Proof: By Euclid
(related to Proposition: Prop. 10.104: Straight Line Commensurable with Apotome of Medial Straight Line)
 For since $AB$ is an apotome of a medial (straight line), let $EB$ be an attachment to it.
 Thus, $AE$ and $EB$ are medial (straight lines which are) commensurable in square only [Prop. 10.74], [Prop. 10.75].
 And let it have been contrived that as $AB$ is to $CD$, so $BE$ (is) to $DF$ [Prop. 6.12].
 Thus, $AE$ [is] also commensurable (in length) with $CF$, and $BE$ with $DF$ [Prop. 5.12], [Prop. 10.11].
 And $AE$ and $EB$ are medial (straight lines which are) commensurable in square only.
 $CF$ and $FD$ are thus also medial (straight lines which are) commensurable in square only [Prop. 10.23], [Prop. 10.13].
 Thus, $CD$ is an apotome of a medial (straight line) [Prop. 10.74], [Prop. 10.75].
 So, I say that it is also the same in order^{1} as $AB$.
 For since as $AE$ is to $EB$, so $CF$ (is) to $FD$ [Prop. 5.12], [Prop. 5.16] [but as $AE$ (is) to $EB$, so the (square) on $AE$ (is) to the (rectangle contained) by $AE$ and $EB$, and as $CF$ (is) to $FD$, so the (square) on $CF$ (is) to the (rectangle contained) by $CF$ and $FD$], thus as the (square) on $AE$ is to the (rectangle contained) by $AE$ and $EB$, so the (square) on $CF$ also (is) to the (rectangle contained) by $CF$ and $FD$ [Prop. 10.21 lem.] [and, alternately, as the (square) on $AE$ (is) to the (square) on $CF$, so the (rectangle contained) by $AE$ and $EB$ (is) to the (rectangle contained) by $CF$ and $FD$].
 And the (square) on $AE$ (is) commensurable with the (square) on on $CF$.
 Thus, the (rectangle contained) by $AE$ and $EB$ is also commensurable with the (rectangle contained) by $CF$ and $FD$ [Prop. 5.16], [Prop. 10.11].
 Therefore, either the (rectangle contained) by $AE$ and $EB$ is rational, and the (rectangle contained) by $CF$ and $FD$ will also be rational [Def. 10.4] , or the (rectangle contained) by $AE$ and $EB$ [is] [medial]bookofproofs$2115, and the (rectangle contained) by $CF$ and $FD$ [is] also medial [Prop. 10.23 corr.] .
 Therefore, $CD$ is the apotome of a medial (straight line), and is the same in order as $AB$ [Prop. 10.74], [Prop. 10.75].
 (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