(related to Proposition: Prop. 8.25: If Ratio of Cube to Number is as between Two Cubes then Number is Cube)

- For let two numbers, $A$ and $B$, have to one another the ratio which the cube number $C$ (has) to the cube number $D$.
- And let $A$ be cube.
- So I say that $B$ is also cube.

- For since $C$ and $D$ are cube (numbers), $C$ and $D$ are (thus) similar solid (numbers) .
- Thus, two numbers fall (between) $C$ and $D$ in mean proportion [Prop. 8.19].
- And as many (numbers) as fall in between $C$ and $D$ in continued proportion, so many also (fall) in (between) those (numbers) having the same ratio as them (in continued proportion) [Prop. 8.8].
- And hence two numbers fall (between) $A$ and $B$ in mean proportion.
- Let $E$ and $F$ (so) fall.
- Therefore, since the four numbers $A$, $E$, $F$, $B$ are in continued proportion, and $A$ is cube, $B$ (is) thus also cube [Prop. 8.23].
- (Which is) the very thing it was required to show.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"