# Proof: By Euclid

• For let equal multiples $K$ and $L$ have been taken of $E$ and $F$ (respectively), and other random equal multiples $M$ and $N$ of $G$ and $H$ (respectively).
• And since $E$ and $F$ are equal multiples of $A$ and $C$ (respectively), and the equal multiples $K$ and $L$ have been taken of $E$ and $F$ (respectively), $K$ and $L$ are thus equal multiples of $A$ and $C$ (respectively) [Prop. 5.3].
• So, for the same (reasons), $M$ and $N$ are equal multiples of $B$ and $D$ (respectively).
• And since as $A$ is to $B$, so $C$ (is) to $D$, and the equal multiples $K$ and $L$ have been taken of $A$ and $C$ (respectively), and the other random equal multiples $M$ and $N$ of $B$ and $D$ (respectively), then if $K$ exceeds $M$ then $L$ also exceeds $N$, and if ($K$ is) equal (to $M$ then $L$ is also) equal (to $N$), and if ($K$ is) less (than $M$ then $L$ is also) less (than $N$) [Def. 5.5] .
• And $K$ and $L$ are equal multiples of $E$ and $F$ (respectively), and $M$ and $N$ other random equal multiples of $G$ and $H$ (respectively).
• Thus, as $E$ (is) to $G$, so $F$ (is) to $H$ [Def. 5.5] .
• Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth then equal multiples of the first (magnitude) and the third will also have the same ratio to equal multiples of the second and the fourth, being taken in corresponding order, according to any kind of multiplication whatsoever.
• (Which is) the very thing it was required to show.

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