polygon HSEOFPGR, and vertex N: but it is also less; which is impossible : therefore the cone, of which the base is the circle ABCD and vertex L, has not to any solid which is less than the cone of which the base is the circle EFGH and vertex N, the triplicate ratio of that which AC has to EG. In the same manner it may be demonstrated, that neither has the cone EFGHN to any * 14.5. solid which is less than the cone ABCDL, the triplicate ratio of that which EG has to AC. Nor can the cone ABCDL have to any solid which is greater than the cone EFGHN, the triplicate ratio of that which AC has to EG. For, if it be possible, let it have it to a greater, viz. to the solid Z: therefore, inversely, the solid Z has to the cone ABCDL, the triplicate ratio of that which EG has to AC: but as the solid Z is to the cone ABCDL, so is the cone EFGHN to some solid, which must be less * than the cone ABCDL, because the solid Z is greater than the cone EFGHN; therefore the cone EFGHN has to a solid which is less than the cone ABCDL the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible: therefore the cone ABCDL has not to any solid greater than the cone EFGHN, the triplicate ratio of that which AC has to EG: and it was demonstrated, that it could not have that ratio to any solid less than the cone EFGHN: therefore the cone ABCDL has to the cone EFGHN, the triplicate ratio of that which AC has to EG, but as the cone is to the cone, so the cylinder to the cylinder; for every cone is the third part of the cylinder upon the same base, and of the same altitude: therefore also the cylinder has to the cylinder, the triplicate ratio of that which AC has to EĞ. Wherefore, similar cones, &c. * 15. 5. + 10. 12. Q. E. D. If a cylinder be cut by a plane parallel to its opposite See N. planes, or bases, it divides the cylinder into two cylin- cy B T Y Let the cylinder AD be cut by the plane GH paral- R A А E V ways: any num * 11. 12. AK G RB, DT, TQ. And because the axes іEB linders PR, RB, BG, are equal : and KH because the axes LN, NE, EK, are D equal to one another, as also the cylinders PR, RB, BG, and that there are as many T T Y axes as cylinders; therefore, whatever multiple the axis KL is of the axis KE, the same multiple is the cylinder PG of the cylinder GB: for the same reason, whatever multiple the axis MK is of the axis KF, the same multiple is the cylinder QG of the cylinder GD: and if the axis KL be equal to the axis KM, the cylinder PG is equal to the cylinder GQ; and if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if less, less: therefore, since there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD; and that of the axis EK and cylinder BG there have been taken any equimultiples whatever, viz. the axis KL and cylinder PG, and of the axis KF and cylinder GD, any equimultiples whatever, viz. the axis KM and cylinder GQ; and since it has been demonstrated, that if the axis KL be greater than the axis KM, the cylinder PG is greater than the cylinder GQ; and if equal, equal; and if less, * 5 Def. 5. less: therefore* as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. Wherefore, if a cylinder, &c. Q. E. D. PROP. XIV. THEOR. See N. Cones and cylinders upon equal bases are to one another as their altitudes, Let the cylinders EB, FD, be upon the equal bases AB, CD: as the cylinder EB to the cylinder FD, so shall the axis GH be to the axis KL. Produce the axis KL to the point N, and make LN equal to the axis GH, and let CM be a cylinder of which the base is CD, and axis LN. Then because the cylinders EB, CM, have the same altitude, they are - to one another as their bases *: but their bases are equal, therefore also the cylinders EB, CM, are equal : and * 11. 12. E * 13. 12. N because the cylinder FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to the cylinder FD, so is * the axis LN to the axis KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; there AL fore as the cylinder EB to the cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB to the cylinder FD, so is * the * 15.5. cone ABG to the cone CDK, because the cylinders are triple * of the cones: therefore also the axis GH is * 10. 12. to the axis KL, as the cone ABG to the cone CDK, and as the cylinder EB to the cylinder FD. Wherefore cones, &c. R. E. D. PROP. XV. THEOP. The bases and altitudes of equal cones and cylinders are See N. 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, ÉO, the cylinders: the bases and altitudes of the cylinders AX, EO shall be reciprocally proportional ; that is, as the base ABCD to the base EFGH so shall the altitude MN be 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*, therefore the base ABCD is * 11. 12. equal * to the base EFGH; and as the base ABCD is * A. 5. 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 MP equal to KL, and X YZFDs through the point P cut the cylinder EO by the plane TYS, parallel to the opposite planes of the cireles EFGH, RO; there R M B * 7.5. * 11. 12. * 13. 12. fore 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 the same ES: but 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 cylinder ES, so * is the altitude MŇ 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 shall be 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 * to KL; and therefore the cylinder AX is equal * to the cylinder EO. 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 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: 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. * A.5. * 11. 12. * A. 5. * 11. 12: to Q. E. D. |