# Proof: By Euclid

• For as many as $A$, $B$, $C$, $D$ are in multitude, let so many of the least numbers, $G$, $H$, $K$, $L$, having the same ratio as $A$, $B$, $C$, $D$, have been taken [Prop. 7.33].
• Thus, the outermost of them, $G$ and $L$, are prime to one another [Prop. 8.3].
• And since $A$, $B$, $C$, $D$ are in the same ratio as $G$, $H$, $K$, $L$, and the multitude of $A$, $B$, $C$, $D$ is equal to the multitude of $G$, $H$, $K$, $L$, thus, via equality, as $A$ is to $B$, so $G$ (is) to $L$ [Prop. 7.14].
• And as $A$ (is) to $B$, so $E$ (is) to $F$.
• And thus as $G$ (is) to $L$, so $E$ (is) to $F$.
• And $G$ and $L$ (are) prime (to one another).
• And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21].
• And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser - that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20].
• Thus, $G$ measures $E$ the same number of times as $L$ (measures) $F$.
• So as many times as $G$ measures $E$, so many times let $H$, $K$ also measure $M$, $N$, respectively.
• Thus, $G$, $H$, $K$, $L$ measure $E$, $M$, $N$, $F$ (respectively) an equal number of times.
• Thus, $G$, $H$, $K$, $L$ are in the same ratio as $E$, $M$, $N$, $F$ [Def. 7.20] .
• But, $G$, $H$, $K$, $L$ are in the same ratio as $A$, $C$, $D$, $B$.
• Thus, $A$, $C$, $D$, $B$ are also in the same ratio as $E$, $M$, $N$, $F$.
• And $A$, $C$, $D$, $B$ are in continued proportion.
• Thus, $E$, $M$, $N$, $F$ are also in continued proportion.
• Thus, as many numbers as have fallen in between $A$ and $B$ in continued proportion, so many numbers have also fallen in between $E$ and $F$ in continued proportion.
• (Which is) the very thing it was required to show.

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:
@Fitzpatrick