Proof: By Euclid

Let $AB$ be a minor (straight line), and $CD$ a rational (straight line).
And let $CE$, equal to the (square) on $AB$, have been applied to the rational (straight line) $CD$, producing $CF$ as breadth.
I say that $CF$ is a fourth apotome.
For let $BG$ be an attachment to $AB$.
Thus, $AG$ and $GB$ are incommensurable in square, making the sum of the squares on $AG$ and $GB$ rational, and twice the (rectangle contained) by $AG$ and $GB$ medial [Prop. 10.76].
And let $CH$, equal to the (square) on $AG$, have been applied to $CD$, producing $CK$ as breadth, and $KL$, equal to the (square) on $BG$, 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 rational.
$CL$ is thus also rational.
And it is applied to the rational (straight line) $CD$, producing $CM$ as breadth.
Thus, $CM$ (is) also rational, and commensurable in length with $CD$ [Prop. 10.20].
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 $FL$ 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 through $N$, parallel to either of $CD$ or $ML$.
Thus, $FO$ and $NL$ are each equal to the (rectangle contained) by $AG$ and $GB$.
And since twice the (rectangle contained) by $AG$ and $GB$ is medial, and is equal to $FL$, $FL$ is thus also medial.
And it is applied to the rational (straight line) $FE$, producing $FM$ as breadth.
Thus, $FM$ is 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]bookofproofs$1095 with twice the (rectangle contained) by $AG$ and $GB$.
And $CL$ (is) equal to the (sum of the squares) on $AG$ and $GB$, and $FL$ equal to twice the (rectangle contained) by $AG$ and $GB$.
$CL$ [is] thus incommensurable with $FL$.
And as $CL$ (is) to $FL$, so $CM$ is to $MF$ [Prop. 6.1].
$CM$ is thus incommensurable in length with $MF$ [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 fourth (apotome) .
For since $AG$ and $GB$ are incommensurable in square, the (square) on $AG$ (is) thus also incommensurable with the (square) on on $GB$.
And $CH$ is equal to the (square) on $AG$, and $KL$ equal to the (square) on $GB$.
Thus, $CH$ is incommensurable with $KL$.
And as $CH$ (is) to $KL$, so $CK$ is to $KM$ [Prop. 6.1].
$CK$ is thus incommensurable in length with $KM$ [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 the (square) on $AG$ is equal to $CH$, and the (square) on $GB$ to $KL$, and the (rectangle contained) by $AG$ and $GB$ to $NL$, $NL$ is thus 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].
The (rectangle contained) by $CK$ and $KM$ is thus 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 the (rectangle contained) by $CK$ and $KM$, equal to the fourth part of the (square) on $MF$, has been applied to $CM$, falling short by a square figure, and divides it into incommensurable (parts), the square on $CM$ is thus greater than (the square on) $MF$ by the (square) on (some straight line) incommensurable (in length) with ($CM$) [Prop. 10.18].
And the whole of $CM$ is commensurable in length with the (previously) laid down rational (straight line) $CD$.
Thus, $CF$ is a fourth apotome [Def. 10.14] .
Thus, the (square) on a minor, and so on ....
∎

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References

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

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)