---
book: 12
number: 4
id: "XII.4"
kind: "theorem"
uses: ["[[book-12/proposition-3]]", "[[book-6/proposition-22]]", "[[book-5/proposition-16]]", "[[book-11/proposition-39]]", "[[book-5/proposition-12]]", "[[book-11/proposition-17]]", "[[book-11/proposition-32]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:12.prop.4"
license: "CC-BY-SA-4.0"
---

# XII.4

*If there be two pyramids of the same height which have triangular bases*, *and each 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.*

## Proof

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; [[book-12/proposition-3|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*; [[book-6/proposition-22|VI. 22]] therefore, alternately, as the triangle *ABC* is to the triangle *DEF*, so is the triangle *LOC* to the triangle *RVF*. [[book-5/proposition-16|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. [[book-11/proposition-39|XI. 39]]; cf. [[book-12/proposition-3|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 [[book-5/proposition-12|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*. 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.

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. [[book-11/proposition-17|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.

Hence also the parallelepipedal solids described from the said prisms are of equal height and are to one another as their bases; [[book-11/proposition-32|XI. 32]] therefore their halves, namely the said prisms, are to one another as the base *LOC* is to the base *RVF*. Q. E. D.