Alternate Ratios of Multiples

Proof
Let the unit $A$ measure any (natural) number $BC$.

Let another (natural) number $D$ measure any other (natural) number $EF$ the same number of times.

We need to show that $A$ measures $D$ the same number of times that $BC$ measures $EF$.


 * Euclid-VII-15.png

We have that $A$ measures $BC$ the same number of times that $D$ measures $EF$.

So as many units as there are in $BC$ there are numbers equal to $D$ in $EF$.

Let $BC$ be divided into the units in it: $BG, GH, HC$.

Let $EF$ be divided into the numbers in it equal to $D$: $EK, KL, LF$.

So the multitude of $BG, GH, HC$ will equal the multitude of $EK, KL, LF$.

We have that $BG = GH = HC$ and $EK = KL = LF$.

So $BG : EK = GH : KL = HC : LF$.

So from, $BG : EK = BG + GH + HC : EK + KL + LF$.

That is, $BG : EK = BC : EF$.

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

So $A : D = BC : EF$.

So $A$ measures $D$ the same number of times that $BC$ measures $EF$.