If two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them.

Let two magnitudes *AB* and *CD* be equimultiples of two magnitudes *E* and *F,* and let *AG* and *CH* subtracted from them be equimultiples of the same two *E* and *F.*

I say that the remainders *GB* and *HD* either equal *E* and *F* or are equimultiples of them.

First, let *GB* equal *E.*

I say that *HD* also equals *F.*

Make *CK* equal to *F.*

Since *AG* is the same multiple of *E* that *CH* is of *F,* while *GB* equals *E,* and *KC* equals *F,* therefore *AB* is the same multiple of *E* that *KH* is of *F.*

But, by hypothesis, *AB* is the same multiple of *E* that *CD* is of *F,* therefore *KH* is the same multiple of *F* that *CD* is of *F.*

Since then each of the magnitudes *KH* and *CD* is the same multiple of *F,* therefore *KH* equals *CD.*

Subtract *CH* from each. Then the remainder *KC* equals the remainder *HD.*

But *F* equals *KC,* therefore *HD* also equals *F.*

Hence, if *GB* equals *E, HD* also equals *F.*

Similarly we can prove that, even if *GB* is a multiple of *E, HD* is also the same multiple of *F.*

Therefore, *if two magnitudes are equimultiples of two magnitudes, and any magnitudes subtracted from them are equimultiples of the same, then the remainders either equal the same or are equimultiples of them.*

Q.E.D.

Its proof depends on a distributivity, namely that multiplication by magnitudes distributes over subtraction of numbers: (*m* – *n*)*a* = *ma* – *na.* Euclid takes 4 as *m* and 3 as *n.* He has two cases since since he doesn’t take 1 to be a number.

This proposition is not used in the rest of the *Elements.*