Alternate Ratios of Equal Fractions

Proof
Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$, and another $D$ be the same aliquot part of another, $EF$, that $A$ is of $BC$.

We need to show that whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$.


 * Euclid-VII-9.png

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

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

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

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

Thus the multitude of $BG, GC$ equals the multitude of $EH, HF$.

We have that $BG = GC$ and $EH = HF$.

Therefore whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part or the same aliquant part is $GC$ of $HF$ also.

So, from and, we have that whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part also, or the same aliquant part, is the sum $BC$ of the sum $EF$.

But $BG = A$ and $EH = D$.

So whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$ also.