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

## Theorem

In the words of Euclid:

*Similar parallelepidedal solids are to one another in the triplicate ratio of their corresponding sides.*

(*The Elements*: Book $\text{XI}$: Proposition $33$)

### Porism

In the words of Euclid:

*From this it is manifest that, if four straight lines be < continuously > proportional, as the first is to the fourth, so will a parallelepidedal solid on the first be to the similar and similarly described parallelepidedal solid on the second, inasmuch as the first has to the fourth and the ratio triplicate of that which it has to the second.*

(*The Elements*: Book $\text{XI}$: Proposition $33$ : Porism)

## Proof

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 Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

- $AE : EK = AG : GK$
- $GE : EL = GK : KL$
- $HE : EM = QE : KM$

Therefore:

- $AG : GK = GK : KL = QE : LM$

- $AG : GK = AB : EO$
- $GK : KL = OE : QL$
- $QE : KM = QL : KP$

Therefore:

- $AB : EO = EO : QL = QL : KP$

So from Book $\text{V}$ Definition $10$: Triplicate Ratio:

- the parallelepipeds $AB$ has to the parallelepipeds $KP$ the ratio triplicate of that which $AE$ has to $EO$.

But from Proposition $1$ of Book $\text{VI} $: Areas of Triangles and Parallelograms Proportional to Base:

- $AB : EO = AG : GK = AE : EK$

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 $CF$.

$\blacksquare$

## Historical Note

This proof is Proposition $33$ of Book $\text{XI}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 3*(2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions