PROPOSITION 13.
To construct a pyramid,
to comprehend it in a given sphere,
and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
Let the diameter
AB of the given sphere be set out, and let it be cut at the point
C so that
AC is double of
CB; let the semicircle
ADB be described on
AB, let
CD be drawn from the point
C at right angles to
AB, and let
DA be joined; let the circle
EFG which has its radius equal to
DC be set out, let the equilateral triangle
EFG be inscribed in the circle
EFG, [
IV. 2] let the centre of the circle, the point
H, be taken, [
III. 1] let
EH,
HF,
HG be joined; from the point
H let
HK be set up at right angles to the plane of the circle
EFG, [
XI. 12] let
HK equal to the straight line
AC be cut off from
HK, and let
KE,
KF,
KG be joined.
Now, since
KH is at right angles to the plane of the circle
EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle
EFG. [
XI. Def. 3]
But each of the straight lines
HE,
HF,
HG meets it: therefore
HK is at right angles to each of the straight lines
HE,
HF,
HG.
And, since
AC is equal to
HK, and
CD to
HE, and they contain right angles, therefore the base
DA is equal to the base
KE. [
I. 4]
For the same reason each of the straight lines
KF,
KG is also equal to
DA; therefore the three straight lines
KE,
KF,
KG are equal to one another.
And, since
AC is double of
CB, therefore
AB is triple of
BC.
But, as
AB is to
BC, so is the square on
AD to the square on
DC, as will be proved afterwards.
Therefore the square on
AD is triple of the square on
DC.
But the square on
FE is also triple of the square on
EH, [
XIII. 12] and
DC is equal to
EH; therefore
DA is also equal to
EF.
But
DA was proved equal to each of the straight lines
KE,
KF,
KG; therefore each of the straight lines
EF,
FG,
GE is also equal to each of the straight lines
KE,
KF,
KG; therefore the four triangles
EFG,
KEF,
KFG,
KEG are equilateral.
Therefore a pyramid has been constructed out of four equilateral triangles, the triangle
EFG being its base and the point
K its vertex.
It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
For let the straight line
HL be produced in a straight line with
KH, and let
HL be made equal to
CB.
Now, since, as
AC is to
CD, so is
CD to
CB, [
VI. 8, Por.] while
AC is equal to
KH,
CD to
HE, and
CB to
HL, therefore, as
KH is to
HE, so is
EH to
HL; therefore the rectangle
KH,
HL is equal to the square on
EH. [
VI. 17]
And each of the angles
KHE.
EHL is right; therefore the semicircle described on
KL will pass through
E also. [cf.
VI. 8,
III. 31.]
If then,
KL remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, it will also pass through the points
F,
G, since, if
FL,
LG be joined, the angles at
F,
G similarly become right angles; and the pyramid will be comprehended in the given sphere.
For
KL, the diameter of the sphere, is equal to the diameter
AB of the given sphere, inasmuch as
KH was made equal to
AC, and
HL to
CB.
I say next that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid
For, since
AC is double of
CB, therefore
AB is triple of
BC; and,
convertendo,
BA is one and a half times
AC.
But, as
BA is to
AC, so is the square on
BA to the square on
AD.
Therefore the square on
BA is also one and a half times the square on
AD.
And
BA is the diameter of the given sphere, and
AD is equal to the side of the pyramid.
Therefore the square on the diameter of the sphere is one and a half times the square on the side of the pyramid. Q. E. D.
LEMMA.
It is to be proved that, as
AB is to
BC, so is the square on
AD to the square on
DC.
For let the figure of the semicircle be set out, let
DB be joined, let the square
EC be described on
AC, and let the parallelogram
FB be completed.
Since then, because the triangle
DAB is equiangular with the triangle
DAC, as
BA is to
AD, so is
DA to
AC, [
VI. 8,
VI. 4] therefore the rectangle
BA,
AC is equal to the square on
AD. [
VI. 17]
And since, as
AB is to
BC, so is
EB to
BF, [
VI. 1] and
EB is the rectangle
BA,
AC, for
EA is equal to
AC, and
BF is the rectangle
AC,
CB, therefore, as
AB is to
BC, so is the rectangle
BA,
AC to the rectangle
AC,
CB.
And the rectangle
BA,
AC is equal to the square on
AD, and the rectangle
AC,
CB to the square on
DC, for the perpendicular
DC is a mean proportional between the segments
AC,
CB of the base, because the angle
ADB is right. [
VI. 8, Por.]
Therefore, as
AB is to
BC, so is the square on
AD to the square on
DC. Q. E. D.