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.