until it reached the trace of A'; and so we would have the surface of the lune less than its trace, which is contrary to our first assumption (657). ..the two angles cannot be unequal.

DEFINITIONS.

679. The Supplement of a Sect is the sect by which it differs from a half-line.

680. One-quarter of a line is called a Quadrant.

618. A Spherical Polygon is a closed figure in the sphere formed by sects.

682. A Spherical Triangle is a convex spherical polygon of three sides.

683. Symmetrical Spherical Polygons are those in which the sides and angles of the one are respectively equal to those of

the other, but arranged in the reverse order. If one end of a sect were pivoted within one polygon, and one end of another sect pivoted within the symmetrical polygon, and the two sects revolved so as to pass over the equal parts at the same time, one sect would move clockwise, while the other moved counterclockwise.

D

fixed

684. Two points are symmetrical with respect to line, called the axis of symmetry, when this axis bisects at right angles the sect joining the two points.

Any two figures are symmetrical with respect to an axis when every point of one has its symmetrical point on the

other.

THEOREM III.

685. The perimeter of a convex spherical polygon wholly contained within a second spherical polygon is less than the perimeter of the second.

G

B

Let ABFGA be a convex spherical polygon wholly contained in

Produce the sides AG and GF to meet the containing perimeter at

K and L respectively.

By 661, a sect is the smallest path between its end points,

:. BF BCLF, and LG < LKG, and KA < KDEA;

<

:. BFGA< BCLGA < BCKA < BCDEA.

THEOREM IV.

686. The sum of the sides of a convex spherical polygon is less than a line.

Let ABCDEA be the polygon, and let its sides BA and BC be produced to meet again at M. Since the polygon is convex, the BCD < st. 4, and also ☀ BAE < st. ; therefore ABCDEA lies wholly within ABCMA; therefore, by 685, the perimeter of ABCDEA <the perimeter of ABCMA, that is, less than a line.

SYMMETRY WITHOUT CONGRUENCE.

687. If two figures have central symmetry in a plane, either can be made to coincide with the other by turning it in the plane through a straight angle. This holds good when for "plane" we substitute "sphere."

If two figures have axial symmetry in a plane, they can be made to coincide by folding the plane over along the axis, but not by any sliding in their plane. That is, we must use the

third dimension of space, and then their congruence depends on the property of the plane that its two sides are indistinguishable, so that any piece will fit its trace after being turned over. This procedure, folding along a line, can have no place in a strictly two-dimensional geometry; and, were we in tridimensional spherics, we could say, that from the outside a sphere is convex, while from the inside it is concave, and that a piece of it, after being turned over, will not fit its trace, but only touch it at one point. So figures with axial symmetry, according to 684, on a sphere cannot be made to coincide; and the word symmetrical is henceforth devoted entirely to such.

THEOREM V.

688. Two spherical triangles having two sides and the included angle of one equal respectively to two sides and the included angle of the other, are either congruent or symmetrical.

B'B

(1) If the parts given equal are arranged in the same order, as in DEF and ABC, then the triangle DEF can be moved in the sphere until it coincides with ABC.

(2) If the parts given equal are arranged in reverse order, as in DEF and A'B'C', by making a triangle symmetrical to A'B'C', as ABC, we get the equal parts arranged in the same order as in DEF, which proves DEF congruent to any triangle symmetrical to A'B'C'.

THEOREM VI.

689. If two spherical triangles have two sides of the one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first will be greater than the third side of the second.

HYPOTHESIS. AB = DE, and AC:

=

EF, but BAC > ¥ EDF.

CONCLUSION. BC > EF.

PROOF. When AB coincides with DE, if the points C and F are on the same side of the line AB, then, since ☀ BAC > × EDF, the side DF will lie between BA and AC, and DF must stop either within the triangle ABC, on BC, or after cutting BC.

When F is within the triangle, by 685, BC + CA > EF + FD; but AC = DF,

:. BC> EF.

When F lies on BC, then EF is but a part of BC. In case DF cuts BC, call their intersection point G. Then BG + GF > EF, and GC + GD> AC,

but FD = AC,

.. BC + FD > AC + EF;

.. BC> EF.

If, when one pair of equal sides coincide, the other pair lie on opposite sides of the line of coincidence, the above proof will show the third side of a triangle symmetrical to the first to be greater than the third side of the second triangle, and therefore the third side of the first greater than the third side of the second.