Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms

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 :
 * $EH \parallel DB$

For the same reason:
 * $HK \parallel AB$

Therefore $HEBK$ is a parallelogram.

Therefore from :
 * $HK = EB$

But $EB = EA$.

Therefore:
 * $AE = HK$

But also $AH = HD$.

Thus:
 * $EA = KH$
 * $AH = HD$

and:
 * $\angle EAH = KHD$

Therefore from :
 * $\triangle AEH = \triangle HKD$

For the same reason:
 * $\triangle AHG = \triangle HLD$

From :
 * $\angle EHG = \angle KDL$

We have:
 * $EH = KD$
 * $HG = DL$

and:
 * $\angle EHG = KDL$

Therefore from :
 * $\triangle EHG = \triangle KDL$

For the same reason:
 * $\triangle AEG = \triangle HKL$

Therefore from :
 * 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 :
 * $\triangle ADB$ is equiangular with $\triangle DHK$

and their sides are proportional.

Therefore by :
 * $\triangle ADB$ 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 :
 * $\angle BAC = \angle KHL$

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 :
 * $(1): \quad$ 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.