R L E -M Y DEMONSTRATION. Let the cylinder AD be cut by the plane GH parallel to the opposite planes 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 of the cylinder AD. Let AEFC be the parallelogram in any position of it, by the revolution of which about the straight 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, and GK equal to EA, the straight line from the center of the circle AB: for the same reason, each of the straight lines 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 another; therefore the line GH is the circumference of a circle of which the center is the point K (6): 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 shown, that the cylinder AH is to the cylinder HC, as the axis EK is to the axis KF. a) XI. 16. (b) I. Def. 15. (c) XII. 11. V. Def. 5. 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 points L, N, X, M: therefore the common sections of these planes with the cylinder produced, are circles, the centers 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. And because the axes LN, NE, EK are all equal, therefore the cylinders PR, RB, BG are to one another as their bases (c): but their bases are equal, and therefore the cylinders PR, RB, BG are equal: and 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 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, less; therefore (d), as the axis EK is to the axis KF, so is the cylinder BG to the cylinder GD. PROPOSITION XIV. THEOREM.-If cones and cylinders are upon equal bases, they are to one another as their altitudes. DEMONSTRATION. Let the cylinders EB, FD be upon the equal bases AB, CD; as the cylinder EB is to the cylinder FD, so shall the axis GH be to the axis KL. E H F N 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 (a): but their bases are equal, therefore also the cylinders EB, CM are equal: and because the cylinder FM is cut by the plane CD parallel to its opposite planes, as the cylinder CM is to the cylinder FD, so is the axis LN to the axis KL (6): but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB is to the cylinder FD, so is the axis GH to the axis KL: and as the cylinder EB is to the cylinder FD, so is the cone ABG to the cone CDK (c), because the cylinders are triple of the cones (d): therefore also the axis GH is to the axis KL, as the cone ABG is to the cone CDK, and as the cylinder EB is to the cylinder FD. (a) XII. 11. PROPOSITION XV. THEOREM [1.]-If cones and cylinders are equal, their bases and altitudes are reciprocally proportional; [2.] and if the bases and altitudes be reciprocally proportional, the cones and cylinders are equal to one another. DEMONSTRATION [1]. 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 shall be reciprocally proportional; that is, as the base ABCD is to the base EFGH, so shall the altitude MN be to the altitude KL. L X Y N E K M F B (a) XII. 11. (b) V. A. (c) V. 7. (d) XII. 13. 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), therefore the base ABCD is equal to the base EFGH (6); and 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 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 the same ES (c): but as the cylinder AX is to the cylinder ES, so is the base ABCD to the base EFGH (a); for the cylinders AX, ES are of the same altitude; and as the cylinder EO is to the cylinder ES, so is the altitude MN to the altitude MP (d), because the cylinder EO is cut by the plane TYS parallel to its opposite planes; therefore as the base ABCD is 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 is 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. [2] 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 is 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 (6); and therefore the cylinder AX is equal to the cylinder EO (a). 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 (b). Then, the same construction being made as before, because as the base ABCD is 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 is to the cylinder ES (a); and as the altitude MN is 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. PROPOSITION XVI. PROBLEM. In the greater of two given circles (ABCD, EFGH) that have the same center (K), to inscribe a polygon of an even number of equal sides, that shall not meet the lesser circle (EFGH). E H K GM (a) III. 16, Cor. SOLUTION. Through the center K draw the straight line BD, and from the point G, where it meets the circumference of the lesser circle, draw GA at right angles to BD, and produce it to C; therefore AC touches the circle EFGH (a): then, if the circumference BAD be bisected, and the half of it be again bisected, and so on, there must at length remain a circumference less than AD (b): 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 that 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; so that if straight lines equal to LD be applied in the circle ABCD from the point L around to N, there shall be inscribed in the circle a polygon of an even number of equal sides not meeting the lesser circle. LEMMA II. THEOREM.-If two trapeziums ABCD, EFGH be inscribed in the circles, the centers 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 center of the circle in which the greater sides are, is greater than the straight line LE, drawn from the center to the circumference of the other circle. DEMONSTRATION. If it be possible, let KA be not greater than LE; then KA must be either equal to it, or less than it. 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 (a); 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. P; But let KA be less than LE, and make LM equal to KA; and from the center L and distance LM describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, and join MN, NO, OP, PM which are respectively parallel to and less than EF, FG, GH, HE (6): then because EH is greater than MP, AD is greater than MP; and the circles ABCD, MNOP are equal; therefore the circumference AD is greater than MP: for the same reason, the circumference BC is greater than NO: and because the |