Subtraction of Multiples of Divisors obeys 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 the same aliquant part of the (natural) number $CD$ that $AE$ subtracted is of $CF$ subtracted.

We need to show that $EB$ is also the same aliquant part of the remainder $FD$ that the whole $AB$ is of the whole $CD$.


 * Euclid-VII-8.png

Let $GH = AB$.

Then whatever aliquant part $GH$ is of $CD$, the same aliquant part also is $AE$ of $CF$.

Let $GH$ be divided into the aliquant parts of $CD$, namely $GK + KH$, and $AE$ into the aliquant parts of $CF$, namely $AL + LE$.

Thus the multitude of $GK, KH$ will be equal to the multitude of $AL, LE$.

We have that whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $AL$ of $CF$.

We also have that $CD > CF$.

Therefore $GK > AL$.

Now let $GM = AL$.

Then whatever aliquot part $GK$ is of $CD$, the same aliquot part also is $GM$ of $CF$.

Therefore from Subtraction of Divisors obeys Distributive Law the remainder $MK$ is of the same aliquot part of the remainder $FD$ that the whole $GK$ is of the whole $CD$.

Again, we have that whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $EL$ of $CF$.

We also have that $CD > CF$.

Therefore $HK > EL$.

Let $KN = EL$.

Then whatever aliquot part $KH$ is of $CD$, the same aliquot part also is $KN$ of $CF$.

Therefore from Subtraction of Divisors obeys Distributive Law the remainder $NH$ is of the same aliquot part of the remainder $FD$ that the whole $KH$ is of the whole $CD$.

But $MK$ was proved to be the same aliquot part of $FD$ that $GK$ is of $CD$.

Therefore $MK + NH$ is the same aliquant part of $DF$ that $HG$ is of $CD$.

But $MK + NH = EB$ and $HG = BA$.

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