Join AC, BD, AD; and let AD meet the plane KL in the point X and join EX, XF. Be cause the two parallel planes KL, CH MN are cut by the plane EBDX, the common sections BD, EX are parallel (VII. 15). For the same reason, because the two parallel planes GH, KL are cut by the plane ACD, AC and XF are parallel; and because EX is parallel to BD, one side of the triangle ABD, (V. 2), AE: EB :: AX : XD; again, because XF is parallel to AC, (V. 2), AX: XD:: CF: FD; ... (IV. 15), AE: EB:: CF: FD. Proposition 17. Theorem.—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 stand at right angles to the plane CD; any plane through AB will also be perpendicular to CD. H A B G D Let HF be a plane passing through AB, and let EF be the common section of the two planes. In the plane CD draw BG at right angles to EF. Then because AB is at right angles to CD, it is at right angles to EF and BG, which it meets in that plane; but BG is also at right angles to EF. Hence AB, BG are drawn at right angles to the common section of the two planes, one in each plane, and they are at right angles to each other; therefore (VII., Def. 7) the planes HF, CD are at right angles to each other. Proposition 18. Theorem.-If two planes cutting each other 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 is perpendicular to the plane ADC. B 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 AB and ADC, and because the plane AB is at right angles to ADC, DE is at right angles to the plane AB, and therefore to DB in that plane. For the same reason, DF is at right angles to DB. Therefore DB being at right angles to DE, DF at their point of intersection is at right angles to the plane ADC (VII. 4). Proposition 19. A F E C Theorem. Any two angles of a triedral angle are together greater than the third. Let the solid angle at A be contained by BAD, DAC, CAB; any two of these are together greater than the third. If they are all three equal, it is evident that any two of them are greater than the third If not, let BAC' be that angle which is not less than either of the other two, and is greater than one of them, DAB. Make (I. 5), in the plane, which passes through BA, AC, the angle BAE equal to the angle DAB; take any point, D, in AD, and make A AE equal to AD, and through E draw BEC, cutting AB, AC A in the points B, C, and join BD, CD. And because DA is equal to AE, and AB is common to the two triangles ABD, ABE, and also the angle DAB is equal to the angle EAB, therefore DB is equal to BE. And because BD, DC are greater than BC, and one of them, BD, has been proved equal to BE, the remainder, DC, is greater than the remainder EC, therefore the angle DAC is greater (I. 23) than the angle EA C, and by construction DAB is equal BAE; therefore the angles DAB, DAC are together greater than BAE, EAC-that is, than BAC; but BAC is not less than either of the others; therefore BAC with either of them is greater than the other. Proposition 20. Theorem.-The plane angles which contain any solid angle are together less than four right angles. Let A be a solid angle 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 A be cut by another plane, and let the section of them by this plane be the rectilineal figure BCDEF. Let M=the sum of all the angles at A; B A E D S=the sum of the angles at the bases of the triangles; R= one right angle. Because the angle at B is triedral, the two angles ABC, ABF are together greater (VIII. 10) than CBF. In the same way all the angles at the bases of the triangles having a common vertex at A, are greater than the angles of BCDEF, or S> P. Now (I. 30), the angles of ABC are equal to two right angles, and the same is true of the other triangles; hence, Theorem.-If the plane angles which constitute one triedral angle be equal, each to each, to the plane angles which constitute another, the faces of the two angles are equally inclined to each other. Let A and B be two triedral angles, of which the plane angles CAD, FBG, as also DAE, GBH, and CAE, FBH are equal to each other; then the planes containing the equal angles are equally inclined to each other. In AC take any point K, and draw, in the planes CAD, CAE, KD, KE at right angles to AC; then DKE will be (Def. 7) the C K B inclination of those planes; make BL equal to AK, and draw GL, LH in the planes FBG, FBH, at right angles to BF; then GLH will be the inclination of those planes. Because in the triangles AKD, BLG, AK is equal to BL, the angle KAD to the angle LBG (Hyp.), and the angles AKD, BLG, right angles, there A B angles ADE, BGH, the two sides DA, AE are equal to GB, BH, and their included angles are equal; therefore (I. 7), DE is equal to GH. Lastly, in the triangles DKE, GLH, because the three sides of one are equal to the three sides of the other, the angles DKE, GLH are equal (I. 4) to each other; but these angles are the inclination of the planes DAC, CAE, and the planes GBF, FBH, to each other; therefore the planes are equally inclined. And in the same way we may prove it of the other faces. EXERCISES. 1. To pass a plane perpendicular to two parallel lines. 2. If two planes, which intersect, contain two lines which are parallel to each other, one in each plane, the common section of the planes will be parallel to the lines. Pass a plane perpendicular to the parallel lines. 3. To pass a plane through a given line perpendicular to a given plane. 4. If a line be parallel to one plane and perpendicular to another, the planes are at right angles to each other. 5. The perpendicular is the shortest line drawn from a point to a plane. Oblique lines drawn at equal distances from the perpendicular are equal; and of two oblique lines unequally distant from the perpendicular, the more remote is the greater. |