Multiplication of Numbers is Right Distributive over Addition

Theorem

 * If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.

That is:
 * $ma + na + pa + \cdots = \left({m + n + p + \cdots }\right) a$

Proof
Let a first magnitude, $AB$, be the same multiple of a second, $C$, that a third, $DE$, is of a fourth, $F$.

Let a fifth, $BG$, be the same multiple of $C$ that a sixth, $EH$, is of $F$.
 * Euclid-V-2.png

We need to show that $AG = AB + BG$ is the same multiple of $C$ that $DH = DE + EH$ is of $F$.

We have that $AB$ is the same multiple of $C$ that $DE$ is of $F$.

It follows that as many magnitudes as there are in $AB$ equal to $C$, so many also are there in $DE$ equal to $F$.

For the same reason, as many as there are in $BG$ equal to $C$, so many also are there in $EH$ equal to $F$.

So as many as there are in the whole $AG$ equal to $C$, so many also are there in the whole $DH$ equal to $F$.

Therefore the sum of the first and fifth, $AG$, is the same multiple of the second, $C$, that the sum of the third and sixth, $DH$ is of the fourth, $F$.

Also see

 * Real Multiplication Distributes over Addition
 * Multiplication of Numbers is Left Distributive over Addition