Equality of Ratios is Transitive

Theorem

In the words of Euclid:

Ratios which are the same with the same ratio are also the same with one another.

(The Elements: Book $\text{V}$: Proposition $11$)

That is:

$A : B = C : D, C : D = E : F \implies A : B = E : F$

Proof

Let $A : B = C : D$ and $C : D = E : F$.

Of $A, C, E$ let equimultiples $G, H, K$ be taken.

Of $B, D, F$ let other arbitrary equimultiples $L, M, N$ be taken.

We have that:

$A : B = C : D$

$G, H$ are equimultiples of $A, C$

$L, M$ are equimultiples of $B, D$

So:

$G > L \implies H > M$

$G = L \implies H = M$

$G < L \implies H < M$

Also, we have that:

$C : D = E : F$

$H, K$ are equimultiples of $C, E$

$M, N$ are equimultiples of $D, F$

So:

$H > M \implies K > N$

$H = M \implies K = N$

$H < M \implies K < N$

Also, we have that:

$G, K$ are equimultiples of $A, E$

$L, N$ are equimultiples of $B, F$

Therefore $A : B = E : F$.

$\blacksquare$

Historical Note

This proof is Proposition $11$ 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