At a given point in a given straight line, to make a solid 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 solid angle contained by the three plane angles EDC, EDF, FDC : it is required to make at the point A in the straight line AB á solid angle equal to the solid angle D. In the straight line DF take any point F, 11. 11. from which draw* FG perpendicular to the plane EDC, meeting that plane in G, and join DG: at the point A, in the straight line AB, 23. 1. 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 * 12. 11. to DG, and from the point K erect* KH at right angles to the plane BAL, and make KH equal to GF, and join AH. The solid angle at A which is contained by the three plane angles BAL, BAH, HAL shall be equal to the solid angle at D contained by the three plane angles EDC, EDF, FDC. 11. * 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 3 Def angles with every straight line meeting it in that plane: therefore each of the angles FGD, FGE is a right angle. For the same reason, HKA, HKB are right angles. And because KA, AB are equal to GD, DE, each to each, and that they contain equal angles, therefore the base BK is equal to the 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, each to each, 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, each to #4.1. *4.1. * each; and the base HB is equal to the base FE; therefore the angle BAH is equal to the angle * 8. 1. EDF. For the same reason, the angle HAL is equal to the angle FDC: because if AL and DC be made equal, and KL, HL, GC, FC be joined; since the whole angle BAL is equal to the whole EDC, and the parts of them BAK, EDG are, 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, each to each, and contain equal angles, the base KL is equal to the base * 4. 1. GC; and KH is equal to GF; so that LK, KH are equal to CG, GF, each to each; and they contain right angles; therefore the base HL is + Def A * 11. equal to the base FC: again, because HA, AL +4. 1. are equal to FD, DC, each to each, and the base HL to the base FC, the angle HAL is equal* *8.1. to the angle FDC. Therefore, because the three plane angles BAL, BAH, HAL, which contain the solid angle at A, are equal to the three plane angles EDC, EDF, FDC, which contain the solid angle at D, each to each, and are situated in the same order, the solid angle at A is equal* *B 11. to the solid angle at D. Therefore at a given point in a given straight line a solid angle has been made equal to a given solid angle contained by three plane angles. Which was to be done. PROP. XXVII. THEOR. To describe from a given straight line a solid parallelopiped similar and similarly situated to one given. Let AB be the given straight line, and CD the given solid parallelopiped. It is required from AB to describe a solid parallelopiped similar and similarly situated to CD. At the point A of the given straight line AB *26. 11. make* a solid angle equal to the solid 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 12. 6. HAB to FCE: and as EC to CG, so make * BA * 12. 6. to AK; and as GC to CF, so make* KA to AH; *22.5. wherefore, ex æquali, * as EC to CF, so is BA to AH: complete the parallelogram BH, and the solid AL AL shall be similar and similarly, situated to CD. Because, as EC to GC, so BA to AK, the sides about the equal angles ECG, BAK, are proportionals; therefore the parallelogram BK +1 Def. is similar to EG. For the same reason, the parallelogram KH is similar to GF, and HB to FE; wherefore three parallelograms of the solid AL are similar to three of the solid CD: and * *24. 11. the three opposite ones in each solid are equal and similar to these, each to each. Also, because the plane angles which contain the solid angles of the figures are equal, each to each, and situated in the same order, the solid angles are *B. 11. equal,* each to each. Therefore the solid AL *11 Def. is similar to the solid CD. Wherefore from a 11. given straight line AB a solid parallelopiped ÅL has been described similar and similarly situated to the given one CD. Which was to be done. PROP. XXVIII. THEOR. If a solid parallelopiped be cut by a plane passing through the diagonals of two of the opposite planes; it shall be cut into two equal parts. Let AB be a solid parallelopiped, and DE, CF the diagonals of the opposite 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, FE are* parallel; wherefore the diagonals CF, DE, are in the plane in which the parallels are, and are themselves pa-G rallels and the plane CDEF shall cut the solid AB into two D equal parts. B *9.11. 16. 11. F H E Because the triangle CGF is A equal to the triangle CBF, and the triangle *34. 1. DAE to DHE; and that the parallelogram CA is equal and similar to the opposite one BE, *24.11. 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 to the prism contained by the two *C. 11. triangles CBF, DHE, and the three parallelograms BE, CH, EC; because they are contained by the same number of equal and similar planes, alike situated, and none of their solid angles are contained by more than three plane angles. Therefore the solid AB is cut into two equal parts by the plane CDEF. Q. E. D. N. B. The insisting straight lines of a parallelopiped, ⚫ mentioned in the next and some following propositions, are 'the sides of the parallelograms betwixt the base and the opposite plane parallel to it,' PROP. XXIX. THEOR. Solid parallelopipeds 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 parallelopipeds 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 shall be equal to the solid AK. First let the parallelograms DG, HN, which are opposite to the base AB, have a common side HG. Then, because the solid AH is cut by the plane AGHC passing through the diagonals, AG, CH, of the opposite planes ALGF, 28. 11. CBHD, AH is cut into two equal parts* by the plane AGHC; therefore the solid AH is double of the prism which is contained betwixt the triangles ALG, CBH: for the same reason, because the solid AK is cut by the plane LGHB, through the diago A N nals LG, BH of the opposite planes ALNG, CBKH, the solid AK is double of the same prism which is contained betwixt the triangles ALG, +6 Ax. CBH: therefore the solid AH is equal to the solid AK. Next let the parallelograms DM, EN, oppo site to the base, have no common side. Then, |