which EG has to AC. but as the solid Z is to the cone ABCDL, Book XII. so is the cone EFGHN to some solid, which must be less i than the cone ABCDL, because the solid Z is greater than the cone EFGHN. i. 14. s. 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 folid 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 k the cylinder to the cylinder, for every cone is the third part of k. 15.5. 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. IF: 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 I plane GH parallel to the opposite planes 0 P AB, CD, meeting the axis E F in the point K, and let the line GH be the common section of the plane GH and R N S the surface of the cylinder AD. let AEFC be the parallelogram, in any position of it, by the revolution of which А E B about the straight line EF the cylinder AD is described ; and let GK be the common section of the plane GH, and G K H the plane AEFC. and because the parallel planes AB, GH are cut by the C F D plane AEKG, AE, KG, their common fections with it, are parallel*; where- T X Y a. 16. 18. fore AK is a parallelogram, and GK e Q qual to EA the straight line from the M center of the circle AB. for the same reason, Book XII, reason, each of the straight lines drawn from the point K to the line NGH 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 another. therefore the line GH is the circumference of b.15.Def.s.a circle bof which the center is the point K. therefore the plane GH divides the cylinder AD into the cylinders AH, GD; for they are the fame 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 E F both ways; and take any number of straight lints 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 O L P А LN, NE, EK are all equal, therefore c. 11. 12. the cylinders PR, RB, BG are o to one G K H F Y V M M cylinders; therefore whatever multiple the axis KL is of the axis KE, the fame multiple is the cylinder PG of the cylinder GB. for the same reason, whatever multiple the axis MK is of the axis KF, the fame 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. since therefore there are four magnitudes, viz. the axes EK, KF, and the cylinders BG, GD, and that of the axis EK and cylinder BG there has been taken any equimulples whatever, viz. the axis KL and cylinder PG; and of the axis KF E KF and cylinder GD, any equimultiples whatever, viz. the axis KM Book xii. and cylinder GQ_; and it has been demonstrated if the axis KL be in greater than the axis KM, the cylinder PG is greater than the cylinder GQ, and if equal, equal; and if lefs, less. therefore d the axis d.s. Def. o EK is to the axis KF, as the cylinder BG to the cylinder GD: Wherefore if a cylinder, &c. Q. E. D. 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. as the cylinder EB to the cylinder FD, so is the axis GH 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. and because the cylinders EB, CM have the fame altitude, they are to one another as their bases“. but their bases are equal, there. a. st. ito fore also the cylinders EB, CM F K are equal. and because the cylinder FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM to the cylinder E FD, so is the axis LN to the b. 13. 14. axis KL. but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH. therefore as the cylinder EB to A M N the cylinder FD, fo is the axis GH to the axis KL. and as the cylinder EB to the cylinder FD, so is the cone ABG to the cone CDK, because the cylinders are c. 15. $. triple d of the cones. therefore also the axis GH is to the axis KL, d. 10.12. as the cone ABG to the cone CDK, and the cylinder EB to the cylinder FD. Wherefore cones, &c. Q. E. D. LID н В Book XII. See N. PRO P. XV. THE OR. HE bafes and altitudes of equal cones and cylinders, 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 bafes and altitudes of the cylinders AX, EO are reciprocally proportional; that is, as the base ABCD to the base CFGH, fo 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 alío ecual, and cones and cylinders of the fame alti8. 11. 12. tude being to one another as their bases a, therefore the base ABCD 8. A. 5. is equal o in the lase EFGH; and as the base ABCD is to the bale EFGH, fo is the al- N R L P S D H E K M P, cut the cylinder B. 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 c. 7. 5. is equal to the cylinder EO, as AX is to the cylinder ES, fois the cylinder EO to the same ES. but as the cylinder AX to the cylinder ES, fo * is the base ABCD to the base EFGH; for the cr. linder's AX, ES are of the same altitude; and as the cylinder EO d. 13. 12. to the cylinder ES, fod is the altitude MN to the altitude MP, he: caue A cause the cylinder EO is cut by the plane TYS parallel to its op- Book xli. posite 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 altitudė 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 b to KL, and therefore the b. A. cylinder AX is equal to the cylinder EO. a 11.is 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 b than KL; then, the fame 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 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. therefore the cylinder AX is to the cylinder ES, as the cylinder EO is to the fine ES. whence the cylinder AX is equal to the cylinder EO. and the same reasoning holds in cones. Q. E. D. To describe in the greater of two circles that have the same center, a polygon of an even number of equal fides, that shall not meet the lesser circle. Let ABCD, EFGH be two given circles having the same center 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 letter circle. Through the center K draw the straight line BD, and from the point G, where it meets the circumference of the leger circle, draw |