# Proof: By Euclid

• For let one number $C$ fall between the two numbers $A$ and $B$ in mean proportion.
• I say that $A$ and $B$ are similar plane numbers. • For let the least numbers, $D$ and $E$, having the same ratio as $A$ and $C$ have been taken [Prop. 7.33].

• Thus, $D$ measures $A$ as many times as $E$ (measures) $C$ [Prop. 7.20].
• So as many times as $D$ measures $A$, so many units let there be in $F$.
• Thus, $F$ has made $A$ (by) multiplying $D$ [Def. 7.15] .
• Hence, $A$ is plane, and $D$, $F$ (are) its sides.
• Again, since $D$ and $E$ are the least of those (numbers) having the same ratio as $C$ and $B$, $D$ thus measures $C$ as many times as $E$ (measures) $B$ [Prop. 7.20].
• So as many times as $E$ measures $B$, so many units let there be in $G$.
• Thus, $E$ measures $B$ according to the units in $G$.
• Thus, $G$ has made $B$ (by) multiplying $E$ [Def. 7.15] .
• Thus, $B$ is plane, and $E$, $G$ are its sides.
• Thus, $A$ and $B$ are (both) plane numbers.
• So I say that (they are) also similar.
• For since $F$ has made $A$ (by) multiplying $D$, and has made $C$ (by) multiplying $E$, thus as $D$ is to $E$, so $A$ (is) to $C$ - that is to say, $C$ to $B$1 [Prop. 7.17].
• Again, since $E$ has made $C$, $B$ (by) multiplying $F$, $G$, respectively, thus as $F$ is to $G$, so $C$ (is) to $B$ [Prop. 7.17].
• And as $C$ (is) to $B$, so $D$ (is) to $E$.
• And thus as $D$ (is) to $E$, so $F$ (is) to $G$.
• And, alternately, as $D$ (is) to $F$, so $E$ (is) to $G$ [Prop. 7.13].
• Thus, $A$ and $B$ are similar plane numbers.
• For their sides are proportional [Def. 7.21] .
• (Which is) the very thing it was required to show.

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### References

1. This part of the proof is defective, since it is not demonstrated that $F\times E = C$. Furthermore, it is not necessary to show that $D/E=A/C$, because this is true by hypothesis (translator's note).