the same altitude, they are to one another as their bases: Book XII. But their bases are equal, therefore also the cylinders EB, CM are equal. And because a 11. 12. to the axis KL: And as the A cylinder EB to the cylinder FD, so is the cone ABG to c 15. 5. the cone CDK, because the cylinders are tripled the cones: d 10. 12. PROP. XV. THEOR. The bases and altitudes of equal cones and cylin- See N. ders are reciprocally proportional; and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. Let the circles ABCD, EFGH, the diameters of which are AC, EG, be the bases, and KL, MN the axes, as also the altitudes, of equal cones and cylinders; and let ALC, ENG be the cones, and AX, EO the cylinders: The bases and altitudes of the cylinders AX, EO are reciprocally proportional; that is, as the base ABCD to the base EFGH, so is the altitude MN, to the altitude KL. Either the altitude MN is equal to the altitude KL, or these altitudes are not equal. First, Let them be equal; and the cylinders AX, EO being also equal, and cones and cylinders of the same altitude being to one another as their bases, a 11. 12. therefore the base ABCD is equal to the base EFGH; and b A. 5. as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL. But let the altitudes KL, MN be unequal, and MN the greater of the two, and from MN take N R Book XII. MP equal to KL, and through the point P cut the cylinder EO by the plane TYS, parallel to the opposite planes of the circles EFGH, RO; therefore the common section of the plane TYS and the cylinder EO is a circle, and consequently ES is a cylinder the base of which is the circle EFGH, and altitude MP: And because the cylinder AX is equal to the cylinder EO, as AX is to the cylinder ES, so is the cylinder EO to c 7. 5. a 11. 12. c the same ES. But X Y H C E G K M B F as the cylinder AX to the cylinder ES, so is the base ABCD to the base EFGH: for the cylinders AX, ES are of the same altitude; and as the cylinder EO to the cylind 13. 12. der ES, sod is the altitude MN to the altitude MP, because the cylinder EO is cut by the plane TYS, parallel to its opposite planes. Therefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude MP: But MP is equal to the altitude KL; wherefore as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; that is, the bases and altitudes of the equal cylinders AX, EO are reciprocally proportional. But let the bases and altitudes of the cylinders AX, EO, be reciprocally proportional, viz. the base ABCD to the base EFGH, as the altitude MN to the altitude KL: The cylinder AX is equal to the cylinder EO. First, Let the base ABCD be equal to the base EFGH: then because as the base ABCD is to the base EFGH, so is the altitude MN to the altitude KL: MN is equal e to KL, f11. 12. and therefore the cylinder AX is equal to the cylinder EO. e A. 5. e But let the bases ABCD, EFGH be unequal, and let ABCD be the greater; and because as ABCD is to the base EFGH, so is the altitude MN to the altitude KL; therefore MN is greater than KL. Then, the same construction being made as before, because as the base ABCD to the base EFGH, so is the altitude MN to the altitude KL; and because the altitude KL is equal to the altitude MP; therefore the base ABCD is f to the base EFGH, as the cylinder AX to the cylinder ES; and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES: Book XII. Therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the same ES: Whence the cylinder AX is equal to the cylinder EO: and the same reasoning holds in cones. Q. E. D. PROP. XVI. PROB. To describe in the greater of two circles that have the same centre, a polygon of an even number of equal sides, that shall not meet the less circle. Let ABCD, EFGH be two given circles having the same centre K; it is required to inscribe in the greater circle ABCD, a polygon of an even number of equal sides, that shall not meet the less circle. Through the centre K draw the straight line BD, and from the point G, where it meets the circumference of the less circle, draw GA at right angles to BD, and produce it to C; therefore AC touches the circle EFGG: Then, if the circum- a 16. 3. ference BAD be bisected, and the half of it be again bisected, A H E K GM D F N and so on, there must at length remain a circumference less b b Lemma. than AD: Let this be LD; and from the point L draw LM perpendicular to BD, and produce it to N; and join LD, DN. Therefore LD is equal to DN; and because LN is parallel to AC, and AC touches the circle EFGH; therefore LN does not meet the circle EFGH: And much less shall the straight lines LD, DN, meet the circle EFGH: B So that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be described in the circle a polygon of an even number of equal sides not meeting the less circle. Which was to be done. S LEMMA II. If two trapezium's ABCD, EFGH be inscribed in the circles, the centres of which are the points K, L; and if the sides AB, DC be parallel, as also EF, HG; and the other four sides AD, BC, EH, FG, be all equal to one another; but the side AB greater than EF, and DC greater than HG. The straight line KA from the centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle. If it be possible, let KA be not greater than LE; then KA must be either equal to it or less. First, Let KA be equal to LE: Therefore, because in two equal circles, AD, BC, in the one, are equal to EH, FG in the other, the circumferences AD, BC are equal to the circumferences EH, FG: but because the straight lines AB, DC are respectively greater than EF, GH, the circumferences AB, DC are greater than EF, HG: Therefore, the whole circumference ABCD is greater than the whole EFGH; but it is also equal to it, which is impossible: Therefore the straight line KA is not equal to LE. But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle a 2. 6. MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM, which are respectively parallel to, and less than EF, FG, GH, HE: Then, because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore the circum ference AD is greater than MP; for the same reason, the Book XII. circumference BC is greater than NO; and because the straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN: Therefore the circumference AB is greater than MN; and, for the same reason, the circumference DC is greater than PO: Therefore the whole circumference ABCD is greater than the whole MNOP; but it is likewise equal to it, which is impossible: Therefore KA is not less than LE: nor is it equal to it; the straight line KA must therefore be greater than LE. Q. E. D. COR. And if there be an isosceles triangle, the sides of which are equal to AD, BC, but its base less than AB the greater of the two sides AB, DC; the straight line KA may, in the same manner, be demonstrated to be greater than the straight line drawn from the centre to the circumference of the circle described about the triangle. PROP. XVII. PROB. To describe in the greater of two spheres which See N. have the same centre, a solid polyhedron, the superficies of which shall not meet the less sphere. Let there be two spheres about the same centre A; it is required to describe in the greater a solid polyhedron, the superficies of which shall not meet the less sphere. Let the spheres be cut by a plane passing through the centres; the common sections of this plane with the spheres will be circles; because the sphere is described by the revolution of a semicircle about the diameter remaining unmoveable; so that in whatever position the semicircle be conceived, the common section of the plane in which it is with the superficies of the sphere is the circumference of a circle; and this is a great circle of the sphere, because the diameter of the sphere, which is likewise the diameter of the circle, is greater a 15. 3. than any straight line in the circle or sphere: Let then the circle made by the section of the plane with the greater sphere be BCDE, and with the less sphere be FGH; and draw the two diameters BD, CE, at right angles to one another: and in BCDE, the greater of the two circles, describe b a polygon b 16. 12. of an even number of equal sides not meeting the less circle FGH; and let its sides, in BE the fourth part of the circle, be BK, KL, LM, ME; join KA, and produce it to N; and from A draw AX at right angles to the plane of the circle |