Proof: By Euclid

Let $A$ and $B$ be two numbers (which are) prime to one another, and let the (numbers) $C$ and $D$ fall in between them in continued proportion.
And let the unit $E$ be set out.
I say that, as many numbers as have fallen in between $A$ and $B$ in continued proportion, so many (numbers) will also fall between each of $A$ and $B$ and the unit in continued proportion.
For let the least two numbers, $F$ and $G$, which are in the ratio of $A$, $C$, $D$, $B$, have been taken [Prop. 7.33].
And the (least) three (numbers), $H$, $K$, $L$.
And so on, successively increasing by one, until the multitude of the (least numbers taken) is made equal to the multitude of $A$, $C$, $D$, $B$ [Prop. 8.2].
Let them have been taken, and let them be $M$, $N$, $O$, $P$.
So (it is) clear that $F$ has made $H$ (by) multiplying itself, and has made $M$ (by) multiplying $H$.
And $G$ has made $L$ (by) multiplying itself, and has made $P$ (by) multiplying $L$ [Prop. 8.2 corr.] .
And since $M$, $N$, $O$, $P$ are the least of those (numbers) having the same ratio as $F$, $G$, and $A$, $C$, $D$, $B$ are also the least of those (numbers) having the same ratio as $F$, $G$ [Prop. 8.2], and the multitude of $M$, $N$, $O$, $P$ is equal to the multitude of $A$, $C$, $D$, $B$, thus $M$, $N$, $O$, $P$ are equal to $A$, $C$, $D$, $B$, respectively.
Thus, $M$ is equal to $A$, and $P$ to $B$.
And since $F$ has made $H$ (by) multiplying itself, $F$ thus measures $H$ according to the units in $F$ [Def. 7.15] .
And the unit $E$ also measures $F$ according to the units in it.
Thus, the unit $E$ measures the number $F$ as many times as $F$ (measures) $H$.
Thus, as the unit $E$ is to the number $F$, so $F$ (is) to $H$ [Def. 7.20] .
Again, since $F$ has made $M$ (by) multiplying $H$, $H$ thus measures $M$ according to the units in $F$ [Def. 7.15] .
And the unit $E$ also measures the number $F$ according to the units in it.
Thus, the unit $E$ measures the number $F$ as many times as $H$ (measures) $M$.
Thus, as the unit $E$ is to the number $F$, so $H$ (is) to $M$ [Prop. 7.20].
And it was shown that as the unit $E$ (is) to the number $F$, so $F$ (is) to $H$.
And thus as the unit $E$ (is) to the number $F$, so $F$ (is) to $H$, and $H$ (is) to $M$.
And $M$ (is) equal to $A$.
Thus, as the unit $E$ is to the number $F$, so $F$ (is) to $H$, and $H$ to $A$.
And so, for the same (reasons), as the unit $E$ (is) to the number $G$, so $G$ (is) to $L$, and $L$ to $B$.
Thus, as many (numbers) as have fallen in between $A$ and $B$ in continued proportion, so many numbers have also fallen between each of $A$ and $B$ and the unit $E$ in continued proportion.
(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"

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Proof: By Euclid

References

Adapted from (subject to copyright, with kind permission)