PROP. XX. THEOR. Book XI. IF a folid angle be contained by three plane angles, any see N. two of them are greater than the third. Let the folid angle at A be contained by the three plane angles BAC, CAD, DAB. Any two of them are greater than the third. If the angles BAC, CAD, DAB be all equal, it is evident that any two of them are greater than the third. But if they are not, let BAC be that angle which is not less than either of the other two, and is greater than one of them DAB; and at the point A in the straight line AB, make, in the plane which paffes through BA, AC, the angle BAE equal to the a 23. 1. angle DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the D A b 4. 1. EC C are greater than CB, and one of them C 20. 1. BD has been proved equal to BE a part of CB, therefore the other DC is greater than the remaining part EC. And because DA is equal to AE, and AC common, but the bafe DC greater than the base EC; therefore the angle DAC is greater than d 25. I. the angle EAC; and, by the construction, the 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 less 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. XXI. THEOR. VERY folid angle is contained by plain angles First, Let the folid angle at A be contained by three plane These three together are less than Take 03 Book XI. 2 20, II. b 32.1. Take in each of the straight 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 plain angles CBA, ABD, DBC, any two of them are greater than the third; 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 equal to two right angles: Therefore the fix angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles: And because the three angles of each of the triangles B ABC, ACD, ADB are equal to two D A right angles, therefore the nine angles of these three triangles viz. the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DBA, BAD are equal to fix right angles: Of these 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 A F C E D Therefore all the angles at the bases of the triangles are toge ther 32. Ι. ther greater than all the angles of the polygon: And because Book XI. 2: all the angles of the triangles are together equal to twice as many right angles as there are triangles; that is, as there are b 32. 1. fides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise equal to twice as many right angles as there are fides in the polygon; c r. Cor. 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 bases 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 angle at A, are less than four right angles. Therefore every folid angle, &c. Q. E. D. PROP. XXII. THEOR. 1 IF every two of three plain angles be greater than the see N. third, and if the straight lines which contain them be all equal; a triangle may be made of the straight lines that join the extremities of those equal straight lines. Let ABC, DEF, GHK be three plane angles, whereof every two are greater than the third, and are contained by the equal straight lines AB, BC, DE, EF, GH, HK; if their extremities be joined by the straight lines AC, DF, GK, a triangle may be made of three straight lines equal to AC, DF, GK; that is, every two of them are together greater than the third. If the angles at B, E, H are equal; AC, DF, GK are also equal, and any two of them greater than the third: Buta 4. I. if the angles are not all equal, let the angle ABC be not less than either of the two at E, H; therefore the straight line AC is not less than either of the other two DF, GKb; and 64. or 24.1. it is plain that AC, together with either of the other two, must be greater than the third: Alfo DF with GK are greater than AC: For, at the point B in the straight line AB make the c 23. 1. angle 04 Book XI. angle ABL equal to the angle GHK, and make BL equal to ~ one of the straight lines AB, BC, DE, EF, GH, HK, and join AL, LC: Then because AB, BL are equal to GH, HK, and the angle ABL to the angle GHK, the base AL is equal to the base GK: And because the angles at E, H are greater than the angle ABC, of which the angle at H is equal to ABL; therefore the remaining angle at E is greater than the angle LBC: d 24. 1. e 20. 1. f 22. I. A L C FGK And because the two fides LB, BC are equal to the two DE, EF, and that the angle DEF is greater than the angle LBC, the base DF is greater than the base LC: And it has been proved that GK is equal to AL; therefore DF and GK are greater than AL and LC: But AL and LC are greater than AC, much more then are DF and GK greater than AC. Wherefore every two of these straight lines AC, DF, GK are greater than the third; and therefore a triangle may be made f the fides of which shall be equal to AC, DF, GK. QE. D. See N. T O make a folid angle which shall be contained by three given plane angles, any two of them being greater than the third, and all three together less than four right angles. Let the three given plane angles be ABC, DEF, GHK, any two of which are greater than the third, and all of them together less than four right angles. It is required to make a folid angle contained by three plane angles equal to ABC, DEF, GHK, each to each. From From the straight lines containing the angles, cut off AB, Book XI. BC, DE, EF, GH, HK all equal to one another; and join AC, DF, GK: Then a triangle may be made of three straight a 22. II. lines equal to AC, DF, GK. Let this be the triangle LMN b, b 22. 1. fo that AC be equal to LM, DF to MN, and GK to LN; and about the triangle LMN describe a circle, and find its center c 5.4 X, which will either be within the triangle, or in one of its fides, or without it. First, Let the center X be within the triangle, and join LX, MX, NX: AB is greater than LX: If not, AB must either be equal to, or less than LX; first, let it be equal: Then because AB is equal to LX, and that AB is also equal to BC, and LX to XM, AB and BC are equal to LX and XM, each to each; and the base AC is, by construction, equal to the bafe LM: wherefore the angle ABC is equal to the angle LXMd: For the fame reason, the angle DEF is equal to the d 8. 1. angle MXN, and the angle GHK to the angle NXL: Therefore the three angles ABC, DEF, GHK are equal to the three angles LXM, MXN, NXL: But the three angles LXM, MXN, NXL are equal to four right angles; therefore alfo the three angles ABC, DEF, GHK are equal to four right angles: But, by the hypothesis, they are lefs than four right angles; which is M R L X N abfurd; therefore AB is not equal to LX: But neither can AB be less than LX: For, if possible, let it be less, and upon the straight line LM, on the fide of it on which is the center X, describe the triangle LOM, the fides LO, OM, of which are equal to AB, BC; and because the base LM is equal to the 1 bafe e2. Cor. 15. I. |