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The demonstration of this proposition is entirely similar to that

of art. 42.

THEOREM.

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.

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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

AHG DFE = 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 IG BK, and IB = GK.

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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.

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 11. 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.

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Besides, if we exclude from the definition triangles, the sides and angles of which are so great, it is because the resolution of them, or the determination of their parts, reduces itself always to that of triangles contained in the definition. Indeed, it will be readily perceived, that if we know the angles and sides of the triangle ABC, we shall know immediately the angles and sides of the triangle of the same name, which is the remainder of the surface of the hemisphere.

Fig. 236.

THEOREM.

493. The lunary surface AMBNA (fig. 236) is to the surface of the sphere as the angle MAN of this surface is to four right angles, or as the arc MN, which measures this angle, is to the circumference.

Demonstration. Let us suppose in the first place, that the arc MN is to the circumference MNPQ in the ratio of two entire numbers, as 5 to 48, for example. The circumference MNPQ may be divided into 48 equal parts, of which MN will contain 5; then joining the pole A and the points of division by as many quadrants, we shall have 48 triangles in the surface of the hemisphere AMNPQ, which will be equal among themselves, since they have all their parts equal. The entire sphere then will contain 96 of these partial triangles, and the lunary surface AMBNA will contain 10 of them; therefore the lunary surface is to that of the sphere as 10 is to 96, or as 5 is to 48, that is, as the arc MN is to the circumference.

If the arc MN is not commensurable with the circumference, it may be shown by a course of reasoning, of which we have already had many examples, that the lunary surface is always to that of the sphere as the arc MN is to the circumference. 494. Corollary 1. Two lunary surfaces are to each other as their respective angles.

495. Corollary II. We have already seen that the entire surface of the sphere is equal to eight triangles having each three right angles (491); consequently, if the area of one of these triangles be taken for unity, the surface of the sphere will be represented by eight. This being supposed, the lunary surface, of which the angle is A, will be expressed by 24, the angle A being estimated by taking the right angle for unity; for we have 2A:8:: A : 4. Here are then two kinds of units; one for

angles, this is the right angle; the other for surfaces, this is the spherical triangle, of which all the angles are right angles and the sides quadrants.

496. Scholium. The spherical wedge comprehended by the planes AMB, ANB, is to the entire sphere, as the angle A is to four right angles. For the lunary surfaces being equal, the spherical wedges will also be equal; therefore two spherical wedges are to each other as the angles formed by the planes which comprehend them.

THEOREM.

497. Two symmetrical spherical triangles are equal in sarface. Demonstration. Let ABC, DEF (fig. 237), be two symmetri- Fig. 237. cal triangles, that is, two triangles which have their sides equal, namely, AB = DE, AC = DF, CB = EF, and which at the same time do not admit of being applied the one to the other; we say that the surface ABC is equal to the surface DEF.

Let P be the pole of the small circle which passes through the three points A, B, C*; from this point draw the equal arcs PA, PB, PC (464); at the point F make the angle DFQ = ACP, the arc FQ CP, and join DQ, EQ.

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The sides DF, FQ, are equal to the sides AC, CP; the angle DFQ ACP; consequently the two triangles DFQ, ACP, are equal in all their parts (480); therefore the side DQ = AP, and the angle DQF=APC.

In the proposed triangles DFE, ABC, the angles DFE, ACB, opposite to the equal sides DE, AB, being equal (481), if we subtract from them the angles DFQ, ACP, equal, by construction, there will remain the angle QFE equal to PCB. Moreover the sides QF, FE, are equal to the sides PC, CB; consequently the two triangles FQE, CPB, are equal in all their parts; therefore the side QE = PB, and the angle FQE = CPB.

If we observe now that the triangles DFQ, ACP, which have the sides equal each to each, are at the same time isosceles, we

* The circle, which passes through the three points A, B, C, or which is circumscribed about the triangle ABC, can only be a small circle of the sphere; for, if it were a great one, the three sides AB, BC, AC, would be situated in the same plane, and the triangle ABC would be reduced to one of its sides.

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