Bisect the arcs EF, FG, GH, HE, in the points O, P, R, S, and join EO, OF, &c. Then each of the triangles EOF, FPG, GRH, HSE, is greater than half of the segment of the circle in which it is. Upon each of these triangles form a pyramid having the same vertex with the cone; each of these pyramids is greater than the half of the segment of the cone in which it is: and thus dividing each of these arcs into two equal parts, and from the points of division drawing straight lines to the extremities of the arcs, and upon each of the triangles thus made, describing pyramids having the same vertex with the cone, and so on, there must at length remain (XII. lem. 1.) some segments of the cone which are together less than Z. Let these be the segments upon EO, OF, FP, PG, GR, RH, HS, SE: therefore the remainder of the cone, viz., the pyramid of which the base is the polygon EOFPGRHS, and its vertex the same with that of the cone, is greater than X. In the circle BD describe the polygon ATBYCVDQ similar to EOFPGRHS, and upon it form a pyramid having the same vertex with the cone AL. Then, because (XII. 1.) the square of AC is to the square of EG, as the polygon ATBYCVDQ to the polygon EOFPGRHS; and (XII. 2.) as the square of AC to the square of EG, so is the circle BD to the circle FH; therefore (V. I.) the circle BD is to the circle FH, as the polygon ATBYCVDQ to the polygon EOFPGRHS. But as the circle BD to the circle FH, so is the cone AL to X; and (XII. 6.) the polygon ATBYCVDQ to the polygon EOFPGRHS, as the pyramid of which the base is the first of these polygons, and vertex L, to the pyramid of which the base is the other polygon, and its vertex N : therefore, as the cone AL to X, so is the pyramid of which the base is the polygon ATBYCVDQ, and vertex L, to the pyramid the base of which is the polygon EOFPGRHS, and vertex N. But the cone AL is greater than the pyramid contained in it; therefore (V. 14.) the solid X is greater than the pyramid in the cone EN: but it is less, as was shown; which is absurd. Therefore BD is not to FH, as the cone AL to any solid which is less than the cone EN. In the same manner, it may be demonstrated that FH is not to BD, as the cone EN to any solid less than the cone AL. Nor can BD be to FH, as AL to any solid greater than EN. For, if it be possible, let it be so to the solid I, which is greater than En. Therefore, by inversion, as FH to BD, so is I to AL. But as I to AL, so is EN to some solid, which (V. 14.) must be less than AL, because I is greater than EN. Therefore, as FH is to BD, so is EN to a solid less than AL, which was shown to be impossible. BD, therefore, is not to FH, as AL is to any solid greater than EN: and it has been demonstrated that neither is BD to FH, as AL to any solid less than EN. Therefore BD is to FH, as the cone AL to the cone EN. But (V. 15.) as the cone is to the cone, so is the cylinder to the cylinder, because the cylinders are triple (XII. 10.) of the cones, each of each. Therefore, as the circle BD to the circle FH, so are the cylinders upon them of the same altitude. Cones and cylinders, therefore, &c. PROP. XII. THEOR. Similar cones and cylinders have to one another the triplicate ratio of that which the diameters of their bases have. If there be similar cones and cylinders of which the bases are the circles ABCD, EFGH; the diameters of the bases AC, EG; and the axes of the cones and cylinders KL, MN: the first cone has to the second, and the first cylinder to the second, the triplicate ratio of that which AC has to EG. For if the cone ABCDL have not to the cone EFGHN the triplicate ratio of that which AC has to EG, ABCDL will bave the triplicate of that ratio to some solid which is less or greater than EFGHN. First, let it have it to a less, viz., to the solid X. Make the same construction as in the preceding proposition, and it may be demonstrated in the same manner as in that proposition, that the pyramid, of which the base is the polygon EOFPGRHS, and vertex N, is greater than X. Describe also in the circle BD the polygon ATBYCVDQ similar to EOFPGRHS, upon which form a pyramid having the same vertex with the cone ; and let LAQ, NES be two of the triangles containing those pyramids ; and join KQ, MS. Then, in the triangles AKL, QKL, KA, KQ are equal, KL common, and the angles AKL, QKL equal, being (XI. def. 21.) right angles : therefore (I. 4.) LA, LQ are equal. For the same reason, NE, NS are equal ; and therefore AL: LQ :: EN : NS. Also (XI. def. 27.) because the cones are similar, AC: EG :: KL: MN; whence (V. 15.) AK : EM :: KL: MN, and, alternately, AK : KL : : ÈM : MN: therefore (VI. 6.) since the angles AKL, EMN are equal, the triangles AKL, EMN are equiangular; and (VI. 4.) LA: AK::NE: EM. Again, the angles AKQ, EMS, being each the same part of four right angles at the centres K, M, are equal ; and AK : KQ :: EM : MS: therefore (VI. 6.) the triangles AKQ, EMS are equiangular ; and (VI. 4.) AK: AQ :: EM : ES. But, as has been proved, LA : AK :: NE: EM: therefore, ex æquo, LA : AQ:: NE: ES; or since QL is equal to AL, and NS to NE, QL :QA :: SN : SE. Hence, (VI. 5.) the triangles LQA, NSE, having the sides about all their angles proportionals, are equiangular and similar: and therefore the pyramids AKQL, EMSN are similar, because (XI. B.) their solid angles are equal to one another, and they are contained by the same number of similar planes: and (XII. 8.) AKQL has to EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from D, V, C, &c., to K and from H, R, G, &c., to M, and pyramids be described