Subtraction of Divisors obeys Distributive Law/Proof 1

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

Proof
Let $AB$ be that part of the (natural) number $CD$ which $AE$ subtracted is of $CF$ subtracted.

We need to show that the remainder $EB$ is also the same part of the number $CD$ which $AE$ subtracted is of $CF$ subtracted.


 * Euclid-VII-7.png

Whatever part $AE$ is of $CF$, let the same part $EB$ be of $CG$.

Then from, whatever part $AE$ is of $CF$, the same part also is $AB$ of $GF$.

But whatever part $AE$ is of $CF$, the same part also is $AB$ of $CD$, by hypothesis.

Therefore, whatever part $AB$ is of $GF$, the same pat is it of $CD$ also.

Therefore $GF = CD$.

Let $CF$ be subtracted from each.

Therefore $GC = FD$.

We have that whatever part $AE$ is of $CF$, the same part also is $EB$ of $CG$

Therefore whatever part $AE$ is of $CF$, the same part also is $EB$ of $FD$.

But whatever part $AE$ is of $CF$, the same part also is $AB$ of $CD$.

Therefore the remainder $EB$ is the same part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.