Proof: By Euclid
(related to Proposition: Prop. 10.103: Straight Line Commensurable with Apotome)
 For since $AB$ is an apotome, let $BE$ be an attachment to it.
 Thus, $AE$ and $EB$ are rational (straight lines which are) commensurable in square only [Prop. 10.73].
 And let it have been contrived that the (ratio) of $BE$ to $DF$ is the same as the ratio of $AB$ to $CD$ [Prop. 6.12].
 Thus, also, as one is to one, (so) all [are] to all [Prop. 5.12].
 And thus as the whole $AE$ is to the whole $CF$, so $AB$ (is) to $CD$.
 And $AB$ (is) commensurable in length with $CD$.
 $AE$ (is) thus also commensurable (in length) with $CF$, and $BE$ with $DF$ [Prop. 10.11].
 And $AE$ and $BE$ are rational (straight lines which are) commensurable in square only.
 Thus, $CF$ and $FD$ are also rational (straight lines which are) commensurable in square only [Prop. 10.13].
 $CD$ is thus an apotome.
 So, I say that (it is) also the same in order as $AB$.
 Therefore, since as $AE$ is to $CF$, so $BE$ (is) to $DF$, thus, alternately, as $AE$ is to $EB$, so $CF$ (is) to $FD$ [Prop. 5.16].
 So, the square on $AE$ is greater than (the square on) $EB$ either by the (square) on (some straight line) commensurable, or by the (square) on (some straight line) incommensurable (in length) with ($AE$).
 Therefore, if the (square) on $AE$ is greater than (the square on) $EB$ by the (square) on (some straight line) commensurable (in length) with ($AE$) then the square on $CF$ will also be greater than (the square on) $FD$ by the (square) on (some straight line) commensurable (in length) with ($CF$) [Prop. 10.14].
 And if $AE$ is commensurable in length with a (previously) laid down rational (straight line) then so (is) $CF$ [Prop. 10.12], and if $BE$ (is commensurable), so (is) $DF$, and if neither of $AE$ or $EB$ (are commensurable), neither (are) either of $CF$ or $FD$ [Prop. 10.13].
 And if the (square) on $AE$ is greater [than (the square on) $EB$] by the (square) on (some straight line) incommensurable (in length) with ($AE$) then the (square) on $CF$ will also be greater than (the square on) $FD$ by the (square) on (some straight line) incommensurable (in length) with ($CF$) [Prop. 10.14].
 And if $AE$ is commensurable in length with a (previously) laid down rational (straight line), so (is) $CF$ [Prop. 10.12], and if $BE$ (is commensurable), so (is) $DF$, and if neither of $AE$ or $EB$ (are commensurable), neither (are) either of $CF$ or $FD$ [Prop. 10.13].
 Thus, $CD$ is an apotome, and (is) the same in order^{1} as $AB$ [Def. 10.11] , [Def. 10.12] , [Def. 10.13] , [Def. 10.14] , [Def. 10.15] , [Def. 10.16] .
 (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"
Footnotes