Book 13
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Porism. From this it is manifest that, when the side of the cube is cut in extreme and mean ratio, the greater segment is the side of the dodecahedron. Q. E. D.
PROPOSITION 18.
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To set out the sides of the five figures and to compare them with one another.
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Let AB, the diameter of the given sphere, be set out, and let it be cut at C so that AC is equal to CB, and at D so that AD is double of DB; let the semicircle AEB be described on AB, from C, D let CE, DF be drawn at right angles to AB, and let AF, FB, EB be joined.
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Then, since AD is double of DB, therefore AB is triple of BD.
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Convertendo, therefore, BA is one and a half times AD.
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But, as BA is to AD, so is the square on BA to the square on AF, [V. Def. 9, VI. 8] for the triangle AFB is equiangular with the triangle AFD; therefore the square on BA is one and a half times the square on AF.
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But the square on the diameter of the sphere is also one and a half times the square on the side of the pyramid. [XIII. 13]
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And AB is the diameter of the sphere; therefore AF is equal to the side of the pyramid.
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Again, since AD is double of DB, therefore AB is triple of BD.
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But, as AB is to BD, so is the square on AB to the square on BF; [VI. 8, V. Def. 9] therefore the square on AB is triple of the square on BF.
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But the square on the diameter of the sphere is also triple of the square on the side of the cube. [XIII. 15]
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And AB is the diameter of the sphere; therefore BF is the side of the cube.
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And, since AC is equal to CB, therefore AB is double of BC.
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But, as AB is to BC, so is the square on AB to the square on BE; therefore the square on AB is double of the square on BE.
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But the square on the diameter of the sphere is also double of the square on the side of the octahedron. [XIII. 14]
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And AB is the diameter of the given sphere; therefore BE is the side of the octahedron.
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Next, let AG be drawn from the point A at right angles to the straight line AB, let AG be made equal to AB, let GC be joined, and from H let HK be drawn perpendicular to AB.
