prisms described about EFGH exceed the same pyramid Book Hk ABCD. Therefore the pyramid EFGH is greater than the sum of the prisms described about it, which is impossible. Therefore the pyramids ABCD, EFGH are not unequal, that is, they are equal to each other. Therefore, pyramids, &c. Q. E. D. PROP. XV. THEOR. EVERY prism having a triangular base may be divided into three pyramids which have triangular bases, and are equal to one another. Let there be a prism of which the base is the triangle ABC, and let DEF be the triangle opposite to the base; the prism ABCDEF may be divided into three equal pyramids having triangular bases. Join AE, EC, CD. Because ABED is a parallelogram, of which AE is the diameter, the triangle ADE is equala to a 34. 1. the triangle ABE; therefore the F Eb14.3.Sup. pyramid of which the base is the triangle ADE, and vertex the point C, is equal to the pyramid D of which the base is the triangle ABE, and vertex the point C. But the pyramid of which the base is the triangle ABE, and vertex the point C, that is, the pyramid ABCE, is equal to the pyramid DEFCb, for they have equal bases, viz. the triangles ABC, DFE, and the same altitude, viz. the altitude of the prism ABCDEF. Therefore the three pyramids ADEC, AВЕС, C B DFEC are equal to one another. But the pyramids ADEC, ABEC, DFEC make up the whole prism ABCDEF; therefore the prism ABCDEF is divided into three equal pyra mids. Wherefore, every prism, &c. Q. E. D. Suppl. c 1. Cor. 8.3. Sup. COR. 1. From this it is manifest that every pyramid is the third part of a prism which has the same base and the same altitude; for if the base of the prism be any other figure but a triangle, it may be divided into prisms having triangular bases. COR. 2. Pyramids of equal altitudes are to one another as their bases; because the prisms upon the same bases, and of the same altitude, are to one another as their bases. PROP. XVI. THEOR. IF from any point in the circumference of the base of a cylinder a straight line be drawn perpendicular to the plane of the base, it will be wholly in the cy. lindric superficies. Let ABCD be a cylinder, the circle AEB the base, DFC the circle opposite to the base, and GH the axis; from E, any point in the circumference AEB, let EF be drawn perpendicular to the plane of the circle AEB; the straight line EF is in the superficies of the cylinder. Let F be the point in which EF meets the plane DFC opposite to the base; join EG and FH; and let AGHD be the a Def. 14. rectanglea by the revolution of scribed. Because GH is at right angles to the straight line GA, which by its revolution describes the circle AEB, it is at right angles to all the straight lines in the plane of that circle which meet it in G, and is therefore at right angles to the b 4. 2. Sup. plane of the circle AEBb. But EF is at right angles to the same plane; c 6. 2. Sup. therefore EF and GH are parallels, and in the same plane. The plane through GH, EF cuts the parallel D H C planes AEB, DFC in the straight lines EG, FH, therefore Book III. EG is parallel to FHd. Therefore the figure EGHF is a parallelogram; and because EGH is a right angle, it is a rect- d14.2.Sup. angle; and because EG is equal to AG, it is equal to the rectangle AH. Therefore when, in the revolution of the rectangle AH, the straight line AG coincides with EG, the two rectangles AH, EH will coincide, and the straight line AD will coincide with the straight line EF. But AD is always in the superficies of the cylinder, for it describes the superficies; therefore EF is also in the superficies of the cylinder. Therefore, if from any point, &c. Q. E. D. PROP. XVII. THEOR. A CYLINDER and a parallelepiped having equal bases and altitudes are equal to each other. Let ABCD be a cylinder, and EF a parallelepiped, having equal bases and altitudes; the cylinder ABCD is equal to the parallelepiped EF. If not, let them be unequal; and, first, let the cylinder be less than the parallelepiped. From the parallelepiped EF let there be cut off, by a plane PQ parallel to NF, a part EQ Suppl. equal to the cylinder ABCD. In the circle AGB inscribe the polygon AGKBLM, which shall differ from the circle by a 1. Cor. a space less than the parallelogram PHa; and cut off from 4. 1. Sup. the parallelogram EH a part OR equal to the polygon AGKBLM. The point R will fall between P and N. On the polygon AGKBLM let an upright prism AGBCD be constituted of the same altitude with the cylinder, which will 616.3. Sup. be less than the cylinder, because it is within it. Through the point R let a plane RS be drawn parallel to NF, and it will cut off the parallelepiped ES equal to the prism AGBCD, because its base and altitude are equal to the base and alti8. 3. Sup. tude of the prism. But the prism AGBCD is less than the cylinder ABCD, and the cylinder ABCD is equal to the parallelepiped EQ, by hypothesis; therefore ES is less than EQ; and it is also greater, which is impossible. Therefore the cylinder ABCD is not less than the parallelepiped EF. In the same manner it may be proved that it is not greater than EF. Therefore they are equal. Therefore, a cylinder, &c. Q. E. D. c 2. Cor. PROP. XVIII. THEOR. IF a cone and a cylinder have the same base and the same altitude, the cone is the third part of the cylinder. Let the cone ABCD and the cylinder BFKG have the same base BCD, and the same altitude, viz. the perpendicular from the point A upon the plane BCD; the cone ABCD is the third part of the cylinder BFKG. If not, let the cone ABCD be the third part of another cylinder LMNO, having the same altitude as the cylinder BFKG; let the bases BCD, LIM be unequal; and, first, let BCD be greater than LIM. Because the circle BCD is greater than the circle LIM, a polygon may be inscribed in a 4. 1. Sup. BCD, which will differ from it less than LIM does, and therefore will be greater than LIM. Let this be the polygon BECFD; and upon BECFD let there be constituted the pyramid ABECFD, and the prism BCFKHG. Because the polygon BECFD is greater than the circle LIM, the prism BCFKHG is greater than the cylinder LMNO, which has the same altitude. But the pyramid Book III. ABECFD is the third part of the prismb BCFKHG, and therefore is greater than the third part of the cylinder LMNO. b15.3.Sup. Now the cone ABECFD is, by hypothesis, the third part of the cylinder LMNO; therefore the pyramid ABECED is greater than the cone ABCD; and it is also less, because it is inscribed in the cone, which is impossible. Therefore the cone ABCD is not less than the third part of the cylinder BFKG. In the same manner, by circumscribing a polygon about the circle BCD, it may be shown that the cone ABCD is not greater than the third part of the cylinder BFKG. Therefore it is equal to a third part of the cylinder BFKG. Therefore, if a cone, &c. Q. E. D PROP. XIX. THEOR. IF a hemisphere and a cone have equal bases and altitudes, a series of cylinders may be inscribed in the hemisphere, and a series of cylinders may be described about the cone, having the same altitudes as the former series, such that the sum of the two serieses of cylinders shall differ from the sum of the hemisphere and the cone by a solid less than any given solid. i |