:300px 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 {{EuclidPropLink|book = VI|prop = 2|title = Parallel Line in Triangle Cuts Sides Proportionally}}: :$EH \parallel DB$ For the same reason: :$HK \parallel AB$ Therefore $HEBK$ is a parallelogram. Therefore from {{EuclidPropLink|book = I|prop = 34|title = 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 {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle AEH = \triangle HKD$ For the same reason: :$\triangle AHG = \triangle HLD$ From {{EuclidPropLink|book = XI|prop = 10|title = 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 {{EuclidPropLink|book = I|prop = 4|title = Triangle Side-Angle-Side Equality}}: :$\triangle EHG = \triangle KDL$ For the same reason: :$\triangle AEG = \triangle HKL$ Therefore from {{EuclidDefLink|XI|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 {{EuclidPropLink|book = I|prop = 29|title = Parallelism implies Equal Corresponding Angles}}: :$\triangle ADB$ is equiangular with $\triangle DHK$ and their sides are proportional. Therefore by {{EuclidDefLink|VI|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 {{EuclidPropLink|book = XI|prop = 10|title = 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 {{EuclidPropLink|book = XI|prop = 39|title = 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. {{qed}} {{Euclid Note|3|XII}}