¶The 28. Proposition. To proue that an Octohedron is quadruple sesquialter to a Cube inscri∣bed in it.
SVppose that the Octohedron geuen be ABCDEF: and let the cube inscribed in it be GHIK, VQRS. Then I say, that the Octohedron is quadruple sesquialter to the cube inscribed in it. Forasmuch as the lines drawen from the centre of the Octohedron, or of the Sphere which containeth it, vnto the centres of the bases of the Octohedron, are pro∣ued equall, by the 21. of the fiuetenth: and the angles of the cube are set in the centres of those bases, by the 4. of the fiuetenth: it followeth, that the selfe same right lines are drawen from one and the selfe same centre of the cube and of the Octohedron: for they haue eche one and the selfe same centre, by the Corollary of the 21. of the fiuetenth. Let that centre be the point T. Wherefore the base BDFC, which cutteth the Octohedron into two equall and quadrilater pyra∣mids, by the Corollary of the 14. of the thirtenth, shall also cut the cube into two equall partes, by the Corollary of the 39. of the eleuenth. For it passeth by the centre T, by that which was demonstra∣ted in the 14. of the thirtenth. And forasmuch as the base of the cube is in the 4. centres G, H, I, K, of the bases of the pyramis ABDFC, a plaine LNOM, extended by those pointes, shall be parallel to the plaine BDFC, by that which was demonstrated in the 4. of the fiuetenth, and shall cut the pyra∣mis in the pointes L, N, O, M: and the lines LN, BD, and NO, DF, shall be parallels, so also shall the lines OM, FC, and LM, BC: and the square GHIK of the cube shall be inscribed in the square LNOM, by the same. Wherefore the square LNOM is duple to the square GHIK, by the 47. of the first. From the solide angle A, let there be drawen to the plaine superficies BDFC, a perpen∣dicular, which let fall vppon it in the point T, and let the same perpendicular be AT, cutting the plaine LNOM in the point P. And it shall also be a perpendicular to the plaine LNOM, by the Corollary of the 14. of the eleuenth. Againe from the angle BAD of the triangle ADB, let there be drawen by the centre H of the triangle, to the base a line AHX. Wherefore the line AX is ses∣quialter to the line AH, by the Corollary of the 12. of the thirtenth. Wherefore the line AH is duple to the line HX. But the other lines AB, AD, AF, AC, and the perpendicular APT, are cut like vnto the line AHX, by the 17. of the eleuenth: Wherefore the line AP is double to the line PT. Wherefore the line AP is the altitude of the cube, for the line PT is the halfe thereof.