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, = { 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. 232.

Fig. 233.

the greater; take BO=AC, and join OC. The two sides BO, BC, are equal to the two AC, BC; and the angle OBC contained by the first is equal to the angle ACB contained by the second. Consequently the two triangles have their other parts equal (480), namely OCB = ABC; but the angle ABC is, by hypothesis, equal to ACB; whence OCB is equal to ACB, which is impossible; AB then cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles B, C, are equal.

484. Scholium. It is evident, from the same demonstration, that the angle BAD = DAC, and the angle BDA = ADC. Consequently the two last are right angles; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts.

THEOREM.

16

485. In any spherical triangle ABC (fig. 232), if the angle A is greater than the angle B, the side BC opposite to the angle A will be greater than the side AC opposite to the angle B; conversely, if the side BC is greater than AC, the angle A will be greater than the angle B.

Demonstration. 1. Let the angle A> B; make the angle BAD = B, and we shall have AD = DB (483); but

AD+DCAC;

in the place of AD substitute DB, and we shall have DB + DC or BC > AC.

2. If we suppose BC > AC, we say that the angle BAC will be greater than ABC. For, if BAC were equal to ABC, we should have BC AC; and if BAC were less than ABC, it would follow, according to what has just been demonstrated, that BC<AC, which is contrary to the supposition, therefore the angle BAC is greater than ABC.

THEOREM.

17

486. If the two sides AB, AC (fig. 233), of the spherical triangle ABC are equal to the two sides DE, DF, of the triangle DEF described upon an equal sphere, if at the same time the angle A is greater than the angle D, we say that the third side BC of the first triangle will be greater than the third side EF of the second.

The demonstration of this proposition is entirely similar to that of art. 42.

487. If two triangles described upon the same sphere or upon equal spheres are equiangular with respect to each other, they will also be equilateral with respect to each other.

Demonstration. Let A, B, be the two given triangles, P, Q, their polar triangles. Since the angles are equal in the triangles A, B, the sides will be equal in the polar triangles P, Q, (476); but, since the triangles P, Q, are equilateral with respect to each other, they are also equiangular with respect to each other (482); and, the angles being equal in the triangles P, Q, it follows that the sides are equal in their polar triangles A, B. Therefore the triangles A, B, which are equiangular with respect to each other, are at the same time equilateral with respect to each other.

This proposition may be demonstrated without making use of polar triangles in the following manner.

Let ABC, DEF (fig. 234), be two triangles equiangular with Fig. 234. respect to each other, having A=D, B = E, C=F; we say that the sides will be equal, namely, AB = DE, AC=DF, BC= EF.

Produce AB, AC, making AG = DE, AH= DF; join GH, and produce the arcs BC, GH, till they meet in I and K.

The two sides AG, AH, are, by construction, equal to the two DF, DE, the included angle GAH = BAC=EDF, consequently the triangles AGH, DEF, are equal in all their parts (480); therefore the angle AGH= DEF = ABC, and the angle

AHGDFE=ACB.

In the triangles IBG, KBG, the side BG is common, and the angle IGB = GBK; and, since IGB + BGK is equal to two right angles, as also GBK + IBG, it follows that BGK = IBG. Consequently the triangles IBG, GBK, are equal (481); therefore IGBK, and IB = GK.

In like manner, since the angle AHG = ACB, the triangles ICH, HCK, have a side and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other; consequently they are equal; therefore IH= CK, and HK = IC.

Now, if from the equals BK, IG, we take the equal CK, IH, the remainders BC, GH, will be equal. Besides, the angle BCA = AHG, and the angle ABC=AGH. Whence the triangles ABC, AHG, have a side and the two adjacent angles of the one respectively equal to a side and the two adjacent angles of the other; consequently they are equal. But the triangle DEF is equal in all its parts to the triangle AHG; therefore it is also equal to the triangle ABC, and we shall have AB = DE, AC = DF, BC EF; hence, if two spherical triangles are equiangular with respect to each other, the sides opposite to the equal angles will be equal.

488. Scholium. This proposition does not hold true with regard to plane triangles, in which from the equality of the angle we can only infer the proportionality of the sides. But it is easy to account for the difference in this respect between plane and spherical triangles. In the present proposition, as well as those of articles 480, 481, 482, 486, which relate to a comparison of triangles, it is said expressly that the triangles are described upon the same sphere or upon equal spheres. Now similar arcs are proportional to their radii; consequently upon equal spheres two triangles cannot be similar without being equal. It is not therefore surprising that equality of angles should imply equality of sides.

It would be otherwise, if the triangles were described upon unequal spheres; then, the angles being equal, the triangles would be similar, and the homologous sides would be to each other as the radii of the spheres.

THEOREM.

19.

489. The sum of the angles of every spherical triangle is less than six and greater than two right angles.

Demonstration. 1. Each angle of a spherical triangle is less than two right angles (see the following scholium); therefore the sum of the three angles is less than six right angles.

2. The measure of each angle of a spherical triangle is equal to the semicircumference minus the corresponding side of the polar triangle (476); therefore the sum of the three angles has for its measure three semicircumferences minus the sum of the sides of the polar triangle. Now this last sum is less than a

circumference (461); consequently, by subtracting it from three semicircumferences the remainder will be greater than a semicircumference, which is the measure of two right angles; therefore the sum of the three angles of a spherical triangle is greater than two right angles.

490. Corollary 1. The sum of the angles of a spherical triangle is not constant like that of a plane triangle; it varies from two right angles to six, without the possibility of being equal to either limit. Thus, two angles being given, we cannot thence determine the third.

491. Corollary II. A spherical triangle may have two or three right angles, also two or three obtuse angles.

If the triangle ABC (fig. 235) has two right angles B and C, Fig. 235. the vertex A will be the pole of the base BC (467); and the sides AB, AC, will be quadrants.

If the angle A also is a right angle, the triangle ABC will have all its angles right angles, and all its sides quadrants. The triangle having three right angles is contained eight times in the surface of the sphere; this is evident from figure 236, if we suppose the arc MN equal to a quadrant.

492. Scholium. We have supposed in all that precedes, conformably to the definition art. 442, that spherical triangles always have their sides less each than a semicircumference; then it follows that the angles are always less than two right angles. For the side AB (fig. 224) is less than a semicircumfe- Fig. 224. rence as also AC; these arcs must both be produced in order to meet in D. Now the two angles ABC, CBD, taken together, are equal to two right angles; therefore the angle ABC is by itself less than two right angles.

We will remark, however, that there are spherical triangles of which certain sides are greater than a semicircumference, and certain angles greater than two right angles. For, if we produce the side AC till it becomes an entire circumference ACE, what remains, after taking from the surface of the hemisphere the triangle ABC, is a new triangle, which may also be designated by ABC, and the sides of which are AB, BC, AEDC. We see then, that the side AEDC is greater than the semicircumference AED; but, at the same time, the opposite angle B exceeds two right angles by the quantity CBD.