5.11].
And, thus, as $HK$ (is) to $KF$, so $CD$ is to $DB$ [Prop. 5.11].
And the (square) on $CD$ (is) commensurable with the (square) on on $DB$ [Prop. 10.36].
The (square) on $HK$ is thus also commensurable with the (square) on on $KF$ [Prop. 6.22], [Prop. 10.11].
And as the (square) on $HK$ is to the (square) on $KF$, so $HK$ (is) to $KE$, since the three (straight lines) $HK$, $KF$, and $KE$ are proportional [Def. 5.9] .
$HK$ is thus commensurable in length with $KE$ [Prop. 10.11].
Hence, $HE$ is also commensurable in length with $EK$ [Prop. 10.15].
And since the (square) on $A$ is equal to the (rectangle contained) by $EH$ and $BD$, and the (square) on $A$ is rational, the (rectangle contained) by $EH$ and $BD$ is thus also rational.
And it is applied to the rational (straight line) $BD$.
Thus, $EH$ is rational, and commensurable in length with $BD$ [Prop. 10.20].
And, hence, the (straight line) commensurable (in length) with it, $EK$, is also rational [Def. 10.3] , and commensurable in length with $BD$ [Prop. 10.12].
Therefore, since as $CD$ is to $DB$, so $FK$ (is) to $KE$, and $CD$ and $DB$ are (straight lines which are) commensurable in square only, $FK$ and $KE$ are also commensurable in square only [Prop. 10.11].
And $KE$ is rational.
Thus, $FK$ is also rational.
$FK$ and $KE$ are thus rational (straight lines which are) commensurable in square only.
Thus, $EF$ is an apotome [Prop. 10.73].
And the square on $CD$ is greater than (the square on) $DB$ either by the (square) on (some straight line) commensurable, or by the (square) on (some straight line) incommensurable (in length) with ($CD$).
Therefore, if the square on $CD$ is greater than (the square on) $DB$ by the (square) on (some straight line) commensurable (in length) with [$CD$] then the square on $FK$ will also be greater than (the square on) $KE$ by the (square) on (some straight line) commensurable (in length) with ($FK$) [Prop. 10.14].
And if $CD$ is commensurable in length with a (previously) laid down rational (straight line), (so) also (is) $FK$ [Prop. 10.11], [Prop. 10.12].
And if $BD$ (is commensurable), (so) also (is) $KE$ [Prop. 10.12].
And if neither of $CD$ or $DB$ (is commensurable), neither also (are) either of $FK$ or $KE$.
And if the square on $CD$ is greater than (the square on) $DB$ by the (square) on (some straight line) incommensurable (in length) with ($CD$) then the square on $FK$ will also be greater than (the square on) $KE$ by the (square) on (some straight line) incommensurable (in length) with ($FK$) [Prop. 10.14].
And if $CD$ is commensurable in length with a (previously) laid down rational (straight line), (so) also (is) $FK$ [Prop. 10.11], [Prop. 10.12].
And if $BD$ (is commensurable), (so) also (is) $KE$ [Prop. 10.12].
And if neither of $CD$ or $DB$ (is commensurable), neither also (are) either of $FK$ or $KE$.
Hence, $FE$ is an apotome whose terms, $FK$ and $KE$, are commensurable (in length) with the terms, $CD$ and $DB$, of the binomial, and in the same ratio.
And ($FE$) has the same order as $BC$ [Def. 10.5] , [Def. 10.6] , [Def. 10.7] , [Def. 10.8] , [Def. 10.9] , [Def. 10.10] .
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!

Github:
non-Github:: @Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

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

◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)