Book 13
14
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.
14
For the same reason LH is also equal to HE; therefore the triangle LEH is equilateral.
14
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.
14
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.
14
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.
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.
