# Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms

## Theorem

In the words of Euclid:

*Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another, similar to the whole and having triangular bases, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.*

(*The Elements*: Book $\text{XII}$: Proposition $3$)

## Proof

Let $ABCD$ be a tetrahedron whose base is $ABC$ and whose apex is $D$.

It is to be demonstrated that $ABCD$ can be divided into two equal tetrahedra which are similar to $ABCD$ and two equal prisms which form more than half $ABCD$.

Let $AB, BC, CA, A, DB, DC$ be bisected at the points $E, F, G, H, K, L$.

Let $HE, EG, GH, HK, KL, LH, KF, FG$ be joined.

We have that:

- $AE = EB$
- $AH = DH$

From Proposition $2$ of Book $\text{VI} $: Parallel Transversal Theorem:

- $EH \parallel DB$

For the same reason:

- $HK \parallel AB$

Therefore $HEBK$ is a parallelogram.

Therefore from Proposition $34$ of Book $\text{I} $: Opposite Sides and Angles of Parallelogram are Equal:

- $HK = EB$

But $EB = EA$.

Therefore:

- $AE = HK$

But also $AH = HD$.

Thus:

- $EA = KH$
- $AH = HD$

and:

- $\angle EAH = KHD$

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

- $\triangle AEH = \triangle HKD$

For the same reason:

- $\triangle AHG = \triangle HLD$

- $\angle EHG = \angle KDL$

We have:

- $EH = KD$
- $HG = DL$

and:

- $\angle EHG = KDL$

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

- $\triangle EHG = \triangle KDL$

For the same reason:

- $\triangle AEG = \triangle HKL$

Therefore from Book $\text{XI}$ Definition $10$: Similar Equal Solid Figures:

- the pyramid of which $AEG$ is the base and $H$ the apex is equal and similar to the pyramid of which $HKL$ is the base and $D$ the apex.

We have that $HK$ is parallel to $AB$, which is one of the sides of $\triangle ADB$.

From Proposition $29$ of Book $\text{I} $: Parallelism implies Equal Corresponding Angles:

- $\triangle ADB$ is equiangular with $\triangle DHK$

and their sides are proportional.

Therefore by Book $\text{VI}$ Definition $1$: Similar Rectilineal Figures:

- $\triangle ADB$ is similar to $\triangle DHK$.

For the same reason:

- $\triangle DBC$ is similar to $\triangle DKL$

and:

- $\triangle ADC$ is similar to $\triangle DLH$

- $\angle BAC = \angle KHL$

We also have:

- $BA : AC = KH : HL$

and so:

- $\triangle ABC = \triangle HKL$

Therefore the pyramid of which $ABC$ is the base and $D$ the apex is similar to the pyramid of which $HKL$ is the base and $D$ the apex.

But we have that the pyramid of which $AEG$ is the base and $H$ the apex is equal and similar to the pyramid of which $HKL$ is the base and $D$ the apex.

Therefore each of the pyramids $AEGH$ and $HKLD$ is similar to the pyramid $ABCD$.

We have that $BF = FC$.

We also have that the parallelogram $EFGH$ is double the triangle $GFC$.

- $(1): \quad$ the prism contained by $\triangle BKF$, $\triangle EHG$ and the three parallelograms $EBFG, EBKH, HKFG$ is equal to the prism contained by $\triangle GFC$, $\triangle HKL$ and the three parallelograms $KFCL, LCGH, HKFG$.

By joining the straight lines $EF$ and $EK$, it can be seen that the prism in which $EBFG$ is the "base" and the straight line $HK$ is opposite it is greater than pyramid $EBFK$.

But the pyramid $EBFK$ equals pyramid $AEHG$ as they are contained by equal and similar planes.

Thus each of the two prisms in $(1)$ above is greater than each of the pyramids $AEGH$ and $HKLD$.

Thus these two prisms together are greater than the two pyramids $AEGH$ and $HKLD$ together,

But the whole pyramid $ABCD$ consists of these two prisms and these two pyramids.

$\blacksquare$

## Historical Note

This proof is Proposition $3$ of Book $\text{XII}$ 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 $\text{XII}$. Propositions