Magnitudes with Rational Ratio are Commensurable

Proof
Let $A$ and $B$ be magnitudes which have to one another the ratio which the number $D$ has to the number $E$.

Let $A$ be divided into as many equal parts as there are units in $D$.

Let $C$ be equal to one of those parts.

Let $F$ be made up of as many magnitudes equal to $C$ as there are units in $E$.

Since:
 * there are in $A$ as many magnitudes equal to $C$ as there are units in $D$

then:
 * whatever part the unit is of $D$, the same part is $C$ of $A$.

From Ratios of Fractions in Lowest Terms:
 * $\dfrac C A = \dfrac 1 D$

and so from the porism to Ratios of Equal Magnitudes:
 * $\dfrac A C = \dfrac D 1 = D$

Since:
 * there are in $F$ as many magnitudes equal to $C$ as there are units in $E$

then from Ratios of Fractions in Lowest Terms:
 * $\dfrac C F = \dfrac 1 E$

From Equality of Ratios Ex Aequali:
 * $\dfrac A F = \dfrac D E$

But:
 * $\dfrac D E = \dfrac A B$

and so from Equality of Ratios is Transitive:
 * $\dfrac A B = \dfrac A F$

Therefore $A$ has the same ratio to each of the magnitudes $B$ and $F$.

Therefore from Magnitudes with Same Ratios are Equal:
 * $B = F$

But $C$ measures $F$.

Therefore $C$ also measures $B$.

Further, $C$ also measures $A$.

Therefore $C$ measures both $A$ and $B$.

Therefore $A$ is commensurable with $B$.