PROPOSITION 14.
To construct an octahedron and comprehend it in a sphere,
as in the preceding case;
and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.
Let the diameter
AB of the given sphere be set out, and let it be bisected at
C; let the semicircle
ADB be described on
AB, let
CD be drawn from
C at right angles to
AB, let
DB be joined; let the square
EFGH, having each of its sides equal to
DB, be set out, let
HF,
EG be joined, from the point
K let the straight line
KL be set up at right angles to the plane of the square
EFGH [
XI. 12], and let it be carried through to the other side of the plane, as
KM; from the straight lines
KL,
KM let
KL,
KM be respectively cut off equal to one of the straight lines
EK,
FK,
GK,
HK, and let
LE,
LF,
LG,
LH,
ME,
MF,
MG,
MH be joined.
Then, since
KE is equal to
KH, and the angle
EKH is right, therefore the square on
HE is double of the square on
EK. [
I. 47]
Again, since
LK is equal to
KE, and the angle
LKE is right, therefore the square on
EL is double of the square on
EK. [
id.]
But the square on
HE was also proved double of the square on
EK; therefore the square on
LE is equal to the square on
EH; therefore
LE is equal to
EH.
For the same reason
LH is also equal to
HE; therefore the triangle
LEH is equilateral.
Similarly we can prove that each of the remaining triangles of which the sides of the square
EFGH are the bases, and the points
L,
M the vertices, is equilateral; therefore an octahedron has been constructed which is contained by eight equilateral triangles.
It is next required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is double of the square on the side of the octahedron.
For, since the three straight lines
LK,
KM,
KE are equal to one another, therefore the semicircle described on
LM will also pass through
E.
And for the same reason, if,
LM 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,
H, and the octahedron will have been comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For, since
LK is equal to
KM, while
KE is common, and they contain right angles, therefore the base
LE is equal to the base
EM. [
I. 4]
And, since the angle
LEM is right, for it is in a semicircle, [
III. 31] therefore the square on
LM is double of the square on
LE. [
I. 47]
Again, since
AC is equal to
CB,
AB is double of
BC.
But, as
AB is to
BC, so is the square on
AB to the square on
BD; therefore the square on
AB is double of the square on
BD.
But the square on
LM was also proved double of the square on
LE.
And the square on
DB is equal to the square on
LE, for
EH was made equal to
DB.
Therefore the square on
AB is also equal to the square on
LM; therefore
AB is equal to
LM.
And
AB is the diameter of the given sphere; therefore
LM is equal to the diameter of the given sphere.
Therefore the octahedron has been comprehended in the given sphere, and it has been demonstrated at the same time that the square on the diameter of the sphere is double of the square on the side of the octahedron. Q. E. D.