Proof: By Euclid
(related to Proposition: Prop. 10.066: Straight Line Commensurable with Binomial Straight Line is Binomial and of Same Order)
 For since $AB$ is a binomial (straight line), let it have been divided into its (component) terms at $E$, and let $AE$ be the greater term.
 $AE$ and $EB$ are thus rational (straight lines which are) commensurable in square only [Prop. 10.36].
 Let it have been contrived that as $AB$ (is) to $CD$, so $AE$ (is) to $CF$ [Prop. 6.12].
 Thus, the remainder $EB$ is also to the remainder $FD$, as $AB$ (is) to $CD$ [Prop. 6.16], [Prop. 5.19 corr.] 2.
 And $AB$ (is) commensurable in length with $CD$.
 Thus, $AE$ is also commensurable (in length) with $CF$, and $EB$ with $FD$ [Prop. 10.11].
 And $AE$ and $EB$ are rational.
 Thus, $CF$ and $FD$ are also rational.
 And as $AE$ is to $CF$, (so) $EB$ (is) to $FD$ [Prop. 5.11].
 Thus, alternately, as $AE$ is to $EB$, (so) $CF$ (is) to $FD$ [Prop. 5.16].
 And $AE$ and $EB$ [are] [commensurable in square]bookofproofs$2082 only.
 Thus, $CF$ and $FD$ are also commensurable in square only [Prop. 10.11].
 And they are rational.
 $CD$ is thus a binomial (straight line) [Prop. 10.36].
 So, I say that it is the same in order as $AB$.
 For the square on $AE$ is greater than (the square on) $EB$ by the (square) on (some straight line) either commensurable or 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 (some previously) laid down rational (straight line) then $CF$ will also be commensurable (in length) with it [Prop. 10.12].
 And, on account of this, $AB$ and $CD$ are each first binomial (straight lines) [Def. 10.5]  that is to say, the same in order.
 And if $EB$ is commensurable (in length) with the (previously) laid down rational (straight line) then $FD$ is also commensurable (in length) with it [Prop. 10.12], and, again, on account of this, ($CD$) will be the same in order as $AB$.
 For each of them will be second binomial (straight lines) [Def. 10.6] .
 And if neither of $AE$ and $EB$ is commensurable (in length) with the (previously) laid down rational (straight line) then neither of $CF$ and $FD$ will be commensurable (in length) with it [Prop. 10.13], and each (of $AB$ and $CD$) is a third (binomial straight line) [Def. 10.7] .
 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 $CF$ is also 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 the (previously) laid down rational (straight line) then $CF$ is also commensurable (in length) with it [Prop. 10.12], and each (of $AB$ and $CD$) is a fourth (binomial straight line) [Def. 10.8] .
 And if $EB$ (is commensurable in length with the previously laid down rational straight line) then $FD$ (is) also (commensurable in length with it), and each (of $AB$ and $CD$) will be a fifth (binomial straight line) [Def. 10.9] .
 And if neither of $AE$ and $EB$ (is commensurable in length with the previously laid down rational straight line) then also neither of $CF$ and $FD$ is commensurable (in length) with the laid down rational (straight line), and each (of $AB$ and $CD$) will be a sixth (binomial straight line) [Def. 10.10] .
 Hence, a (straight line) commensurable in length with a binomial (straight line) is a binomial (straight line), and the same in order.^{1}
 (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