Proportional Magnitudes are Proportional Alternately

Theorem

In the words of Euclid:

If four magnitudes be proportional, they will also be proportional alternately.

That is:

$a : b = c : d \implies a : c = b : d$

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 $D$.

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 $E : F = G : H$.

$E > G \implies F > H$

$E = G \implies F = H$

$E < G \implies F < H$

Now:

$E, F$ are equimultiples of $A, B$

$G, H$ are equimultiples of $C, D$.

$A : C = B : D$

$\blacksquare$