Multiples of Divisors obey Distributive Law/Proof 1

Theorem
In modern algebraic language:
 * $a = \dfrac m n b, c = \dfrac m n d \implies a + c = \dfrac m n \left({b + d}\right)$

Proof
Let the (natural) number $AB$ be parts of the (natural) number $C$.

Let the (natural) number $DE$ be the same parts of another (natural) number $F$ that $AB$ is of $C$.

We need to show that $AB + DE$ is the same parts of $C + F$.


 * Euclid-VII-6.png

We have that whatever parts $AB$ is of $C$, $DE$ is also the same parts as $F$.

It follows that as many parts of $C$ as there are in $AB$, so many parts of $F$ are there also in $DE$.

Let $AB$ be divided into the parts of $C$, namely $AG, GB$, and $DE$ into the parts of $F$, namely $DH, HE$.

Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.

We have that whatever part $AG$ is of $C$, the same part is $DH$ of $F$ also.

Therefore, from, whatever part $AG$ is of $C$, the same part also is $AG + DH$ of $C + F$ also.

For the same reason, whatever part $GB$ is of $C$, the same part also is $GB + HE$ of $C + F$.

Therefore whatever parts $AB$ is of $C$, the same parts also is $AB + DE$ of the $C + F$