Proof: By Euclid

Let the two rational (straight lines) $A$ and $B$, (which are) commensurable in square only, be laid down.
And let $C$ - in mean proportion3 (straight line) to $A$ and $B$ - have been taken [Prop. 6.13].
And let it be contrived that as $A$ (is) to $B$, so $C$ (is) to $D$ [Prop. 6.12].
And since the rational (straight lines) $A$ and $B$ are commensurable in square only, the (rectangle contained) by $A$ and $B$ - that is to say, the (square) on $C$ [Prop. 6.17] - is thus medial [Prop. 10.21].

fig027e

Thus, $C$ is medial [Prop. 10.21].
And since as $A$ is to $B$, [so] $C$ (is) to $D$, and $A$ and $B$ [are] [commensurable in square]bookofproofs$2082 only, $C$ and $D$ are thus also commensurable in square only [Prop. 10.11].
And $C$ is medial.
Thus, $D$ is also medial [Prop. 10.23].
Thus, $C$ and $D$ are medial (straight lines which are) commensurable in square only.
I say that they also contain a rational (area).
For since as $A$ is to $B$, so $C$ (is) to $D$, thus, alternately, as $A$ is to $C$, so $B$ (is) to $D$ [Prop. 5.16].
But, as $A$ (is) to $C$, (so) $C$ (is) to $B$.
And thus as $C$ (is) to $B$, so $B$ (is) to $D$ [Prop. 5.11].
Thus, the (rectangle contained) by $C$ and $D$ is equal to the (square) on $B$ [Prop. 6.17].
And the (square) on $B$ (is) rational.
Thus, the (rectangle contained) by $C$ and $D$ [is] also rational.
Thus, (two) medial (straight lines, $C$ and $D$), containing a rational (area), (which are) commensurable in square only, have been found.¹
(Which is) the very thing it was required to show.

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$C$ and $D$ have lengths $\rho^{1/4}$ and $\rho^{3/4}$ times that of $A$, respectively, where the length of $B$ is $\rho^{1/2}$ times that of $A$, for some rational number $\rho$ (translator's note) ↩