the lines LM, LO which it meets in that plane. Therefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH (4. def. 2. Sup.). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But because KN and LO are parallel, being the common sections of the parallel planes AB and CD with. a third plane, the interior angle NKM is equal to the exterior angle OLM (21. 1.); that is, the inclination of the plane AB to the plane EH, is aqual to the inclination of the plane CD to the same plane EH. PROP. XVI. THEOR. If two straight lines be cut by parallel planes, they must be cut in the same ratio. D: 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, the common sections EX, BD, are parallel (14. 2. Sup.). 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 (2. 6.) AX to XD. Again, because XF is parallel to AC, aside of the triangle ADC, AX to XD, so is CF to FD and it was proved that AX is to XD, as AE to EB: Therefore (11.5.), as AE to EB, so is CF to FD. PROP. XVII. K M THEOR. H If a straight line be at right angles to a plane, every plane which passes through that line is at right angles to the first mentioned plane. Let the straight line AB be at right angles to the plane CK; every plane which passes through AB is 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 DE at right angles to CE: And because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line meeting it in that plane (1. def. 2. Sup.); and consequently it is perpendicular to CE: Wherefore ABF is a right angle; But GFB is likewise a right angle; therefore AB is parallel (20. 1.) to FG.. And AB is at right angles to the plane CK: therefore FG is also at right angles to the same D G A H K plane (7. 2. Sup.). But 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 (def. 2. 2.); 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. PROP. XVIII. THEOR. F B E If two planes cutting 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 BD be the common section of the first two; BD is perpendicular to the plane ADC. From D in the plane ADC, draw DE perpendicular to AD, and DF to DC. Because DE is perpendicular to AD, the common section of the planes AB and ADC; and because the plane AB is at right angles to ADC, DE is at right angles to the plane AB (def. 2. 2. Sup.), and therefore also to the straight line BD in that plane (def. 1. 2. Sup.). For the same reason, DF is at right angles to DB. Since BD is therefore at right angles to both the lines DE and DF, it is at right angles to the plane in which DE and DF are, that is, to the plane ADC (4. 2. Sup.). PROP. XIX. THEOR. B AE E Two straight lines not in the same plane being given in position, to draw a straight line perpendicular to them both. Let AB and CD be the given lines, which are not in the same plane; it is required to draw a straight line which shall be perpendicular both to AB and CD. In AB take any point E, and through E draw EF parallel to CD, and let EG be drawn perpendicular to the plane which passes through EB, BF (10. 2. Sup.). Through AB and EG let a plane pass, viz. GK, and let this plane meet CD in H; from H draw HK perpendicular to AB; and HK is the line required. Through H, draw HG parallel to AB. Then, since HK and GE, which are in the same plane, are both at right angles to the straight line AB, they are parallel to one another. And because the lines HG, HD are parallel to the lines EB, EF, each to each, the plane GHD is parallel to the plane (13. 2. Sup.) BEF; and therefore EG, which is perpendicular to the plane BEF, is perpendicular also to the plane (Cor. 13. 2. Sup.) GHD. Therefore HK, which is parallel to GE, is also perpendicular to the plane GHD (7. 2. Sup.), and it is therefore perpendicular to HD (def. 1. 2. Sup.), which is in that plane, and it is also perpendicular to AB; therefore HK is drawn perpendicular to the two given lines, AB and CD. PROP. XX. THEOR. If a solid angle be contained by three plane angles, any two of these angles 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. any two If the angles BAC, CAD, DAB be all equal, it is evident that 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 D 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 (Prop. 9. 1.) 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 to the two triangles ABD, ABE, and also the angle DAB equal to the angle EAB; therefore the base DB is equal (1. 1.) to the base BE. And because BD, DC are greater (13. 1.) than CB, and one of them BD has been proved equal 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 (16. 1.) than the angle EAC; and, by the construction, B E 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. But BAC is not less than either of the angles DAB, DAC; therefore BAC, with either of them, is greater than the other. PROP. XXI. THEOR. The plane angles which contain any solid angle are together less than four right angles. B Let A be a solid ang e contained by any number of plane angles BAC, CAD, DAE, EAF, FAB; these together are less than four right angles. Let the planes which contain the solid angle at A be cut by another plane, and let the section of them by that plane be the rectilineal figure BCDEF. And because the solid angle at B is contained by three plane angles CBA, ABF, FBC, of which any two are greater (20. 2. Sup.) than the third, the angles CBA, ABF are greater than the angle FBC For the same reason, the two 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 figure BCDEF therefore all the angles at the bases of the triangles are together greater than all the angles of the figure: and because all the angles of the triangles are together equal to twice as many right angles as there are triangles (25. 1.); that is, as there are sides in the figure BCDEF; and because all the angles of the figure, together with four right angles, are likewise equal to twice as many right angles as there are sides in the figure (26. 1.); therefore all the angles of the triangles are equal to all the angles of the rectilineal figure, together with four right angles. But all the angles at the bases of the triangles are greater than all the angles of the rectilineal, as has been proved. Wherefore, the remaining angles of the triangles, viz. those at the vertex, which contain the solid angle at A, are less than four right angles, : Otherwise. D E Let the sum of all the angles at the bases of the triangles S; the sum of all the angles of the rectilineal figure BCDEF=Σ; the sum of the plane angles at A=X, and let R= a right angle. Then, because S+X= twice (25. 1.) as many right angles as there are triangles, or as there are sides of the rectilineal figure BCDEF, and as +4R is also equal to twice as many right angles as there are sides of the same figure; therefore S+X=+4R. But because of the three plane angles which contain a solid angle, any two are greater than the third, $75; and therefore X4R; that is, the sum of the plane angles which contain the solid angle at A is less than four right angles. SCHOLIUM. A It is evident, that when any of the angles of the figure BCDEF is exterior, like the angle at D, in the annexed figure, the reasoning in the above proposition does not hold, because the solid angles at the base are not all contained by plane angles, of which two belong to the triangular planes, having their common vertex in A, and the third is an interior angle of the rectilineal figure, or base. Therefore, it cannot be concluded that S is necessarily great B D er than . This proposition, therefore, is subject to a limitation, which is farther explained in the notes on this Book. |