Relative Sizes of Elements in Perturbed Proportion

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

Then:
 * $a > c \implies d > f$
 * $a = c \implies d = f$
 * $a < c \implies d < f$

Proof
Let there be three 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.

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

Let $A > C$.

Then we need to show that $D > F$.


 * Euclid-V-21.png

We have that $A > C$.

So from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.

But $A : B = E : F$, and $C : B = E : D$

So from Relative Sizes of Proportional Magnitudes $E : F > E : D$.

But from Relative Sizes of Magnitudes on Unequal Ratios $F < D$ and so $D > F$.

Similarly we can prove that $A = C \implies D = F$ and $A < C \implies D < F$.