lel, for the fame reafon, because the two parallel planes GH, KL Book XI. a. 16. II. H b. 2. 6. C. II. 5. PROP. XVIII. THEOR. I' Fa ftraight line be at right angles to a plane, every plane which paffes thro' it fhall be at right angles to that plane. Let the ftraight line AB be at right angles to the plane CK. every plane which paffes through AB fhall be at right angles to the plane CK. D GAH Let any plane DE pass through AB, and let CE be the common fection of the planes DE, CK; take any point F in CE, from which draw FG in the plane DE at right angles to CE. and because AB is perpendicular to the plane CK, therefore it is alfo perpendicular to every ftraight line in that plane meeting it 2. and confequently it is perpendicular to CE, wherefore ABF is a right angle; but GFB is like C K a.3.Def 11 FBE wife a right angle; therefore AB is parallel to FG. and AB is at b. 28. 1. right angles to the plane CK; therefore FG is alfo at right angles to the fame plane c. but one plane is at right angles to another plane c. 8. 11. when the straight lines drawn in one of the planes, at right angles d.4.Def.11. Book XI. to their common fection, are also at right angles to the other plane ; and any straight line FG in the plane DE, which is at right angles to CE the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pafs through AB are at right angles to the plane CK. Therefore if a ftraight line, &c. Q. E. D. I a F two planes cutting one another be each of them perpendicular to a third plane; their common fection shall be perpendicular to the fame plane. Let the two planes AB, BC be each of them perpendicular to a third plane, and let BD be the common fection of the first two, BD is perpendicular to the third plane. B If it be not, from the point D draw, in the plane AB, the ftraight line DE at right angles to AD the common section of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD the common section of the plane BC with the third plane. and because the plane AB is perpendicular to the third plane, and DE is drawn in the plane AB at right angles to AD their common fection, DE is perpendicular to the third 2.4.Def.11. plane 2. in the fame manner, it may be proved that DF is perpendicular to the third plane. wherefore from the point D two ftraight lines ftand at right angles to the third plane, upon the fame fide of it, b. 13. 11. which is impoffible b. therefore from the A EF D point D there cannot be any straight line at right angles to the third plane, except BD the common fection of the planes AB, BC. BD therefore is perpendicular to the third plane. Wherefore if two planes, &c. Q. E, D. I PROP. XX. THEOR. Book XI. F a folid angle be contained by three plane angles, See N. any 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. a 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 passes through BA, AC, the angle BAE equal to the angle DAB; and make AE a. 23. I. equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. and because DA is equal to AE, and AB is common, the two DA, AB are equal to the two EA, AB, and the angle DAB is equal to the angle EAB. therefore the bafe DB is equal b to the bafe BE. and becaufe BD, DCB d D A b. 4. I. E C are greater than CB, and one of them BD has been proved equal c. 20. 1. 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 base DC greater than the base EC; therefore the angle DAC is greater than the angle EAC; and, by the d. 25. 2o conftruction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater 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 VERY folid angle is contained by plane angles First, Let the folid angle at A be contained by three plane angles BAC, CAD, DAB. these three together are less than four right angles. Book XI. b. 32. I. 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 a. 20. 11. 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 b. 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 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. B Next, Let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are lefs than four right angles. Let the planes in which the angles are be cut by a plane, and let the common fections of it with those a planes be BC, CD, DE, EF, FB. and A F E D is one of the angles of the polygon BCDEF. therefore all the angles at the bafes of the triangles are together greater than all the angles of the polygon. and because all the angles of the triangles Book XI. are together equal to twice as many right angles as there are triangles ; that is, as there are fides in the polygon BCDEF; and b. 32. 1. 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 ; therefore all the angles of the triangles are equal c. 1. Cor. to all the angles of the polygon together with four right angles. 32. I. 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 lefs than four right angles. Therefore every folid angle, &c. Q. E. D. F every two of three plane 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 ftraight 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 equala, a. 4. In and any two of them greater than the third. but 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 lefs than either of the |