# Proof: By Euclid

• 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.

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### References

#### Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"