right angles to the plane AEBF, is at right angles to the common section of the two planes, it is (Def. 2. 2. Sup.) also at right angles to the plane AEBF, and therefore (Cor. 1. 2.) C is the pole of the circle AEBF; and if CG be produced to D, D is the other pole of the circle AEBF.

A

F

G

B

E

In the same manner, by drawing GE in the plane AEBF, perpendicular to AB, and producing it to F, it has shewn that E and F are the poles of the circle ACBD. Therefore, the poles of each of these circles are in the circumference of the other.

Again, If C be one of the poles of the circle AEBF, the great circle ACBD which passes through C, is at right angles to the circle AEBF. For, CG being drawn from the pole to the centre of the circle AEBF is at right angles (Cor. 1. 2.) to the plane of that circle; and therefore, every plane passing through CG (17. 2. Sup.) is at right angles to the plane AEBF; now, the plane ACBD passes through CG. Therefore, &c. Q. E. D.

COR. 1. If of two great circles, the first passes through the poles of the second, the second also passes through the poles of the first. For, if the first passes through the poles of the second, the plane of the first must be at right angles to the plane of the second, by the second part of this proposition; and therefore, by the first part of it, the circumference of each passes through the poles of the other.

COR. 2. All great circles that have a common diameter have their poles in the circumference of a circle, the plane of which is perpendicular to that diameter.

PROP. V.

In isosceles spherical triangles the angles at the base are equal.

Let ABC be a spherical triangle, having the side AB equal to the side 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 to the plane AEG, which D

A

E

B

passes through AE, EG (4. 2. Sup.); and therefore, every planet that passes through DE is at right angles to the plane AEG (17. 2. Sup.); wherefore, the plane DBC is at right angles to the plane AEG. For the same reason, the plane DBC is at right angles to the plane AFG, and therefore AG, the common section of the planes AFG, AEG is at right angles (18. 2. Sup.) to the plane DBC, and the angles AGE, AGF are consequently right angles.

But since 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 side AD common, therefore the other sides are equal, viz. AE to AF, (26. 1.), and DE to DF. Again, be cause 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 square of AG, the remaining squares of GE and GF are equal, and GE is therefore equal to GF. Wherefore, in the triangles AFG, AEG, the side GF is equal to the side GE, and AF has been proved to be equal to AE, and the base AG is common, therefore, the angle AFG is equal to the angle AEG (8. 1.). But the angle AFG is the angle which the plane ADC makes with the plane DBC (4. def. 2. Sup.) because FA and FG, which are drawn in these planes, are at right angles to DF, the common section of the planes. The angle AFG (3. def.) is therefore equal to the spherical angle ACB; and, for the same reason, the angle AEG is equal to the spherical angle ABC. But the angles AFG, AEG are equal. Therefore the spherical angles ACB, ABC are also equal. Q. E. D.

PROP. 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

D

G

E

B

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, (3. def.), 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.

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 DE and DA; 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 common, and also AF equal to AE, the angle ADF is equal to the angle ADE; therefore also the arches AC and AB, which are the measures of the angles ADF and ADE, are equal to one another; and the triangle ABC is isosceles. Q. E. D.

PROP. 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.

The solid angle at D is contained by three plane angles ADB, ADC, BDC; any two of which, ADB, BDC are greater (20. 2. Sup.) than the third ADC; and therefore any two of the arches AB, AC, BC, which measure these angles, as AB and BC, must also be greater than the third AC. Q. E. D.

D

PROP. VIII.

A

The three sides of a spherical triangle are less than the circumference of a great circle.

Let ABC be a spherical triangle as before, the three sides AB, BC, AC are less than the circumference of a great 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. 2. Sup.) therefore the sides AB, BC, AC, which are the measures of these angles, are together less than four quadrants described with the radius AD, that is, than the circumference of a great circle. Q. E. D.

1

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, (6.), and therefore AD, DC are equal to BC; but AD, DC are greater than AC (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 Prop. 19. 1. Elem. Q. E. D.

PROP. X.

B

A

D

"

According as the sum of two of the sides of a spherical triangle is greater than a semicircle, equal to it, or less, each of the interior angles at the base is greater than the exterior and opposite angle at the base, equal to it, or less; and also 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 any of the two sides as AB, and the base AC, till they meet again in D; then, the arch 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 (5.) the angle D, or the angle A will be equal to the angle BCD, that is, the interior angle at the base equal to the exterior and opposite.

A

B

2. If AB, BC together be greater than a semicircle, that is greater than ABD, BC will be greater than BD; and therefore (9.), the angle D, that is the angle A, is greater than the angle BCD.

3. In the same manner it is shewn, if AB, BC together be less than

a semicircle, that the angle A is less than the angle BCD,

Now, 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. 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 arches 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, viz. 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 circle AC passes through A, EF will pass through the pole of AC (1. Cor. 4.) 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.

F

N

M

B

K

D

G

E

H

And since F, E are the poles of AL, AM, the arches FL and EM (2.) 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, (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 and BH together, that is, NH and 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 shewn that the measures of the angles DEF, EFD are the supplements of the sides AB, AC, in the triangle ABC, Q. E. D.