Proof: By Euclid

(related to Proposition: Prop. 10.097: Square on Apotome applied to Rational Straight Line)

Let $AB$ be an apotome, 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 first apotome.
For let $BG$ be an attachment to $AB$.
Thus, $AG$ and $GB$ are rational (straight lines which are) commensurable in square only [Prop. 10.73].
And let $CH$, equal to the (square) on $AG$, and $KL$, (equal) to the (square) on $BG$, have been applied to $CD$.
Thus, 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 $FL$ is thus equal to twice the (rectangle contained) by $AG$ and $GB$ [Prop. 2.7].
Let $FM$ have been cut in half at point $N$.
And let $NO$ have been drawn through $N$, parallel to $CD$.
Thus, $FO$ and $LN$ are each equal to the (rectangle contained) by $AG$ and $GB$.
And since the (sum of the squares) on $AG$ and $GB$ is rational, and $DM$ is equal to the (sum of the squares) on $AG$ and $GB$, $DM$ is thus rational.
And it has been applied to the rational (straight line) $CD$, producing $CM$ as breadth.
Thus, $CM$ is rational, and commensurable in length with $CD$ [Prop. 10.20].
Again, since twice the (rectangle contained) by $AG$ and $GB$ is medial, and $FL$ (is) equal to twice the (rectangle contained) by $AG$ and $GB$, $FL$ (is) thus a medial (area) .
And it is applied to the rational (straight line) $CD$, producing $FM$ as breadth.
$FM$ is thus rational, and incommensurable in length with $CD$ [Prop. 10.22].
And since the (sum of the squares) on $AG$ and $GB$ is rational, and twice the (rectangle contained) by $AG$ and $GB$ medial, the (sum of the squares) on $AG$ and $GB$ is thus incommensurable with twice the (rectangle contained) by $AG$ and $GB$.
And $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, and $FL$ to twice the (rectangle contained) by $AG$ and $GB$.
$DM$ is thus incommensurable with $FL$.
And as $DM$ (is) to $FL$, so $CM$ is to $FM$ [Prop. 6.1].
$CM$ is thus incommensurable in length with $FM$ [Prop. 10.11].
And both are 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 first (apotome) .
For 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 $BG$, and $NL$ 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, the (rectangle contained) by $CK$ and $KM$ is equal to the (square) on $NM$ - that is to say, to the fourth part of the (square) on $FM$ [Prop. 6.17].
And since the (square) on $AG$ is commensurable with the (square) on on $GB$, $CH$ [is] also commensurable with $KL$.
And as $CH$ (is) to $KL$, so $CK$ (is) to $KM$ [Prop. 6.1].
$CK$ is thus commensurable (in length) with $KM$ [Prop. 10.11].
Therefore, since $CM$ and $MF$ are two unequal straight lines, and the (rectangle contained) by $CK$ and $KM$, equal to the fourth part of the (square) on $FM$, has been applied to $CM$, falling short by a square figure, and $CK$ is commensurable (in length) with $KM$, 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 $CM$ is commensurable in length with the (previously) laid down rational (straight line) $CD$.
Thus, $CF$ is a first apotome [Def. 10.15] .
Thus, the (square) on an apotome, applied to a rational (straight line), produces a first apotome as breadth.
(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"

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)