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. 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 Pupon 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.

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 A 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.

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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 arc DE, described from the point A as a pole, and comprehended between the sides AB, AC, produced if necessary.

5

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.

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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.

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.

THEOREM.

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.

Demonstration. Let the sides AB, AC, be produced, if necessary, till they meet EF in G and H; since the point A is the pole of the arc GH, the angle A will have for its measure the arc GH. But the arc EH is a quadrant, as also GF, since E is the pole of AH, and F the pole of AG (465); consequently EH+GF is equal to a semicircumference. But EH+GF is the same as EF + GH; therefore the arc GH, which measures the angle A, is equal to a semicircumference minus the side EF; likewise the angle B has for its measure circ. — DF, and the angle C, circ. - DE.

This property must be reciprocal between the two triangles, since they are described in the same manner, the one by means of the other. Thus we shall find that the angles D, E, F, of the triangle DEF have for their measure respectively circ. BC, circ. — AC, circ. - AB. Indeed, the angle D, for example, has for its measure the arc MI; but

MI + BC= MC + BI = ↓ circ.; therefore the arc MI, the measure of the angle D, = 1⁄2 circ. — BC, and so of the others.

477. Scholium. It may be remarked that, beside the triangle DEF, three others may be formed by the intersection of the three arcs DE, EF, DF. But the proposition applies only to the central triangle, which is distinguished from the three others by this, that the two angles A, D, are situated on the same side of BC, the two B, E, on the same side of AC, and the two C, F, on the same side of AB.

Different names are given to the triangles ABC, DEF; we shall call them polar triangles.

478. The triangle ABC (fig. 229) being given, if, from the pole Fig. 229 A and with the distance AC, an arc of a small circle DEC be described, if, also from the pole B and with the distance BC, the arc DFC be described, and from the point D where the arcs DEC, DFC, cut each other, the arcs of great circles AD, DB, be drawn; we say that of the triangle ADB thus formed the parts will be equal to those of the triangle ACB.

Demonstration. For, by construction the side AD = AC, DB=BC, and AB is common; therefore the two triangles will

Fig. 230.

have the sides equal, each to each. We say, moreover, that the angles opposite to the equal sides are equal.

Indeed, if the centre of the sphere be supposed in O, we can suppose a solid angle formed at the point O by the three plane angles AOB, AOC, BOC; we can suppose likewise a second solid angle formed by the three plane angles AOB, AOD, BOD. And since the sides of the triangle ABC are equal to those of the triangle ADB, it follows that the plane angles, which form one of the solid angles, are equal to the plane angles which form the other solid angle, each to each. But the planes of any two angles in the one solid have the same inclination to each other as the planes of the homologous angles in the other (359); conse quently the angles of the spherical triangle DAB are equal to those of the triangle CAB, namely, DAB = BAC, DBA = CBA, and ADB ACB; therefore the sides and the angles of the triangle ADB are equal to the sides and angles of the triangle ACB.

479. Scholium. The equality of these triangles does not depend upon an absolute equality, or equality by superposition, for it would be impossible to make them coincide by applying the one to the other, at least except they should happen to be isosceles. The equality then under consideration is that which we have already called equality by symmetry, and for this reason, we shall call the triangles ACB, ADB, symmetrical triangles.

480. Two triangles situated on the same sphere, or on equal spheres, are equal in all their parts, when two sides and the included angle of the one are equal to two sides and the included angle of the other, each to each.

Demonstration. Let the side AB = EF (fig. 230), the side AC = EG, and the angle BAC = FEG, the triangle EFG can be placed upon the triangle ABC, or upon the triangle symmetrical with it ABD, in the same manner as two plane triangles are applied, when they have two sides and the included angle of the one respectively equal to two sides and the included angle of the other (36). Therefore all the parts of the triangle EFG will be equal to those of the triangle ABC, that is, beside the three parts which were supposed equal we shall have the side BC= FG, the angle ABC EFG, and the angle ACB= EGF.