Fig. 220.

Fig. 221.

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

455. 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.

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 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 111. 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.

459. In any spherical triangle ABC (fig. 222) either side is less Fig 222. 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,

Fig. 224.

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 AN<AM. 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.

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) ; ABAC we shall have AB+ AC+BC <.?! is, less than the circumference of a great ch

A

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

to the same radius AO; therefore the angle FAG is equal to the angle of the planes OAB, OAC (349), which is that of the arcs AB, AC, and which is designated by BAC.

In like manner, if the arc AD is equal to a quadrant, and also AE, the lines OD, OE, will be perpendicular to AO, and the angle DOE will be equal to the angle of the planes AOD, AOE; therefore the arc DE is the measure of the angle of these planes, or the measure of the angle CAB.

472. Corollary. The angles of spherical triangles may be compared with each other by means of the arcs of great circles, described from their vertices as poles and comprehended between their sides; thus it is easy to make an angle equal to a given angle.

473. Scholium.

The angles opposite to each other at the Fig. 238. vertex, as ACO, BCN (fig. 238), are equal; for each is equal to the angle formed by the two planes ACB, OCN (350).

Fig. 227.

It will be perceived also that in the meeting of two arcs ACB, OCN, the two adjacent angles ACO, OCB, taken together, are equal to two right angles.

THEOREM.

G

474. The triangle ABC (fig. 227) being given, if, from the points A, B, C, as poles, the arcs EF, FD, DE, be described, forming the triangle DEF, reciprocally the points D, E, F, will be the poles of the sides BC, AC, AB.

Demonstration. The point A being the pole of the arc EF, the distance AE is a quadrant; the point C being the pole of the arc DE, the distance CE is likewise a quadrant; consequently the point E is distant a quadrant from each of the points A, C; therefore it is the pole of the arc AC (467). It may be shown, in the same manner, that D is the pole of the arc BC, and F that of the arc AB.

475. Corollary. Hence the triangle ABC may be described by means of DEF, as DEF is described by means of ABC.

476. The same things being supposed as in the preceding theorem, Fig. 227. each angle of one of the triangles ABC, DEF (fig. 227), will have for its measure a semicircumference minus the side opposite in the other triangle.