Proof: By Euclid

Let $AB$ be the square root of (the sum of) two medial (areas), and (let) $CD$ (be) commensurable (in length) with $AB$.
We must show that $CD$ is also the square root of (the sum of) two medial (areas).

fig066e

For since $AB$ is the square root of (the sum of) two medial (areas), let it have been divided into its (component) straight lines at $E$.
Thus, $AE$ and $EB$ are incommensurable in square, making the sum of the [squares] on them medial, and the (rectangle contained) by them medial, and, moreover, the sum of the (squares) on $AE$ and $EB$ incommensurable with the (rectangle) contained by $AE$ and $EB$ [Prop. 10.41].
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 $FD$.
Hence, the sum of the squares on $CF$ and $FD$ is also medial, and the (rectangle contained) by $CF$ and $FD$ (is) medial, and, moreover, the sum of the squares on $CF$ and $FD$ (is) incommensurable with the (rectangle contained) by $CF$ and $FD$.
Thus, $CD$ is the square root of (the sum of) two medial (areas) [Prop. 10.41].
(Which is) the very thing it was required to show.
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