PROP. XVII. THEOR. If two straight lines be cut by parallel planes, they shall be cut in the same ratio. Let the straight lines AB, CD be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D; as AE is to EB, so shall CF be to FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, 16. 11. the common sections EX, BD, are parallel: for the same reason, because the two parallel planes GH, KL are cut by the plane AXFC, the common sections AC, XF are parallel: and because EX is parallel to BD, a side of the triangle ABD, as AE to EB, so is AX to XD. Again, because XF is parallel to AC, a side of the triangle * 2.6. * ADC, as AX to XD, so is CF to FD: and it was 11. 5. proved that AX is to XD, as AE to EB; therefore, as AE to EB, so is CF to FD. straight lines, &c. Q. E. D. Wherefore, if two PROP. XVIII. THEOR. If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane. Let the straight line AB be at right angles to the plane CK; every plane 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 ing it and consequently it is perpendicular to CE: wherefore ABF is a right angle; but GFB is likewise a right angle: therefore AB is parallel to FG: and * 28. 1. AB is at right angles to the plane CK; therefore FG 11. is also at right angles to the same plane. But one * 8. 11. plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also* at right angles *4 Def. 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 pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q. E. D. PROP. XIX. THEOR. If two planes which cut one another be each of them perpendicular to a third plane; their common section is perpendicular to the same 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 shall be perpendicular to the third plane. If it be not, from the point D, in the plane AB, draw the straight line DE at right angles to AD the *11. 1. common section of the plane AB with the third plane; and in the plane BC draw DF at right angles to CD B 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 section, *4 Def. DE is perpendicular to the third plane. In the same manner it may 11. be proved that DF is perpendicular to the third plane: wherefore, from the point D two straight lines stand at right angles to the third plane, upon the same side 13. 11. of it, which is impossible; 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: therefore BD is perpendicular to the third plane. Wherefore, if two planes, &c. Q. E. D. PROP. XX. THEOR. If a solid angle be contained by three plane angles, 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 shall be 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 23. 1. through BA, AC, the angle BAE equal to the angle 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 because DA is equal to AE, and AB is common, the two DA, D 1 And because * 4. 1. AB are equal to the two EA, AB, each to each, and the angle DAB is equal to the angle EAB: therefore the base DB is equal to the base BE. * BD, DC are greater than CB, and one of them BD 20. 1. has been proved equal to BE, a part of CB, therefore the other DC is greater than the remaining part EC. * 5 Ax. 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 * 25. 1. the construction, the angle DAB is equal to the angle BAE; wherefore the angles DAB, DAC are together greater than BAE, EAC, that is, than the angle BAC: * 4 Ax. 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. PROP. XXI. THEOR. Every solid 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 angles BAC, CAD, DAB: these three together shall be 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, ABD, DBC, any two of them are greater** 20. 11. 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 equal to two right B * D C * 32. 1. angles: therefore the six 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 six right angles: of these the six angles CBA, ACB, ACD, CDA, ADB, DBA are greater than two right angles: therefore the remaining three angles BAC, DAC, BAD, which contain the solid 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 shall be 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 those planes be BC, CD, DE, EF, FB. And because the solid angle at B is contained by three plane angles CBA, ABF, *20. 11. FBC, of which any two are greater* than the third, the angles CBA, ABF are greater than the angle FBC: for the same reason, the two B 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 because all the angles of the tri32. 1. angles are together equal to twice as many right angles* as there are triangles; that is, as there are sides in the polygon BCDEF; and that all the angles of the polygon, together with four right angles, are likewise |