Proof: By Euclid
(related to Proposition: Prop. 10.071: Sum of Rational Area and Medial Area gives rise to four Irrational Straight Lines)
 Let $AB$ be a rational (area), and $CD$ a medial (area) .

I say that the square root of area $AD$ is either binomial, or first bimedial, or major, or the square root of a rational plus a medial (area) .

For $AB$ is either greater or less than $CD$.
 Let it, first of all, be greater.
 And let the rational (straight line) $EF$ be laid down.
 And let (the rectangle) $EG$, equal to $AB$, have been applied to $EF$, producing $EH$ as breadth.
 And let (the recatangle) $HI$, equal to $DC$, have been applied to $EF$, producing $HK$ as breadth.
 And since $AB$ is rational, and is equal to $EG$, $EG$ is thus also rational.
 And it has been applied to the [rational] (straight line) $EF$, producing $EH$ as breadth.
 $EH$ is thus rational, and commensurable in length with $EF$ [Prop. 10.20].
 Again, since $CD$ is medial, and is equal to $HI$, $HI$ is thus also medial.
 And it is applied to the rational (straight line) $EF$, producing $HK$ as breadth.
 $HK$ is thus rational, and incommensurable in length with $EF$ [Prop. 10.22].
 And since $CD$ is medial, and $AB$ rational, $AB$ is thus incommensurable with $CD$.
 Hence, $EG$ is also incommensurable with $HI$.
 And as $EG$ (is) to $HI$, so $EH$ is to $HK$ [Prop. 6.1].
 Thus, $EH$ is also incommensurable in length with $HK$ [Prop. 10.11].
 And they are both rational.
 Thus, $EH$ and $HK$ are rational (straight lines which are) commensurable in square only.
 $EK$ is thus a binomial (straight line), having been divided (into its component terms) at $H$ [Prop. 10.36].
 And since $AB$ is greater than $CD$, and $AB$ (is) equal to $EG$, and $CD$ to $HI$, $EG$ (is) thus also greater than $HI$.
 Thus, $EH$ is also greater than $HK$ [Prop. 5.14].
 Therefore, 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 the greater (of the two components of $EK$) $HE$ is commensurable (in length) with the (previously) laid down (straight line) $EF$.
 $EK$ is thus a first binomial (straight line) [Def. 10.5] .
 And $EF$ (is) rational.
 And if an area is contained by a rational (straight line) and a first binomial (straight line) then the square root of the area is a binomial (straight line) [Prop. 10.54].
 Thus, the square root of $EI$ is a binomial (straight line).
 Hence the square root of $AD$ is also a binomial (straight line).
 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 the greater (of the two components of $EK$) $EH$ is commensurable in length with the (previously) laid down rational (straight line) $EF$.
 Thus, $EK$ is a fourth binomial (straight line) [Def. 10.8] .
 And $EF$ (is) rational.
 And if an area is contained by a rational (straight line) and a fourth binomial (straight line) then the square root of the area is the irrational (straight line) called major [Prop. 10.57].
 Thus, the square root of area $EI$ is a major (straight line).
 Hence, the square root of $AD$ is also major.
 And so, let $AB$ be less than $CD$.
 Thus, $EG$ is also less than $HI$.
 Hence, $EH$ is also less than $HK$ [Prop. 6.1], [Prop. 5.14].
 And the square on $HK$ is greater than (the square on) $EH$ either by the (square) on (some straight line) commensurable (in length) with ($HK$), or by the (square) on (some straight line) incommensurable (in length) with ($HK$).
 Let it, first of all, be greater by the square on (some straight line) commensurable in length with ($HK$).
 And the lesser (of the two components of $EK$) $EH$ is commensurable in length with the (previously) laid down rational (straight line) $EF$.
 Thus, $EK$ is a second binomial (straight line) [Def. 10.6] .
 And $EF$ (is) rational.
 And if an area is contained by a rational (straight line) and a second binomial (straight line) then the square root of the area is a first bimedial (straight line) [Prop. 10.55].
 Thus, the square root of area $EI$ is a first bimedial (straight line).
 Hence, the square root of $AD$ is also a first bimedial (straight line).
 And so, let the square on $HK$ be greater than (the square on) $HE$ by the (square) on (some straight line) incommensurable (in length) with ($HK$).
 And the lesser (of the two components of $EK$) $EH$ is commensurable (in length) with the (previously) laid down rational (straight line) $EF$.
 Thus, $EK$ is a fifth binomial (straight line) [Def. 10.9] .
 And $EF$ (is) rational.
 And if an area is contained by a rational (straight line) and a fifth binomial (straight line) then the square root of the area is the square root of a rational plus a medial (area) [Prop. 10.58].
 Thus, the square root of area $EI$ is the square root of a rational plus a medial (area) .
 Hence, the square root of area $AD$ is also the square root of a rational plus a medial (area) .
 Thus, when a rational and a medial area are added together, four irrational (straight lines) arise (as the square roots of the total area)  either a binomial, or a first bimedial, or a major, or the square root of a rational plus a medial (area) .
 (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"