Book XI. multiple foever the base LF is of the base AF, the fame multiple is the folid LV of the folid AV. for the same reason, whatever multiple the base NF is of the base HF, the fame multiple is the folid NV of the folid ED. and if the bafe LF be equal to the base NF, c. C. 11. the folid 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 folid NV; and if lefs, lefs. fince then there are four magnitudes, viz. C the two bafes AF, FH, and the two folids AV, ED, and of the bafe AF and folid AV, the bafe LF and folid 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 bafe LF is greater than the bafe FN, the folid LV is greater than the folid NV; and if equal, equal; and if less, d. 5.Def. 5. lefs. Therefore as the base AF is to the bafe FH, fo is the folid AV to the solid ED. Wherefore if a folid, &c. Q. E. D. See N. 2. II. II. A d PROP. XXVI. PROB. T a given point in a given ftraight line, to make a folid angle equal to a given folid angle contained by three plane angles. Let AB be a given straight line, A a given point in it, and Da 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, from which draw FG perpendicular to the plane EDC, meeting that plane in G'; b. 23. 1. join DG, and at the point A in the ftraight 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 at right angles to the Book XI. plane BAL; and make KH equal to GF, and join AH. then the C. 12. II. folid angle at A which is contained by the three plane angles BAL, BAH, HAL is equal to the folid angle at D contained by the three plane angles EDC, EDF, FDC. Take the equal ftraight lines AB, DE, and join HB, KB, FE, d e e. GE. and because FG is perpendicular to the plane EDC, it makes right angles with every straight line' meeting it in that plane, d.3.Def.it. 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 bafe EG. and KH is equal • 4. I. 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 bafe DF; and AB is equal to DE; therefore HA, AB are equal to FD, DE, and the bafe HB is equal to the bafe FE; therefore the angle BAH is equal to the angle EDF. for the fame reason, the angle HAL is equal to the B angle FDC. because if AL and DC be made equal, and KL, HL, 4.4. e f. 8. t. GC, FC be joined, fince the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, by the conftruction, 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 bafe 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 bafe FC. again, because HA, AL are equal to FD, DC, and the base HL to the bafe FC, the angle HAL is equal 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 folid angle at D, each to each, and are fituated in the fame order; the folid angle at A is equal to the folid angle at D. Therefore at a given g. B. 11. f Book XI. 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 be done. T O defcribe from a given ftraight line a folid parallelepiped fimilar, and fimilarly fituated to one given. Let AB be the given straight line, and CD the given folid parallelepiped. It is required from AB to defcribe a folid parallelepiped fimilar, and fimilarly fituated to CD. At the point A of the given straight line AB make a a folid angle equal to the folid angle at C, and let BAK, KAH, HAB 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 aequali, as EC to CF, fo is BA to AH. complete the parallelogram BH, and the folid AL. and because, as EC to CG, fo GF, and HB to FE. wherefore three parallelograms of the folid AL are fimilar to three of the folid CD; and the three oppofitę d. 24. 11. ones in each folid are equal and fimilar to these, each to each, d alfo, because the plane angles which contain the folid angles of the figures are equal, each to each, and fituated in the fame order, e. B. 11. the folid angles are equal, each to each. Therefore the folid f.11.Def.11 AL is fimilar to the folid CD. wherefore from a given straight line AB a solid parallelepiped AL has been described similar, and fimilarly fituated to the given one CD. Which was to be done. f PROP. XXVIII. THEOR. Book XI. Fa folid parallelepiped be cut by a plane paffing See N. thro' the diagonals of two of the oppofite planes; it fhall be cut into two equal parts. Let AB be a folid parallelepiped, and DE, CF the diagonals of the oppofite parallelograms AH, GB, viz. those which are drawn betwixt the equal angles in each. and because CD, FE are each of them parallel to GA, and not in the same plane with it, CD, EF are parallel ; wherefore the diagonals CF, DE are in the plane in a. 9. 11. which the parallels are, and are themfelves parallels . and the plane CDEF a C B b. 16. 11. F C. 34. I. H A D E Because the triangle CGF is equal © to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal and fimilar to the oppofite one BE; and the parallelogram GE to CH. therefore the prism contained by the two triangles CGF, DAE, and the three parallelograms CA, GE, EC, is equal e to the prism contained by the two triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the fame number of equal and fimilar planes, alike fituated, and none of their solid angles are contained by more than three plane angles. Therefore the folid AB is cut into two equal parts by the plane CDEF. Q. E. D. N. B. The infifting ftraight lines of a parallelepiped, men⚫tioned in the next and some following Propofitions, are the fides ⚫ of the parallelograms betwixt the base and the oppofite plane ' parallel to it.' d. 24. II. SOLI OLID parallelepipeds upon the fame bafe, and of the See N. same altitude, the infifting straight lines of which are terminated in the fame ftraight lines in the plane oppofite to the bafe, are equal to one another. Book XI. Let the folid parallelepipeds AH, AK be upon the fame base AB, and of the fame altitude, and let their infifting ftraight lines See the AF, AG, LM, LN; CD, CE, BH, BK be terminated in the fame figures be low. ftraight lines FN, DK. the folid AH is equal to the folid AK. First, Let the parallelograms DG, HN which are oppofite to the bafe AB have a common fide HG. then because the folid AH is cut by the plane AGHC paffing thro' the diagonals AG, CH of the oppofite planes ALGF, CBHD, AH is cut into two equal a. 28. 11. parts a by the plane AGHC. I. therefore the folid AH is double Ꭰ H K the oppofite planes ALNG, CBKH, the folid AK is double of the fame prism which is contained by the triangles ALG, CBH. Therefore the folid AH is equal to the folid AK. b But let the parallelograms DM, EN oppofite to the base have no common fide. then because CH, CK are parallelograms, CB is b. 34. t. equal to each of the oppofite fides DH, EK; wherefore DH is equal to EK. add, or take away the common part HE; then DE is equal to HK. wherefore also the triangle CDE is equal to the triangle BHK. and the parallelogram DG is equal to the parallelogram HN. for the same reason, the triangle AFG is equal to e. 24. 11. the triangle LMN, and the parallelogram CF is equal to the paral c. 38. 1. d. 36. 1. c K D E H K A N B I lelogram BM, and CG to BN; for they are oppofite. Therefore the prism which is contained by the two triangles AFG, CDE, and f. C. II. the three parallelograms AD, DG, GC is equal to the prism contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prifm LMN, BHK be |