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.] [p. 475]

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. [p. 476]

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.

[p. 478]