Equality of Ratios Ex Aequali

Theorem

 * 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.

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:
 * $A : C = D : F$


 * Euclid-V-22.png

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, $A : C = D : F$.