# Proof: By Euclid

• Let the equal multiples $G$, $H$, and $K$ have been taken of $A$, $B$, and $D$ (respectively), and the other random equal multiples $L$, $M$, and $N$ of $C$, $E$, and $F$ (respectively).
• And since $G$ and $H$ are equal multiples of $A$ and $B$ (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as $A$ (is) to $B$, so $G$ (is) to $H$.
• And, so, for the same (reasons), as $E$ (is) to $F$, so $M$ (is) to $N$.
• And as $A$ is to $B$, so $E$ (is) to $F$.
• And, thus, as $G$ (is) to $H$, so $M$ (is) to $N$ [Prop. 5.11].
• And since as $B$ is to $C$, so $D$ (is) to $E$, also, alternately, as $B$ (is) to $D$, so $C$ (is) to $E$ [Prop. 5.16].
• And since $H$ and $K$ are equal multiples of $B$ and $D$ (respectively), and parts have the same ratio as similar multiples [Prop. 5.15], thus as $B$ is to $D$, so $H$ (is) to $K$.
• But, as $B$ (is) to $D$, so $C$ (is) to $E$.
• And, thus, as $H$ (is) to $K$, so $C$ (is) to $E$ [Prop. 5.11].
• Again, since $L$ and $M$ are equal multiples of $C$ and $E$ (respectively), thus as $C$ is to $E$, so $L$ (is) to $M$ [Prop. 5.15].
• But, as $C$ (is) to $E$, so $H$ (is) to $K$.
• And, thus, as $H$ (is) to $K$, so $L$ (is) to $M$ [Prop. 5.11].
• Also, alternately, as $H$ (is) to $L$, so $K$ (is) to $M$ [Prop. 5.16].
• And it was also shown (that) as $G$ (is) to $H$, so $M$ (is) to $N$.
• Therefore, since $G$, $H$, and $L$ are three magnitudes, and $K$, $M$, and $N$ other (magnitudes) of equal number to them, (being) in the same ratio taken two by two, and their proportion is perturbed, thus, via equality, if $G$ exceeds $L$ then $K$ also exceeds $N$, and if ($G$ is) equal (to $L$ then $K$ is also) equal (to $N$), and if ($G$ is) less (than $L$ then $K$ is also) less (than $N$) [Prop. 5.21].
• And $G$ and $K$ are equal multiples of $A$ and $D$ (respectively), and $L$ and $N$ of $C$ and $F$ (respectively).
• Thus, as $A$ (is) to $C$, so $D$ (is) to $F$ [Def. 5.5] .
• Thus, if there are three magnitudes, and others of equal number to them, (being) in the same ratio taken two by two, and (if) their proportion is perturbed, then they will also be in the same ratio via equality.
• (Which is) the very thing it was required to show.

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