453. Corollary 1. If the cutting plane pass through the centre of the sphere, the radius of the section will be the radius of the sphere; therefore all great circles are equal to each other. 454. Corollary 11. Two great circles always bisect each other; for the common intersection, passing through the centre, is a diameter. 455. Corollary III. Every great circle bisects the sphere and its surface; for, if having separated the two hemispheres from each other, we apply the base of the one to that of the other, turning the convexities the same way, the two surfaces will coincide with each other; if they did not, there would be points in these surfaces unequally distant from the centre. 456. Corollary iv. The centre of a small circle and that of the sphere are in the same straight line perpendicular to the plane of the small circle. 457. Corollary v. Small circles are less according to their distance from the centre of the sphere; for, the greater the distance CO, the smaller the chord AB, the diameter of the small circle AMB. 458. Corollary vi. Through two given points on the surface of a sphere an arc of a great circle may be described; for the two given points and the centre of the sphere determine the position of a plane. If, however, the two given points be the extremities of a diameter, these two points and the centre would be in a straight line, and any number of great circles might be made to pass through the two given points. THEOREM. 459. In any spherical triangle ABC (fig. 222) either side is less Fig than the sum of the other two. Demonstration. Let O be the centre of the sphere, and let the radii OA, OB, OC, be drawn. If the planes AOB, AOC, COB, be supposed, these planes will form at the point O a solid angle, and the angles AOB, AOC, COB, will have for their measure the sides AB, AC, BC, of the spherical triangle ABC (123). But each of the three plane angles, which form the solid angle, is less than the sum of the two others (356); therefore either side of the triangle ABC is less than the sum of the other two. THEOREM. 460. The shortest way from one point to another on the surface of a sphere is the arc of a great circle which joins the two given points. Demonstration. Let ANB (fig. 223) be the arc of a great circle which joins the two given points A and B, and let there be without this arc, if it be possible, a point M of the shortest line between A and B. Through the point M draw the arcs of great circles MA, MB, and take BN=MB. According to the preceding theorem the arc ANB is less than AM+MB; taking from one BN, and from the other its equal BM, we shall have ANAM. Now the distance from B to M, whether it be the same as the arc BM, or any other line, is equal to the distance from B to N; for, by supposing the plane of the great circle BM to turn about the diameter passing through B, the point M may be reduced to the point N, and then the shortest line from M to B, whatever it may be, is the same as that from N to B; consequently the two ways from A to B, the one through M and the other through N, have the part from M to B equal to that from N to B. But the first way is, by hypothesis, the shortest; consequently the distance from A to M is less than the distance from A to N, which is absurd, since the arc AM is greater than AN; whence no point of the shortest line between A and B can be without the arc ANB; therefore this line is itself the shortest that can be drawn between its extremities. Fig. 224. THEOREM. 461. The sum of the three sides of a spherical triangle is less than the circumference of a great circle. Demonstration. Let ABC (fig. 224) be any spherical triangle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be the semicircumferences of great circles, since two great circles always bisect each other (454); but in the triangle BCD the side BC < BD + CD (459); adding to each AB+ AC we shall have AB + AC +BC < ABD+ACD, that is, less than the circumference of a great circle. THEOREM. 462. The sum of the sides of any spherical polygon is less than the circumference of a great circle. Demonstration. Let there be, for example, the pentagon ABCDE (fig. 225); produce the sides AB, DC, till they meet Fig. 225. in F; since BC is less than BF + CF, the perimeter of the pentagon ABCDE is less than that of the quadrilateral AEDF. Again produce the sides AE, FD, till they meet in G, and we shall have ED<EG + GD; consequently the perimeter of the quadrilateral AEDF is less than that of the triangle AFG; but this last is less than the circumference of a great circle (461); therefore for a still stronger reason the perimeter of the polygon ABCDE is less than this same circumference. 463. Scholium. This proposition is essentially the same as that of art. 357; for if O be the centre of the sphere, we can suppose at the point O a solid angle formed by the plane angles AOB, BOC, COD, &c., and the sum of these angles must be less than four right angles, which does not differ from the proposition enunciated above; but the demonstration just given is different from that of art. 357; it is supposed in each that the polygon ABCDE is convex, or that no one of the sides produced would cut the figure. THEOREM. 