Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms

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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

Euclid-XII-3.png

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 Line in Triangle Cuts Sides Proportionally:

$EH \parallel DB$

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 Equality:

$\triangle AEH = \triangle HKD$

For the same reason:

$\triangle AHG = \triangle HLD$

From Proposition $10$ of Book $\text{XI} $: Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles:

$\angle EHG = \angle KDL$

We have:

$EH = KD$
$HG = DL$

and:

$\angle EHG = KDL$

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

$\triangle EHG = \triangle KDL$

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 $\triangle DLH$

From Proposition $10$ of Book $\text{XI} $: Two Lines Meeting which are Parallel to Two Other Lines Meeting contain Equal Angles:

$\angle BAC = \angle KHL$

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 $GFC$.

From Proposition $39$ of Book $\text{XI} $: Prisms of equal Height with Parallelogram and Triangle as Base:

$(1): \quad$ 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.


Sources