Proof: By Euclid

Let $AB$ be the square root of a rational plus a medial (area) , and let $CD$ be commensurable (in length) with $AB$.
We must show that $CD$ is also the square root of a rational plus a medial (area) .

fig066e

Let $AB$ have been divided into its (component) straight lines at $E$.
$AE$ and $EB$ are thus incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational [Prop. 10.40].
And let the same construction have been made as in the previous (propositions).
So, similarly, we can show that $CF$ and $FD$ are also incommensurable in square, and that the sum of the (squares) on $AE$ and $EB$ (is) commensurable with the sum of the (squares) on $CF$ and $FD$, and the (rectangle contained) by $AE$ and $EB$ with the (rectangle contained) by $CF$ and $FD$.
And hence the sum of the squares on $CF$ and $FD$ is medial, and the (rectangle contained) by $CF$ and $FD$ (is) rational.
Thus, $CD$ is the square root of a rational plus a medial (area) [Prop. 10.40].
(Which is) the very thing it was required to show.
∎

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