If a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order.

Let a first magnitude *A* have to a second *B* the same ratio as a third *C* to a fourth *D,* and let equimultiples *E* and *F* be taken of *A* and *C,* and *G* and *H* other, arbitrary, equimultiples of *B* and *D.*

I say that *E* is to *G* as *F* is to *H.*

Take equimultiples *K* and *L* of *E* and *F,* and other, arbitrary, equimultiples *M* and *N* of *G* and *H.*

Since *E* is the same multiple of *A* that *F* is of *C,* and equimultiples *K* and *L* of *E* and *F* have been taken, therefore *K* is the same multiple of *A* that *L* is of *C.* For the same reason *M* is the same multiple of *B* that *N* is of *D.*

And, since *A* is to *B* as *C* is to *D,* and equimultiples *K* and *L* have been taken of *A* and *C,* and other, arbitrary, equimultiples *M* and *N* of *B* and *D,* therefore, if *K* is in excess of *M,* then *L* is in excess of *N*; if it is equal, equal; and if less, less.

And *K* and *L* are equimultiples of *E* and *F,* and *M* and *N* are other, arbitrary, equimultiples of *G* and *H,* therefore *E* is to *G* as *F* is to *H.*

Therefore, *if a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order.*

Q.E.D.

The statement of the proposition says that

Note how Euclid uses the definition to prove that the two ratios *pa* : *qb* and *pc* : *qd* are the same. (Here, *a* and *b* are magnitudes of one kind, and *c* and *d* are magnitudes of another kind, but *p* and *q* are numbers.)
We are given *a* : *b* = *c* : *d.* That means for any numbers *m* and *n* that

We have to prove that *pa* : *qb* = *pc* : *qd* for any numbers *p* and *q.* That means, we have to prove that for any *m* and *n,*

But that’s just a special case of the given relation