Ratio Equals its Multiples

Theorem
That is:
 * $a : b \implies ma = mb$

Proof
Let $AB$ be the same multiple of $C$ that $DE$ is of $F$.


 * Euclid-V-15.png

So as many magnitudes as there are in $AB$ equal to $C$, so many are there also in $DE$ equal to $F$.

Let $AB$ be divided into the magnitudes $AG, GH, HB$ equal to $C$.

Let $DE$ be divided into the magnitudes $DK, KL, LE$ equal to $F$.

Then the number of magnitudes $AG, GH, GB$ is the same as the number of magnitudes in $DK, KL, LE$.

We have that $AG = GH = HB$ and $DK = KL = LE$.

So from Ratios of Equal Magnitudes it follows that $AG : DK = GH : KL = HB : LE$.

Then from Sum of Components of Equal Ratios $AG : DK = AB : DE$.

But $AG = C$ and $DK = F$.