Book XI. PROP. XXVII. PROB. See N. To describe from a given straight line a solid parallelopiped similar, and similarly situated, to one given. 2 26. 11. 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. a At the point A of the given straight line AB, make a 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 b 12. 6. GCF, and HAB to FCE: And as EC to CG, so makeb BA to AK; and as GC to CF, so makeb KA to AH; 22.5. wherefore, ex æqualis, as EC to CF, so is BA to AH: Complete the parallelogram BH, and the solid AL: And because, as EC to 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 the three opposite 24. 11. 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 situB. 11. ated in the same order, the solid angles are equale, each to f11 Def. 11. each. Therefore the solid AL is similarf 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. BOOK XI. PROP XXVIII. THEOR. If a solid parallelopiped be cut by a plane passing See N. through the diagonals of two of the opposite planes; it shall be cut in 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 cach: And because CD, FE, are each of them parallel to GA, and not in the same plane with it, CD, FE, are parallel; where- 9. 11. fore the diagonals CF, DE, are in the plane in which the parallels are, and are themselves parallels: And the plane CDEF shall cut the solid AB into two equal parts. Because the triangle CGF is equal to the triangle CBF, and the triangle DAE to DHE; and that the parallelogram CA is equal and similar to the opposite one BE; and G D B 16. 11. F H 34. 1. 24.11. E 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 C. 11. 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 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 See N. the same altitude, the insisting straight lines of which are terminated in the same straight lines in the planc opposite to the base, are equal to one another. BOOK XI. See the Let the solid parallelopipeds AH, AK, be upon the same base AB, and of the same altitude, and let their insisting figures be 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. low. 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 * 28. 11. ALGF, CBHD, AH is cut into two equal parts a by the plane AGHC: Therefore the solid AH is double of the prism which is contained be- D K twixt the triangles ALG, G CBH: For the same rea son, because the solid AK is N BH of the opposite planes A e D ALNG, CBKH, the solid AK is double of the same prism which is contained betwixt the triangles ALG, ČBH. Therefore the solid AH is equal to the solid AK. But, let the parallelograms DM, EN, opposite to the base, have no common side: Then, because CH, CK, are 34. 1. parallelograms, CB is equal 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: €38. 1. Wherefore also the triangle CDE is equal to the triangle * 36. 1. BHK: And the parallelogram DG is equal to the parallelogram HN: For the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is 24. 11. equale to the parallelogram BM, and CG to BN; for they K D E H K H E A are opposite. Therefore the prism which is contained by the two triangles AFG, CDE, and the three parallelograms fC. 11. 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 prism LMN, BHK, be taken Boox XI. 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 AFG, CDE, 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 See N. 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 LBHM is parallel to the op posite plane ACDF, and that the plane LBHM is that in which are the parallels LB, MHPQ, in which 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: 4 Book XI! 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; there fore the solid CP is a parallelopiped: But the solid CM, of which the base is ACBL, to which FDHM is the opposite 229. 11. parallelogram, is equal to the solid CP, of which the base Kdy N is the parallelogram ACBL, to which ORQP is the one opposite; because they are upon the same base, and their insisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ, are in the same straight lines FR, MQ: And the solid CP is equal to the solid CN: for they are upon the same base ACBL, and their insisting straight lines AO, AG, LP, LN; CR, CE, BQ, BK, are in the same straight lines ON, RK: Therefore the solid CM is equal to the solid CN. Wherefore solid parallelopipeds, &c. Q. E. D. PROP. XXXI. THEOR. See N. SOLID parallelopipeds, which are upon equal bases, and of the same altitude, are equal to one another. Let the solid parallelopipeds AE, CF, be upon equal bases AB, CD; and be of the same altitude; the solid AE is equal to the solid CF. First, Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same |