464. If the diameter DE (fig. 220) be drawn perpendicular to Fig. 220. the plane of the great circle AMB, the extremities D and E of this diameter will be the poles of the circle AMB, and of every small circle FNG which is parallel to it. Demonstration. DC, being perpendicular to the plane AMB, is perpendicular to all the straight lines CA, CM, CB, &c., drawn through its foot in this plane (325); consequently all the arcs DA, DM, DB, &c., are quarters of a circumference. The same may be shown with respect to the arcs EA, EM, EB, &c., whence the points D, E, are each equally distant from all the points in the circumference of the circle AMB; therefore they are the poles of this circle (441). Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; consequently it passes through the centre O of the circle FNG (456); whence, if DF, DN, DG, be drawn, these oblique lines will be equally distant from the perpendicular DO, and will be equal (329). But, the chords being equal, the arcs are equal; consequently all the arcs DF, DN, DG, &c., are equal; therefore the point D is the pole of the small circle, FNG, and for the same reason the point E is the other pole. 465. Corollary 1. Every arc DM, drawn from a point in the arc of a great circle AMB to its pole, is the fourth part of the circumference, which for the sake of conciseness we shall call a quadrant; and this quadrant at the same time makes a right angle with the arc AM. For the line DC being perpendicular to the plane AMC, every plane DMC, which passes through the line DC, is perpendicular to the plane AMC (351); therefore the angle of these planes, or, according to the definition art. 442, the angle AMD is a right angle. 466. Corollary II. In order to find the pole of a given arc AM, draw the indefinite arc MD perpendicular to AM, take MD equal to a quadrant, and the point D will be one of the poles of the arc MD; or rather, draw to the two points A, M, the arcs AD, MD, perpendicular each to AM, the point of meeting D of these two arcs will be the pole required. 467. Corollary III. Conversely, if the distance of the point D from each of the points A, M, is equal to a quadrant, we say that the point D will be the pole of the arc AM, and that, at the same time, the angles DAM, AMD, will be right angles. For, let C be the centre of the sphere, and let the radii CA, CD, CM, be drawn. Since the angles ACD, MCD, are right angles, the line CD is perpendicular to the two straight lines CA, CM; whence it is perpendicular to their plane (325); therefore the point D is the pole of the arc AM; and consequently the angles DAM, AMD, are right angles. 468. Scholium. By means of poles, arcs may be traced upon the surface of a sphere as easily as upon a plane surface. We see, for example, that by turning the arc DF, or any other line of the same extent about the point D, the extremity F will describe the small circle FNG; and, by turning the quadrant DFA about the point D, the extremity A will describe the arc of a great circle AM. If the arc AM is to be produced, or if only the points A, M, are given, through which this arc is to pass, we determine, in the first place, the pole D by the intersection of two arcs described from the points A, M, as centres, with an extent equal to a quadThe pole D being found, we describe from the point D, as a centre, and with the same extent, the arc AM or the continuation of it. rant. If it is required to let fall a perpendicular from a given point P upon a given arc AM, we produce this arc to S, so that the distance PS shall be equal to a quadrant; then from the pole S and with the distance PS we describe the arc PM, which will be the perpendicular arc required. THEOREM. 469. Every plane perpendicular to the radius at its extremity is a tangent to the sphere. Demonstration. Let FAG (fig. 226) be a plane perpendicular Fig. 226. to the radius AO at its extremity; if we take any point M in this plane and join OM, AM, the angle OAM will be a right angle, and thus the distance OM will be greater than OA; consequently the point M is without the sphere; and, as the same might be shown with respect to every other point of the plane FAG, it follows that this plane has only the point 4 in common with the surface of the sphere; therefore it is a tangent to this surface (440). 470. Scholium. It may be shown, in like manner, that two spheres have only one point common, and are consequently tangents to each other, when the distance of their centres is equal to the sum or the difference of their radii; in this case, the centres and the point of contact are in the same straight line. THEOREM. 471. The angle BAC (fig. 226), which two arcs of great circles Fig. 226, make with each other, is equal to the angle FAG formed by the tangents of these arcs at the point A; it has also for its measure the urc DE, described from the point A as a pole, and comprehended between the sides AB, AC, produced if necessary. Demonstration. For the tangent AF, drawn in the plane of the arc AB, is perpendicular to the radius AO (110); and the tangent AG, drawn in the plane of the arc AC, is perpendicular Geom. 21 |