PROPOSITION 18.-Theorem. If a straight line be at right angles to a plane, every plane which passes through it is at right angles to that plane. Let the straight line AB be at right angles to the plane CK. Then every plane which passes through AB is at right angles to the plane CK. therefore G and consequently, B Construction. Let any plane DE pass through AB, and let CE be the common section of the planes DE and CK. Take any point Fin CE, and from F draw FG in the plane DE at right angles to CE (I. 11). Demonstration. Because AB is perpendicular to the plane CK, 1. It is perpendicular to CE meeting it in that plane (XI. Def. 3), 2. ABF is a right angle; but GFB is likewise a right angle (constr.); therefore 3. AB is parallel to FG (I. 28). But AB is at right angles to the plane CK; therefore 4. FG is also at right angles to the same plane (XI. 8). Because one 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 to the other plane (XI. Def. 4); 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 5. The plane DE is at right angles to the plane CK. In like manner, it may be proved that all planes which pass through AB are at right angles to the plane CK. Therefore, if a straight line, &c. Q.E.D. U PROPOSITION 19.-Theorem. 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 and BC be each of them perpendicular to a third plane, and let BD be the common section of the first two. Then BD shall be perpendicular to the third plane. E Construction. If it be not, from the point D, draw, in the plane AB, the straight line DE at right angles to AD, the common section of the plane AB with the third plane (I. 11); and in the plane BC draw DF at right angles to CD, the common section of the plane BC with the third plane. Demonstration. 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; therefore 1. DE is perpendicular to the third plane (XI. Def. 4). In the same manner, it may be proved that DF is perpendicular to the third plane. But 2. From the point D two straight lines are drawn at right angles to the third plane, upon the same side of it (XI. 13), 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 and BC; therefore 3. BD is perpendicular to the third plane. Wherefore, if two planes, &c. Q.E.D. PROPOSITION 20.-Theorem. 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, and DAB. Then any two of them are greater than the third. Construction. If the angles BAC, CAL, and 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. At the point A in the straight line AB, make, in the plane BA and AC, the angle BAE equal to the angle DAB (I. 23). Make AE equal to AD, and through E draw BC cutting AB and AC in the points B and C. Join DB and DC. Demonstration. Because DA is equal to AE, and AB is common, the two DA and AB are equal to the two EA and AB, each to each; and the angle DAB is equal to the angle EAB; therefore 1. The base DB is equal to the base BE (I. 4) ; and because BD and DC are greater than CB (I. 20); and the one BD has been proved equal to BE, a part of CB, therefore 2. The other DC is greater than the remaining part EC (I. Ax. 5). Because DA is equal to AE, and AC common, but the base DC is greater than the base EC; therefore 3. The angle DAC is greater than the angle EAC (I. 25) ; but the angle DAB is equal to the angle BAE (constr.); therefore 4. The angles DAB and DAC are together greater than the angles BAE and EAC; that is, than the angle BAC (I. Ax. 4); but BAC is not less than either of the angles DAB and DAC; therefore 5. BAC with either of them is greater than the third. Wherefore, if a solid angle, &c. Q.E.D. PROPOSITION 21.-Theorem. The plane angles by which every solid angle is contained, are together less than four right angles. First, let the solid angle at A be contained by three plane angles BAC, CAD, and DAB. Then these three together are less than four right angles. B D Construction. Take in each of the straight lines AB, AC, and AD, any points B, C, and D; and join BC, CD, and DB. Demonstration. Because the solid angle at B is contained by the three plane angles CBA, ABD, and DBC; any two of them are greater than the third (XI. 20); therefore 1. The angles CBA and ABD are greater than the angle DBC. For the same reason, the angles BCA and ACD are greater than the angle DCB; and the angles CDA and ADB greater than the angle BDC; therefore 2. The six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than the three angles DBC, BCD, and CDB; but the angles DBC, BCD, and CDB are equal to two right angles (I. 32); therefore 3. The six angles CBA, ABD, BCA, ACD, CDA, and ADB are greater than two right angles. Because the three angles of each of the triangles ABC, ACD, and ADB are equal to two right angles; therefore 4. The nine angles of these three triangles-viz., the angles CBA, BAC, ACB, ACD, CDA, DAC, ADB, DRA, and BAD are equal to six right angles; but of these, the six angles CBA, ACB, ACD, CDA, ADB, and DBA are greater than two right angles; therefore 5. The remaining three angles BAC, CAD, and DAB, 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, and FAB. Then these angles are together less than four right angles. Construction. Let the plane angles which form the solid angle be cut by a plane at any distance from its vertex, and let the common sections of it with those plane angles be BC, CD, DE, EF, and FB. Demonstration. Because the solid angle at B is contained by three plane angles CBA, ABF, and FBC, of which any two are greater than the third (XI. 20); therefore 1. The angles CBA and ABF are greater than the angle FBC. For the same reason, the two plane angles at each of the points C, D, E, and F-viz., those 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 2. All the angles at the bases of the triangles are together greater than all the angles of the polygon. Because all the angles of the triangles are together equal to twice as many right angles as there are triangles (I. 32);-that is, as there |