Proof: By Euclid
(related to Proposition: Prop. 10.027: Construction of Components of First Bimedial)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes