PROPOSITION 15.
To construct a cube and comprehend it in a sphere,
like the pyramid; and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.
Let the diameter
AB of the given sphere be set out, and let it be cut at
C so that
AC is double of
CB; let the semicircle
ADB be described on
AB, let
CD be drawn from
C at right angles to
AB, and let
DB be joined; let the square
EFGH having its side equal to
DB be set out, from
E,
F,
G,
H let
EK,
FL,
GM,
HN be drawn at right angles to the plane of the square
EFGH, from
EK,
FL,
GM,
HN let
EK,
FL,
GM,
HN respectively be cut off equal to one of the straight lines
EF,
FG,
GH,
HE, and let
KL,
LM,
MN,
NK be joined; therefore the cube
FN has been constructed which is contained by six equal squares.
It is then required to comprehend it in the given sphere, and to prove that the square on the diameter of the sphere is triple of the square on the side of the cube.
For let
KG,
EG be joined.
Then, since the angle
KEG is right, because
KE is also at right angles to the plane
EG and of course to the straight line
EG also, [
XI. Def. 3] therefore the semicircle described on
KG will also pass through the point
E.
Again, since
GF is at right angles to each of the straight lines
FL,
FE,
GF is also at right angles to the plane
FK; hence also, if we join
FK,
GF will be at right angles to
FK; and for this reason again the semicircle described on
GK will also pass through
F.
Similarly it will also pass through the remaining angular points of the cube.
If then,
KG remaining fixed, the semicircle be carried round and restored to the same position from which it began to be moved, the cube will be comprehended in a sphere.
I say next that it is also comprehended in the given sphere.
For, since
GF is equal to
FE, and the angle at
F is right, therefore the square on
EG is double of the square on
EF.
But
EF is equal to
EK; therefore the square on
EG is double of the square on
EK; hence the squares on
GE,
EK, that is the square on
GK [
I. 47], is triple of the square on
EK.
And, since
AB is triple of
BC, while, as
AB is to
BC, so is the square on
AB to the square on
BD, therefore the square on
AB is triple of the square on
BD.
But the square on
GK was also proved triple of the square on
KE.
And
KE was made equal to
DB; therefore
KG is also equal to
AB.
And
AB is the diameter of the given sphere; therefore
KG is also equal to the diameter of the given sphere.
Therefore the cube 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 triple of the square on the side of the cube. Q. E. D.