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Proposition 4

If there are two pyramids of the same height with triangular bases, and each of them is divided into two pyramids equal and similar to one another and similar to the whole, and into two equal prisms, then the base of the one pyramid is to the base of the other pyramid as all the prisms in the one pyramid are to all the prisms, being equal in multitude, in the other pyramid.
Let there be two pyramids of the same height with triangular bases ABC and DEF the points G and H the vertices, 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 the base ABC is to the base DEF as all the prisms in the pyramid ABCG to all the prisms, being equal in multitude, in the pyramid DEFH.

Since BO equals OC, and AL equals 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 CO, and EF double FV, therefore BC is to CO as EF is to FV.
And on BC and CO are described the similar and similarly situated rectilinear figures ABC and LOC, and on EF and FV the similar and similarly situated figures DEF and RVF, therefore the triangle ABC is to the triangle LOC as the triangle DEF is to the triangle RVF. | VI.22
Therefore, alternately the triangle ABC is to the triangle DEF as the triangle LOC is to the triangle RVF. But the triangle LOC is to the triangle RVF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite. | V.16 Lemma below
Therefore the triangle ABC is to the triangle DEF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite.
But the said prisms are to one another as the prism with the parallelogram KBOL the base and the straight line PM opposite is to the prism with the parallelogram QEVR the base and the straight line ST opposite. | XI.39
Therefore the two prisms, that with the parallelogram KBOL the base and PM opposite, and that with the triangle LOC the base and PMN opposite, are to the prisms with QEVR the base and the straight line ST opposite and with the triangle RVF the base and STU opposite in the same ratio. | V.12
Therefore the base ABC is to the base DEF as the said two prisms are to the said two prisms.
And similarly, if the pyramids PMNG and STUH are divided into two prisms and two pyramids, then the base PMN is to the base STU as the two prisms in the pyramid PMNG are to the two prisms in the pyramid STUH.
But the base PMN is to the base STU as the base ABC is to the base DEF, for the triangles PMN and STU equal the triangles LOC and RVF respectively.
Therefore the base ABC is to the base DEF as the four prisms are to the four prisms. And similarly, if we divide the remaining pyramids into two pyramids and into two prisms, then the base ABC is to base the DEF as all the prisms in the pyramid ABCG are to all the prisms, being equal in multitude, in the pyramid DEFH.

### Lemma But that the triangle LOC is to the triangle RVF as the prism with the triangle LOC the base and PMN opposite is to the prism with the triangle RVF the base and STU opposite, we must prove as follows.
In the same figure draw perpendiculars from G and H to the planes ABC and DEF. These are, of course, equal since the pyramids are of equal height by hypothesis. | XI.11
Now, since the two straight lines GC and the perpendicular from G are cut by the parallel planes ABC and PMN, therefore they are 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 is also bisected by the plane PMN. For the same reason the perpendicular from H to the plane DEF is also bisected by the plane STU.
And the perpendiculars from G and H to the planes ABC and DEF are equal, therefore the perpendiculars from the triangles PMN and STU to the planes ABC and DEF are also equal.
Therefore the prisms with the triangles LOC and RVF the bases, and PMN and STU opposite, are of equal height.
Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases. Therefore their halves, namely the said prisms, are to one another as the base LOC is to the base RVF. | XI.32 XI.28
Therefore, if there are two pyramids of the same height with triangular bases, and each of them is divided into two pyramids equal and similar to one another and similar to the whole, and into two equal prisms, then the base of the one pyramid is to the base of the other pyramid as all the prisms in the one pyramid are to all the prisms, being equal in multitude, in the other pyramid.
Q.E.D.

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This proposition is subordinate to the next, XII.5, in which two pyramids with triangular bases and the same height are shown to be proportional to their bases. Its proof proceeds by partitioning each of the two original pyramids into the two- pyramid-two-prism division of the previous proposition, then doing the same partition to the two smaller pyramids, then to the four even smaller pyramids, until a sufficiently small part of each original pyramid remains in whatever tiny pyramids there are while a sufficiently large part of each is composed of various sized prisms.

This proposition, at least in the last paragraph, considers that situation and concludes that the base of the first pyramid is to the second as the union of the various sized prisms in the first pyramid is to the union of the various sized prisms in the second pyramid. This is the crucial step in the proof of XII.5.

Book XII Introduction - Proposition XII.3 - Proposition XII.5.

© 1996
D.E.Joyce
Clark University