Proof: By Euclid
(related to Proposition: Prop. 10.032: Constructing Medial Commensurable in Square II)
 Let three rational (straight lines), $A$, $B$ and $C$, commensurable in square only, be laid out such that the square on $A$ is greater than (the square on $C$) by the (square) on (some straight line) commensurable (in length) with ($A$) [Prop. 10.29].
 And let the (square) on $D$ be equal to the (rectangle contained) by $A$ and $B$.
 Thus, the (square) on $D$ (is) medial.
 Thus, $D$ is also medial [Prop. 10.21].
 And let the (rectangle contained) by $D$ and $E$ be equal to the (rectangle contained) by $B$ and $C$.
 And since as the (rectangle contained) by $A$ and $B$ is to the (rectangle contained) by $B$ and $C$, so $A$ (is) to $C$ [Prop. 10.21 lem.] , but the (square) on $D$ is equal to the (rectangle contained) by $A$ and $B$, and the (rectangle contained) by $D$ and $E$ to the (rectangle contained) by $B$ and $C$, thus as $A$ is to $C$, so the (square) on $D$ (is) to the (rectangle contained) by $D$ and $E$.
 And as the (square) on $D$ (is) to the (rectangle contained) by $D$ and $E$, so $D$ (is) to $E$ [Prop. 10.21 lem.] .
 And thus as $A$ (is) to $C$, so $D$ (is) to $E$.
 And $A$ (is) commensurable in square [only] with $C$.
 Thus, $D$ (is) also commensurable in square only with $E$ [Prop. 10.11].
 And $D$ (is) medial.
 Thus, $E$ (is) also medial [Prop. 10.23].
 And since as $A$ is to $C$, (so) $D$ (is) to $E$, and the square on $A$ is greater than (the square on) $C$ by the (square) on (some straight line) commensurable (in length) with ($A$), the square on $D$ will thus also be greater than (the square on) $E$ by the (square) on (some straight line) commensurable (in length) with ($D$) [Prop. 10.14].
 So, I also say that the (rectangle contained) by $D$ and $E$ is medial.
 For since the (rectangle contained) by $B$ and $C$ is equal to the (rectangle contained) by $D$ and $E$, and the (rectangle contained) by $B$ and $C$ (is) medial [for $B$ and $C$ are [rational][bookofproofs$2083] ([straight lines][bookofproofs$645] which are) [commensurable in square][bookofproofs$2082] only] [[Prop. 10.21]]bookofproofs$2115, the (rectangle contained) by $D$ and $E$ (is) thus also medial.
 Thus, two medial (straight lines), $D$ and $E$, commensurable in square only, (and) containing a medial (area) , have been found such that the square on the greater is larger than the (square on the) lesser by the (square) on (some straight line) commensurable (in length) with the greater.^{1}
 So, similarly, (the proposition) can again also be demonstrated for (some straight line) incommensurable (in length with the greater), provided that the square on $A$ is greater than (the square on) $C$ by the (square) on (some straight line) incommensurable (in length) with ($A$)^{2} [Prop. 10.30].
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes