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