Proof: By Euclid
(related to Proposition: Prop. 10.038: Second Bimedial is Irrational)
 For let the two medial (straight lines), $AB$ and $BC$, commensurable in square only, (and) containing a medial (area) , be laid down together [Prop. 10.28].

I say that $AC$ is irrational.

For let the rational (straight line) $DE$ be laid down, and let (the rectangle) $DF$, equal to the (square) on $AC$, have been applied to $DE$, making $DG$ as breadth [Prop. 1.44].
 And since the (square) on $AC$ is equal to (the sum of) the (squares) on $AB$ and $BC$, plus twice the (rectangle contained) by $AB$ and $BC$ [Prop. 2.4], so let (the rectangle) $EH$, equal to (the sum of) the squares on $AB$ and $BC$, have been applied to $DE$.
 The remainder $HF$ is thus equal to twice the (rectangle contained) by $AB$ and $BC$.
 And since $AB$ and $BC$ are each medial, (the sum of) the squares on $AB$ and $BC$ is thus also medial.^{1}
 And twice the (rectangle contained) by $AB$ and $BC$ was also assumed (to be) medial.
 And $EH$ is equal to (the sum of) the squares on $AB$ and $BC$, and $FH$ (is) equal to twice the (rectangle contained) by $AB$ and $BC$.
 Thus, $EH$ and $HF$ (are) each medial.
 And they were applied to the rational (straight line) $DE$.
 Thus, $DH$ and $HG$ are each rational, and incommensurable in length with $DE$ [Prop. 10.22].
 Therefore, since $AB$ is incommensurable in length with $BC$, and as $AB$ is to $BC$, so the (square) on $AB$ (is) to the (rectangle contained) by $AB$ and $BC$ [Prop. 10.21 lem.] , the (square) on $AB$ is thus incommensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.11].
 But, the sum of the squares on $AB$ and $BC$ is commensurable with the (square) on on $AB$ [Prop. 10.15], and twice the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.6].
 Thus, the sum of the (squares) on $AB$ and $BC$ is incommensurable with twice the (rectangle contained) by $AB$ and $BC$ [Prop. 10.13].
 But, $EH$ is equal to (the sum of) the squares on $AB$ and $BC$, and $HF$ is equal to twice the (rectangle) contained by $AB$ and $BC$.
 Thus, $EH$ is incommensurable with $HF$.
 Hence, $DH$ is also incommensurable in length with $HG$ [Prop. 6.1], [Prop. 10.11].
 Thus, $DH$ and $HG$ are rational (straight lines which are) commensurable in square only.
 Hence, $DG$ is irrational [Prop. 10.36].
 And $DE$ (is) rational.
 And the rectangle contained by irrational and rational (straight lines) is irrational [Prop. 10.20].
 The area $DF$ is thus irrational, and (so) the square root [of it] is irrational [Def. 10.4] .
 And $AC$ is the square root of $DF$.
 $AC$ is thus irrational  let it be called a second bimedial (straight line).
 (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