Proof: By Euclid

For let the two straight lines, $AB$ and $BC$, incommensurable in square, (and) fulfilling the prescribed (conditions), be laid down together [Prop. 10.35].
I say that $AC$ is irrational.

fig041e

Let the rational (straight line) $DE$ be laid out, and let (the rectangle) $DF$, equal to (the sum of) the (squares) on $AB$ and $BC$, and (the rectangle) $GH$, equal to twice the (rectangle contained) by $AB$ and $BC$, have been applied to $DE$.
Thus, the whole of $DH$ is equal to the square on $AC$ [Prop. 2.4].
And since the sum of the (squares) on $AB$ and $BC$ is medial, and is equal to $DF$, $DF$ is thus also medial.
And it is applied to the rational (straight line) $DE$.
Thus, $DG$ is rational, and incommensurable in length with $DE$ [Prop. 10.22].
So, for the same (reasons), $GK$ is also rational, and incommensurable in length with $GF$ - that is to say, $DE$.
And since (the sum of) the (squares) on $AB$ and $BC$ is incommensurable with twice the (rectangle contained) by $AB$ and $BC$, $DF$ is incommensurable with $GH$.
Hence, $DG$ is also incommensurable (in length) with $GK$ [Prop. 6.1], [Prop. 10.11].
And they are rational.
Thus, $DG$ and $GK$ are rational (straight lines which are) commensurable in square only.
Thus, $DK$ is irrational, and that (straight line which is) called binomial [Prop. 10.36].
And $DE$ (is) rational.
Thus, $DH$ is irrational, and its square root is irrational [Def. 10.4] .
And $AC$ (is) the square root of $HD$.
Thus, $AC$ is irrational - let it be called the square root of (the sum of) two medial (areas).
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!