on the triangles having the same vertices with the cones, it may be demonstrated that each pyramid in the first cone has to each in the other, taking them in the same order, the triplicate ratio of that which AK has to EM ; that is, which AC bas to EG. But (V. 12.) as one antecedent is to its cons onsequent, so are all the ante. cedents to all the consequents; therefore as the pyramid AKQL to the pyramid EMSN, so is the whole pyramid DQATBYCVL, to the whole HSEOFPGRN: and therefore also the first of these two last named pyramids has to the other the triplicate ratio of that which AC has to FG. But, by the hypothesis, the cone ABCDL has to X, the triplicate ratio of that which AC has to EG; therefore, as the cone ABCDL is to X, so is the pyramid DQATBYCVL, to the pyramid HSEOFPGRN. But the same cone is greater than the pyramid contained in it; therefore (V. 14.) X is greater than the pyramid HSEOFPGRN; but it is also less ; which is impossible. Therefore the cone ABCDL has not to any solid which is less than EFGHN, 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 solid which is less than ABCDL, the triplicate ratio of that which EG has to AC. Nor can ABCDL have to any solid which is greater than 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, Z has to ABCDL, the triplicate ratio of that which EG has to AC. But as Z is to ABCDL, so is EFGHN to some solid, which (V. 14.) must be less than ABCDL, because Z is greater than EFGHN. Therefore EFGHN has to a solid which is less than ABCDL, the triplicate ratio of that which EG has to AC, which was demonstrated to be impossible. ABCDL, therefore, has not to any solid greater than 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 EFGHN. Therefore ABCDL has to EFGHN, the triplicate ratio of that which AC has to EG. But (V. 15.) as the cone is to the cone, so is the cylinder to the cylinder; for (XII. 10.) every cone is a 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 EG: wherefore similar cones, &c. PROP. XIII. THEOR. р R N E B G H с D If a cylinder be cut by a plane parallel to its bases; the plane divides the cylinder into two cylinders, which are proportional to their axes. Let the cylinder AD be cut by the plane GH parallel to the bases AB, CD, meeting the axis EF in K, and let the line GH be the common section of the plane GH and the surface of the cylinder. Let AEFC be the rectangle in any position of it, by the revolution of which about EF the cylinder AD is described ; and let GK be the common section of the plane GH, and the plane AF. Then, because the parallel planes AB, GH are cut by the plane AEKG, A E, KG, their common sections with it, are (XI. 16.) parallel; wherefore AK is a parallelogram, and GK equal to EA, the radius of the circle AB. For the same reason, each of the straight lines drawn from the point K to the line GA may be proved to be equal to the radius of the circle AB: they are therefore all equal to one another ; and (I. def. 30.) the line GH is the circumference of a circle, of which the centre is K. Therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same that would be described by the revolution of the rectangles AK, GF, about EK, KF: and it is now to be shown that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF. Produce EF both ways; and take any number of straight lines EN, NL, each equal to EK, and any number FX, XM, each equal to FK; and let planes parallel to AB, CD, pass through L, N, X, M. The common sections of these planes with the cylinder continued, are circles the centres of which are the points L, N, X, M, as was proved of the plane GH; and these planes cut off the cylinders PR, RB, DT, TQ. Again, because the axes LN, NE, EK are all equal : therefore the cylinders PR, RB, BG, being to one another (XII. 11.) as their bases, are equal. But because the axes LN, NE, EK are equal to one another, as also the cylinders PR, RB, BG, and that there are as many axes T X V M м 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 KL be equal to KM, PG is equal to GQ; if KL be greater than KM, PG is greater than QG; and if less, less. Since therefore, of EK and BG there have been taken any like multiples whatever, viz., KL and PG; and of KF and GD, any like multiples whatever, viz., KM and GQ; and since it has been demonstrated, that if KL be greater than KM, PG is greater than GQ; if equal, equal; and if less, Jess: therefore (V. def. 5.) the axis EK is to the axis KF, as the cylinder BG to the cylinder GD: wherefore, if a cylinder, &c. PROP. XIV. THEOR. 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: EB is to FD, as the axis GH to the axis KL. Produce KL making LN equal to 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 (XII. 11.) as their bases : but the bases are equal ; therefore also EB, CM are equal. Again, (XII. 13.) because the cylinder FM is cut by the plane CD parallel to its bases, CM is to FD, as LN to KL. But CM is equal to EB, and the axis LN to GH: therefore EB is to FD, as GH to KL: and (V. 15.) as EB to FD, so is the cone ABG to the cone CDK, because (XII. 10.) the cylinders are triple of the cones. Therefore also GH is to KL, as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c. K G E L D A B H M м N PROP. XV. THEOR. Tue bases and altitudes of equal cones and cylinders are reciprocally proportional ; and (2.) if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. 1. Let the circles BD, FH, 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 BD to FH, so is MN to KL. The altitudes MN, KL are either equal or not. First, let |