Let ABC be a spherical triangle; and from the points A, B, and C, as poles, let the great circles FE, ED, DF, be described, intersecting one another in F, D, and E; the sides of the triangle FED are the supplements of the measures of the angles A, B, C; namely, FE of the angle BAC, DE of the angle ABC, and DF of the angle ACB. And again, AC is the supplement of the angle DFE, AB of the angle FED, and BC of the angle EDF. Let AB produced meet DE, EF, in G, M; let AC meet FD, FE, in K, L; and let BC meet FD, DE, in N, H. Since A is the pole of FE, and the LM: circle AC passes through A, EF will pass through the pole of AC (Sp. Ge. I. Cor. 7); and since AC passes through C, the pole of FD, FD will k a c hu pass through the pole of AC; there- U fore the pole of AC is in the point F, E in which the arcs DF, EF, intersect each other. In the same manner, D is the pole of BC, and E the pole of · And since F, E, are the poles of AL, AM, the arcs FL and EM are quadrants, and FL, EM, together, that is, FE and ML together, are equal to a semicircle. But since A is the pole of ML, ML is the measure of the angle BAC (Sp. Ge. 3), consequently FE is the supplement of the measure of the angle BAC. In the same manner, ED, DF, are the supplements of the measures of the angles ABC, BCA. Since likewise CN, BH, are quadrants, CN, BH, together, that is, NH, BC, together, are equal to a semicircle; and since D is the pole of NH, NH is the measure of the angle FDE, therefore the measure of the angle FDE is the supplement of the side BC. In the same manner, it is shown that the measures of the angles DEF, EFD, are the supplements of the sides AB, AC, in the triangle ABC. Schol.The triangles ABC, DEF, are called polar ttiangles. ." !!! PROPOSITION XII. The three angles of a spherical triangle are greater than two right angles, and less than six right angles. The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the E supplemental triangle DEF, are (Sp. Ge. 11) equal to three semicircles; N but the three sides of the triangle FDE are (Sp. Ge. 8) less than two semicircles; therefore the measures of the angles A, B, C, are greater than *HTA a semicircle ; and hence the angles [ A, B, C, are greater than two right D G angles. And because all the external and internal angles of any triangle are equal to six right angles; therefore, all the internal angles are less than six right angles. PROPOSITION XIII. If to the circumference of a great circle, from a point which is not the pole of it, arcs of great circles be drawn; the greatest of these arcs is that which passes through the pole of the first-mentioned circle, and the supplement of it is the least ; and of the others, that which is nearer to the greatest is greater than that which is more remote. Let ADB be the circumference of a great circle, of which the pole is H, and let C be any other point; through C and H let the semicircle ACB be drawn meeting the circle ADB in A and B; and let the arcs CD, CE, CF, also be described. From C draw CG perpendicular to AB, and then, because the circle AHCB which passes through H, the pole of the circle ADB, is at right angles to ADB, CG is perpendicular to the plane ADB. Join GĎ, GE, GF, CD, CE CF,CA, CB. Of all the straight lines drawn from G to the circumference ADB, GA is the greatest, and GB the least (Pl. Ge. III. 7); and GD, which is nearer to GA, is greater than GE, which is more remote. But the triangles CGA, CGD, are right angled at G, and they have the common side CG; therefore the squares of CG, GA, together, that is, the square of CA is greater than the squares of CG, GD, together, that is, than the square of CD; therefore CA is greater than CD, and the arc CA than the arc CD. In the same manner, since GD is greater than GE, and GE than GF, it is shown that CD A D лв is greater than CE, and CE than D E CF, and, consequently, the arc CD greater than the arc CE, and the arc CE greater than the arc CF. Also, because AG is the greatest, and GB the.least, of all the lines drawn from G, CA is the greatest, and CB the least, of all the lines drawn from C, and therefore the arc CA is the greatest, and CB, its supplement, the least of all the arcs drawn through C. PROPOSITION XIV. In a right-angled spherical triangle the sides are of the same affection with the opposite angles ; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles. Let ABC be a spherical triangle right angled at A, any side AB will be of the same affection with the opposite angle ACB. Case 1. Let AB be less than a quadrant. Let AE be a quadrant, and EC an arc of a great circle passing through E, C. Since A is a right angle, and AE a quadrant, E is the pole of the great circle AC, and ECA a right angle ; but ECA is greater than BCA, therefore BCA is less than a right angle. Case 2. Let AB be greater than a quadrant; make AE equal to a quadrant, and let a great circle pass through C, E. ECA is a right angle as before, and BCA is greater than ECA, that is, greater than a right angle. PROPOSITION XV. If the two sides of a right-angled spherical triangle be of the same affection, the hypotenuse will be less than a quadrant; and if they be of different affection, the hypotenuse will be greater than a quadrant. Let ABC (last figure) be a right-angled spherical triangle; if the two sides AB, AC, be of the same or of different affection, the hypotenuse BC will be less or greater than a quadrant. Case 1. Let AB, AC, be each less than a quadrant. Let AE, AG, be quadrants ; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but (Sp. Ge. 13) CE is greater than CB, since CB is farther off from CGD than CE. In the same manner, it is shown that CB, in the triangle CBD, where the two sides CD, BD, are each greater than a quadrant, is less than CE, that is, less than a quadrant. Case 2. Let AC be less, and AB greater than a quadrant; then the hypotenuse BC will be greater than a quadrant; for, let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But CB is greater than CE (Sp. Ge. 13), since AC passes through the pole of ABD. Cor. 1. Hence, conversely, if the hypotenuse of a right angled triangle be greater or less than a quadrant, the sides will be of different or the same affection. Cor. 2.-Since (Sp. Ge. 14) the angles of a right-angled spherical triangle have the same affection with the opposite sides, therefore, according as the hypotenuse is greater or less than a quadrant, the angles will be of different or of the same affection. PROPOSITION XVI. In any spherical triangle, if the perpendicular upon the base from the opposite angle fall within the triangle, the angles at the base are of the same affection; and if the perpendicular fall without the triangle, the angles at the base are of different affection. Let ABC be a spherical triangle, and let the arc CD be drawn from C perpendicular to the base AB. 1. Let CD fall within the triangle; then since ADC, BDC, are right-angled spherical triangles, the angles A, B, must each be of the same affection with CD (Sp. Ge. 14). 2. Let CD fall without the triangle; then (Sp. Ge. 14) the angle B is of the same affection with CD; and the angle CAD is of the same affection with CD; therefore the angles CAD and B are of the same affection, and the angles CAB and B of different affections. CoR.--Hence, if the angles A and B be of the same af fection, the perpendicular will fall within the base; for, if it did not, A and B would be of different affection, And if the angles A and B be of opposite affection, the perpendicular will fall without the triangle; for, if it did not, the angles A and B would be of the same affection, contrary to the supposition. PROPOSITION XVII. If to the base of a spherical triangle a perpendicular be drawn from the opposite angle, which either falls within the triangle, or is the nearest of the two that fall without; the least of the segments of the base is adjacent to the least of the sides of the triangle, or to the greatest, according as the sum of the sides is less or greater than a semicircle. Let ABEF be a great circle of a sphere, H its pole, and GHD any circle passing through H, which therefore is perpendicular to the circle ABEF. Let A and B be two points in the circle ABEF on opposite sides of the point D, and let D be nearer to A than to B, and let C be any point in the circle GHD, between H and D. Through the points A and C, B and C, let the arcs AC and BC be drawn, and let them be produced till they meet the circle ABEF in the points E and F; then the arcs ACE, BCF, are semicircles. Also ACB, ACF, CFE, ECB, are four spherical triangles |