XV. The axis of a cylinder is the fixed straight line about which the parallelogram revolves./ XVI. The bases of a cylinder are the circles described by the two revolving opposite sides of the parallelogram. ✓ XVII. Similar cones and cylinders are those which have their axes and the diameters of their bases proportionals. / PROP. I. THEOR. IF two solids be contained by the same number of equal and similar planes, similarly situated, and if the inclination of any two contiguous planes in one solid be the same with the inclination of the two equal and similarly situated planes in the other, the solids are equal and similar. Let AG and KQ be two solids contained by the same number of equal and similar planes, similarly situated, so that the plane AC is similar and equal to the plane KM, the plane AF to the plane KP, BG to LQ, GD to QN, DE to NO, and FH to PR; and let the inclination of the plane AF to the plane AC be the same with that of the plane KP to the plane KM, and so of the rest; the solid KQ is equal and similar to the solid AG. Book III. Suppl. Let the solid KQ be applied to the solid AG, so that the bases KM and AC, which are equal and similar, may coina 8. Ax. 1. cide, the point N coinciding with the point D, K with A, L with B, and M with C. Because the plane KM coincides with the plane AC, and, by hypothesis, the inclination of KR to KM is the same with the inclination of AH to AC, the plane KR will be upon the plane AH, and will coincide with it, because they are similar and equala, and because their equal sides KN and AD coincide. And in the same manner it may be shown that the other planes of the solid KQ coincide with the other planes of the solid AG, each with each. Wherefore the solids KQ and AG do wholly coincide, and are equal and similar to each other. Therefore, if two solids, &c. Q. E. D. PROP. II. THEOR. IF a solid be contained by six planes, of which two and two are parallel, the opposite planes are similar and equal parallelograms. Let the solid CDGH be contained by the parallel planes AC and GF, BG and CE, FB and AE; its opposite planes are similar and equal parallelograms. Because the two parallel planes BG, CE are cut by the a14.2.Sup. plane AC, their common sections AB, CD are parallela; and because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are pa rallela. Hence AC is a parallelo- A gram. In like manner it may be proved that each of the figures CE, FG, GB, BF, AE is a parallelogram. Join AH, DF. Because AB is parallel to DC, and BH to CF, the two straight lines AB, BH, which meet each other, are parallel to DC, CF, which meet each other; therefore the angle ABH is equal to the angle b 9.2. Sup. DCFb. Because AB, BH are equal to DC, CF, and the G C F D E angle ABH is equal to the angle DCF, the base AH is equal Book III. to the base DFc, and the triangle ABH to the triangle DCF. For the same reason the triangle AGH is equal to the trian- c 4. 1. gle DEF. Therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner it may be proved that the parallelogram AC is equal and similar to the parallelogram GF, and the parallelogram AE to BF. Therefore, if a solid, &c. Q. E. D. PROP. III. THEOR. (IF a solid parallelepiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to each other as their bases. Let the solid parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the solids ABFV, EGCD; as the base AEFY is to the base EHCF, so is the solid ABFV to the solid EGCD. Produce AH both ways, and take any number of straight lines HM, MN, each equal to EH, and any number AK, KL, each equal to EA; and complete the parallelograms LO, KY, HQ, MS, and the solids LP, KR, HU, MT. Be cause the straight lines LK, KA, AE are equal, and also the straight lines KO, AY, EF, which make equal angles with LK, KA, AE, the parallelograms LO, KY, AF are equal a 36. 1. and similarb; and likewise the parallelograms KX, KB, AG; b Def. 1. 6. and also the parallelograms LZ, KP, AR, because they are Suppl. opposite planes. For the same reason the parallelograms EC, HQ, MS are equal; and the parallelograms HG, HI, c 2.3. Sup. IN; and also HD, MU, NT. Therefore three planes of the solid LP are equal and similar to three planes of the solid KR, and also to three planes of the solid AV. But the three planes opposite to these three are equal and similar to theme in the several solids; therefore the solids LP, KR, AV are contained by equal and similar planes. Because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ to XP is equal to that of KP d15.2.Sup. to PB, or of AR to BVd; and the same is true of the other contiguous planes; therefore the solids LP, KR, and AV are e 1. 3. Sup. equal to one another. For the same reason the three solids ED, HU, MT are equal to one another. Therefore whatever multiple the base LF is of the base AF, the same multiple is the solid LV of the solid AV. For the same reason, whatever multiple the base NF is of the base HF, the same multiple is the solid NV of the solid ED. Therefore if the base LF be equal to the base NF, the solid LV is equale to the solid NV; and if the base LF be greater than the base NF, the solid LV is greater than the solid NV; and if less, less. Now there are four magnitudes, viz. the two bases AF, FH, and the two solids AV, ED; and of the base AF and solid AV the base LF and solid LV are any equimultiples whatever; and of the base FH and solid ED the base FN and solid NV are any equimultiples whatever; and it has been proved that if the base LF is greater than the base FN, the solid LV is greater than the solid NV; and if equal, f 5. Def. 5. equal; and if less, less; thereforef as the base AF is to the base FH, so is the solid AV to the solid ED. Wherefore, if a solid, &c. Q. E. D. g 1. 6. COR. Because the parallelogram AF is to the parallelogram FH as YF to FCs, the solid AV is to the solid ED as YF to FC. PROP. IV. THEOR. IF a solid parallelepiped be cut by a plane passing through the diagonals of two of the opposite planes, it will be cut into two equal prisms. J Let AB be a solid parallelepiped, and DE, CF the diago- Book III. nals of the opposite parallelograms AH, GB, viz. those which are drawn between the equal angles in each. Because CD, FE are parallel to GA, though not in the same plane with it, they are parallela to each other; wherefore the diagonals CF, DE are in the plane in which the parallels are, and are themselves parallelb. The plane CDEF will cut the solid AB into two equal parts. The triangle CGF is equal to the triangle CBF, and the triangle DAE to DHE; and the parallelogram CA is equald and similar to the opposite one BE, and the parallelogram GE to CH; therefore the corresponding planes which contain the prisms CAE, CBE are equal and similar, each to each. They are also equally inclined to each other, because the planes AC, EB are parallel, and also AF, BD, and they are cut by the plane CE. Therefore the prism CAE is equal to the prism CBE, e 1.3. Sup. and the solid AB is cut into two equal prisms by the plane CDEF. Therefore, if a solid parallelepiped, &c. Q. E. D. N. B. The insisting straight lines of a parallelepiped, mentioned in the following propositions, are the sides of the parallelograms between the base and the plane parallel to it. PROP. V. THEOR. SOLID parallelepipeds upon the same base and of the same altitude, the insisting straight lines of which are terminated in the same straight lines in the plane opposite to the base, are equal to one another. Let the solid parallelepipeds AH, AK be upon the same base AB, and of the same altitude, and let their insisting straight lines AF, AG, LM, LN be terminated in the same straight line FN, and CD, CE, BH, BK be terminated in the same straight line DK; the solid AH is equal to the solid AK. |