Let AB and CD be similar parallelepipedal solids, and let AE be the side corresponding to CF.
I say that the solid AB has to the solid CD the ratio triplicate of that which AE has to CF.
Produce EK, EL, and EM in a straight line with AE, GE, and HE. Make EK equal to CF, EL equal to FN, and EM equal to FR. Complete the parallelogram KL and the solid KP.
Now, since the two sides KE and EL equal the two sides CF and FN, while the angle KEL equals the angle CFN, for the angle AEG also equals the angle CFN because AB and CD are similar solids, therefore the parallelogram KL equals and is similar to the parallelogram CN. For the same reason the parallelogram KM equals and is similar to CR, and EP equals and is similar to DF.
Therefore three parallelograms of the solid KP equal and are similar to three parallelograms of the solid CD. But the former three parallelograms equal and are similar to their opposites, and the latter three equal and are similar to their opposites, therefore the whole solid KP equals and is similar to the whole solid CD.
Complete the parallelogram GK, and complete the solids EO and LQ on the parallelograms GK and KL as bases with the same height as that of AB.
Then since the solids AB and CD are similar, therefore AE is to CF as EG is to FN, and as EH is to FR. And CF equals EK, FN equals EL, and FR equals EM, therefore AE is to EK as GE is to EL, and as HE is to EM.
But AE is to EK as AG is to the parallelogram GK, therefore GE is to EL as GK is to KL, and HE is to EM as QE is to KM. Therefore the parallelogram AG is to GK as GK to is KL, and as QE is to KM.
But AG is to GK as the solid AB is to the solid EO, GK is to KL as the solid OE is to the solid QL, and QE is to KM as the solid QL is to the solid KP, therefore the solid AB is to EO as EO is to QL, and as QL is to KP.
But, if four magnitudes are continuously proportional, then the first has to the fourth the ratio triplicate of that which it has to the second, therefore the solid AB has to KP the ratio triplicate of that which AB has to EO.
But AB is to EO as the parallelogram AG is to GK, and as the straight line AE is to EK, hence the solid AB also has to KP the ratio triplicate of that which AE has to EK.
But the solid KP equals the solid CD, and the straight line EK equals CF, therefore the solid AB has also to the solid CD the ratio triplicate of that which the corresponding side of it, AE, has to the corresponding side CF.
Therefore, Similar parallelepipedal solids are to one another in the triplicate ratio of their corresponding sides.