A DEFINITIONS.. I. Great circle of the sphere is any circle on the superficies of the sphere of which the plane paffes through the centre of the sphere, and of which the centre therefore is the same with the centre of the sphere. II. The pole of a great circle of the sphere is a point in the superficies of the sphere, from which all straight lines drawn to the circumference of the circle are equal. A spherical angle is that which on the fuperficies of a sphere is contained by two arches of great circles, and is the fame with the inclination of the planes of these great circles. IV. A spherical triangle is a figure upon the superficies of a sphere comprehended by three arches of three great circles, each of which is less than a femicircle. PROP. PROP. I. PROP. V. NY two great circles of a sphere bisect one another. AN For, as they have the same centre, their common section is a diameter of both, and therefore bisects both. PROP. II. : THE arch of a great circle between the pole and the circumference of another great circle is a quadrant. Let ABC be a great circle, and D its pole; if a great circle CD pass through D, and meet ABC in C, the arch DC is a quadrant. Let the great circle CD meet ABC again in A, and let AC be the common section of the great circles, which will pass through E the centre of the sphere: Join DE, DA, DC: By def. 2. DA, DC are equal, A E C and AE, EC are also equal, and DE is common; therefore B (8. 1.) the angles DEA, DEC are equal; wherefore the arches DA, DC are equal, and confequently each of them is a qua drant. Q. E. D. Cor. The circle ABC has two poles, one on each fide of its plane, which are the extremities of a diameter of the sphere perpendicular to the plane ABC; and no other point but these two can be a pole of the circle ABC. PROP. I PRO P. III. F the pole of a great circle be the fame with the intersection of other two great circles; the arch of the first mentioned circle intercepted between the other two, is the meafure of the spherical angle which the same two circles make with one another. Let the great circles BA, CA on the superficies of a sphere, of which the centre is D, interfect one another in A, and let BC be an arch of another great circle, of which the pole is A; BC is the measure of the spherical angle BAС. Join AD, DB, DC; fince A is the pole of BC, AB, AC are quadrants, and the angles ADB, ADC are right angles; therefore (4. def. 7.) the angle CDB is the inclination of the planes of the circles AB, AC, and is (def. 3. Sp.. T.) equal to the spherical angle BAC. Q.E. D. A A COR. 1. If through the point A, two quadrants AB, AC, be drawn, the point A will be the pole of B the great circle BC, passing through their extremities B, C. D C E Join AC, and draw AE, a straight line to any other point E in BC; join DE: Since AC, AB are quadrants, the angles ADB, ADC are right angles, and AD will be perpendicular to the plane of BC: Therefore the angle ADE is a right angle, and AD, DC are equal to AD, DE each to each; therefore AE, AC are equal, and A is the pole of BC, by def. 2. 3 PROP. IF PROP. IV. F there be two great circles of a sphere, of which the first passes through the poles of the second, the fecond alfo passes through the poles of the first. Let ACBD and AEBF be two circles, the one of which ACBD paffes through C and D, the poles of the other AEBF; the circle AEBF passes through the poles of the circle ACBD. Let G be the centre of the sphere; join CG, GD, which will be in the same straight line: Draw AGB the diameter, which is the common feetion of the circles ACBD, AEBF; and in the plane of the circle AEBF draw from the point G the ftraight line EGF perpendicular to AB. F G Then, because Cand Dare the poles of the circle AEBF, the arch intercepted between either of them and any point in the circumference AEBF, is a quadrant (Prop. 3.), and therefore A B E D the straight line CD is perpendicular to the plane of the circle AEBF; wherefore also the plane of the circle ACBD, which paffes through CD, is perpendicular (17.7.) to the plane AEBF. And fince GE in the plane AEBF, is perpendicular to AB, the common section of the two planes, it is perpendicular to the plane (2. def. 7.) ACBD. The arch of a great circle, therefore, intercepted between the point E and any point of the circumference ACBD is a quadrant, and therefore, (Prop. 3.) the point E is the pole of the circle ACBD. For the fame reason, F is the other pole of the circle ACBD; and E and Fare in the circumference of the circle AEBF. Q. E. D. Nifofceles spherical triangles the angles at the base are equal. Let ABC be a spherical triangle, having the fide AB equal to the fide AC; the spherical angles ABC and ACB are equal. Let D be the centre of the sphere; join DB, DC, DA, and from A on the straight lines DB, DC, draw the perpendiculars AE, AF; and from the points E and F draw in the plane DBC the straight lines EG, FG perpendicular to DB and DC, meeting one another in G: Join AG. Because DE is at right angles to each of the straight lines AE, EG, it is at right angles and therefore AG, the com mon section of the planes A C F G E B AFG, AEG is at right angles (18.7.) to the plane DBC, and the angles AGE, AGF are consequently right angles. But, fince the arch AB is equal to the arch AC, the angle ADB is equal to the angle ADC. Therefore the triangles ADE, ADF, have the angles EDA, FDA equal, as also the angles AED, AFD, which are right angles; and they have the fide AD common, therefore the other fides are equal, viz. AE to AF, (26. 1.) and DE to DF. Again, because the angles AGE, AGF are right angles, the squares on AG and GE are equal to the square of AE; and the squares of AG and GF to the square of AF. But the squares of AE and AF are equal, therefore the squares of AG and GE are equal to the squares of AG and GF, and taking away the common fquare |