Multiples of Terms in Equal Ratios/Euclid's Proof

Proof
Let a first magnitude $A$ have to a second magnitude $B$ the same ratio as a third $C$ to a fourth $D$.

Let equimultiples $E, F$ be taken of $A, C$, and let different equimultiples $G, H$ be taken of $B, D$.

We need to show that $E : G = F : H$.


 * Euclid-V-4.png

Let equimultiples $K, L$ be taken of $E, F$ and other arbitrary equimultiples $M, N$ be taken of $G, H$.

We have that $E$ is the same multiple of $A$ that $L$ is of $C$.

So from Proposition $3$: Associative Law of Multiplication, $K$ is the same multiple of $A$ that $L$ is of $C$.

For the same reason, $M$ is the same multiple of $B$ that $N$ is of $D$.

We have that:
 * $A$ is to $B$ as $C$ is to $D$
 * of $A, C$ equimultiples $K, L$ have been taken
 * of $B, D$ other equimultiples $M, N$ have been taken.

So from the definition of equality of ratios:
 * if $K$ is in excess of $M$, $L$ is also in excess of $N$
 * if $K$ is equal to $M$, $L$ is equal to $N$
 * if $K$ is less than $M$, $L$ is less than $N$

But $K, L$ are equimultiples of $E, F$.

Therefore as $E$ is to $G$, so is $F$ to $H$.