Book XI. the figure LQ, and the straight line CG with MQ, and the point G with the point Q: Since, therefore all the planes and fides of the solid figure AG coincide with the planes and sides of the folid figure KQ, AG is equal and fimilar to KQ: And, in the same manner, any other solid figures whatever contained by the same number of equal and fimilar planes, alike situated, and having none of their folid angles contained by more than three plane angles, may be proved to be equal and similar to one another. Q. E. D. Sce N. IF PROP. XXIV. THEOR. a folid be contained by fix planes, two and two of which are parallel; the oppofite planes are fimilar and equal parallelograms. Let the folid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE: Its opposite planes are fimilar and equal parallelograms. Because the two parallel planes, BG, CE, are cut by the a 16. 11. plane AC, their common sections AB, CD, are parallel. Again, because the two parallel planes BF, AE, are cut by the plane AC, their common sections AD, BC, are parallel: And AB is parallel to CD; therefore AC is a parallelogram. In like manner, it may be proved that each of the figures CE, FG, GB, BF, and are not in the fame plane with C 4. 1. b. 10. 11. tain equal angles; the angle ABH is therefore equal to the angle DCF: And because AB, BH, are equal to DC, CF, and the angle ABH equal to the angle DCF; therefore the base AH is equal to the base DF, and the triangle ABH to the trid 34. 1. angle DCF: And the parallelogram BG is double d of the triangle ABH, and the parallelogram CE double of the triangle DCF; therefore the paralelogram BG is equal and fimilar to the parallelogram CE. In the same manner it may be proved, that the parallelogram AC is equal and fi 1 milar to the parallelogram GF, and the parallelogram AE to Book XI. BF. Therefore, if a folid, &c. Q. E. D. I F PROP. XXV. THEOR. a folid parallelepiped be cut by a plane parallel to two See N. of its oppofite planes; it divides the whole into two solids, the base of one of which shall be to the base of the other, as the one solid is to the other. Let the folid parallelepiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the two solids ABFV, EGCD; as the base AEFY of the first is to the base EHCF of the other, so is the folid ABFV to the folid 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: Then, because the straight lines LK, KA, AE are all equal, the parallelograms LO, KY, AF are equal: And likewise the parallelograms KX, a 36. 1. KB, AG; as also the parallelograms LZ, KP, AR, because b 24. 11. they are opposite planes: For the fame reason, the paral P lelograms EC, HQ, MS are equal; and the parallelograms HG, HI, IN, as also HD, MU, NT: Therefore three planes of the folid I.P, are equal and fimilar to three planes of the folid KR, as also to three planes of the folid AV: But the three planes opposite to these three are equal and fimilar to them in the several folids, and none of their solid angles are contained by more than three plane angles: Therefore the three folids LP, KR, AV are equal to one another: For the fame reason, c C. II. the three folids ED, HU, MT are equal to one another: Therefore Book XI. fore what multiple soever the base LF is of the bafe AF, the fame multiple is the folid 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: And if the base LF be equal to the base NF, the solid LV is equal to the folid NV; and if the base LF be greater than the base NF, the folid LV is greater than the solid NV; and if less, less: Since then there are four magnitudes, viz. the two bases AF, FH, and the two folids 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 folid ED, the base FN and folid NV are any equimultiples whatever; and it has been proved, that if the base LF is greater than the base FN, the folid LV is greater than the folid NV; and if equal, equal, and if less, less. d s. def. 5. Therefored as the base AF is to the base FH, so is the folid AV to the folid ED. Wherefore, if a folid, &c. Q. E. D. T'a PROP. XXVI. PROB. a given point in a given straight line, to make a folid angle equal to a given solid angle contained by three plane angles. Let AB be a given straight line, A a given point in it, and D a given folid angle contained by the three plane angles EDC, EDF, FDC: It is required to make at the point A in the straight line AB a folid angle equal to the folid angle D. In the straight line DF take any point F, froni which draw a 11. II. FG perpendicular to the plane EDC, meeting that plane in G; join DG, and at the point A in the straight line AB make the angle BAL equal to the angle EDC, and in the plane BAL make the angle BAK equal to the angle EDG; then make AK equal to DG, and from the point K erect KH b23. 1. C 12. II. C at at right angles to the plane BAL; and make KH equal to Book XI. GF, and join AH: Then the folid angle at A, which is contain-W ed by the three plane angles BAL, BAH, HAL, is equal to the folid angle at D contained by the three plane augles EDC, EDF, FDC. Take the equal straight lines AB, DE, and join HB, KB, FE, GE: And because FG is perpendicular to the plane EDC, it makes right anglesd with every straight line meeting it in d 3. def. 11. that plane: Therefore each of the angles FGD, FGE is a right angle: For the fame reason, HKA HKB are right angles: And because KA, AB are equal to GD, DE, each to each, and contain equal angles, therefore the base BK is equal to the e 4. 1. base EG: And KH is equal to GF, and HKB, FGE, are right angles, therefore HB is equal to FE: Again, because AK, KH are equal to DG, GF, and contain right angles, the base AH is equal to the base DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the base HB is e qual to the base FE, A therefore the angle BAH is equal f to the angle EDF: For the fame reason, the D f 8. 1. e FC be joined, fince the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG ar, by the construction, equal; therefore the remaining angle KAL is equal to the remaining angle GDC: And because KA, AL are equal to GD, DC, and contain equal angles, the base KL is equal to the base GC: and KH is equal to GF, fo that LK, KH are equal to CG, GF, and they contain right angles; therefore the base HL is equal to the base FC: Again, because HA, AL are equal to FD, DC, and the base HL to the base FC, the angle HAL is equalf to the angle FDC: 'Therefore, because the three plane angles BAL, BAH, HAL, which contain the folid angle at A, are equal to the three plane angles EDC, EDF, FDC, which contain the fold angle at D, each to each, and are fituated in the same order, the solid angle at A is equal to the folid angle at D. Therefore, at a given point in a given straight line, a folid angle has been made equal to a given folid angle contained by three plane angles. Which was to g. B II. be done. Book XI. a 26. 11. b 12. 6. C 22. 5. 1 PROP. XXVII. PRO B. Telepiped fimilar, fimilarly fituated to one given. O describe from a given straight line a solid paralle Let AB be the given straight line, and CD the given solid parallelepiped. It is required from AB to describe a solid parallelepiped fimilar, and fimilarly fituated to CD. At the point A of the given straight line AB, make a solid angle equal to the folid angle at C, and let BAК, КАН, НАВ be the three plane angles which contain it, so that BAK be equal to the angle ECG, and KAH to GCF, and HAB to FCE: And as EC to CG, so make BA to AK; and as GC to CF, fo make KA to AH; wherefore, ex aequalis, as EC to CF, so is BA to AH: Complete the parallelogram BH, and the folid AL: And because, as EC to L CG, so BA to AK, equal angles ECG, F BAK are proporti- parallelogram BK K G is fimilar to EG. B d 24. II. For the fame rea- A fon, the parallelo gram KH is fimilar to GF, and HB to FE. Wherefore three parallelograms of the solid AL are similar to three of the folid CD; and the three opposite ones in each folid are equal d and fimilar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each to each, and fituated in the fame order, the solid angles are equal, each to each. Therefore the folid AL is fimilar to the f11. def. 11. folid CD. Wherefore from a given straight line AB a folid p. rallelepiped AL has been defcribed similar, and fimilarly situated to the given one CD. Which was to be done. e B. II. PROF. |