Alternate Ratios of Equal Fractions

Theorem

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

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

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


 * Euclid-VII-9.png

We have that whatever part $A$ is of $BC$, the same 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 part or parts $BG$ is of $EH$, the same part or the same parts is $GC$ of $HF$ also.

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

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

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