Book 13
14
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.
14
But the square on LM was also proved double of the square on LE.
14
And the square on DB is equal to the square on LE, for EH was made equal to DB.
14
Therefore the square on AB is also equal to the square on LM; therefore AB is equal to LM.
14
And AB is the diameter of the given sphere; therefore LM is equal to the diameter of the given sphere.
14
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.
PROPOSITION 15.
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.
15
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.
