Equality of Ratios in Perturbed Proportion

Theorem
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 $A : C = D : F$.