smaller than the BCA. But that is contrary to the hypothesis.

Hence the two sides cannot be unequal; or in other words, they are equal.

Exercises.

Q. E. D.

-1. Construct a right spherical triangle, having

given the sides adjacent to the right angle.

2. The same, having given the hypothenuse and one side.

3. The same, having given the hypothenuse and an oblique angle.

4. The same, having given an oblique angle and the adjacent base.

5. The same, having given an oblique angle and the side opposite. Discuss thoroughly.

6. Show that two of the angles of a spherical triangle may each be 90°; and that when they are such, the sides opposite them will be 90°.

7. Show that the three angles of a spherical triangle may each be 90°; that when they are such, each side of the spherical triangle will be 90°; and that the area of the spherical triangle will be one-eighth of the surface of the sphere.

8. Show that if two spherical triangles can be brought to have a common side, and two other sides cross each other, the sum of the sides that cross is greater than the sum of the sides that do not cross; i.e.

B C

D

A

E

FIG. 280.

9. Show that if two spherical triangles have a common side AC, and the vertex D, of one, lies within the other,

AB+BC>AD + DC.

10. Show that if two spherical triangles

have two sides in one equal to two sides in A

the other, and the included angles are un

B

E

FIG. 281.

equal, the third sides will be unequal, and the greater side will be opposite the greater angle.

11. Show that the sum of the sides of a convex* spherical polygon of any number of sides is less than a great circle.

polygon were separated into spherical triangles by means of diagonals, some of the spherical triangles would be exterior to the polygon, and the polygon would not be convex. Therefore the spherical polygon lies entirely within the hemisphere. Produce AE to A', ED to H, and DC to K.

AFA' = AEA' = ÁÈ + ÉÀ',

EA' + A'H>ED + DH,

DH+HK>DC+ CK,

CK + KB> CB,

AB = BA,

AFA' + EA' + Á1H + DH + HK + CK + KB + AB>ÁÈ + ÉA' + ED + DH + DC + CK + CB + BA;

AFA' + A'H + HK + KB + AB > AE + ED + DC + CB+BÂ; Great circle > AE + ED + DC + CB + BA.

Q. E. D.

*NOTE. A convex spherical polygon is one in which none of the angles are greater than 180°.

CHAPTER XI.

125. Definitions. We have seen that a single plane separates space into two parts; two planes separate it into four parts, and themselves intersect in a straight line.

In general, three planes separate space into eight parts, called triedrals; have three lines of intersection; and have one, and only one, common point.

Fig. (a) represents the three planes so placed as to be perpendicular to each other. In Fig. (b) they are oblique. Particular cases may arise; for instance:

1. The third plane may be parallel to the line of intersection of the first and second. The three lines of intersection will then be parallel.

In this case space is separated into seven parts.

2. The third plane may pass through the line of intersection of the first and second. In this case space is separated into six parts.

FIG. 285.

3. The three planes may be parallel to each other. In this case they are sometimes said to intersect at infinity. They separate space into four parts.

When three planes form eight triedrals, the common point (I in the figure) is called the ver

tex of each.

The lines of intersection of the planes form the edges of the triedrals.

IJ, IK, and IH are edges.

The angles that the edges make with each other are called the facial angles.

H

FIG. 286.

K

The angles that the planes make with each other (as previously noted) are called diedrals.

126. THEOREM. If the vertex of a triedral be taken as the centre of a sphere of any radius, the intersections will form a spherical triangle, the sides of which will be the measures of the facial angles; and the angles of which will be the measures of the diedrals.

With I as a centre and ID as a radius describe the sphere.

The DE is the measure of the

▲ DIE; the EF is the measure

J

E

G

H

FIG. 287.

of the EIF, and the FD is the measure of the FID.

If at any of the vertices of the

spherical triangle, as F, the measure of the diedral be drawn (§ 106) and be FO and FG, they will (§ 116) make the same angle as do the arcs FE and FD.

Q. E. D.

Exercises. 1. Show that if two facial angles of a triedral are equal, the diedrals opposite them will be equal.

Let IJ, IH, and IK represent the edges of the triedral* of which LEID = LEIF. Conceive a sphere with I as a centre and any radius, say IE, intersecting the planes of the triedral in ED, DF, and EF.

E

F

N

K

M

H

FIG. 288.

Since ZEID = ZEIF, ED = ÉF. :. (§ 124) ≤ EDF = / EFD. Hence from the theorem, the diedrals J-HI-K and J-KI-H are equal.

2. Show that if two diedrals of a triedral are equal, the facial angles opposite them will be equal.

3. Show that in any triedral the sum of any two facial angles will be greater than the third.

4. Show that the facial angles and the diedrals of a triedral and its vertical triedral will be equal.

5. Show that the sum of the facial angles of a triedral will be between 0° and 360°.

6. Show that the sum of the diedrals of a triedral will be between two and six right angles.

7. Show that two triedrals having the three facial angles of one equal to the three facial angles of the other, and similarly arranged, are superimposable.

127. Definition. If two triedrals can be so placed that the edges of one are perpendicular to the faces of the other, they are said to be supplementary.

Let the two triedrals be so placed that their vertices coincide; VK being perpendicular to the face AVC, VE perpendicular to the face AVB, and VD perpendicular to the face BVC.

*NOTE. The plane MN is introduced simply for the purpose of enabling the beginner to more readily comprehend the figure.