G A Ha 16. 11. EX, BD, are parallel *. For the fame reason, because the two Beok XI. parallel planes GH, KL are cut by the plane AXFC, the common fections AC, XF are parallel: And because EX is parallel to BD, a fide of the triangle ABD, as AE to EB, fo is b AX to XD. Again, because XF is parallel to AC, a fide of the triangle ADC, as AX to XD, K fo is CF to FD: And it was proved that AX is to XD, as AE to EB; Therefore, as AE to EB, fo is CF to FD. M Wherefore, iftwo ftraight lines, &c. Q. E. D. b 2.6. L E F C II. 5. N B D PROP. XVIII. THEOR. [F a ftraight line be at right angles to a plane, every plane which passes through it fhall be at right angles to that plane. Let the straight line AB be at right angles to a plane CK; every plane which paffes through AB fhall be at right angles to the plane CK. D G A H K a 3. def. 11. 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 straight line in that plane meeting it a: And confequently it is perpendicular to CE: Wherefore ABF is a right angle; but GFB is likewise a right angle; therefore AB is parallel b to FG. And b 28. I. AB is at right angles to the plane CK; therefore FG is also at right angles to the fame plane. But one plane is at right an- c 8. 11. gles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right Book XI. angles to the other planed; and any straight line FG in the plane DE, which is at right angles to CE the common section d4. def. 11. 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 straight line, &c. Q. E. D. IF PROP. XIX. THEOR. F two planes cutting one another be each of them perpendicular to a third plane; their common fection fhall 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. 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 fection 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 a 4. def. 11. third plane a. 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, which is im13. 11. poffible b: Therefore, from the point D there cannot be any ftraight line at right angles to the third plane, except B EF D 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. PROP. IF PROP. XX. THEOR. Book XI. 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. D 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 lefs 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 angle a 23. i. DAB; and make AE equal to AD, and through E draw BEC cutting AB, AC in the points B, C, and join DB, DC. And becaufe 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 base DB is equal b to the base BE. And because BD, DC B are greater than CB, and one of b 4. t. E C them BD has been proved equal to BE a part of CB, therefore c 20. 1. 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 bafe EC; therefore the angle DAC is greater than the angle EAC; and, by the construction, the angle DAB d 25. 1. 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 VERY folid angle is contained by plane angles which 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. 03 Take Book XI. a 20. II. b. 32. I. 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 plane angles CBA, 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 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, B ADB are equal to two right angles, D A 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. Next, let the folid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; thefe together are less than four right angles. Let the planes in which the angles are, be cut by a plane, and let the common section of it with thofe a planes be BC, CD, DE, EF, FB: And C A F E D fore all the angles at the bafes of the triangles are together greater greater than all the angles of the polygon: And because all the Book XI. angles of the triangles are together equal to twice as many right angles as there are triangles b; that is, as there are fides b 32. I. 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 ; there- c 1. Cor. fore all the angles of the triangles are equal to all the angles 32. 1. 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 the remaining angles of the triangles, viz. those at the vertex, which contain the folid angle at A, are less than_four right angles. There fore every folid angle, &c. Q. E. D. PROP. XXII. THEOR. I F every two of thrée 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 ftraight 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 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 also equal a, and any two of them greater than the third: But if a 4. 1. the angles are not all equal, let the angle ABC be not less than either of the two at E, H; therefore the ftraight line AC 24 I. is not lefs than either of the other two DF, GK ; and it is b 4. Cor. plain that AC, together with either of the other two, must be greater than the third: Also DF with GK are greater than AC: For, at the point B in the ftraight line AB make the c 23. I, 04 -angle |