Ratio Equals its Multiples
Theorem
In the words of Euclid:
(The Elements: Book $\text{V}$: Proposition $15$)
That is:
- $a : b \implies ma = mb$
Proof
Let $AB$ be the same multiple of $C$ that $DE$ is of $F$.
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 $DK = F$.
$\blacksquare$
Historical Note
This proof is Proposition $15$ of Book $\text{V}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{V}$. Propositions