Multiples of Alternate Ratios of Equal Fractions

Theorem

 * If a (natural) number be parts of a (natural) number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth.

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

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


 * Euclid-VII-10.png

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

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

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

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

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

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