Proof: By Euclid
(related to Proposition: Prop. 8.21: Numbers between which exist two Mean Proportionals are Similar Solid)
 For let the two numbers $C$ and $D$ fall between the two numbers $A$ and $B$ in mean proportion.

I say that $A$ and $B$ are similar solid (numbers) .

For let the three least numbers $E$, $F$, $G$ having the same ratio as $A$, $C$, $D$ have been taken [Prop. 8.2].
 Thus, the outermost of them, $E$ and $G$, are prime to one another [Prop. 8.3].
 And since one number, $F$, has fallen (between) $E$ and $G$ in mean proportion, $E$ and $G$ are thus similar plane numbers [Prop. 8.20].
 Therefore, let $H$, $K$ be the sides of $E$, and $L$, $M$ (the sides) of $G$.
 Thus, it is clear from the (proposition) before this (one) that $E$, $F$, $G$ are in continued proportion in the ratio of $H$ to $L$, and of $K$ to $M$.
 And since $E$, $F$, $G$ are the least (numbers) having the same ratio as $A$, $C$, $D$, and the multitude of $E$, $F$, $G$ is equal to the multitude of $A$, $C$, $D$, thus, via equality, as $E$ is to $G$, so $A$ (is) to $D$ [Prop. 7.14].
 And $E$ and $G$ (are) prime (to one another), and prime (numbers) are also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser  that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20].
 Thus, $E$ measures $A$ the same number of times as $G$ (measures) $D$.
 So as many times as $E$ measures $A$, so many units let there be in $N$.
 Thus, $N$ has made $A$ (by) multiplying $E$ [Def. 7.15] .
 And $E$ is the (number created) from (multiplying) $H$ and $K$.
 Thus, $N$ has made $A$ (by) multiplying the (number created) from (multiplying) $H$ and $K$.
 Thus, $A$ is solid, and its sides are $H$, $K$, $N$.
 Again, since $E$, $F$, $G$ are the least (numbers) having the same ratio as $C$, $D$, $B$, thus $E$ measures $C$ the same number of times as $G$ (measures) $B$ [Prop. 7.20].
 So as many times as $E$ measures $C$, so many units let there be in $O$.
 Thus, $G$ measures $B$ according to the units in $O$.
 Thus, $O$ has made $B$ (by) multiplying $G$.
 And $G$ is the (number created) from (multiplying) $L$ and $M$.
 Thus, $O$ has made $B$ (by) multiplying the (number created) from (multiplying) $L$ and $M$.
 Thus, $B$ is solid, and its sides are $L$, $M$, $O$.
 Thus, $A$ and $B$ are (both) solid.
 So I say that (they are) also similar.
 For since $N$, $O$ have made $A$, $C$ (by) multiplying $E$, thus as $N$ is to $O$, so $A$ (is) to $C$  that is to say, $E$ to $F$ [Prop. 7.18].
 But, as $E$ (is) to $F$, so $H$ (is) to $L$, and $K$ to $M$.
 And thus as $H$ (is) to $L$, so $K$ (is) to $M$, and $N$ to $O$.
 And $H$, $K$, $N$ are the sides of $A$, and $L$, $M$, $O$ the sides of $B$.
 Thus, $A$ and $B$ are similar solid numbers [Def. 7.21] .
 (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"