Let the sections EFG, PQR be parallel to the bases ABC, KLMN respectively; and let the distances DI, SO of the vertexes D, 8 from the sections be equal. Then ABC: EFG:: DH: DI (13. 12), and KLMN: PQR: :ST: 80. But DH = ST, and DI = SO; therefore DH = ST, and DI =SO (Propor. 44); also the base ABC KLMN, therefore the section EFG PQR. In the same manner it may be proved that any other two sections at equal distances from the vertexes are equal to each other. But the pyramids are supposed to consist of an infinite number of equal and parallel sections or plates; M therefore they are equal. Therefore, pyramids &c. Q. E. D. COR. If pyramids of equal altitudes, and standing on the same plane, be cut by a plane parallel to their bases, the sections and bases will be proportional. PROPOSITION XIV. THEOREM. Pyramids and cones of equal bases and altitudes are equal to one another. Let the pyramid ABCD and the cone KLMS stand on equal bases ABC, KLMN, and have equal altitudes DH, ST; the pyramid is equal to the cone. Let the sections EFG, PQR, be parallel to the bases ABC, KLMN, respectively; and let DI, SO, the distances of the vertexes from the sections EFG, PQR, be equal. Then ABC : EFG: DH: DI2 (11. 12), and KLMN: PQR:: ST: SO (12. 12). But DH ST, and DI SO; therefore DH-ST, and DI SO (Propor. 44). Hence ABC: EFG:: KLMŃ Y = : PQR. But ABC Cor. If a pyramid and a cone of equal altitudes, and standing on the same plane, be cut by a plane par allel to their bases, the sections and the bases will be propor tional. PROPOSITION XV. THEOREM. Every prism having a triangular base may be divided into three pyramids which have triangular bases, and are equal to one another.* Let ABC-DEF be a triangular prism; it may be divided into three equal pyramids having triangular bases. D F E Join AE, EC, CD. Because ABED is a parallelogram, of which AE is the diagonal, the triangle ADE is equal to ABE (34. 1); therefore the pyramid C ADE is equal to the pyramid C-ABE (13. 12). But the pyramid C-ABE is equal to the pyramid C-DEF, for they have equal bases, ABC, DFE, and the altitude of the prism ABCDEF, Therefore the three pyramids ADEC, ABEC, DFEC are equal to one another. But these pyramids compose the whole prism; therefore the prism ABCDEF is divided into three equal pyramids. Wherefore, every prism &c, Q. E. D, A C B Cor. 1. From this it is manifest that every pyramid is the third part of a prism of the same base and altitude; for if the base of the prism be any other figure but a triangle, it may be divided into prisms having triangular bases. * Learners will scarcely understand the demon. of this prop. without models of the three pyramids. Cor. 2. Pyramids of equal altitudes are to one another as their bases; because the prisms on the same bases as the pyramids, and of the same altitude, are to one another as their ba ses (Cor. 7. 12). Cor. 3. Similar pyramids are to one another as the cubes of their homologous sides (10. 10). PROPOSITION XVI. THEOREM. Every cone is a third part of a cylinder of the same base and altitude. Let AFBC be a cone, and AFBDE a cylinder of the same base and altitude; the cone is a third part of the cylinder. E M L K Let GHIK be a pyramid, and GHILM a prism on the same base GHI and of the same altitude, and let the base and altitude of the prism be equal to those of the cylinder. Then the cylinder is equal to the prism (6. 12), and the cone is equal to the pyramid (14. 12). But the pyramid GHIK is a third part of the prism GHILM (Cor. 15. 12), therefore the cone AFBC is a third part of the cylinder AFBDE. Therefore, every cone &c. Q. E. D. F PROPOSITION XVII. THEOREM. H Cones of equal bases are to one another as their altitudes. Let ABC, DEF be cones of equal bases, and of unequal altitudes CG, FH; the cone ABC DEF:: alt. CG: FH. Let ABKI, DEML be cylinders of equal bases, then ABKI: DEML:: alt. CG: FH(Cor.8.12). But the cone ABC is a third of ABKI (16. I 12), and the cone DEF is a F L C M K third of DEML; therefore Cor. Pyramids of equal bases are to one another as their altitudes. For cones and pyramids of equal bases and altitudes are equal to one another (14. 12). PROPOSITION XVIII. THEOREM. Pyramids of equal altitudes are to one another as their bases. Let ABCE, FGHK be two pyramids having equal altitudes, but unequal bases ABCD, FGHI; the pyramid ABCE : FGHK :: base ABCD: FGHI. For prisms of equal altitudes are to one another as their bases (Cor. 7. 12), therefore the prism ABCOP :FGHLMN :: base ABCD : FGHI. But a pyramid is one third of a prism of the same base and altitude (Cor. 15. 12); therefore the pyramid ABCE: FGHK base A ABCD: FGHI. Wherefore, pyramids &c. Q. E. D. P E M N L K H G Cor. Cones of equal altitudes are to one another as their bases. For a cone is equal to a pyramid of the same base and altitude (16. 12). For the fig. on RS: fig. on RT:: RS: RT (R. 6), T therefore the fig. on RS: RS:: fig. on RT: RT (Propor. 36), and the fig. on ST: ST: : fig. on RT :RT; therefore the fig. on RS: RS:: fig. on ST: ST (Propor. 34), therefore the fig on RS + fig. on ST:: RS + ST:: fig. on RT: RT (Propor. 41). But RS+ST RT (B. 2), therefore the fig. on RS + fig. on ST = fig. on RT (Propor. 32). Therefore, in a right &c. Q. E. D. PROPOSITION XIX. THEOREM. EVERY sphere is two thirds of the circumscribing cylinder; that is, of a cylinder having the same altitude and diameter as the sphere. Let ADBCA be a hemisphere, and AIDEBCA a cylinder described about it; the hemisphere is equal to two thirds of the circumscribing cylinder. Let ADB be a semicircle, of which the centre is C; let CD be perp. to AB; let DEBC and DIAC be squares described on DC; draw the diagonal CE, and through any point & in DC draw GO parallel to DE. Let the figure thus constructed revolve about DC; then I the D E B having for its base the circle described by DE, equal to the circle described by CB, which circle is the base of the hemisphere. Let L be the point in which GO meets the semicircle ADB, and suppose CL to be joined; then, in the rotation of the plane DEBC on DC the three lines GO, GL, GR, will describe circular sections of the cylinder, hemisphere, and cone respectively. Since CGL is a right angle, the two circles described with the distances CG, GL, are together equal to the circle described with the distance CL or GO (Lemma.). = GR Now the triangles CDE and CGR are equiangular, therefore CD: CG:: DE: GR. But CD DE, therefore CG (Propor. 32). Therefore the circles described with the distances GR and GL are together equal to the circle described with the distance GO; that is, the circles described by the revolution of GR and GL about the point G are together equal to the circle described by the revolution of GO about G; or the circular sections of the cone and sphere are together equal to the corresponding section of the cylinder. And because this is the case in every parallel position of GO, it follows that the cylinder AE is equal to the cone CEI and the hemisphere ADBA. But the cone CEI is a third part of the cylinder AE having the same base and altitude (17. 12); consequently the hemisphere ADBA is equal to the remaining two thirds of the cylinder AE. Therefore the whole sphere is equal to two thirds |