Multiplication of Numbers is Right Distributive over Addition

Theorem
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