Book XI. Take in each of the ftraight lines AB, AC, AD any points ~ B, C, D, and join BC, CD, DB: Then, because the folid b 32. I. D angle at B is contained by the three plain angles CBA, ABD, a 20. 11. 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 ABC, ACD, ADB are equal to two 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 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. B Next, Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; thefe toge ther are lefs than four right angles. A Let the planes in which the angles are, be cut by a plane, and let the common fections of it with thofe planes be BC, CD, DE, EF, FB: And because the fold angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater than the third, the angles CBA, ABF are greater than the angle FBC: For the fame reafon, the two plane angles at each of the points C, D, E, F, viz. the angles which are at the bafes of the triangles having the common vertex A, are greater than the third angle at the fame point, which is one of the angles of the polygon BCDEF: B F E D Therefore all the angles at the bafes of the triangles are toge ther ther greater than all the angles of the polygon: And because Book XI. 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. I. fides in the polygon BCDEF: and that all the angles of the polygon, together with four right angles, are likewife 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 an- 34. 1° gles 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 thè remaining angles of the triangles, viz. thofe at the vertex, which contain the folid angle at A, are lefs than four right angles. Therefore every folid angle, &c. Q. E. D. PROP. XXII. THEOR. IF every two of three plain angles be greater than the See . third, and if the ftraight lines which contain them be all equal; a triangle may be made of the ftraight lines that join the extremities of thofe 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 ftraight lines AB, BC, DE, EF, GH, HK; if their extremities be joined by the ftraight lines AC, DF, GK, a triangle may be made of three ftraight 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 alfo equal, and any two of them greater than the third: But a 4. 1. if the angles are not all equal, let the angle ABC be not lefs than either of the two at E, H; therefore the ftraight line AC is not lefs than either of the other two DF, GK; and b 4.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 ftraight line AB make the € 23. 1. 0 4 angle 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 bafe AL is equal to the base GK: And becaufe 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; 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, d 24. 1. the bafe DF is greater than the bafe LC: And it has been proved that GK is equal to AL; therefore DF and GK are e 20. 1. greater than AL and LC: But AL and LC are greater f 22. I. than AC; much more than 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. Q. E. D. See N. Th make a folid angle which fhall be contained by three given plane angles, any two of them being greater than the third, and all three together lefs 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 lets 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 ftraight 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 ftraight a 22. 11. E Ада A D F K 22. I lines equal to AC, DF, GK. Let this be the triangle LMN, b fo that AC be equal to LM, DF to MN, and GK to LN; and about the triangle LMN defcribe a circle, and find its centre e 5. 4. X, which will either be within the triangle, or in one of its fides, or without it. R First, Let the centre X be within the triangle, and join LX, MX, NX: AB is greater than IX: If not, AB muft either be equal to, or less than LX; first, let it be equal: Then becaufe AB is equal to LX, and that AB is alfo 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 conftruction, equal to the bafe LM; wherefore the angle ABC is equal to the angle LXM: For the fame reason, the angle DEF is equal to the 1 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 hypothefis, they are lefs than four right angles; which is abfurd; therefore AB is not equal to LX: But neither can AB be M L N lefs than LX: For, if poffible, let it be lefs, and upon the ftraight line LM, on the fide of it on which is the centre X, defcribe the triangle LOM, the fides LO, OM, of which are equal to AB, BC; and because the bafe LM is equal to the e 2. Cor. 15. I. bafe Book XI. ~ d 8. I. f 21. I. † 20. I. L R bafe AC, the angle LOM is equal to the angle ABC And the fame angles ABC, DEF, GHK are less than four right angles; which is abfurd: Therefore AB is not less than LX, and it has been proved that it is not equal to LX; wherefore AB is greater than LX. L R Next, Let the centre X of the circle fall in one of the fides of the triangle, viz. in MN, and join XL: In this cafe alfo AB is greater than LX. If not, AB is either equal to LX, or lefs than it : First, let it be equal to LX: Therefore AB and BC, that is, DE and EF, are equal to MX and XL, that is, to MN But, by the conftruction, MN is equal to DF; therefore DE, EF are equal to DF, which is im poffible +: Wherefore AB is not equal to LX; nor is it lefs; for then, much more, an abfurdity would follow: Therefore AB is greater than LX. M N X But, let the centre X of the circle fall within the triangle LMN, and join LX, MX, NX. In this cafe likewife 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 fame manner, as in the first cafe, 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 allo the angle MXN is greater than DEF. And becaufe DE, EF |