inclination of KR to KM is the same with the inclination of AH to AC, the plane KR will be upon the plane AH, and will coincide with it, because they are similar and equal (8. Ax. 1.), and because their equal sides KN and AD coincide. And in the same manner it is shewn that the other planes of the solid KQ coincide with the other planes of the solid AG, each with each: wherefore the solids KQ and AG do wholly coincide, and are equal and similar to one another. PROP. II. THEOR. If a solid be contained by six planes, two and two of which are parallel, the opposite planes are similar and equal parallelograms. B Let the solid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE: its opposite planes are similar and equal parallelograms. Because the two parallel planes BG, CE, are cut by the plane AC, their common sections AB, CD are parallel (14. 2. Sup.). Again, because the two parallel planes BF, AE are cut by the plane AC, their common sections AD, BC are parallel (14. 2. Sup.): 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, BE, AE is a parallelogram; join AH, DF; and because AB is parallel to DC, and BH to CF; the two A straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another; wherefore, though the first two are not in the same plane with the other two, they contain equal angles (9. 2. Sup.); 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 (4. 1.) to the base DF, and the triangle ABH to the triangle DCF: For the same reason, the triangle AGH is equal to the triangle DEF: and therefore the parallelogram BG is equal and similar to the parallelogram CE. In the same manner, it may be proved, that the parallelogram AC is equal and similar to the parallelogram GF, and the parallelogram AE to BF. PROP. III. THEOR. D F E If a solid parallelopiped be cut by a plane parallel to two of its opposite planes, it will be divided into two solids, which will be to one another as the bases. Let the solid parallelopiped ABCD be cut by the plane EV, which is parallel to the opposite planes AR, HD, and divides the whole into the solids ABFV, EGCD: as the base AEFY to the base EHCF, so is the solid ABFV to the solid 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 LF KR HU, MT; then, because the straight lines LK, KA, AE are all equal, and also the straight lines KO, AY, EF which make equal angles with LK, KA, AE, the parallelograms LO, KY, AF are equal and similar (36. 1. & def. 1. 6.) and likewise the parallelograms KX, KB, AG; as also (2. 3. Sup.) the parallelograms LZ, KP, AR, because they are opposite planes. For the same reason, the parallelograms EC, HQ, MS are equal (36. 1. & def. 1. 6.); and the parallelograms HG, HI, IN, as also (2. 3. Sup.) HD, MU, NT; therefore three planes of the solid LP, are equal and similar to three planes of the solid KR, as also to three planes of the solid AV but the three planes opposite to these three are equal and similar to them (2. 3. Sup.) in the several solids; therefore the solids LP, KR, AV are contained by equal and similar planes. And because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ to XP is equal to that of KP to PB; or of AR to BV (15. 2. Sup.) and the same is true of the other contiguous planes, therefore the solids LP, KR, and AV, are equal to one another (1. 3. Sup.). For the same reason, the three solids, ED, HU, MT are equal to one another; therefore 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 (1. 3. Sup.) 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, 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: There fore (def. 5. 5.) as the base AF is to the base FH, so is the solid A V to the solid ED. COR. Because the parallelogram AF is to the parallelogram FH as YF to FC (16.), therefore the solid AV is to the solid ED as YF to FC E D H M K N PROP. IV. THEOR. If a solid parallelopiped be cut by a plane passing through the diagonals of two of the opposite planes, it will be cut into two equal prisms. 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, though not in the same plane with it, CD, FE are parallel (8. 2. Sup.) wherefore the diagonals CF, DE are in the plane in which the parallels are, and are themselves parallels (14. 2. Sup.); the plane CDEF cuts the solid AB into two equal parts. G B F D H Because the triangle CGF is equal (34. 1.) to the triangle CBF, and the triangle DAE to DHE; and since the parallelogram CA is equal (2. 3. Sup.) and similar to the opposite one BE; and the parallelogram GE to CH: therefore the planes which contain the prisms CAE, CBE, are equal and similar, each to each; and they are also equally inclined to one another, because the planes AC, EB are parallel, as also AF and BD, and they are cut by the plane CE (15. 2. Sup.). Therefore the prism CAE is equal to the prism CBE (1. 3. Sup.), and the solid AB is cut into two equal prisms by the plane CDEF. A D N. B. The insisting straight lines of a parallelopiped, mentioned in the following propositions, are the sides of the parallelograms betwixt the base and the plane parallel to it. PROP. V. THEOR. Solid parallelopipeds upon the same base, and of the same altitude, the in sisting 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 let the insisting lines CD CE, BH, BK be terminated in the same straight line DK; the solid AF is equal to the solid AK. 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 DG is equal (36. 1.) to the parallelogram HN. For the same reason, the triangle AFG is equal to the triangle LMN, and the parallelogram CF is equal (2. 3. Sup.) to the parallelogram BM, and CG to BN; for they are opposite. Therefore the planes which contain the prism DAG are similar and equal to those which contain the prism HLN, each to each › and the contiguous planes are also equally inclined to one another (15. 2. Sup), because that the parallel planes AD and LH, as also AE and LK are cut by the same piane DN: therefore the prisms DAG, HLN are equal (1. 3. Sup.). If therefore the prism LNH be taken from the solid, of which the base is the parallelogram AB, and FDKN the plane opposite to the base; and if from this same solid there be taken the prism AGD, the remaining solid, viz. the parallelopiped AH is equal to the remaining parallelopiped AK. PROP. VI. 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 poin's O, P, Q, R; and join AO, LP, BQ, CR. Because the planes (def. 5. 3. Sup.), LBHM and ACDF are parallel, and because the plane LBHM is that in which are the parallels LB, MHPQ (def. 5. 3. Sup.), and in which also is the figure BLPQ; and because the plane ACDF is that in which are the parallels AC, FDOR, and in which also is the figure CAOR; therefore the figures BLPQ, CAOR, are in parallel planes. In like manner, because the planes ALNG and CBKE are parallel, and the plane ALNG is that in which are the parallels AL, OPGN, and in which also is the figure ALPO; and the plane CBKE is that in which are the parallels CB, RQEK, and in which also is the figure CBQR; therefore the figures ALPO, CBQR, are in parallel planes. But the planes ACBL, ORQP are also parallel; therefore the solid CP is a parallelopiped. Now the solid parallelopiped CM is equal (5. 2. Sup.) to the solid parallelopiped CP, because they are upon the same base, and their insisting straight lines AF, AO, CD, CR; LM, LP, BH, BQ are terminated in the same straight lines FR, MP; and the solid CP is equal (5. 2. Sup.) 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 terminated in the same straight lines ON, RK; Therefore the solid CM is equal to the solid CN. PROP. VII. THEOR. 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. Case 1. Let the insisting straight lines be at right angles to the bases AB, CD, and let the bases be placed in the same plane, and so as that the sides CL, LB, be in a straight line; therefore the straight line LM, which is at right angles to the plane in which the bases are, in the point L, is common (11. 2. Sup.) to the two solids AE, CF; let the other insisting lines of the solids be AG, HK, BE; DF, OP, CN: and first, let the angle ALB be equal to the angle CLD; then AL, LD are in a straightline (14. 1.). Produce OD, HB, and let them meet in Q and complete the solid parallelopiped LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting straight lines: therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is (7. 5.) the base CD to the same LQ and because the solid parallelopiped AR is cut by the plane LMEB, which is parallel to the opposite plancs AK. DR; as the base AB is to the base LQ, so is (3. 3. Sup.) the solid : |