SPHERICAL TRIANGLES. If C = H = r't 4, and c = h, and AF, then if ABC > G, make * ABD = × G, .. ¥ BDA = × H = 4 C = r't,.. B is pole to CDA. 405. Theorem. Of sects joining two symmetrical points to a third, that cutting the axis is the greater. Proof. BA=BC+ CA = BC+CA'>BA'. A FIG. 186. B 77 406. Theorem. If two spherical triangles have two sides of the one equal to two sides of the other, but the included angles unequal, then that third side is the greater which is opposite the greater angle. Proof. Slide the triangles in the sphere until a pair of equal sides coincide and the other pair of equal sides have a common end point. Bisect the angle made by these equal sides. This axis. cuts the third side, which is opposite the greater angle. 407. Inverse. If two triangles have two sides of the one equal to two sides of the other, but the third sides unequal, then of D A A FIG. 187. the angles opposite these third sides that is the greater which is opposite the greater third side. 408. Theorem. The g-line through the poles of two g-lines is the polar of their intersection points. Proof. If A and B are poles of the g-lines a and b, which intersect in P, then AP and BP are quadrants; . AB is the polar of P. 409. Corollary I. The g-line through the poles of two g-lines cuts both at right angles. FIG. 188. B 410. Corollary II. If three g-lines are concurrent, their poles are collinear. 411. Of the sides of a spherical angle, we may call those poles positive from which in the figure these sides would be described from the vertex by a quadrant rotating positively. 412. Theorem. The sect which an angle intercepts on the polar of its vertex equals the sect between the positive poles of its sides. Proof. Slide the quadrant BF along the polar of A until B comes to C. The+pole F of AB will then coincide with the + pole G of AC. 413. The sect joining any point to one pole of a g-line is less than a quadrant if the two points are in the same one of that g-line's hemispheres; greater than a quadrant if they are in different hemispheres. By a pole's hemisphere we mean that one of its g-line's hemispheres in which the pole is. 414. Of a given spherical triangle ABC, the polar is a new triangle A'B'C', where A' is that pole of BC which has A in its hemisphere, and B' that pole of AC which has B in its hemisphere, and C^ that pole of AB which has C in its hemisphere. 415. Theorem. If of two spherical triangles the second is the polar of the first, then the first is the polar of the second. Hypothesis. Let ABC be the polar of A'B'C'. Conclusion. Then A'B'C' is the polar of ABC. Proof. Join A'B and A'C. Since B is pole of A'C', therefore BA' is a quadrant; and since Cis pole of A'B', therefore CA' is a quadrant; .. A' is pole of BC. In like manner, B' is pole of AC, and C' of AB. SPHERICAL TRIANGLES. 79 Moreover, since A has A' in its hemisphere, .. the sect AA' is less than a quadrant, .. A' has A in its hemisphere. D 416. Theorem. In a pair of polar triangles, any angle of either intercepts, on the side of the other which lies opposite to it, a sect which is the supplement of that side. Proof. Let ABC and A'B'C' be two polar triangles. E B FIG. 193. Produce A'B' and A'C' to meet BC at D and E, respectively. Since B is the pole of A'C', therefore BE is a quadrant ; and since C is the pole of A'B', therefore CD is a quadrant; therefore BE+ CD BECDBC + DE, Therefore DE, the sect of BC which A' intercepts, is the supplement of BC. = half-g-line; but 417. Theorem. Two spherical triangles of the same sense, having three angles of the one equal respectively to three angles of the other, are congruent. Proof. Since the given triangles are respectively equiangular their polars are respectively equilateral. For equal angles at the poles of g-lines intercept equal sects on those lines; and these equal sects are the supplements of corresponding sides. Hence these polars, having three sides equal, are respectively equiangular, and therefore the original triangles are respectively equilateral. F 418. Of a convex spherical polygon ABCD..., the polar is a new spherical polygon A'B'C'D' ..., where A' is that pole of BC which has A in its hemisphere, etc. 419. Theorem. The polar of a cenquad is a concentric cenquad. Proof. The g-line HK through the symcenter 0 and 1 to AB is also to CD; and OH OK are the complements of the sects = D A from 0 to poles D' and B' of the sides AB and CD. KB B' F' FIG. 194. C B Hence O is symcenter for B' and D'. In the same way prove O symcenter for A' and C'. 420. Theorem. The opposite sides of a cenquad intersect on the polar of its symcenter. D Proof. O is symcenter for F and F'. 421. Theorem. Any two consecutive vertices of a cenquad and the opposites of the other two are concyclic. A FIG. 195. Proof. The perpendiculars, to the g-line B through O and bisecting BC and DA, from A and B in one of its hemispheres and C and D in the other, are equal. So also the perpendiculars from their opposites D' and C' in the first hemisphere and A' and B' in the second. So A, B, C', D' and A', B', C, D are on equal circles with opposite q-poles. Such circles are called parallels; the co-polar g-line, equator. 422. The perpendiculars erected at the mid points of the sides of a spherical triangle are concurrent in its circumcenter. 423. Theorem. The g-line bisecting two sides of a triangle intersects the third side at a quadrant from its mid point. .. ALD FIG. 196. BMD' [having two angles and an opposite side equal, and the other pair of opposite sides not supplemental]. .. AD BD', .. DC' from D to C' the mid p't of AB is a quadrant. SPHERICAL TRIANGLES. 81 424. Corollary I. The altitudes of a spherical triangle are concurrent in a point called its orthocenter. For, regarding A'B'C' as the triangle, the perpendicular to DC' at C' is the polar of D, and .. 1 to A'B'. Similarly, the perpendicular to BA' at A' is 1 to B'C', etc. So the three altitudes of A'B'C' are concurrent in the cir cumcenter of ABC. 425. Cor. II. The vertices of spherical triangles of the same angle sum on the same base are on a circle co-polar with the g-line bisecting their sides. = == For AO BO, OAB = OBA, & LAB = 4 MBA = { [A + B + C]. Hence AOB is fixed, and .. OC [supplemental to OA]. 426. Theorem. The g-lines through the corresponding vertices of a triangle and its polar are concurrent in the common orthocenter of the two triangles. Proof. For AA' is 1 to BC and B'C', B'K since it passes through their poles. 427. Theorem. The sides of a triangle A intersect the corresponding sides of its polar on the polar of their orthocenter. Proof. For AA' is the polar of the intersection points of BC and B'C'; similarly, BB' is the polar of the intersection points of CA and C'A', etc. Sects from the orthocenter to these intersection points are all quadrants. 428. Theorem. A triangle's in-center is also its polar's circumcenter; and R is complemental to r. .. IA' is B' FIG. 198. Proof. ID to BC contains A'. the complement of r. So is IB' and IC'. |