:600px Let there be two tetrahedra of the same height whose bases are $ABC, DEF$ and whose apices are $G$ and $H$. From {{EuclidPropLink|book = XII|prop = 3|title = Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms}}: :let each of $ABCG$ and $DEFH$ be divided into two equal tetrahedra which are similar to the whole and two equal prisms. It is to be demonstrated that the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of all the prisms in $ABCG$ to all the prisms in $DEFH$. We have that: :$BO = OC$ and: :$AL = LC$ Therefore: :$LO \parallel AB$ and $\triangle ABC$ is similar to $\triangle LOC$. For the same reason, $\triangle DEF$ is similar to $\triangle RVF$. We have that: :$BC = 2 \cdot CO$ and: :$EF = 2 \cdot FV$ Therefore: :$BC : CO = EF : FV$ So from {{EuclidPropLink|book = VI|prop = 22|title = Similar Figures on Proportional Straight Lines}}: :$\triangle LOC : \triangle ABC = \triangle DEF : \triangle RFV$ Therefore from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$\triangle ABC : \triangle DEF = \triangle LOC : \triangle RFV$ But from {{EuclidPropLink|book = XII|prop = 4|qual=Lemma|title = Proportion of Sizes of Tetrahedra divided into Two Similar Tetrahedra and Two Equal Prisms}}: :the ratio of $\triangle LOC$ to $\triangle RFV$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite. Therefore the ratio of $\triangle ABC$ to $\triangle DEF$ equals the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite. But from {{EuclidPropLink|book = XI|prop = 39|title = Prisms of equal Height with Parallelogram and Triangle as Base}}: :the ratio of the prism of which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite to the prism of which $\triangle RFV$ is its base and $\triangle STU$ is its opposite equals the ratio of the prism of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite. Therefore, by {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :the ratio of the two prisms of which the parallelogram $KBOL$ is its "base" and the straight line $PM$ lies opposite and in which $\triangle LOC$ is its base and $\triangle PMN$ is its opposite are to the two prisms of which the the parallelogram to the prism of which $QEVR$ is its "base" and the straight line $ST$ lies opposite and in which $\triangle RFV$ is its base and $\triangle STU$ is its opposite. Similarly from {{EuclidPropLink|book = XII|prop = 3|title = Tetrahedron divided into Two Similar Tetrahedra and Two Equal Prisms}}: :the two tetrahedra $PMNG$ and $STUH$ can be divided into two equal tetrahedra which are similar and two equal prisms. As the ratio of the base $PMN$ is to the base $STU$, so will the ratio of the two prisms in the tetrahedron $PMNG$ be to the two prisms in the tetrahedron $STUH$. But: :$\triangle PMN : \triangle STU = \triangle LOC : \triangle RFV$ Therefore: :$\triangle ABC : \triangle DEF$ equals the ratio of the four prisms in the tetrahedron $ABCG$ to the four prisms in the tetrahedron $DEFH$. Similarly, if the remaining tetrahedra are divided into two tetrahedra and two prisms, then, as $\triangle ABC$ is to $\triangle DEF$, so will all the prisms in $ABCG$ be to all the prisms in $DEFH$. {{qed}} {{Euclid Note|4|XII}}