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 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. And because the axes LŇ, NE, EK, are all equal, therefore the cylinders PR, RB, BG, are to one another as their bases: (xii. 11.) 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 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. PROPOSITION XIV. THEOREM. 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. 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 : (x11. 11.) 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 to the cylinder FD, so is the axis LN to the axis KL: (x11. 13.) but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore 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 cone A BG to the cone CDK, (v.15.) because the cylinders are triple of the cones: (XII. 10.) therefore also the axis GH is to the axis KL, as the cone ABG to the cone CDK, and as the cylinder EB to the cylinder FD. Wherefore cones, &c. Q.E.D. PROPOSITION XV. THEOREM. The bases and altitudes of equal cones and cylinders are reciprocally proportional; and conversely, 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: proportional; 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; as their bases, (x11. 11.) : (v. A.) 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: (v.7.) but as the cylinder AX to the cylinder ES, so is the base ABCD to the base EFGH; (XII. 11.) for the cylinders AX, ES are of the same altitude; and as the cylinder EO to the cylinder ES, so is the altitude MN to the altitude MP, (XII. 13.) 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; (v. A.) and therefore the cylinder AX is equal to the cylinder EO. (XI. 11.) 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. (v. A.) 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 to the base EFGH, as the cylinder AX to the cylinder ES; (XII. 12.) and as the altitude MN to the altitude MP or KL, so is the cylinder EO to the cylinder ES: is to the same ES: and the same reasoning holds in cones. Q.E.D. PROPOSITION XVI. PROBLEM. In the greater of two circles that have the same centre, to inscribe a polygon of an even number of equal sides, that shall not meet the lesser circle. Let ABCD, EFGH be two given circles having the same centre 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 lesser circle. Through the centre K draw the straight line BD, lesser circle, therefore AC touches the circle EFGH: (111. 16. Cor.) 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: (xii. Lem. 1.) 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. Q. E. F. LEMMA II. If two trapeziums ABCD, EFGH be inscribed in the circles, the centres 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 centre of the circle in which the greater sides are, is greater than the straight line LE drawn from the centre to the circumference of the other circle. If it be possible, let KA be not greater than LE; 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; (111. 28.) 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. But let KA be less than LE, and make LM equal to KA, and from the centre L, and distance LM, describe the circle MNOP, meeting the straight lines LE, LF, LG, LH, in M, N, O, P; and join MN, NO, OP, PM, which are respectively parallel to, and less than EF, FG, GH, HE: (VI. 2.) 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 straight line AB is greater than EF, which is greater than MN, much more is AB greater than MN: therefore the circumference AB is greater than MN; and for the same reason, the circumference DC is greater than PO: therefore the whole circumference ABCD is greater than the whole MNOP: but it is likewise equal to it, which is impossible: therefore KA is not less than LE: nor is it equal to it; therefore the straight line KA must be greater than LE. Q. E. D. |