Book XI. plane angles BAC, CAD, DAB. These three together are less than four right angles. Take in each of the straight lines AB, AC, AD any points B, C, D, and join BC, CD, DB: Then because the solid angle at B is contained by the three plane angles CBA, a 20. 11. ABD, DBC, any two of them are greater a than the third; therefore the angles CBA, ABD are greater than the angle DBC: For the same reason, the angles BCA, ACD are greater than the angle DCB; and the angles CDA, ADB, greater than BDC: Wherefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles DBC, BCD, CDB: But the three angles DBC, BCD, CDB are 6 32. 1. equal to two right anglesb: There fore the six angles CBA, ABD, BCA, Next, Let the solid 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 beBC,CD,DE,EF, FB: And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater a than the third, the BE angles CBA, ABF, are greater than the angle FBC: for the same reason, the two plane angles at each of the points C, D, E, F, viz. the angles which are at the bases of the triangles having the common vertex A, are greater than the third angle at the same point, which is one of the angles of the polygon BCDEF : Therefore all the angles at the bases of the triangles are together greater than all the angles of the polygon: And Book XI. because all the angles of the triangles are together equal to twice as many right angles as there are triangles b, that is, 32, 1. as there are sides 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 sides in the polygone; therefore all the angles of the trian- c 1 Cor. gles are equal to all the angles of the polygon together with 32. 1. 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 solid angle at A, are less than four right angles. Therefore every solid angle, &c. Q. E. D. PROP. XXII. THEOR. Ir every two of three plane angles be greater than See N. the 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 the 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 a, and any two of them greater than the third :* 4. 1 But if the angles are not 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; b 4. or and it is plain that AC, together with either of the other 24. 1. two, must be greater than the third: Also DF with GK are greater than AC: For, at the point B, in the straight Book XI. line AB, make the angle ABL equal to the angle GHK, and make BL equal to one of the straight lines AB, BC, * 23. 1. 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: h AAA с K DE, EF, and that the angle DEF is greater than the angle à 24. 1. LBC, the base DF is greaterd 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 20. 1. 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, there* 22.1. fore, a triangle may be made f, the sides of which shall be equal to AC, DF, GK Q. E. D. See N. To make a solid 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 solid angle contained by three plane angles equal to ABC, DEF, GHK, each to each. From the straight lines containing the angles, cut off Book XI. AB, BC, DE, EF, GH, HK, all equal to one another; and join AC, DF, GK: Then a triangle may be made a of. 29. 11. three straight lines equal to AC, DF, GK. Let this be the triangle LMNb, so that AC be equal to LM, DF to MN, - 22. 1. and GK to LN; and about the triangle LMN describe a .5. 4. circle, and find its centre X, which will either be within the triangle, or in one of its sides, or without it. First, Let the centre 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 base LM; wherefore the angle ABC is equal to the angle LXMd. For the same reason, the angle DEF.8. 1. is equal to the angle MXN, and the angle GHK to theangle 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 an- !, glese; therefore also the three 1 • 2 Cor. angles ABC, DEF, GHK are 15. 1. equal to four right angles: But, by the hypothesis, they are less than four right angles, which is absurd; 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 side of it on which is the centre X, describe the triangle LOM, the sides LO, OM of which are equal to AB, BC; and because the base Boox XI. LM is equal to the base AC, the angle LOM is equal to the angle ABCd: And AB, that is, LO, by the hypothesis, d 8. 1. is less than LX; whereforę LO, OM fall within the trian. 1901 LOM, that is, the angle ABC, is L 1 21. 1. greater than the angle LXMf: In the same manner it may be proved X N Next, Let the centre X of the circle fall in one of the sides of the triangle, viz. in MN, R X pios EF, are equal to DF, which is deco + 20. 1. impossible t: Wherefore AB is not equal to LX; nor is it less; for then, much more, an: absurdity would follow : Therefore AB is greater than LX. But, let the centre X of the circle fall without the triangle LMN, and join LX, MX, NX. In this case likewise AB is greater than LX: If not, it is either equal to, or less than LX: First, let it be equal; it may be proved, in the same manner as in the first case, that the angle ABC is equal to the angle MXL, and GHK to LXN; therefore the whole angle MXN is equal to the two angles ABC, GHK: But ABC and GHK are together greater than the angle DEF: therefore also the angle MXN is greater than DEF. And because DE, EF are equal to |