Volumes of Similar Parallelepipeds are in Triplicate Ratio to Length of Corresponding Sides

Proof

 * Euclid-XI-33.png

Let $AB$ and $CD$ be similar parallelepipeds.

Let $AE$ be the side of $AB$ corresponding to the side $CF$ of $CD$.

It is to be demonstrated that the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$.

That is:


 * $AB : CD = AE^3 : CF^3$

Let $EK, EL, EM$ be produced in a straight line with $AE, GE, HE$.

Let $EK = CF, EL = FN, EM = FR$.

Let the parallelogram $KL$ and parallelepiped $KP$ be completed.

We have that the two sides $KE$ and $EL$ equal the two sides $CF$ and $FN$.

Also $\angle KEL = \angle CFN$ because of similarity of $AB$ and $CD$.

Therefore the parallelogram $KL$ is equal and similar to the parallelogram $CN$.

For the same reason:
 * the parallelogram $KM$ is equal and similar to the parallelogram $CR$

and:
 * the parallelogram $EP$ is equal and similar to the parallelogram $DF$.

Therefore from:

and:

the whole parallelepiped $KP$ is equal and similar to the parallelepiped $CD$.

Let the parallelogram $GK$ be completed.

On the parallelograms $GK$ and $KL$ as bases, let the parallelepipeds $EO$ and $LQ$ be constructed with the same height as $AB$.

We have that $AB$ and $CD$ are similar.

So:
 * $AE : CF = EG : FN = EH : FR$

and:
 * $CF = EK$
 * $FN : EL$
 * $FR : EM$

Therefore:
 * $AE : EK = GE : EL = HE : EM$

But from :
 * $AE : EK = AG : GK$
 * $GE : EL = GK : KL$
 * $HE : EM = QE : KM$

Therefore:
 * $AG : GK = GK : KL = QE : LM$

But from :
 * $AG : GK = AB : EO$
 * $GK : KL = OE : QL$
 * $QE : KM = QL : KP$

Therefore:
 * $AB : EO = EO : QL = QL : KP$

So from :
 * the parallelepipeds $AB$ has to the parallelepipeds $KP$ the ratio triplicate of that which $AE$ has to $EO$.

But from :
 * $AB : EO = AG : GK = AE : EK$

Hence:
 * $AB : KP = AE^3 : EK^3$

But $KP = CD$ and $EK = CF$.

Therefore the parallelepiped $AB$ has to the parallelepiped $CD$ the ratio triplicate of that which $AE$ has to $CF$.