Proof: By Euclid

(related to Proposition: Prop. 10.099: Square on Second Apotome of Medial Straight Line applied to Rational Straight Line)

Let $AB$ be the second apotome of a medial (straight line), and $CD$ a rational (straight line).
And let $CE$, equal to the (square) on $AB$, have been applied to $CD$, producing $CF$ as breadth.
I say that $CF$ is a third apotome.
For let $BG$ be an attachment to $AB$.
Thus, $AG$ and $GB$ are medial (straight lines which are) commensurable in square only, containing a medial (area) [Prop. 10.75].
And let $CH$, equal to the (square) on $AG$, have been applied to $CD$, producing $CK$ as breadth.
And let $KL$, equal to the (square) on $BG$, have been applied to $KH$, producing $KM$ as breadth.
Thus, the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$ [and the (sum of the squares) on $AG$ and $GB$ is medial].
$CL$ (is) thus also medial [Prop. 10.15], [Prop. 10.23 corr.] .
And it has been applied to the rational (straight line) $CD$, producing $CM$ as breadth.
Thus, $CM$ is rational, and incommensurable in length with $CD$ [Prop. 10.22].
And since the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, of which $CE$ is equal to the (square) on $AB$, the remainder $LF$ is thus equal to twice the (rectangle contained) by $AG$ and $GB$ [Prop. 2.7].
Therefore, let $FM$ have been cut in half at point $N$.
And let $NO$ have been drawn parallel to $CD$.
Thus, $FO$ and $NL$ are each equal to the (rectangle contained) by $AG$ and $GB$.
And the (rectangle contained) by $AG$ and $GB$ (is) medial.
Thus, $FL$ is also medial.
And it is applied to the rational (straight line) $EF$, producing $FM$ as breadth.
$FM$ is thus rational, and incommensurable in length with $CD$ [Prop. 10.22].
And since $AG$ and $GB$ are commensurable in square only, $AG$ [is] thus incommensurable in length with $GB$.
Thus, the (square) on $AG$ is also incommensurable with the (rectangle contained) by $AG$ and $GB$ [Prop. 6.1], [Prop. 10.11].
But, the (sum of the squares) on $AG$ and $GB$ is commensurable with the (square) on on $AG$, and twice the (rectangle contained) by $AG$ and $GB$ with the (rectangle contained) by $AG$ and $GB$.
The (sum of the squares) on $AG$ and $GB$ is thus incommensurable with twice the (rectangle contained) by $AG$ and $GB$ [Prop. 10.13].
But, $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, and $FL$ is equal to twice the (rectangle contained) by $AG$ and $GB$.
Thus, $CL$ is incommensurable with $FL$.
And as $CL$ (is) to $FL$, so $CM$ is to $FM$ [Prop. 6.1].
$CM$ is thus incommensurable in length with $FM$ [Prop. 10.11].
And they are both rational (straight lines).
Thus, $CM$ and $MF$ are rational (straight lines which are) commensurable in square only.
$CF$ is thus an apotome [Prop. 10.73].
So, I say that (it is) also a third (apotome) .
For since the (square) on $AG$ is commensurable with the (square) on on $GB$, $CH$ (is) thus also commensurable with $KL$.
Hence, $CK$ (is) also (commensurable in length) with $KM$ [Prop. 6.1], [Prop. 10.11].
And since the (rectangle contained) by $AG$ and $GB$ is in mean proportion3 to the (squares) on $AG$ and $GB$ [Prop. 10.21 lem.] , and $CH$ is equal to the (square) on $AG$, and $KL$ equal to the (square) on $GB$, and $NL$ equal to the (rectangle contained) by $AG$ and $GB$, $NL$ is thus also in mean proportion3 to $CH$ and $KL$.
Thus, as $CH$ is to $NL$, so $NL$ (is) to $KL$.
But, as $CH$ (is) to $NL$, so $CK$ is to $NM$, and as $NL$ (is) to $KL$, so $NM$ (is) to $KM$ [Prop. 6.1].
Thus, as $CK$ (is) to $MN$, so $MN$ is to $KM$ [Prop. 5.11].
Thus, the (rectangle contained) by $CK$ and $KM$ is equal to the [ (square) on $MN$ - that is to say, to the] fourth part of the (square) on $FM$ [Prop. 6.17].
Therefore, since $CM$ and $MF$ are two unequal straight lines, and (some area), equal to the fourth part of the (square) on $FM$, has been applied to $CM$, falling short by a square figure, and divides it into commensurable (parts), the square on $CM$ is thus greater than (the square on) $MF$ by the (square) on (some straight line) commensurable (in length) with ($CM$) [Prop. 10.17].
And neither of $CM$ and $MF$ is commensurable in length with the (previously) laid down rational (straight line) $CD$.
$CF$ is thus a third apotome [Def. 10.13] .
Thus, the (square) on a second apotome of a medial (straight line), applied to a rational (straight line), produces a third apotome as breadth.
(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)