Proof: By Euclid

For let the two medial (straight lines), $AB$ and $BC$, commensurable in square only, (and) containing a rational (area), be laid down together.
I say that the whole (straight line), $AC$, is irrational.

fig036e

For since $AB$ is incommensurable in length with $BC$, (the sum of) the (squares) on $AB$ and $BC$ is thus also incommensurable with twice the (rectangle contained) by $AB$ and $BC$ [see previous proposition].
And, via composition, (the sum of) the (squares) on $AB$ and $BC$, plus twice the (rectangle contained) by $AB$ and $BC$ - that is, the (square) on $AC$ [Prop. 2.4] - is incommensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.16].
And the (rectangle contained) by $AB$ and $BC$ (is) rational - for $AB$ and $BC$ were assumed to enclose a rational (area).
Thus, the (square) on $AC$ (is) irrational.
Thus, $AC$ (is) irrational [Def. 10.4] - let it be called a first bimedial (straight line).
(Which is) the very thing it was required to show.
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