Proof: By Euclid
(related to Proposition: Prop. 10.072: Sum of two Incommensurable Medial Areas give rise to two Irrational Straight Lines)
 For $AB$ is either greater than or less than $CD$.
 By chance, let $AB$, first of all, be greater than $CD$.
 And let the rational (straight line) $EF$ be laid down.
 And let $EG$, equal to $AB$, have been applied to $EF$, producing $EH$ as breadth, and $HI$, equal to $CD$, producing $HK$ as breadth.
 And since $AB$ and $CD$ are each medial, $EG$ and $HI$ (are) thus also each medial.
 And they are applied to the rational straight line $FE$, producing $EH$ and $HK$ (respectively) as breadth.
 Thus, $EH$ and $HK$ are each rational (straight lines which are) incommensurable in length with $EF$ [Prop. 10.22].
 And since $AB$ is incommensurable with $CD$, and $AB$ is equal to $EG$, and $CD$ to $HI$, $EG$ is thus also incommensurable with $HI$.
 And as $EG$ (is) to $HI$, so $EH$ is to $HK$ [Prop. 6.1].
 $EH$ is thus incommensurable in length with $HK$ [Prop. 10.11].
 Thus, $EH$ and $HK$ are rational (straight lines which are) commensurable in square only.
 $EK$ is thus a binomial (straight line) [Prop. 10.36].
 And the square on $EH$ is greater than (the square on) $HK$ either by the (square) on (some straight line) commensurable (in length) with ($EH$), or by the (square) on (some straight line) incommensurable (in length with $EH$).
 Let it, first of all, be greater by the square on (some straight line) commensurable in length with ($EH$).
 And neither of $EH$ or $HK$ is commensurable in length with the (previously) laid down rational (straight line) $EF$.
 Thus, $EK$ is a third binomial (straight line) [Def. 10.7] .
 And $EF$ (is) rational.
 And if an area is contained by a rational (straight line) and a third binomial (straight line) then the square root of the area is a second bimedial (straight line) [Prop. 10.56].
 Thus, the square root of $EI$  that is to say, of $AD$  is a second bimedial.
 And so, let the square on $EH$ be greater than (the square) on $HK$ by the (square) on (some straight line) incommensurable in length with ($EH$).
 And $EH$ and $HK$ are each incommensurable in length with $EF$.
 Thus, $EK$ is a sixth binomial (straight line) [Def. 10.10] .
 And if an area is contained by a rational (straight line) and a sixth binomial (straight line) then the square root of the area is the square root of (the sum of) two medial (areas) [Prop. 10.59].
 Hence, the square root of area $AD$ is also the square root of (the sum of) two medial (areas).
 So, similarly, we can show that, even if $AB$ is less than $CD$, the square root of area $AD$ is either a second bimedial or the square root of (the sum of) two medial (areas).
 Thus, when two medial (areas which are) incommensurable with one another are added together, the remaining two irrational (straight lines) arise (as the square roots of the total area)  either a second bimedial, or the square root of (the sum of) two medial (areas).
 A binomial (straight line), and the (other) irrational (straight lines) after it, are neither the same as a medial (straight line) nor (the same) as one another.
 For the (square) on a medial (straight line), applied to a rational (straight line), produces as breadth a rational (straight line which is) also incommensurable in length with (the straight line) to which it is applied [Prop. 10.22].
 And the (square) on a binomial (straight line), applied to a rational (straight line), produces as breadth a first binomial [Prop. 10.60].
 And the (square) on a first bimedial (straight line), applied to a rational (straight line), produces as breadth a second binomial [Prop. 10.61].
 And the (square) on a second bimedial (straight line), applied to a rational (straight line), produces as breadth a third binomial [Prop. 10.62].
 And the (square) on a major (straight line), applied to a rational (straight line), produces as breadth a fourth binomial [Prop. 10.63].
 And the (square) on the square root of a rational plus a medial (area) , applied to a rational (straight line), produces as breadth a fifth binomial [Prop. 10.64].
 And the (square) on the square root of (the sum of) two medial (areas), applied to a rational (straight line), produces as breadth a sixth binomial [Prop. 10.65].
 And the aforementioned breadths differ from the first (breadth), and from one another  from the first, because it is rational  and from one another, because they are not the same in order.^{1}
 Hence, the (previously mentioned) irrational (straight lines) themselves also differ from one another.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes