Parallelepipedal solids which are of the same height are to one another as their bases.

Let *AB* and *CD* be parallelepipedal solids of the same height.

I say that the parallelepipedal solids *AB* and *CD* are to one another as their bases, that is, that the solid *AB* to the solid *CD* as the base *AE* is to the base *CF* .

Apply *FH* equal to *AE* to *FG.* Complete the parallelepipedal solid *GK* with the same height as that of *CD* on *FH* as base.

Then the solid *AB* equals the solid *GK* for they are on equal bases *AE* and *FH* and of the same height.

And, since the parallelepipedal solid *CK* is cut by the plane *DG* which is parallel to opposite planes, therefore the solid *CD* is to the solid *DH* as the base *CF* is to the base *FH.*

But the base *FH* equals the base *AE,* and the solid *GK* equals the solid *AB,* therefore the solid *AB* to the solid *CD* as the base *AE* is to the base *CF.*

Therefore, *parallelepipedal solids which are of the same height are to one another as their bases.*

Q.E.D.