Equality of Ratios is Transitive

Theorem
That is:
 * $A : B = C : D, C : D = E : F \implies A : B = E : F$

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


 * Euclid-V-11.png

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 $A : B = E : F$.