Fig. 220.

444. A spherical polygon is a part of the surface of a sphere terminated by several arcs of great circles.

445. A lunary surface is the part of the surface of a sphere comprehended between two semicircumferences of great circles, which terminate in a common diameter.

446. We shall call a spherical wedge the part of a sphere comprehended between the halves of two great circles, and to which the lunary surface answers as a base.

447. A spherical pyramid is the part of a sphere comprehended between the planes of a solid angle whose vertex is at the centre. The base of the pyramid is the spherical polygon intercepted by these planes.

448. A zone is the part of the surface of a sphere comprehended between two parallel planes, which are its bases. One of these planes may be a tangent to the sphere, in which case the zone has only one base.

449. A spherical segment is the portion of a sphere comprehended between two parallel planes which are its bases. One of these planes may be a tangent to the sphere, in which case the spherical segment has only one base.

450. The altitude of a zone or of a segment is the distance between the parallel planes which are the bases of the zone or segment.

451. While the semicircle DAE (fig. 220), turning about the diameter DE, describes a sphere, every circular sector, as DCF, or FCH, describes a solid, which is called a spherical sector.

THEOREM.

Fig. 221.

452. Every section of a sphere made by a plane is a circle. Demonstration. Let AMB (fig. 221) be a section, made by a plane, of the sphere of which C is the centre. From the point C draw CO perpendicular to the plane AMB, and different oblique lines CM, CM, to different points of the curve AMB which terminates the section.

The oblique lines CM, CM, CB, are equal, since they are radii of the sphere; consequently they are at equal distances from the perpendicular CO (329); whence all the lines OM, OM, OB, are equal; therefore the section AMB is a circle of which the point O is the centre

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 II. 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 tlre 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 Fig. 222, less 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.

Fig. 223.

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 Ñ; 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 GEOM.

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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 11. 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.