Equality of Ratios in Perturbed Proportion

Theorem

 * If there be three magnitudes, and others equal to them in multitude, which taken two and two together are in the same ratio, and the proportion of them be perturbed, they will also be in the same ratio ex aequali.

That is, if:
 * $a : b = e : f$
 * $b : c = d : e$

then:
 * $a : c = d : f$

Proof
Let there be three magnitudes $A, B, C$, and others equal to them in multitude, which taken two and two together are in the same proportion, namely $D, E, F$.

Let the proportion of them be perturbed, that is:
 * $A : B = E : F$
 * $B : C = D : E$

then we need to show that:
 * $A : C = D : F$


 * Euclid-V-23.png

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

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

From Ratio Equals its Multiples it follows that $A : B = G : H$

For the same reason $E : F = M : N$.

We have that $A : B = E : F$

From Equality of Ratios is Transitive $G : H = M : N$.

Next, we have that $B : C = D : E$.

From Proportional Magnitudes are Proportional Alternately $B : D = C : E$.

From Ratio Equals its Multiples $B : D = H : K$.

We also have that $B : D = C : E$ and $H : K = C : E$.

So from Equality of Ratios is Transitive $H : K = C : E$.

From Ratio Equals its Multiples $C : E = L : M$.

We also have that $C : E = H : K$.

So from Equality of Ratios is Transitive $H : K = L : M$.

From Proportional Magnitudes are Proportional Alternately $H : L = K : M$.

But we have $G : H = M : N$.

Thus it follows from Relative Sizes of Elements in Perturbed Proportion that:
 * $G > L \implies K > N$
 * $G = L \implies K = N$
 * $G < L \implies K < N$

We have that:
 * $G, K$ are equimultiples of $A, D$
 * $L, N$ are equimultiples of $C, F$

Therefore $A : C = D : F$.