690. From 688 and 689, by 33, Rule of Inversion, if two spherical triangles have two sides of the one equal respectively to two sides of the other, the included angle of the first is greater than, equal to, or less than, the included angle of the second, according as the third side of the first is greater than, equal to, or less than, the third side of the second. 691. COROLLARY. Therefore, by 688, two spherical triangles having three sides of the one equal respectively to three sides of the other are either congruent or symmetrical. PROBLEM I. 692. To bisect a given spherical angle. Let BAC be the given . In its arms take equal sects AB and AC each less than a quadrant. Join BC. With B as pole, and any sect greater than half BC, but not greater than a quadrant, as a spherical radius, describe the circle EDF. With an equal spherical radius from C as pole describe an arc intersecting the circle EDF at D within the lune BAC. Join DB, DC, DA. DA shall bisect the angle BAC. For the s ABD, ACD, having AD common and AB = AC, and BD = DC (spherical radii of equal circles), are, by 691, symmetrical, :: BAD = ¥ CAD. 693. COROLLARY I. To bisect the re-entrant angle BAC, produce the bisector of the angle BAC. 694. COROLLARY II. To erect a perpendicular to a given line from a given point in the line, bisect the straight angle at that point. With A and B, the extremities of the given sect, as poles, and equal spherical radii greater than half AB, but less than a quadrant, describe arcs intersecting at C. Join AC and BC; and, by 692, bisect the angle ACB, and produce the bisector to meet AB at D. D is the mid point of the sect AB. For, by 688, ACD is symmetrical to BCD. 696. COROLLARY. The angles at the base of an isosceles spherical triangle are equal. PROBLEM III. 697. To draw a perpendicular to a given line from a given point in the sphere not in the line. D GIVEN, the line AB, and point C. Take a point D on the other side of the line AB, and with C as pole, and CD as spherical radius, describe an arc cutting AB in F and G. Bisect the sect FG at H, and join CH. CH shall be 1 AB. For, by 691, As GCH and FCH are symmetrical. THEOREM VII. 698. If two lines be drawn in a sphere, at right angles to a given line, they will intersect at a point from which all sects drawn to the given line are equal. Let AB and CB, drawn at right angles to AC, intersect at B, and meet AC again at A' and C' respectively. Then BA'C' = BAC', and BC'A' = BCA'; (678. The angles contained by the sides of a lune, at their two points of intersection, moreover, are equal.) AC = A'C', Hence, keeping A and C for they have the common supplement AC'. on the line AC, slide ABC until AC comes into coincidence with A'C'. Then, the angles at A, C, A', C', being all right, AB will lie along A'B, and CB along C'B, and hence the figures ABC and A'BC' coincide. .. each of the half-lines ABA' and CBC' is bisected at B. In like manner, any other line drawn at right angles to AC passes through B, the mid point of ABA'. Hence every sect from AC to B is a quadrant. 699. COROLLARY I. A line is a circle whose spherical radius is a quadrant. 700. COROLLARY II. A point which is a quadrant from two points in a line, and not in the line, is its pole. 701. COROLLARY III. Any sect from the pole of a line to the line is perpendicular to it. 702. COROLLARY IV. Equal angles at the poles of lines intercept equal sects on those lines. 703. COROLLARY V. If K be the pole, and FG a sect of any other line, the angles and semilunes ABC and FKG are to one another as AC to FG. 704. We see, from 698, that a spherical triangle may have two or even three right angles. If a spherical triangle ABC has two right angles, B and C, it is called a bi-rectangular triangle. By 698, the vertex A is the pole of BC, and therefore AB and AC are quadrants. 705. The Polar of a given spherical triangle is a spherical triangle, the poles of whose sides are respectively the vertices of the given triangle, and its vertices each on the same side of a side of the given triangle as a given vertex. THEOREM VIII. 706. If, of two spherical triangles, the first is the polar of the second, then the second is the polar of the first. HYPOTHESIS. Let A'B'C' be the polar of ABC. |