Proof: By Euclid
(related to Proposition: Prop. 10.026: Medial Area not greater than Medial Area by Rational Area)
 And let the rational (straight line) $EF$ be laid down.
 And let the rectangular parallelogram $FH$, equal to $AB$, have been applied to to $EF$, producing $EH$ as breadth.
 And let $FG$, equal to $AC$, have been cut off (from $FH$).
 Thus, the remainder $BD$ is equal to the remainder $KH$.
 And $DB$ is rational.
 Thus, $KH$ is also rational.
 Therefore, since $AB$ and $AC$ are each medial, and $AB$ is equal to $FH$, and $AC$ to $FG$, $FH$ and $FG$ are thus each also medial.
 And they are applied to the rational (straight line) $EF$.
 Thus, $HE$ and $EG$ are each rational, and incommensurable in length with $EF$ [Prop. 10.22].
 And since $DB$ is rational, and is equal to $KH$, $KH$ is thus also rational.
 And ($KH$) is applied to the rational (straight line) $EF$.
 $GH$ is thus rational, and commensurable in length with $EF$ [Prop. 10.20].
 But, $EG$ is also rational, and incommensurable in length with $EF$.
 Thus, $EG$ is incommensurable in length with $GH$ [Prop. 10.13].
 And as $EG$ is to $GH$, so the (square) on $EG$ (is) to the (rectangle contained) by $EG$ and $GH$ [Prop. 10.13 lem.] .
 Thus, the (square) on $EG$ is incommensurable with the (rectangle contained) by $EG$ and $GH$ [Prop. 10.11].
 But, the (sum of the) squares on $EG$ and $GH$ is commensurable with the (square) on on $EG$.
 For ($EG$ and $GH$ are) both rational.
 And twice the (rectangle contained) by $EG$ and $GH$ is commensurable with the (rectangle contained) by $EG$ and $GH$ [Prop. 10.6].
 For (the former) is double the latter.
 Thus, the (sum of the squares) on $EG$ and $GH$ is incommensurable with twice the (rectangle contained) by $EG$ and $GH$ [Prop. 10.13].
 And thus the sum of the (squares) on $EG$ and $GH$ plus twice the (rectangle contained) by $EG$ and $GH$, that is the (square) on $EH$ [Prop. 2.4], is incommensurable with the (sum of the squares) on $EG$ and $GH$ [Prop. 10.16].
 And the (sum of the squares) on $EG$ and $GH$ (is) rational.
 Thus, the (square) on $EH$ is irrational [Def. 10.4] .
 Thus, $EH$ is irrational [Def. 10.4] .
 But, (it is) also rational.
 The very thing is impossible.
 Thus, a medial (area) does not exceed a medial (area) by a rational (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"