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IF AB (Fig. 100) be the diameter of a sphere(α) 1.1 divided in D so that AD shall be ⅓ AB, then (having erected the perpendicular DF) BF will be the side of the Tetraëdrum by Prop. 56. and AF the side of the Hexaëdrum by Prop. 58. Cons. 2. and BE or AE (erecting from the center the perpendicular CE) will be the side of the Octaëdrum by Prop. 57. Now if AF be cut in mean and extreme reason in O, you'l have AO the side of the Dodecaëdrum by Prop. 59. Lastly, if you erect BG double of CB, HI will be double of CI, and the □ of HI=4 □ of CI; consequently the □ CH or CB=5 □ CI. Therefore the □ of AB (double of CH) is also=to 5 □ of HI (which is double of CI) therefore HI is the radius of the circle circumscribing the Pentagon of the Icosaëdrum, and IB the side of the Decagon inscribed in the same circle, and HB the side of the Pentagon, and also the side of the Ico∣saëdrum, by Prop. 60.