# Proof: By Euclid

• For since $A$ is greater than $C$, and $B$ (is) another random [magnitude], $A$ thus has a greater ratio to $B$ than $C$ (has) to $B$ [Prop. 5.8].
• And as $A$ (is) to $B$, so $C$ (is) to $D$.
• Thus, $C$ also has a greater ratio to $D$ than $C$ (has) to $B$.
• And that (magnitude) to which the same (magnitude) has a greater ratio is the lesser [Prop. 5.10].
• Thus, $D$ (is) less than $B$.
• Hence, $B$ is greater than $D$.
• So, similarly, we can show that even if $A$ is equal to $C$ then $B$ will also be equal to $D$, and even if $A$ is less than $C$ then $B$ will also be less than $D$.
• Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and the first (magnitude) is greater than the third, then the second will also be greater than the fourth.
• And if (the first magnitude is) equal (to the third then the second will also be) equal (to the fourth).
• And if (the first magnitude is) less (than the third then the second will also be) less (than the fourth).
• (Which is) the very thing it was required to show.

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