what multiple soever 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; and if the base LF, be equal to the base NF, the solid LV is equal (C. 11.) 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 : since then there are four magnitudes, viz, the two bases AF, FH, and the two solids AV, B G I O Y F C Q S 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, equal; and if less, less. Therefore (5. def. 5.) 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. PROP. XXVL PROB. 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 Da 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 a solid angle equal to the solid angle D. In the straight line DF take any point F, from which draw (11. 11.) GF perpendicular to the plane EDC, meeting that plane in G; join DG, and at the point A in the straight line AB make (23. 1.) 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 (12. 11.) KH at right angles to the plane BAL; and make KH equal to GF, and join AH: then the solid angle at A, which is contained by the three plane angles BAL, BAH, HAL, is equal to the solid angle at D contained by the three plane angles 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 angles (3. def. 11.) with every straight line meeting * See Note. A 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 contain equal angles, therefore the base BK is equal (4. 1.) to the base EG: and KH is equal to GF, and HKB, FGE are right angles, therefore HB is equal (4. 1.) 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 equal to the base FE, therefore the angle BAH is equal (8. 1.) to the angle EDF: for the same reason, the angle HAL is equal to the an. gle FDC. Because if AL DL EL and DC be made equal, K I and KL, HL, GC, FC be Η joined, since the whole angle GAL 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, and contain equal angles, the base KL is equal (4. 1.) to the base GC: and KH is equal to GF, so 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 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. PROB. 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, make (26. 11.) 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 HAB to FCE: and as EC to CG, so make (12. 6.) BA to AK: and as GC to CF, so make (12. 6.) KA to AH; wherefore ex æquali (22. 5.) as EC to CF, so is BA to AH; complete the parallelogram D BH, and the solid AL: and because, as EC to CG, so BA to AK, the Η sides about the equal angles ECG, BAK are proportionals: therefore the parallelogram BK is similar to EG. For the same reason, the parallelogram KH is similar to GF, and с Е HB to FE. Wherefore three parallelograms of the solid AL are similar to three of the solid CD; and the three opposite ones in each solid are equal (24. 11.) 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 equal (B. 1 1.) each to each. Therefore the solid AL is similar (11. def. 11.) to the solid CD. Wherefore from a given straight line AB, a solid parallelopiped AL has been described similar and similarly situated to the given one CD. Which was to be done. PROP. XXVIII. THEOR. Ir 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 pa. rallel to GA, and not in the same plane with it, CD, FE are parallel (9. 11.); wherefore the diagonals CF, DE are in the plane in which the parallels are, and are themselves paral- C B lels (16. 11.); and the plane CDEF shall cut the solid AB into two equal parts. Because the triangle CGF is equal (34. 1.) to the triangle CBF, and the triangle DAE, to DHE; and that the parallelogram CA is equal (24. 11.) and similar to the opposite D LIH one BE; and the parallelogram GE to CH: therefore the prism contained by the two triangles CGF, DAE, and the three parallel. ograms CA, GE, EC, is equal (C. 11.) to the prism contained by the two 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 * See Note. 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 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 is 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, CBHD, AH is cut into two equal parts (28. 11.) 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 D K is cut by the plane LGHB through I FAG IN the diagonals LG, BH of the opposite planes ALNG, CBKH, the solid AK is double of the same prism which is contained betwixt the tri- C angles ALG, CBH. Therefore the solid AH is equal to the solid AK. A But let the parallelograms DM, EN opposite to the base have no common side: then, because CH, CK are parallelograms, CB is equal (34. 1.) to each of the opposite sides 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 (38. 1.) to the triangle BHK: and the parallelogram DC is equal (36. 1.) to the parallelogram HN: for the same reason the triangle Η Ε «A K D E Η Κ у м. Лc A AFG is equal to the triangle LMN, and the parallelogram CF is equal (24. 11.) to the parallelogram BN, and CG to BN; for they are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms AD, DG, GC, is equal (C. 11. to the prism contained by the two triangles LMN, BHK, and the three parallelograms BM, MK, KL. If therefore the prism LMNBHK be taken from the solid of which the base is the parallelogram AB, and in which FDKN is the one opposite to it; and if from this same solid there be taken the prism AFGCDE, the remaining solid, viz. the parallelopiped AH, is equal to the remaining parallelopiped AK. Therefore, solid parallelopipeds, &c. Q. E. D. PROP. XXX. THEOR. Solid parallelopipeds upon the same base, and of the same altitude, the insisting straight lines of which are not terminated in the same straight lines in the plane opposite to the base, are equal to one another.* Let the parallelopipeds CM, CN be upon the same base AB, and of the same altitude, but their insisting straight lines AF, AG, LM, LN, CD, CE, BH, BK, not terminated in the same straight lines; the solids CM, CN, are equal to one another. Produce FD, MH, and NG, KE; and let them meet one another in the points O, P, Q, R; and join AO, LP, BQ, CR: and because the plane LBHN is parallel to the opposite plane ACDF, and that the Ν Κ A C plane LBHM is that in which are the parallels LB, MHPQ, in whiclı also is the figure BLPQ, and the plane ACDF is that in which are the parallels AC, FDOR, in which also is the figure CAOR; therefore the figures BLPQ, CAOR are in parallel planes ; in like manner, because the plane ALNG is parallel to the opposite plane CBKE, and that the plane ALNG is that in which are the parallels AL, OPGN, in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, in which also is the figure CBQR; therefore the figures ALPO, CBQR are in parallel planes; and the planes ACBL, ORQP are parallel; therefore the solid CP is a paral * See Note. |