Equality of Ratios Ex Aequali

Theorem

As Euclid defined it:

If there be any number of magnitudes whatever, and others equal to them in multitude, which taken two and two together are in the same ratio, they will also be in the same ratio ex aequali.

(The Elements: Book V: Proposition $22$)

That is, if:

$a : b = d : e$

$b : c = e : f$

then:

$a : c = d : f$

Proof

Let there be any number of magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two together are in the same ratio, so that:

$A : B = D : E$

$B : C = E : F$

Then we need to show that:

$A : C = D : F$

Let equimultiples $G, H$ be taken of $A, D$.

Let other arbitrary equimultiples $K, L$ be taken of $B, E$.

Let other arbitrary equimultiples $M, N$ be taken of $C, F$.

We have that $A : B = D : E$.

So from Multiples of Terms in Equal Ratios $G : K = H : L$.

For the same reason, $K : M = L : N$.

We have that there are three magnitudes $G, K, M$ and others $H, L, N$ which taken two and two together are in the same ratio.

So from Relative Sizes of Successive Ratios it follows that:

$G > M \implies H > N$

$G = M \implies H = N$

$G < M \implies H < N$

We also have that $G, H$ are equimultiples of $A, D$ and that $M, N$ are equimultiples of $C, F$.

So from Book V Definition 5: Equality of Ratios, $A : C = D : F$.

$\blacksquare$

Historical Note

This is Proposition 22 of Book V of Euclid's The Elements.