Proof: By Euclid
(related to Proposition: Prop. 10.067: Straight Line Commensurable with Bimedial Straight Line is Bimedial and of Same Order)
 For since $AB$ is a bimedial (straight line), let it have been divided into its (component) medial (straight lines) at $E$.
 Thus, $AE$ and $EB$ are medial (straight lines which are) commensurable in square only [Prop. 10.37], [Prop. 10.38].
 And let it have been contrived that as $AB$ (is) to $CD$, (so) $AE$ (is) to $CF$ [Prop. 6.12].
 And thus as the remainder $EB$ is to the remainder $FD$, so $AB$ (is) to $CD$ [Prop. 5.19 corr.] 2, [Prop. 6.16].
 And $AB$ (is) commensurable in length with $CD$.
 Thus, $AE$ and $EB$ are also commensurable (in length) with $CF$ and $FD$, respectively [Prop. 10.11].
 And $AE$ and $EB$ (are) medial.
 Thus, $CF$ and $FD$ (are) also medial [Prop. 10.23].
 And since as $AE$ is to $EB$, (so) $CF$ (is) to $FD$, and $AE$ and $EB$ are commensurable in square only, $CF$ and $FD$ are [thus] also commensurable in square only [Prop. 10.11].
 And they were also shown (to be) medial.
 Thus, $CD$ is a bimedial (straight line).
 So, I say that it is also the same in order as $AB$.
 For since as $AE$ is to $EB$, (so) $CF$ (is) to $FD$, thus also as the (square) on $AE$ (is) to the (rectangle contained) by $AEB$, so the (square) on $CF$ (is) to the (rectangle contained) by $CFD$ [Prop. 10.21 lem.] .
 Alternately, as the (square) on $AE$ (is) to the (square) on $CF$, so the (rectangle contained) by $AEB$ (is) to the (rectangle contained) by $CFD$ [Prop. 5.16].
 And the (square) on $AE$ (is) commensurable with the (square) on on $CF$.
 Thus, the (rectangle contained) by $AEB$ (is) also commensurable with the (rectangle contained) by $CFD$ [Prop. 10.11].
 Therefore, either the (rectangle contained) by $AEB$ is rational, and the (rectangle contained) by $CFD$ is rational [and, on account of this, ($AE$ and $CD$) are first bimedial (straight lines)], or (the rectangle contained by $AEB$ is) medial, and (the rectangle contained by $CFD$ is) medial, and ($AB$ and $CD$) are each second (bimedial straight lines) [Prop. 10.23], [Prop. 10.37], [Prop. 10.38].
And, on account of this, $CD$ will be the same in order^{1} as $AB$.
 (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