PROPOSITION 16.-Theorem. If two parallel planes be cut by another plane, their common sections with it are parallels. Let the parallel planes A B and CD be cut by the plane FG, and let their common sections with it be EF and Gł. Then EF is parallel to GH. Construction. For, if they be not parallel, EF and GH shall meet if produced either on the side of F# or EĞ. Let them be produced on the side of FH, and meet in the point K. Demonstration. Because the straight line EFK is in the plane AB (XI. 1), therefore 1. The point K is in the plane AB; for the same reason, the point Ķ is also in the plane CD; wherefore 2. The planes AB and CD being produced, meet one another. But they do not meet, because they are parallel (hyp.) ; therefore 3. The straight lines EF and GH do not meet when produced on the side of FH. In the same manner, it may be proved that EF and GH do not meet when produced on the side of EG. But straight lines which are in the same plane, and do not meet, though produced either way, are parallel (I. Def. 35); therefore 4. EF is parallel to GH. Wherefore, if two parallel planes, &c. Q.E.D. PROPOSITION 17.-Theorem. If two straight lines be cut by parallel planes, they are cut in the same ratio, Let the straight lines AB and CD be cut by the parallel planes GH, KL, and M N in the points A, E, and B; C, F, and D. Then as AE is to EB, so is CF to FD. Construction. Join AC, BD, and AD, and let AD meet the plane KL in the point X. Join EX and XF. Demonstration. Because the two parallel planes KL and MN are cut by the plane BX; therefore 1. The common sections EX and BD are parallel (XI. 16) ; for the same reason, because the two parallel planes GH and KL are cut by the plane XC; therefore 2. The common sections AC and XF are parallel (XI. 16). And because EX is parallel to BD, a side of the triangle ABD; therefore 3. As AE to EB, so is AX to XD (VI. 2). Again, because XF is parallel to AC, a side of the triangle ADC; therefore 4. As AX is to XD, so is CF to FD. But it was proved that AX is to XD, as AE is to EB; therefore 5. As AE is to EB, so is CF to FD (V. 11). Wherefore, if two straight lines, &c. Q.E.D. 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. Construction. Let any plane DE pass through AB, and let CE be the common section of the planes DE and CX. 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, therefore 1. It pis perpendicular to CE meeting it in that plane (XI. and consequently, 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. T 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 ED 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 ĀB 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 (L. 23). Make A E equal to AD, and through E draw BĈ 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); |