Divisors obey Distributive Law/Proof 1

Theorem
In modern algebraic language:
 * $\displaystyle a = \frac 1 n b, c = \frac 1 n d \implies a + c = \frac 1 n \left({b + d}\right)$

Proof
Let the (natural) number $A$ be a part of the (natural) number $BC$.

Let the (natural) number $D$ be the same part of another (natural) number $EF$ that $A$ is of $BC$.

We need to show that $A + D$ is the same part of $BC + EF$.


 * Euclid-VII-5.png

We have that whatever part $A$ is of $BC$, then $D$ is also the same part of $EF$.

Therefore as many numbers as there are in $BC$ equal to $A$, so many numbers are there in $EF$ equal to $D$.

Let $BC$ be divided into the numbers equal to $A$, that is, $BG, GC$.

Let $EF$ be divided into the numbers equal to $D$, that is, $EH, HF$.

Then the multitude of $BG, GC$ will be equal to the multitude of $EH, HF$.

Since $BG = A$ and $EH = D$, then $BG + EH = A + D$.

For the same reason, $GC + HF = A + D$.

Therefore, as many numbers as there are in $BC$ equal to $A$, so many are there also in $BC + EF$ equal to $A + D$.

Therefore, whatever multiple $BC$ is of $A$, the same multiple also is $BC + EF$ of $A + D$.

Therefore, whatever part $A$ is of $BC$, the same part also is $A + D$ of $BC + EF$.