Proof: By Euclid

Let $AB$ be a (straight line) which with a rational (area) makes a medial whole, and (let) $CD$ (be) commensurable (in length) with $AB$.
I say that $CD$ is also a (straight line) which with a rational (area) makes a medial (whole).

fig103e

For let $BE$ be an attachment to $AB$.
Thus, $AE$ and $EB$ are (straight lines which are) incommensurable in square, making the sum of the squares on $AE$ and $EB$ medial, and the (rectangle contained) by them rational [Prop. 10.77].
And let the same construction have been made (as in the previous propositions).
So, similarly to the previous (propositions), we can show that $CF$ and $FD$ are in the same ratio as $AE$ and $EB$, and 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, $CF$ and $FD$ are also (straight lines which are) incommensurable in square, making the sum of the squares on $CF$ and $FD$ medial, and the (rectangle contained) by them rational.
$CD$ is thus a (straight line) which with a rational (area) makes a medial whole [Prop. 10.77].
(Which is) the very thing it was required to show.
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