in F, and draw, in the plane ADB, EG at right angles to the same ED; therefore the rectilineal angle FEG is (6. def. 11.) the inclination of the planes ADC, ADB, and therefore is the same with the spherical angle BAC; from F draw FH perpendicu lar to DB, and from H draw, in the plane ADB, the straight line HG at right angles to HD meeting EG in G, and join GF. Because DE is at right angles to EF and EG, it is perpendicular to the plane FEG, (4. 11.) and therefore the plane FEG, is perpendicular to the plane ADB, in which DE is: (18. 11.) in the same manner the plane FHG is perpendicular to the plane ADB; and therefore GF the common section of the planes FEG, FHG is perpendicular to the plane ADB; (19. 11.) and because the angle FHG is the inclination of the planes BDC, BDA, it is the same with the spherical angle ABC; and the sides AC, CB of the spherical triangle being equal, the angles EDF, HDF, which stand upon them at the centre of the sphere, are equal; and in the triangles EDF, HDF, the side DF is common, and the angles DEF, DHF are right angles; therefore EF, FH are equal; and in the triangles FEG, FHG the side GF is common, and the sides EG, GH will be equal by the 47. 1. and therefore the angle FEG is equal to FHG; (8. 1.) that is, the spherical angle BAC is equal to the spherical angle ABC.

PROP. V. FIG. 3.

IF, in a spherical triangle ABC, two of the angles BAC, ABC be equal, the sides BC, AC opposite to them, are equal.

Read the construction and demonstration of the preceding proposition, unto the words, "and the sides AC, CB," &c. and the rest of the demonstration will be as follows, viz.

And the spherical angles BAC, ABC, being equal, the rectilineal angles FEG, FHG, which are the same with them, are equal; and in the triangles FGE, FGH the angles at G are right angles, and the side FG opposite to two of the equal angles is common; therefore (26. 1.) EF is equal to FH; and in the right angled triangles DEF, DHF the side DF is common; wherefore (47. 1.) ED is equal to DH, and the angles EDF, HDF, are therefore equal, (4. 1.) and consequently the sides AC, BC of the spherical triangle are equal.

PROP. VI. FIG. 4.

ANY two sides of a spherical triangle are greater that the third.

Let ABC be a spherical triangle, any two sides AB, BC will be greater than the other side AC.

Let D be the centre of the sphere; join DA, DB, DC.

The solid angle at D, is contained by three plane angles, ADB, ADC, BDC; and by 20. 11. any two of them ADB, BDC are greater than the third ADC; that is, any two sides AB, BC of the spherical triangle ABC, are greater than the third AC.

PROP. VII. FIG. 4.

THE three sides of a spherical triangle are less than a circle.

Let ABC be a spherical triangle as before, the three sides AB, BC, AC are less than a circle.

Let D be the centre of the sphere: the solid angle at D is contained by three plane angles BDA, BDC, ADC, which together are less than four right angles, (21. 11.) therefore the sides AB, BC, AC together, will be less than four quadrants; that is, less than a circle.

PROP. VIII. FIG. 5.

IN a spherical triangle the greater angle is opposite to the greater side; and conversely.

Let ABC be a spherical triangle, the greater angle A is opposed to the greater side BC.

Let the angle BAD be made equal to the angle B, and then BD, DA will be equal, (5. of this) and therefore AD, DC are equal to BC; but AD, DC are greater than AC, (6. of this), therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as prop. 19. 1. El. Q. E. D.

PROP. IX. FIG. 6.

IN any spherical triangle ABC, if the sum of the sides AB, BC, be greater, equal, or less than a semicircle, the internal angle at the base AC will be greater, equal, or less than the external and opposite BCD; and therefore the sum of the angles A and ACB will be greater, equal, or less than two right angles.

Let AC, AB produced meet in D.

1. If AB, BC be equal to a semicircle, that is, to AD, BC, BD will be equal, that is, (4. of this) the angle D, or the angle A will be equal to the angle BCD.

2. If AB, BC together be greater than a semicircle, that is, greater than ABD, BC will be greater than BD; and therefore (8. of this) the angle D, that is, the angle A, is greater than the angle BCD.

3. In the same manner it is shown, that if AB, BC together be less than a semicircle, the angle A is less than the angle BCD. And since the angles BCD, BCA are equal to two right angles, if the angle A be greater than BCD, A and ACB together will be greater than two right angles. If A be equal to BCD, A and ACB together will be equal to two right angles; and if A be less than BCD, A and ACB will be less than two right angles. Q. E. D.

PROP. X. FIG. 7.

IF the angular points A, B, C of the spherical triangle ABC be the poles of three great circles, these great circles by their intersections will form another triangle FDE, which is called supplemental to the former; that is, the sides FD, DE, EF are the supplements of the measures of the opposite angles C, B, A, of the triangle ABC, and the measures of the angles F, D, E of the triangle FDE, will be the supplements of the sides AC. BC, BA, in the triangle ABC.

Let AB produced meet DE, EF in G, M, and AC meet FD, FE in K, L, and BC meet FD, DE in N, H.

Since A is the pole of FE, and the circle AC passes through A, EF will pass through the pole of AC, (13. 15. 1. Th.) and since AC passes through C, the pole of FD, FD will pass through the pole of AC; therefore the pole of AC is in the point F, in which the arches DF, EF intersect each other. In the same manner D is the pole of BC, and E the pole of AB.

And since F, E are the poles of AL, AM, 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, 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. Q. E. D.

PROP. XI. FIG. 7.

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 supplemental triangle DEF, are (10. of this) equal to three semicircles; but the three sides of the triangle FDE, are (7. of this) less than two semicircles; therefore the measures of the angles A, B, C are greater than a semicircle; and hence the angles A, B, C are greater than two right angles.

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.

PROP. XII. FIG. 8.

IF from any point C, which is not the pole of the great circle ABD, there be drawn arches of great circles CA,

CD, CE, CF, &c. the greatest of these is CA, which passes through H the pole of ABD, and CB the remainder of ACB is the least, and of any others CD, CE, CF, &c. CD, which is nearer to CA, is greater than CE, which is more remote.

Let the common section of the planes of the great circles ACB, ADB be AB: and from C, draw CG perpendicular to AB, which will also be perpendicular to the plane ADB (4. def. 11.); join GD, 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 (7. 3.); and GD which is nearer to GA is greater than GE, which is more remote. 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, the square of CD; and CA is greater than CD, and therefore the arch CA is greater than CD. In the same manner, since GD is greater than GE, and GE than GF, &c. it is shown that CD is greater than CE, and CE than CF, &c. and consequently, the arch CD greater than the arch CE, and the arch CE greater than the arch CF, &c. And since GA is the greatest, and GB the least of all the straight lines drawn from G to the circumference ADB, it is manifest that CA is the greatest, and CB the least of all the straight lines drawn from C to the circumference; and therefore the arch CA is the greatest, and CB the least of all the circles drawn through C, meeting ADB. Q. E. D.

PROP. XIII. FIG. 9.

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 let EC be a great circle passing through E, C. Since A is a right angle, and AE a quadrant, E is the pole of the great cir