Equality of Ratios Ex Aequali

Theorem
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, $A : C = D : F$.