Book 13
13
But the square on FE is also triple of the square on EH, [XIII. 12] and DC is equal to EH; therefore DA is also equal to EF.
13
But DA was proved equal to each of the straight lines KE, KF, KG; therefore each of the straight lines EF, FG, GE is also equal to each of the straight lines KE, KF, KG; therefore the four triangles EFG, KEF, KFG, KEG are equilateral.
13
Therefore a pyramid has been constructed out of four equilateral triangles, the triangle EFG being its base and the point K its vertex.
13
It is next required to comprehend it in the given sphere and to prove that the square on the diameter of the sphere is one and a half times the square on the side of the pyramid.
13
For let the straight line HL be produced in a straight line with KH, and let HL be made equal to CB.
13
Now, since, as AC is to CD, so is CD to CB, [VI. 8, Por.] while AC is equal to KH, CD to HE, and CB to HL, therefore, as KH is to HE, so is EH to HL; therefore the rectangle KH, HL is equal to the square on EH. [VI. 17]
13
And each of the angles KHE. EHL is right; therefore the semicircle described on KL will pass through E also. [cf. VI. 8, III. 31.]
13
If then, KL 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, since, if FL, LG be joined, the angles at F, G similarly become right angles; and the pyramid will be comprehended in the given sphere.
