Then because di divides the parallelogram AE into two equal parts, the pyramid whose base is ABD, and vertex c, is equal to the pyramid whose base is bed and vertex c (VIII. 15.) And since the opposite ends of the prism are equal to each other (VIII. Def. 3.), the pyramid whose base is ABC and vertex D, is equal to the pyramid whose base is der and vertex c (VIII. 15.) But the pyramid whose base is ABC and vertex D, is equal to the pyramid whose base is ABD and vertex c, being both contained by the fame planes. The three pyramids DABC, CBED and card are, therefore, all equal to each other, and consequently the prism FDABE, which is composed of them, is triple the pyramid DABC, as was to be thewn. Cor. Every pyramid is the third part of a prism of the same base and altitude ; since the base of the prism, whatever be its figure, may be divided into triangles, and the whole solid into triangular prisms, and pyramids. SCHOLIUM. Whatever has been demonstrated of the proportionality of prisms, holds equally true of pyramids; the former being always triple the latter. • PROP. 9 PROP. XVII. THEOREM. If a cylinder be cut by a plane parallel ta its base, the section will be a circle, equal to the base. Let AF be a cylinder, and GHK a section parallel to its base ABC; then will Ghk be a circle, cqual to ABC. For let the planes NE, NF pass through the axis of the cylinder LN, and meet the section GK in M, H and K. Then, since the circle Def is equal and parallel to the circle ABC (VIII. Def.8.), the radii LF, LE will be equal and parallel to the radii nc, NB (III. 5. and VII. 12.) And because lines which join the corresponding extremes of equal and parallel lines are themselves parallel (I. 29.), FC, EB will be parallel to LN; or kc, he to MN. In like manner, since the circle GHK is parallel to the circle ABC (by Hyp.), MK, MH will be parallel to NC, NB. And, because the opposite sides of parallelograms are equal (I. 30.), MK will be equal to nc, and mh to NB. But NC, NB are equal to each other, being radii of the same circle ; whence MK, MH are also equal to each other. And the same may be shewn of any other lines, drawn from the point M, to the circumference of the fection GHK; consequently gik is a circle, and equal to ABC, as was to be shewn. PRO P. XVIII. THEOREM. Every cylinder is equal to a prism of an equal base and altitude. 1 Let AH be a cylinder, and DM a prism, standing upon equal bases ACB, der, and having equal altitudes; then will ah be equal to DM. For parallel to the bases, and at equal distances from them, draw the planes onm, and urw. Then, by the last Prop. and Prop. 8, the section onm. is equal to the base ACB, and the section vrw to the base DEF. But the base ACB is equal to the base def, by hypo. thesis; whence the section onm is also equal to the fec And, in the fame manner, it may be thewn, that any other sections, at equal distances from the base, are equal to each other. Since, therefore, every section of the cylinder is equal to its correspondent section in the prism, the folids themfelves, which are composed of those sections, must also be equal. Q. E. D. SCHOLIUM. Whatever has been demonstrated of the proportionality of prisms, holds equally true of cylinders; the former being equal to the latter. PRO P. XIX. THEOREM. If a cone be cut by a plane parallel to its base, the section will be to the base as the squares of their distances from the vertex. D A Let DABC be a cone, and nmp a section parallel to the base ABC, then will nmp be to ABC as the squares of their distances from the vertex, For draw the perpendicular dr; and let the planes CDP, BDP pass through the axis of the cone, and meet the section in o, p, and m. Then fince the section nmp is parallel to the base ABC (by Hyp.), and the planes Bo, co cut them, op will be parallel to pc, and om to PB (VII. 12.) And because the triangles formed by these lines are equiangular, om will be to PB as Do to DP, or as op to PC (VI. 5.) But PB is equal to pc, being radii of the same circle ; wherefore om will also be equal to op (V. 10.) And the same may be shewn of any other lines drawn from the point o to the circumference of the section nmp; whence nmp is a circle. Again, by similar triangles, Ds is to pr as Do to DP, or as om to PB; whence the square of Ds is to the square · of dr as the square of om is to the square of PB (VI. 19.) But the square of om is to the square of PB as the circle nmp is to the circle ÅBC (VIII. 5.); therefore the square of os is to the square of Dr as the circle nmp is to the circle ABC (V.11.) Q. E. D. Cor. If a cone be cut by a plane parallel to its base the section will be a circle. |