Kitap 13
14
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.
14
I say next that it is also comprehended in the given sphere.
14
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]
14
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]
14
Again, since AC is equal to CB, AB is double of BC.
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.
