PROPOSITION 4.
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.
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. [
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. 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. [
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; [
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.