ILL.-In Fig. 333, ABC and A'B'C' represent symmetrical spherical triangles. In these triangles A = A', B = B', C = C',

=

AC A'C', AB = : A'B', and BC B'C'; nevertheless we cannot conceive one triangle superimposed upon the other. Thus, were we to make the attempt by placing A'B' in its equal AB, A' at A, and B' at B, the angle C' would fall on the opposite side of AB from C. Now, we cannot revolve A'C'B' on AB (or its chord), and thus make the two coincide, for this would bring their convexities together. Nor can we make them coincide by reversing A'B'C', and placing B' at A, and A' at B. For, although these two arcs will thus coincide, as the angle B' is not equal to A, B'C' will not fall in AC; and, again, if it did, C' would not fall at C, since B'C' and AC are not equal.

=

But, considering the triangles ABC and A'B'C' in Fig. 334, in which A A', B = B', C = C', AC = A'C', AB = A'B', and BC = B'C', we can readily conceive the latter as superimposed upon the former. [The student should make the application.] Now, the two triangles are equal in each case, as will subsequently appear of the former. Such triangles as those in Fig. 333 are called symmetrically equal, while the latter are said to be equal by superposition.

Fig. 335 represents the same triangles as Fig. 334, and exhibits a complete projection* of the semicircumferences of which the sides of the triangles are

arcs.

A'

The student should become perfectly familiar with it, and be able to draw it readily. Thus, aABb is the projection of the semicircumference of which AB is an arc, aACc of the semicircumference of which AC is an arc, etc., etc.

FIG. 335.

PROPOSITION XVII.

583. Theorem.-Symmetrical spherical triangles are equivalent.

*To understand what is meant by the projection of these lines, conceive a hemisphere with its base on the paper, and represented by the circle abc, and all the arcs raised up from the paper as they would be on the surface of such a hemisphere. Thus, considering the arc aABb, the ends a and b would be in the paper just where they are, but the rest of the arc would be off the paper, as though you could take hold of B and raise it from the paper while a and b remain fixed. The lines in the figure are representations of lines on the surface of such a hemisphere, as they would appear to an eye situated in the axis of the circle abc, and at an infinite distance from it; that is, just as if each point in the lines dropped perpendicularly down upon the paper. Arcs of great circles perpendicular to the base are projected in straight lines passing through the centre, and oblique arcs are projected in ellipses. See Spherical Trigonometry (97-109).

DEM.-Let ABC and A'B'C' be two symmetrical spherical triangles, with AB = A'B', AC A'C'. BC B'C', A = A', B = B', and C = C'; then are they equivalent.

=

=

For, pass circumferences of small circles through the vertices A, B, C and A', B', C', as abc and a'b'c', of which o and o' are the poles. [The student should execute this on the spherical blackboard.] Now, by reason of the mutual equality of the sides, the chord AC chord A'C', chord AB = chord A'B', and chord BC chord B'C', and as the small circles are circumscribed about the equal plane triangles ABC and A'B'C', these circles are equal. Hence, oAo'A' oBo'B' oc = o'C'. The triangle AoB is therefore equal to A'o'B', BoC B'o'C', and AoC A'o'C'. [The student should make the application of these equal triangles.] Hence, ABC is equivalent to A'B'C', as the two are composed of equal parts.

FIG. 336.

If the poles of the small circles fell without the given triangles, ABC would be equivalent to the sum of two of the partial triangles minus the third.

PROPOSITION XVIII.

584. Theorem.-On the same or equal spheres, two spherical triangles having two sides and the included angle

FIG. 337.

FIG. 338.

of the one equal to two sides and the included angle of the other, each to each, are equal, or symmetrical and equivalent.

=

=

DEM.-Let ABC and A'B'C', Fig. 337, be two spherical triangles having AB A'B', AC A'C', and A = A'. In this case, as the parts are similarly arranged, by placing AC in its equal A'C', AB will fall in its equal A'B' (as A = A'), and the two triangles will coincide. Hence, they are equal. Again, let the two triangles be ABC and A'B'C', Fig. 338, in which AB = A'B', AC A'C', and A = A', the parts not being similarly arranged, so that the triangles are incapable of superposition. Thus, if AB is placed in its equal A'B', A at A', and B at B', C and C' will fall ou opposite sides of AB. We may, however, construct ABC, Fig. 337, symmetrical with A'B'C' in this figure, and apply ABC of Fig. 338 to it, and find that they coincide. Now, ABC, Fig. 337, and A'B'C', Fig. 338, are equivalent (583); hence ABC, Fig. 338, is equivalent to A'B'C', Fig. 338.

585. SCH.-This proposition is virtually the same as (446) concerning trie drals. Thus, in Fig. 337, drawing the radii AO, BO, CO, A'O, B'O, and C'O, two

triedrals are formed, having the facial angle AOB = A'OB, AOC = A'OC', the included diedrals equal, and the parts similarly disposed, whence the triedrals are equal. In like manner the triedral O-ABC, Fig. 338, is symmetrical and equivalent to O-A'B'C', Fig. 338. Hence, in either case, all the parts of one spherical triangle are equal to all the parts of the other, each to each.

PROPOSITION XIX.

586. Theorem.-On the same, or on equal spheres, two spherical triangles having two angles and the included side of the one equal to two angles and the included side of the other, each to each, are equal, or symmetrical and equivalent.

