If there be two pyramids of the same height which have triangular bases, and cach of them be divided into two pyramids equal to one another and similar to the whole, and into two equal prisms, then, as the base of the one pyramid is to the base of the other pyramid, so will all the prisms in the one pyramid be to all the prisms, being equal in multitude, in the other pyramid.
设有两个同高的三角锥，底面分别为三角形ABC和DEF，顶点分别为G和H。将每个三角锥分割成两个全等且与原锥相似的小三角锥，以及两个相等的棱柱。则底面ABC与底面DEF之比，等于锥ABCG中所有棱柱之和与锥DEFH中所有棱柱之和（数量相等）之比。
Let there be two pyramids of the same height which have the triangular bases ABC, DEF, and vertices the points G, H, and let each of them be divided into two pyramids equal to one another and similar to the whole and into two equal prisms; [XII. 3] I say that, as the base ABC is to the base DEF, so are all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH, For, since BO is equal to OC, and AL to LC, therefore LO is parallel to AB, and the triangle ABC is similar to the triangle LOC. For the same reason the triangle DEF is also similar to the triangle RVF. And, since BC is double of CO, and EF of FV, therefore, as BC is to CO, so is EF to FV. And on BC, CO are described the similar and similarly situated rectilineal figures ABC, LOC, and on EF, FV the similar and similarly situated figures DEF, RVF; therefore, as the triangle ABC is to the triangle LOC, so is the triangle DEF to the triangle RVF; [VI. 22] therefore, alternately, as the triangle ABC is to the triangle DEF, so is the triangle LOC to the triangle RVF. [V. 16] But, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite; [Lemma following] therefore also, as the triangle ABC is to the triangle DEF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite. But, as the said prisms are to one another, so is the prism in which the parallelogram KBOL is the base and the straight line PM its opposite to the prism in which the parallelogram QEVR is the base and the straight line ST its opposite.
由分割知BO=OC，AL=LC，故LO平行于AB，三角形ABC相似于三角形LOC。同理三角形DEF相似于三角形RVF。
[XI. 39; cf. XII. 3] Therefore also the two prisms, that in which the parallelogram KBOL is the base and PM its opposite, and that in which the triangle LOC is the base and PMN its opposite, are to the prisms in which QEVR is the base and the straight line ST its opposite and in which the triangle RVF is the base and STU its opposite in the same ratio [V. 12] Therefore also, as the base ABC is to the base DEF, so are the said two prisms to the said two prisms. And similarly, if the pyramids PMNG, STUH be divided into two prisms and two pyramids, as the base PMN is to the base STU, so will the two prisms in the pyramid PMNG be to the two prisms in the pyramid STUH. But, as the base PMN is to the base STU, so is the base ABC to the base DEF; for the triangles PMN, STU are equal to the triangles LOC, RVF respectively. Therefore also, as the base ABC is to the base DEF, so are the four prisms to the four prisms. And similarly also, if we divide the remaining pyramids into two pyramids and into two prisms, then, as the base ABC is to the base DEF, so will all the prisms in the pyramid ABCG be to all the prisms, being equal in multitude, in the pyramid DEFH.
因BC=2CO，EF=2FV，故BC:CO=EF:FV。由相似图形面积比等于对应边平方比，得三角形ABC:三角形LOC=三角形DEF:三角形RVF，从而ABC:DEF=LOC:RVF。
Q. E. D. LEMMA. But that, as the triangle LOC is to the triangle RVF, so is the prism in which the triangle LOC is the base and PMN its opposite to the prism in which the triangle RVF is the base and STU its opposite, we must prove as follows. For in the same figure let perpendiculars be conceived drawn from G, H to the planes ABC, DEF; these are of course equal because, by hypothesis, the pyramids are of equal height.
由引理，三角形LOC:三角形RVF等于以LOC为底、PMN为对面的棱柱与以RVF为底、STU为对面的棱柱之比。又由XI.39，这些棱柱与以平行四边形KBOL为底、PM为对面的棱柱等成比例，故两棱柱之和与另两棱柱之和的比等于ABC:DEF。
Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC, PMN, they will be cut in the same ratios. [XI. 17] And GC is bisected by the plane PMN at N; therefore the perpendicular from G to the plane ABC will also be bisected by the plane PMN. For the same reason the perpendicular from H to the plane DEF will also be bisected by the plane STU. And the perpendiculars from G, H to the planes ABC, DEF are equal; therefore the perpendiculars from the triangles PMN, STU to the planes ABC, DEF are also equal. Therefore the prisms in which the triangles LOC, RVF are bases, and PMN, STU their opposites, are of equal height.
类似地，将剩余小锥继续分割，每次所得棱柱之和的比始终等于ABC:DEF。重复此过程，最终所有棱柱之和的比等于ABC:DEF。