(related to Proposition: Prop. 8.19: Between two Similar Solid Numbers exist two Mean Proportionals)

- Let $A$ and $B$ be two similar solid numbers, and let $C$, $D$, $E$ be the sides of $A$, and $F$, $G$, $H$ (the sides) of $B$.
- And since similar solid (numbers) are those having proportional sides [Def. 7.21] , thus as $C$ is to $D$, so $F$ (is) to $G$, and as $D$ (is) to $E$, so $G$ (is) to $H$.
- I say that two numbers fall (between) $A$ and $B$ in mean proportion, and (that) $A$ has to $B$ a cubed ratio with respect to (that) $C$ (has) to $F$, and $D$ to $G$, and, further, $E$ to $H$.

- For let $C$ make $K$ (by) multiplying $D$, and let $F$ make $L$ (by) multiplying $G$.
- And since $C$, $D$ are in the same ratio as $F$, $G$, and $K$ is the (number created) from (multiplying) $C$, $D$, and $L$ the (number created) from (multiplying) $F$, $G$, [thus] $K$ and $L$ are similar plane numbers [Def. 7.21] .
- Thus, there exits one number in mean proportion to $K$ and $L$ [Prop. 8.18].
- Let it be $M$.
- Thus, $M$ is the (number created) from (multiplying) $D$, $F$, as shown in the theorem before this (one).
- And since $D$ has made $K$ (by) multiplying $C$, and has made $M$ (by) multiplying $F$, thus as $C$ is to $F$, so $K$ (is) to $M$ [Prop. 7.17].
- But, as $K$ (is) to $M$, (so) $M$ (is) to $L$.
- Thus, $K$, $M$, $L$ are in continued proportion in the ratio of $C$ to $F$.
- And since as $C$ is to $D$, so $F$ (is) to $G$, thus, alternately, as $C$ is to $F$, so $D$ (is) to $G$ [Prop. 7.13].
- And so, for the same (reasons), as $D$ (is) to $G$, so $E$ (is) to $H$.
- Thus, $K$, $M$, $L$ are in continued proportion in the ratio of $C$ to $F$, and of $D$ to $G$, and, further, of $E$ to $H$.
- So let $E$, $H$ make $N$, $O$, respectively, (by) multiplying $M$.
- And since $A$ is solid, and $C$, $D$, $E$ are its sides, $E$ has thus made $A$ (by) multiplying the (number created) from (multiplying) $C$, $D$.
- And $K$ is the (number created) from (multiplying) $C$, $D$.
- Thus, $E$ has made $A$ (by) multiplying $K$.
- And so, for the same (reasons), $H$ has made $B$ (by) multiplying $L$.
- And since $E$ has made $A$ (by) multiplying $K$, but has, in fact, also made $N$ (by) multiplying $M$, thus as $K$ is to $M$, so $A$ (is) to $N$ [Prop. 7.17].
- And as $K$ (is) to $M$, so $C$ (is) to $F$, and $D$ to $G$, and, further, $E$ to $H$.
- And thus as $C$ (is) to $F$, and $D$ to $G$, and $E$ to $H$, so $A$ (is) to $N$.
- Again, since $E$, $H$ have made $N$, $O$, respectively, (by) multiplying $M$, thus as $E$ is to $H$, so $N$ (is) to $O$ [Prop. 7.18].
- But, as $E$ (is) to $H$, so $C$ (is) to $F$, and $D$ to $G$.
- And thus as $C$ (is) to $F$, and $D$ to $G$, and $E$ to $H$, so (is) $A$ to $N$, and $N$ to $O$.
- Again, since $H$ has made $O$ (by) multiplying $M$, but has, in fact, also made $B$ (by) multiplying $L$, thus as $M$ (is) to $L$, so $O$ (is) to $B$ [Prop. 7.17].
- But, as $M$ (is) to $L$, so $C$ (is) to $F$, and $D$ to $G$, and $E$ to $H$.
- And thus as $C$ (is) to $F$, and $D$ to $G$, and $E$ to $H$, so not only (is) $O$ to $B$, but also $A$ to $N$, and $N$ to $O$.
- Thus, $A$, $N$, $O$, $B$ are in continued proportion in the aforementioned ratios of the sides. So I say that $A$ also has to $B$ a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side - that is to say, with respect to (that) the number $C$ (has) to $F$, or $D$ to $G$, and, further, $E$ to $H$.
- For since $A$, $N$, $O$, $B$ are four numbers in continued proportion, $A$ thus has to $B$ a cubed ratio with respect to (that) $A$ (has) to $N$ [Def. 5.10] .
- But, as $A$ (is) to $N$, so it was shown (is) $C$ to $F$, and $D$ to $G$, and, further, $E$ to $H$.
- And thus $A$ has to $B$ a cubed ratio with respect to (that) a corresponding side (has) to a corresponding side - that is to say, with respect to (that) the number $C$ (has) to $F$, and $D$ to $G$, and, further, $E$ to $H$.
- (Which is) the very thing it was required to show.∎

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