Hence these polars, being, by 691, congruent or symmetrical, are respectively equiangular, and therefore the original spherical triangles are respectively equilateral.

THEOREM XIV.

713. An exterior angle of a spherical triangle is greater than, equal to, or less than, either of the interior opposite angles, according as the medial from the other interior opposite angle is less than, equal to, or greater than, a quadrant.

Let ACD be an exterior angle of the ABC. By 695, bisect AC at E. Join BE, and produce to F, making EF = BE. Join FC. λ ABE CFE, ↑

(688. Spherical triangles having two sides and the included angle equal are congru

If, now, the medial BE be a quadrant, BEF is a half-line, and, by 676, F lies on BD;

..DCE coincides with FCE,

If the medial BE be less than a quadrant, BEF is less than a halfline, and F lies between AC and CD;

:. ¥ DCA > & FCE,

X DCA > BAC.

And if BE be greater than a quadrant, BEF is greater than a halfline, and F lies between CD and AC produced;

Thus, according as BE is greater than, equal to, or less than, a quadrant, the exterior X ACD is less than, equal to, or greater than, the interior opposite X BAC.

714. From 713, by 33, Rule of Inversion,

According as the exterior angle ACD is greater than, equal to, or less than, the interior opposite angle BAC, the medial BE is less than, equal to, or greater than, a quadrant.

THEOREM XV.

715. Two spherical triangles having the angles at the base of the one equal to the angles at the base of the other, and the sides opposite one pair of equal angles equal, are either congruent or symmetrical, provided that in neither triangle is the third angular point a quadrant from any point in that half of its base not adjacent to one of the sides equal by hypothesis.

FIRST, if the parts given equal lie clockwise in the two spherical triangles, as in ABC and DEF, where we suppose B = * E, X C × F, and AB — DE, make DE coincide with AB; then EF will lie along BC, and DF must coincide with AC. For if it could take any other position, as AG, it would make a ↑ AGC with exterior ¥ AGB = interior opposite C, and therefore, by 714, with medial AH a quadrant, which is contrary to our hypothesis.

SECOND, if the equal parts lie in one spherical triangle clockwise, in the other counter-clockwise, as in A'B'C' and λ DEF, then λ DEF ≈ λ ABC, which is symmetrical to ▲ A'B'C'.

THEOREM XVI.

716. If a sect drawn in the sphere from a point perpendicular to a line be less than a quadrant, it is the smallest which can be drawn in the sphere from the point to the line; of others, that which is nearer the perpendicular is less than that which is more remote; also to every sect drawn on one side of the perpendicular there can be drawn one, and only one, sect equal on the other side.

Let A be the point, BD the line, AB 1 BD, and AD nearer than AE to AB.

Produce AB to C, making BC= AB. Join CD, CE.

Then AD = CD.

(688. Spherical triangles having two sides and the included angle equal are congruent or symmetrical.)

But, by 661,

AD + CD or 2AD > AC or 2AB,

Also, by 685,

..

AD > AB.

AE+ CE or 2AE > AD + CD or 2AD.

Again, make BF = BD. Join AF.

As before, AF AD, and we have already shown that no two sects on the same side of the perpendicular can be equal.

717. COROLLARY. The greatest sect that can be drawn from A to BD is the supplement of AB.

THEOREM XVII.

718. If two spherical triangles have two sides of the one equal to two sides of the other, and the angles opposite one pair of equal sides equal, the angles opposite the other pair are either equal or supplemental.

FIRST, given the equal parts AB = DE, AC = DF, and X B = E arranged clockwise in the two spherical triangles.

If A = XD, the spherical triangles are congruent.

If A is not = D, Suppose A > X D. XBX E, side EF will

one must be the greater.

Make DE coincide with AB. Then, since lie along BC; since A> D, side DF will lie between AB and AC, as at AG.

Now, the two angles at G are supplemental; but one is F and the other * C, because ↑ CFG is isosceles.

SECOND, if the parts given equal lie clockwise in one spherical triangle and counter-clockwise in the other, as in λ A'B'C' and λ DEF, then, taking the ABC symmetrical to A'B'C', the above proof shows F equal or supplemental to ₫ C, that is, to ¥ C'.

THEOREM XVIII.

719. Symmetrical isosceles spherical triangles are congruent.

B B

A

丸

For, since four sides and four angles are equal, the distinction between clockwise and counter-clockwise is obliterated.

THEOREM XIX.

720. The locus of a point from which the two sects drawn to two given points are equal is the line bisecting at right angles the sect joining the two given points.

For every point in this perpendicular bisector, and no point out of

it, possesses the property.