Alternate Ratios of Multiples

Theorem

 * If an unit measure any (natural) number, and another (natural) number measure any other (natural) number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth.

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 $EF$.