equal. Therefore the spherical angles ACB, ABC, are also equal.

PROPOSITION VI.

If the angles at the base of a spherical triangle be equal, the triangle is isosceles.

Let ABC be a spherical triangle having the angles ABC, ACB, equal to one another; the sides AC and AB are also 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.

Then, it may be proved, as was done in the last proposition, that AG is at right angles to the plane BCD, and that therefore the angles AGF, AGE, are right angles, and also that the angles AFG, AEG, are equal to the angles which the planes DAC, DAB, make with the plane DBC. But because the spherical angles ACB, ABC, are equal, the angles which the planes DAC, DAB, make with the plane DBC, are equal (Sp. Ge. Def. 7), and therefore the angles AFG, AEG, are also equal. The triangles AGE, AGF, have therefore two angles of the one equal to two angles of the other, and they have also the side AG common; wherefore they are equal, and the side AF is equal to the side AE.

C

Again, because the triangles ADF, ADE, are right angled at F and E, the squares of DF and FA are equal to the square of DA, that is, to the squares of A DE and EA; now, the square of AF is equal to the square of AE, therefore the square of DF is equal to the square of DE, and the side DF to the side DE. Therefore in the triangles DAF, DAE, because DF is equal to DE, and DA D common, and also AF equal to AE, the angle ADF is equal to the angle ADE; therefore, also, the arcs AC and AB, which are the measures of the angles ADF and ADE, are equal to one another; and the triangle ABC is isosceles.

E

G

B

PROPOSITION VII.

Any two sides of a spherical triangle are greater than the third.

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

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

D

The solid angle at D is contained by three plane angles ADB, ADC, BDC (So. Ge. II. 1); any two of which 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.

PROPOSITION VIII.

A

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 (So. Ge. II. 2); therefore the sides AB, BC, AC, together, will be less than four quadrants, that is, less than a circle.

PROPOSITION IX.

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 (Sp. Ge. 6), and therefore AD, DC, are equal to BC; but AD, DC, are greater than AC 3

D

A.

(Sp. Ge. 7); therefore BC is greater than AC, that is, the greater angle A is opposite to the greater side BC. The converse is demonstrated as in Pl. Ge. I. 19.

PROPOSITION X.

In a spherical triangle, according as the sum of two of the sides is greater than a semicircle, equal to it, or less, the interior angle at the base is greater than the exterior and opposite angle at the base, equal to it, or less; and the sum of the two interior angles at the base greater than two right angles, equal to two right angles, or less than two right angles.

Let ABC be a spherical triangle, of which the sides are AB and BC; produce the side AB and the base AC till they meet again in D; then, the arc ABD is a semicircle, and the spherical angles at A and D are equal, because each of them is the inclination of the circle ABD to the circle ACD.

1. If AB, BC, be equal to a semicircle, that is, to AD, BC will be equal to BD, and therefore (Sp. Ge. 5) 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, A greater than ABD, BC will be

B

greater than BD; and therefore (Sp. Ge. 9) the angle D, that is, the angle A, is greater than the angle BCD.

3. In the same manner, it is shown if AB, BC, together be less than a semicircle, that 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 that BCD, A and ACB will be less than two right angles.

PROPOSITION XI.

If the angular points of a spherical triangle be made the poles of three great circles, these three circles by their intersections will form a triangle which is said to be supplemental to the former; and the two triangles are such, that the sides of the one are the supplements of the arcs which measure the angles of the other.

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.

N

F

M

B

A

C

L

Since A is the pole of FE, and the 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 pass through the pole of AC; therefore the pole of AC is in the point F, in which the arcs DF, EF, intersect each other. same manner, D is the pole of BC, and E the pole of AB.

D

G

H

In the

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 triangles.

PROPOSITION XII.

The three angles of a spherical triangle are greater than two right angles, and less than six right angles.

F

N

M

B

The measures of the angles A, B, C, in the triangle ABC, together with the three sides of the supplemental triangle DEF, are (Sp. Ge. 11) equal to three semicircles; 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 K a semicircle; and hence the angles A, B, C, are greater than two right angles.

D

A

H

HE

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 I 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 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 (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