Proportional Magnitudes are Proportional Alternately

Theorem
That is:
 * $a : b = c : d \implies a : c = b : d$

Proof
Let $A, B, C, D$ be four proportional magnitudes, so that as $A$ is to $B$, then so is $C$ to $D$.

We need to show that as $A$ is to $C$, then $B$ is to $D$.


 * Euclid-V-16.png

Let equimultiples $E, F$ be taken of $A, B$.

Let other arbitrary equimultiples $G, H$ be taken of $C, D$.

We have that $E$ is the same multiple of $A$ that $F$ is of $B$.

So from Ratio Equals its Multiples we have that $A : B = E : F$

But $A : B = C : D$.

So from Equality of Ratios is Transitive it follows that $C : D = E : F$.

Similarly, we have that $G, H$ are equimultiples of $C, D$.

So from Ratio Equals its Multiples we have that $C : D = G : H$

So from Equality of Ratios is Transitive it follows that $E : F = G : H$.

But from Relative Sizes of Components of Ratios:
 * $E > G \implies F > H$
 * $E = G \implies F = H$
 * $E < G \implies F < H$

Now $E, F$ are equimultiples of $A, B$, and $G, H$ are equimultiples of $C, D$.

Therefore from :
 * $A : C = B : D$