Kitap 13
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For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined.
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Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC.
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Therefore the circumference BE is a sixth part of the circumference of the circle; therefore the straight line BE belongs to a hexagon; therefore it is equal to the radius DE. [IV. 15, Por.]
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And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE.
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But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47] therefore the squares on AB, BE are quadruple of the square on BE.
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Therefore, separando, the square on AB is triple of the square on BE.
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But BE is equal to DE; therefore the square on AB is triple of the square on DE.
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Therefore the square on the side of the triangle is triple of the square on the radius. Q. E. D.
PROPOSITION 13.
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To construct a pyramid, to comprehend it in a 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.
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Let the diameter AB of the given sphere be set out, and let it be cut at the point C so that AC is double of CB; let the semicircle ADB be described on AB, let CD be drawn from the point C at right angles to AB, and let DA be joined; let the circle EFG which has its radius equal to DC be set out, let the equilateral triangle EFG be inscribed in the circle EFG, [IV. 2] let the centre of the circle, the point H, be taken, [III. 1] let EH, HF, HG be joined; from the point H let HK be set up at right angles to the plane of the circle EFG, [XI. 12] let HK equal to the straight line AC be cut off from HK, and let KE, KF, KG be joined.
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Now, since KH is at right angles to the plane of the circle EFG, therefore it will also make right angles with all the straight lines which meet it and are in the plane of the circle EFG. [XI. Def. 3]
