Multiples of Alternate Ratios of Equal Fractions

Proof
Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$, and another $DE$ be the same aliquant part of another, $F$.

We need to show that whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part or aliquot part is $C$ of $F$.


 * Euclid-VII-10.png

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

So as many aliquant parts of $C$ as there are in $AB$, the same aliquant parts also is $DE$ of $F$.

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

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

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

From, whatever aliquot part $AG$ is of $DH$, the same aliquot part or the same aliquant part is $C$ of $F$ also.

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

So from and, whatever  aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part also, or the same aliquot part, is $C$ of $F$.