L to the bafe BE. And be-Book VII. b 4. I. C 20. I. fore the bafe DB is equal caufe BD, DC are greater c than CB, and one of them BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And becaufe DA is equal to AE, and AC common, but the base DC greater than the bafe EC; therefore the angle DAC is greater d than the angle EAC; and, by the conftruction, the d 25. 1. angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC. But BAC is not lefs than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. Wherefore, if a folid angle, &c. Q. E. D. E PROP. XX. THEOR. VERY folid angle is contained by plane angles First, Let the solid angle at A be contained by three plane angles BAC, CAD, DĂB. These three together are lefs than four right angles. D Take in each of the ftraight lines AB, AC, AD any points B, C, D, and join BC, CD, DB: then, because the folid angle at B is contained by the three plane angles CBA, ABD, DBC, any two of them are greater a than the third; a 19. 7. therefore the angles CBA, ABD are greater than the angle DBC: for the fame reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB greater than BDC: Wherefore the fix angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB: but the three angles DBC, BCD, CDB are Β ́ equal to two right angles b: Therefore the fix angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles and because the ༩ three b 23. I. Book VII. three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of thefe three triangles, viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to fix right angles: Of thefe, the fix angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles : therefore the remaining three angles BAC, DAC, BAD, which contain the folid angle at A, are less than four right angles. Next, Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common sections of it with those planes be BC, CD, DE, EF, FB: and because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any 19. 7. two are greater a than the third, the angles CBA, ABF are great- B A F D E the triangles are together greater than all the angles of the polygon and because all the angles of the triangles are together equal to twice as many right angles as there are trib 32. 1. angles; that is, as there are fides in the polygon BCDEF; and because all the angles of the polygon, together with four right angles, are likewife equal to twice as many right angles c 1. cor. as there are fides in the polygon; therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles. But all the angles at the bafes of the triangles are greater than all the angles of the polygon, as has been proved. Wherefore, the remaining angles of the triangles, viz. those at the vertex, which contain the folid 32. I. angle angle at A, are less than four right angles. Therefore every Book VII. folid angle, &c. Q. E. D. PROP. XXI. THE OR. F two folids be contained by the fame number of equal and fimilar planes, fimilarly fituated, and if the inclination of any two contiguous planes in the one folid be the fame with the inclination of the two equal, and fimilarly fituated planes in the other, the folids themselves are equal and fimilar. Let AG and KQ be two folids contained by the fame number of equal and fimilar, planes, fimilarly fituated, so that the plane AC is fimilar and equal to the plane KM, the plane AF to the plane KP; BG to LQ, GD to QN, DE to NO, and FH to PR. Let alfo the inclination of the plane AF to the plane AC be the fame with that of the plane KP to the plane KM, and fo of the reft; the folid KQ is equal and fimilar to the folid AG. Let the folid KQ be applied to the folid AG, fo that the bafes KM and AC, which H are equal and G R fimilar, may coincidea, the with B, and e PROP. Book VII. 1 1 IF F a folid be contained by fix planes, two and two of which are parallel, the oppofite planes are fimilar and equal parallelograins. Let the folid CDGH be contained by the parallel planes AC, GF; BG, CE; FB, AE: its oppofite planes are fimilar and equal parallelograms. B Because the two parallel planes BG, CE, are cut by the a 14. 7. plane AC, their common fections AB, CD, are parallel a. Again, because the two parallel planes BF, AE, are cut by the plane AC, their common fections AD, BC are parallel a; 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, BF, AE is a parallelogram: Join AH, DF; and because AB is parallel to DC, and BH to CF; the two straight lines AB, BH, which meet one another, are parallel to DC and CF, which meet one another; wherefore, tho' the first two are not in the fame plane with the other two, they contain eb97. qual angles b; the angle ABH is therefore equal to the angle DCF. A H G F D E And because AB, BH, are equal to DC, CF, and the angle ABH equal to the angle DCF; therefore the bafe AH is £ 4. 1. equal to the bafe DF, and the triangle ABH to the triangle DCF: For the fame reason, the triangle AGH is equal to the triangle DEF; and therefore the parallelogram BG is equal and fimilar to the parallelogram ČE., In the fame manner it may be proved, that the parallelogram AC is equal and fimilar to the parallelogram, GF, and the parallelogram AE to BF. Therefore, if a solid, &c. Q. E. D. PROP, IF a folid parallelepiped be cut by a plane parallel to two of its oppofite planes, it will be divided into two folids, which will be to one another as their bafes. Let the folid parallelepiped ABCD be cut by the plane EV, which is parallel to the oppofite planes AR, HD, and divides the whole into the folids ABFV, EGCD; as the base AEFY to the base EHCF, fo is the folid ABFV to the folid 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 folids LP, KR, HU, MT: then, because Book VII. the ftraight lines LK, KA, AE are all equal, and alfo the straight lines KO, AY, EF, which make equal angles with LK, KA, AE, the parallelograms LO, KY, AF are equal and fimilar a: and likewise the parallelograms KX, a 20 6. KB, AG; as alfo b the parallelograms LZ, KP, AR, b 22. 7. because they are oppofite planes. For the fame reason, the parallelograms EC, HQ, MS, are equal a; and the parallelograms HG, HI, IN, as alfo b HD, MÚ, NT; therefore three planes of the folid LP, are equal and fimilar to three planes of the folid KR, as alfo to three planes of the folid AV: but the three planes oppofite to these three are equal and fimilar to them b in the several folids; therefore the folids LP, KR, AV are contained by equal and fimilar planes. And because the planes LZ, KP, AR are parallel, and are cut by the plane XV, the inclination of LZ |