a. 16. 11. lela. for the same reason, because the two parallel plancs GH, KL Book XI. are cut by the plane AXFC, the common sections AC, XF H are parallel. and because EX A is parallel to BD, a fide of the G triangle ABD, as AE to EB, b. 2. 6. fo is 6 AX to XD. again, bem cause XF is parallel to AC, a L fide of the triangle ADC, as E F N D X C. II. 5. (F a straight line be at right angles to a plane, every to that plane, Let the straight line AB be at right angles to the plane CK. every planę which passes through AB shall be at right angles to the plane CK. Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG in the plane D G A A H DE at right angles to CE. and because AB is perpendicular to the plane CK, therefore it is also K perpendicular to every straight line in that plane meeting it a. a. 3.Def 114 and consequently it is perpendicular to CE, wherefore ABF is a right angle; but GFB is like c F B E wife a right angle; therefore AB is parallel to FG. and AB is at b. 28. 3. right angles to the plane CK; therefore FG is also at right angles to the same 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 Book XI. to their common section, are also at right angles to the other plane d; and any straight line FG in the plane DE, which is at 4.4.Def.11. 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 pass through AB are at right angles to the plane CK. Therefore if a straiglit line, &c. Q. E. D. T two planes cutting one another be each of them perpendicular to a third plane; their common section 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 section of the first two, BD is perpendicular to the third plane. If it be not, freen the point D draw, in the plane AB, the B section, DE is perpendicular to the third 2.4.Def. 11. plane , in the same maner, it may be proved that DF is perpendicular to the D A C b. 13. 11. which is impossible b. therefore from the point D there cannot be any straight line at right angles to the third plane, except BD the common section of the planes AB, BC. BD therefore is perpendicular to the third plane. Wherefore if two planes, &c. Q. E, D. Book XI. PROP. XX. THEOR. I F a solid angle be contained by three plane angles, See N. any two of them are greater than the third. Let the solid 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 passes through BA, AC, the angle BAE equal to the angle DAB; and make AE a. 23. 1. equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and D 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 A A the angle DAB is equal to the angle EAB. therefore the base DB is equal to the base BE. and because BD, DCB Е с 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 d than the anglé EAC; and, by the d. 25. i. construction, 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 less than either of the angles DAB, DAC, therefore BAC with either of them is greater than the other. Wherefore if a solid angle, &c. Q. E. D. E VERY folid angle is contained by plane angles which together are less than four right angles. First, Let the solid angle at A be contained by three plane ane gles BAC, CAD, DAB. these three together are less than four right angles. Book XI. 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, ABD, DBC, any two of 2. 20. 11. them are greater a than the third; therefore the angles CBA, ABD are greater than the angle DBC. for the same reason, the angles D are equal to two right angles b. there- A B с 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. 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 А E D the third angle at the fame 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 because all the angles of the triangles Book XI. are together equal to twice as many right angles as there are triangles b; that is, as there are sides in the polygon BCDEF; and b. 32. I. 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. I. Cor. to all the angles of the polygon together with four right angles. 32. 3. 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 solid angle, &c. Q. E. D. every two of three plane angles be greater than the See N. 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, a. 4. I. and any two of them greater than the third. but if the angles are A K |