Proof: By Euclid

Let $AB$ be a (straight line) which with a medial (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 medial (area) makes a medial whole.

fig103e

For let $BE$ be an attachment to $AB$.
And let the same construction have been made (as in the previous propositions).
Thus, $AE$ and $EB$ are (straight lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, further, the sum of the squares on them incommensurable with the (rectangle contained) by them [Prop. 10.78].
And, as was shown (previously), $AE$ and $EB$ are commensurable (in length) with $CF$ and $FD$ (respectively), and the sum of the squares on $AE$ and $EB$ 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 $FD$.
Thus, $CF$ and $FD$ are also (straight lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them medial, and, further, the sum of the [squares] on them incommensurable with the (rectangle contained) by them.
Thus, $CD$ is a (straight line) which with a medial (area) makes a medial whole [Prop. 10.78].
(Which is) the very thing it was required to show.
∎

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