# 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.

## 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:

$AE = EB$
$AH = DH$
$EH \parallel DB$

For the same reason:

$HK \parallel AB$

Therefore $HEBK$ is a parallelogram.

$HK = EB$

But $EB = EA$.

Therefore:

$AE = HK$

But also $AH = HD$.

Thus:

$EA = KH$
$AH = HD$

and:

$\angle EAH = KHD$
$\triangle AEH = \triangle HKD$

For the same reason:

$\triangle AHG = \triangle HLD$
$\angle EHG = \angle KDL$

We have:

$EH = KD$
$HG = DL$

and:

$\angle EHG = KDL$
$\triangle EHG = \triangle KDL$

For the same reason:

$\triangle AEG = \triangle HKL$
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 $\triangle ADB$.

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

and their sides are proportional.

$\triangle ADB$ 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 theorem is Proposition $3$ of Book $\text{XII}$ of Euclid's The Elements.