26.

Triangles standing opposite to one another on the sphere are equivalent in surface.

By opposite triangles we here understand such as are made on both sides of the center by the intersections of the sphere with planes; in such triangles, therefore, the sides and angles are in contrary order.

In the opposite triangles ABC and A'B'C' (Fig. 14, where one of them must be looked upon as represented turned about), we have the sides ABA'B', BC= B'C', CA C'A', and the corresponding angles

FIG. 14.

at the points A, B, C are likewise equal to those in the other triangle at the points A', B', C'.

Through the three points A, B, C, suppose a plane passed, and upon it from the center of the sphere a perpendicular dropped whose prolongations both ways cut both opposite triangles in the points D and D' of the sphere. The distances of the first D from the points ABC, in arcs of great circles on the sphere, must be equal (Theorem 12) as well to each other as also to the distances D'A', D'B', D'C', on the other triangle (Theorem 6), consequently the isosceles triangles about the points. D and D' in the two spherical triangles ABC and A'B'C' are congruent. In order to judge of the equivalence of any two surfaces in general, I take the following theorem as fundamental:

Two surfaces are equivalent when they arise from the mating or separating of equal parts.

27. A three-sided solid angle equals the half sum of the surface angles less a right-angle.

In the spherical triangle ABC (Fig. 15), where each side <, desig. nate the angles by A, B, C; prolong the side AB so that a whole circle ABA'B'A is produced; this divides the sphere into two equal parts.

In that half in which is the triangle ABC, prolong now the other two sides through their common intersection point C until they meet the circle in A' and B'.

Α'

FIG. 15.

In this way the hemisphere is divided into four triangles, ABC, ACB', B'CA', A'CB, whose size may be designated by P, X, Y, Z. It is evi. dent that here P+X=B, P+Z= A.

The size of the spherical triangle Y equals that of the opposite triangle ABC', having a side AB in common with the triangle P, and whose third angle C' lies at the end-point of the diameter of the sphere which goes from C through the center D of the sphere (Theorem 26). Hence it follows that

P+Y= C, and since P+X+Y+Z

π, therefore we have also

P (A+B+C — π).

We may attain to the same conclusion in another way, based solely upon the theorem about the equivalence of surfaces given above. (Theorem 26.)

In the spherical triangle ABC (Fig. 16), halve the sides AB and BC, and through the mid-points D and

E draw a great circle; upon this let
fall from A, B, C the perpendiculars
AF, BH, and CG. If the perpendic F
ular from B falls at H between D and
E, then will of the triangles so made
BDH AFD, and BHE EGC (The A

orems 6 and 15), whence follows that

D

H

FIG. 16.

the surface of the triangle ABC equals that of the quadrilateral AFGC

(Theorem 26).

If the point H coincides with the middle point E of the side BC (Fig.

B 17), only two equal right-angled triangles, ADF

C

and BDE, are made, by whose interchange the equivalence of the surfaces of the triangle ABC and the quadrilateral AFEC is established.

If, finally, the point H falls outside the triangle ABC (Fig. 18), the perpendicular CG goes, in consequence, through the triangle, and so we go over from the triangle ABC to the quadrilateral AFGC by adding the

FIG. 17.

EBH.

triangle FAD DBH, and then taking away the triangle CGE Supposing in the spherical quadrilateral AFGC a great circle passed through the points A and G, as also through F and C, then will their arcs between AG and FC equal one another (Theorem 15), consequently also the triangles FAC and ACG be congruent (Theorem 15), and the angle FAC equal the angle ACG.

Hence follows, that in all the preceding cases, the sum of all three angles of the spherical triangle equals the sum of the two equal angles in the quadrilateral which are not the right angles.

Therefore we can, for every spherical triangle, in which the sum of the three angles is S, find a quadrilateral with equivalent surface, in which are two right angles and two equal perpendicular sides, and where the two other angles are each S.

Let now ABCD (Fig. 19) be the spherical quadrilateral, where the sides AB DC are perpendicular to BC, and the angles A and D each S.

=

BC until they cut one another in E, and
EF and let fall upon the prolongation
Bisect the whole arc BG and join the

Prolong the sides AD and further beyond E, make DE of BC the perpendicular FG. mid-point H by great-circle-arcs with A and F.

The triangles EFG and DCE are congruent (Theorem 15), so FG DC AB.

The triangles ABH and HGF are likewise congruent, since they are right angled and have equal perpendicular sides, consequently AH and AF pertain to one circle, the arc AHF π, ADEF likewise: angle HAD

HFES BAH = 1S - HFG =

†S—HAD−7+1; consequently, angle HFE

π, the

S-HFE-EFG (S-7); or what

is the same, this equals the size of the lune AHFDA, which again is equal to the quadrilateral ABCD, as we easily see if we pass over from the one to the other by first adding the triangle EFG and then BAH and thereupon taking away the triangles equal to them DCE and HFG. Therefore (S-) is the size of the quadrilateral ABCD and at the same time also that of the spherical triangle in which the sum of the three angles is equal to S.

28. If three planes cut each other in parallel lines, then the sum of the three surface angles equals two right angles.

Let AA', BB' CC' (Fig. 20) be three parallels made by the intersection of planes (Theorem 25). Take upon them at random three

points A, B, C, and suppose through these a plane passed, which consequently will cut the planes of the parallels along the straight lines AB, AC, and BC. Further, pass through the line AC and any point D on the BB', another plane, whose intersection with the two planes of the parallels AA' and BB', CC' and BB' produces the two lines AD and DC, and whose inclination to the third plane of the parallels AA' and CC' we will designate by w.

The angles between the three planes in which the parallels lie will be designated by X, Y, Z, respectively at the lines AA', BB', CC'; finally call the linear angles BDC a, ADC=b, ADB C.

About A as center suppose a sphere described, upon which the intersections of the straight lines AC, AD AA' with it determine a spherical triangle, with the sides p, q, and r. Call its size. Opposite the side q lies the angle w, opposite r lies X, and consequently opposite p lies the angle +2a-w-X, (Theorem 27).

π

In like manner CA, CD, CC' cut a sphere about the center C, and determine a triangle of size ẞ, with the sides p', q', r', and the angles, w opposite q', Z opposite r', and consequently +23-w-Z opposite p'.

Finally is determined by the intersection of a sphere about D with the lines DA, DB, DC, a spherical triangle, whose sides are l, m, n, and the angles opposite them w+Z-23, w+X-2a, and Y. Consequently its size (X+Y+Z-)—a—B+w.

Decreasing w lessens also the size of the triangles a and ẞ, so that +ẞ-w can be made smaller than any given number.