Book XII. than the cone EFGHN. First, let it have it to a less, viz. to the folid X. make the same construction as in the preceding Proposition, and N II. L; and let NES be one of the triangles containing the pyramid upon the polygon EOFPGRHS of which the vertex is N; and join KQ; MS. because then the cone ABCDL is similar to the cone EFGHN, 2. 24. Def. AC is a to EG, as the axis KL to the axis MN; and as AC to EG, fobis AK to EM; therefore as AK to EM, so is KL to MN; and, alternately, AK to KL, as EM to MN. and the right angles AKL, EMN are equal; therefore, the sides about these equal angles being Com 6. 6. proportionals, the triangle AKL is similar to the triangle EMN, again, because AK is to KQ , as EM to MS, and that these fides b. 15. 5. d. 22. 5. are about equal angles AKQ, EMS, because these angles are, Book XII. each of them, the fame part of four right angles at the centers K, M; therefore the triangle AKQ is similar to the triangle EMS. c. 6.6. and because it has been shewn that as AK to KL, so is EM to MN, and that AK is equal to KQ, and EM to MS, as OK to KL, so is SM to MN; and therefore, the sides about the right angles OKL, SMN being proportionals, the triangle LKQ is similar to the triangle NMS. and because of the fimilarity of the triangles AKL, EMN, as LA is to AK, so is NE to EM; and by the similarity of the triangles AKQ, EMS, as K A to AQ, fo ME to ES; ex aequali d, LA is to AQ, as NE to ES. again, because of the similarity of the triangles LOK, NSM, as LQ to OK, so NS to SM; and from the similarity of the triangles KAQ, MES, as KQ to QA, fo MS to SE; ex aequalid, LQ is to QA, as NS to SE. and it was proved that QA is to AL, as SE to EN; therefore, again, ex aequali, as QL to LA, fo is SN to NE. wherefore the triangles LQA, NSE, having the sides about all their angles proportionals, are equiangular and fimilar to one another. and therefore the pyramid of which e. 5. 6. the base is the triangle AKQ, and vertex L, is similar to the pyramid the base of which is the triangle EMS, and vertex N, because their solid angles are equal to one another, and they are contained f. B. II. by the same number of similar planes. but similar pyramids which have triangular bases have to one another the triplicate & ratio of that 8. 8. 12. which their homologous fides have; therefore the pyramid AKOL has to the pyramid EMSN the triplicate ratio of that which AK has to EM. In the same manner, if straight lines be drawn from the points D, V, C, Y, B, T to K, and from the points H, R, G, P, F, O to M, and pyramids be erected upon the triangles having the fame 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 the side AK has to the side EM; that is, which AC has to EG. but as one antecedent to its consequent, so are all the antecedents to all the consequents h; ho 520 5. therefore as the pyramid AKOL to the pyramid EMSN, so is the whole pyramid the base of which is the polygon DQATBYCV, and vertex L, to the whole pyramid of which the base is the polygon HISEOFPGR, and vertex N. wherefore also the first of these two last named pyramids has to the otherthe triplicate ratio of that which AC has to EG. but, by the Hypothesis, the cone of which the base is the circle ABCD, and vertex L has to the solid X, the triplicate Book XII. ratio of that which AC has to EG; therefore as the cone of which the base is the circle ABCD, and vertex L, is to the folid X, so N i. 14. S. folid 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 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 folid 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, fo is the cone EFGHN to some solid, which must be less i than the cone Book XII. 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 folid 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, fo kk. 15. S. 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 EG. Wherefore similar cones, &c. Q. E. D. IFS (F a cylinder be cut by a plane parallel to its opposite See N. planes, or bases; it divides the cylinder into two cylinders, one of which is to the other as the axis of the first to the axis of the other. Let the cylinder AD be cut by the plane GH parallel to the opposite planes 0 AB, CD, meeting the axis EF in the point K, and let the line GH be the common section of the plane GH and the surface R of the cylinder AD. let AEFC be the parallelogram, in any position of it, by the revolution of which about the straight A line EF the cylinder AD is described ; and let GK be the common section of the plane GH, and the plane AEFC. and because the parallel planes AB, GH are cut by the plane AEKG, AE, KG, their common sections with it, are parallel a ; wherefore AK is a parallelogram, T and GK equal to EA the straight line from the center of the circle AB. for ve the same reason, each of the straight lines Book XII. drawn from the point K to the line GH may be proved to be equal to those which are drawn from the center of the circle AB to its circumference, and are therefore all equal to one anoB.15 Def.1. ther. therefore the line GH is the circumference of a circle 6 of which the center is the point K. therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the same which would be described by the revolution of the parallelograms AK, GF about the straight lines EK, KF. and it is to be shewn that the cylinder AH is to the cylinder HC, as the axis EK to the axis KF. Produce the axis 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 the 0 -L SP R in А. G D Y E B с to one КОН AAA |