If a number is parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth.

Let the number *AB* be parts of the number *C,* and another number *DE* be the same parts of another number *F.*

I say that, alternately, *C* is the same parts or part of *F* that *AB* is of *DE.*

Since *DE* is the same parts of *F* as *AB* is of *C,* therefore *F* is the same parts of *DE* as *C* is of *AB.*

Divide *AB* into the parts of *C,* namely *AG* and *GB,* and divide *DE* into the parts of *F,* namely *DH* and *HE.* Then the multitude of *AG* and *GB* equals the multitude of *DH* and *HE.*

Now since *DH* is the same part of *F* as *AG* is of *C,* therefore, alternately, *C* is the same part or the same parts of *F* as *AG* is of *DH.*

For the same reason, *C* is the same part or the same parts of *F* as *GB* is of *HE,* so that, in addition, *C* is the same part or the same parts of *F* as *AB* is of *DE.*

Therefore, *if a number is parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth.*

Q.E.D.

The sample value taken for *m/n* in the proof is 2/3.