DEM.-Using the same triangles as in the preceding proposition, the student should be able to make the application directly, when the parts are similarly disposed; and when not similarly disposed, he should be able to show that ABC, of Fig. 338, can be applied to ABC, Fig. 337, symmetrical with A'B'C', Fig. 338.

587. SCH.-This proposition is also virtually the same as (447) concerning triedrals. Let the student point out the identity.

PROPOSITION XX.

588. Theorem.-On the same, or on equal spheres, if two spherical triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides are unequal, and the greater third side belongs to the triangle having the greater included angle. Conversely, If the two sides are equal, each to each, and the third sides unequal, the angles included by the equal sides are unequal, and the greater belongs to the triangle having the greater third side.

A'B',

DEM.-In the triangles ABC and A'B'C', let AB = AC = A'C', and A > A'; then is BC B'C'. For, join the vertices with the centre, forming the two triedrals O-ABC and O-A'B'C'. In these triedrals AOB A'OB', AOC = A'OC', being measured by equal arcs; and C-AO-B > C'-A'O-B', having the same measures as A and A' (558). Hence COB > C'OB′ (449). Therefore CB, the measure of COB, is greater than C'B', the measure of C'OB'.

B

FIG 339.

In like manner, the same sides of the triangles, and consequently the same facial angles of the triedrals, being granted equal, and

BC > B'C', A > A'. For, BC being greater than B'C', COB > C'OB'; whence B-AO-C > B'-A'O-C′ (450), or A is greater than A'.

PROPOSITION XXI.

589. Theorem.—On the same, or on equal spheres, two spherical triangles having the sides of the one respectively equal to the sides of the other, or the angles of the one respectively equal to the angles of the other, are equal, or symmetrical and equivalent.

DEM.—The sides of the triangles being equal, the facial angles of the triedrals at the centre are equal, whence the triedrals are equal or symmetrical (451). Consequently the angles of the triangles are equal, and the triangles are equal, or symmetrical and equivalent.

Again, the triangles being mutually equiangular, the triedrals have their diedrals mutually equal; whence the triedrals are equal or symmetrical (452). Therefore, the sides of the triangles are mutually equal, and the triangles are equal, or symmetrical and equivalent. (See Figs. 333, 334.)

PROPOSITION XXII.

590. Theorem.-On spheres of different radii, mutually equiangular triangles are similar (not equal).

bc: BC

FIG. 340.

DEM.-Let O be the common centre of two unequal spheres; and let ABC be a spherical triangle on the surface of the outer. Draw the radii AO, BO, and CO, constructing the triedral O-ABC. Now, the intersections of these faces with the surface of the inner sphere will constitute a triangle which is mutually equiangular with ABC. Thus, A ɑ, Bb, and C = c, since in each case the corresponding diedrals are the same. From the similar sectors aOb, AOB, we have ab: AB :: αO : AO; and, in like manner, ac : AC :: aO: AO. Whence, ab : AB: ac AC. So, also, ab: AB :: ¿0 : BO, and Thus we see that ABC and also their sides proportional:

60: BO; whence, ab : AB :: bc: BC. abc, having their angles equal each to each, have therefore they are similar.

POLAR OR SUPPLEMENTAL TRIANGLES.

591. One triangle is polar to another when the vertices of one are the poles of the sides of the other. Such triangles are also

called supplemental, since the angles of one are the supplements of the sides opposite in the other, as will appear hereafter.

PROPOSITION XXIII.

592. Problem.-Having a spherical triangle given, to draw its polar.

SOLUTION.-Let ABC be the given triangle.* From A as a pole, with a quadrant strike an arc, as C'B'. From B as a pole, with a quadrant strike the arc and from C, the arc A'B'. Then is A'B'C' poiar to ABC.

C'A';

593. COR.-If one triangle is polar to another, conversely, the latter is polar to the former; i. e., the relation is reciprocal.

FIG. 341.

2

Thus, A'B'C' being polar to ABC; reciprocally, ABC is polar to A'B'C'; that is, A' is the pole of CB, B' of AC, and C' of AB. For every point in A'B' is at a quadrant's distance from C, and every point in A'C' is at a quadrant's distance from B. Hence, A' is at a quadrant's distance from the two points Cand B of CB, and is therefore its pole. [In like manner the student should show that B' is the pole of AC, and C' of AB.]

594. SCH.-Ву producing each of the arcs struck from the vertices of the given triangles sufficiently, four new triangles will be formed, viz., A'B'C', QC'B', PC'A', and RA'B'. Only the first of these is called polar to the given triangle. It is easy to observe the relation of any of the parts of any one of the other three triangles to the parts of the polar. Thus, QC' = : 180° — b', QB' 180° - c', QC'B' = 180° ·

=

= 180° — C'B'A', and Q = A′ = 180° appear hereafter.

B'C'A', QB'C'

P

A

2

BB

*This should be executed on a sphere. Few students get clear ideas of polar triangles without it. Care should be taken to construct a variety of triangles as the given triangle, since the polar triangle does not always lie in the position indicated in the figure here given. Let the given triangle have one side considerably greater than 90°, another somewhat less. and the third quite small. Also, let each of the sides of the given triangle be greater than 90°.