PROPOSITION 3.
Any pyramid which has a triangular base is divided into two pyramids equal and similar to one another,
similar to the whole and having triangular bases,
and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
Let there be a pyramid of which the triangle
ABC is the base and the point
D the vertex; I say that the pyramid
ABCD is divided into two pyramids equal to one another, having triangular bases and similar to the whole pyramid, and into two equal prisms; and the two prisms are greater than the half of the whole pyramid.
For let
AB,
BC,
CA,
AD,
DB,
DC be bisected at the points
E,
F,
G,
H,
K,
L, and let
HE,
EG,
GH,
HK,
KL,
LH,
KF,
FG be joined.
Since
AE is equal to
EB, and
AH to
DH, therefore
EH is parallel to
DB. [
VI. 2]
For the same reason
HK is also parallel to
AB.
Therefore
HEBK is a parallelogram; therefore
HK is equal to
EB. [
I. 34]
But
EB is equal to
EA; therefore
AE is also equal to
HK.
But
AH is also equal to
HD; therefore the two sides
EA,
AH are equal to the two sides
KH,
HD respectively, and the angle
EAH is equal to the angle
KHD; therefore the base
EH is equal to the base
KD. [
I. 4]
Therefore the triangle
AEH is equal and similar to the triangle
HKD.
For the same reason the triangle
AHG is also equal and similar to the triangle
HLD.
Now, since two straight lines
EH,
HG meeting one another are parallel to two straight lines
KD,
DL meeting one another, and are not in the same plane, they will contain equal angles. [
XI. 10]
Therefore the angle
EHG is equal to the angle
KDL.
And, since the two straight lines
EH,
HG are equal to the two
KD,
DL respectively, and the angle
EHG is equal to the angle
KDL, therefore the base
EG is equal to the base
KL; [
I. 4] therefore the triangle
EHG is equal and similar to the triangle
KDL.
For the same reason the triangle
AEG is also equal and similar to the triangle
HKL.
Therefore the pyramid of which the triangle
AEG is the base and the point
H the vertex is equal and similar to the pyramid of which the triangle
HKL is the base and the point
D the vertex. [
XI. Def. 10]
And, since
HK has been drawn parallel to
AB, one of the sides of the triangle
ADB, the triangle
ADB is equiangular to the triangle
DHK, [
I. 29] and they have their sides proportional; therefore the triangle
ADB is similar to the triangle
DHK. [
VI. Def. 1]
For the same reason the triangle
DBC is also similar to the triangle
DKL, and the triangle
ADC to the triangle
DLH.
Now, since the two straight lines
BA,
AC meeting one another are parallel to the two straight lines
KH,
HL meeting one another, not in the same plane, they will contain equal angles. [
XI. 10]
Therefore the angle
BAC is equal to the angle
KHL.
And, as
BA is to
AC, so is
KH to
HL; therefore the triangle
ABC is similar to the triangle
HKL.
Therefore also the pyramid of which the triangle
ABC is the base and the point
D the vertex is similar to the pyramid of which the triangle
HKL is the base and the point
D the vertex.
But the pyramid of which the triangle
HKL is the base and the point
D the vertex was proved similar to the pyramid of which the triangle
AEG is the base and the point
H the vertex.
Therefore each of the pyramids
AEGH,
HKLD is similar to the whole pyramid
ABCD.
Next, since
BF is equal to
FC, the parallelogram
EBFG is double of the triangle
GFC.
And since, if there be two prisms of equal height, and one have a parallelogram as base, and the other a triangle, and if the parallelogram be double of the triangle, the prisms are equal, [
XI. 39] therefore the prism contained by the two triangles
BKF,
EHG, and the three parallelograms
EBFG,
EBKH,
HKFG is equal to the prism contained by the two triangles
GFC,
HKL and the three parallelograms
KFCL,
LCGH,
HKFG.
And it is manifest that each of the prisms, namely that in which the parallelogram
EBFG is the base and the straight line
HK is its opposite, and that in which the triangle
GFC is the base and the triangle
HKL its opposite, is greater than each of the pyramids of which the triangles
AEG,
HKL are the bases and the points
H,
D the vertices, inasmuch as, if we join the straight lines
EF,
EK, the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is greater than the pyramid of which the triangle
EBF is the base and the point
K the vertex.
But the pyramid of which the triangle
EBF is the base and the point
K the vertex is equal to the pyramid of which the triangle
AEG is the base and the point
H the vertex; for they are contained by equal and similar planes.
Hence also the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is greater than the pyramid of which the triangle
AEG is the base and the point
H the vertex.
But the prism in which the parallelogram
EBFG is the base and the straight line
HK its opposite is equal to the prism in which the triangle
GFC is the base and the triangle
HKL its opposite, and the pyramid of which the triangle
AEG is the base and the point
H the vertex is equal to the pyramid of which the triangle
HKL is the base and the point
D the vertex.
Therefore the said two prisms are greater than the said two pyramids of which the triangles
AEG,
HKL are the bases and the points
H,
D the vertices.
Therefore the whole pyramid, of which the triangle
ABC is the base and the point
D the vertex, has been divided into two pyramids equal to one another and into two equal prisms, and the two prisms are greater than the half of the whole pyramid. Q. E. D.