COR. 2.-Only one great circle can pass through the same two points, in the surface of the sphere, that are not diametrically opposite.

For the plane of the great circle must pass through these two points and through the centre of the sphere (Sp. Ge. Def. 3), and only one plane can do so (So. Ge. I. 2, Cor. 2). COR. 3.-Any two sides of a spherical triangle being pro

duced, intersect again at the distance of a semicircle. COR. 4.-The two poles of any circle, its centre, and the centre of the sphere, are always in the same straight line, and that straight line is perpendicular to the plane of the circle. COR. 5.-And, therefore, if a line or plane be perpendicular to a circle of the sphere, and pass through one of these points, it will pass through the other three. Or, if it pass through two of them, it will be perpendicular to the circle, and also pass through the remaining two. COR. 6.-Hence, two great circles, whose planes are perpendicular, pass through each other's poles; and conversely.

COR. 7.-And, if one great circle pass through a pole of another, the latter will pass through the poles of the former.

COR. 8.-All parallel circles have the same axis and the same poles; and conversely.

Schol. It appears by the fourth and fifth corollaries, that, of these five conditions of passing through the two poles of a circle of the sphere, through the centre of the circle, through the centre of the sphere, and of being perpendicular to the plane of the circle-if a straight line or a plane fulfil any two, it will also satisfy the other three.

PROPOSITION II.

Each pole of any circle of the sphere is equally distant on the surface from every point in its circumference.

Let ABCD be any circle of the sphere, whose centre is F, and axis GFH. Its poles G, H, are each of them equally distant from its circumference.

For, let the sphere be cut by planes passing through G, H,

D

D

F

G

C

B

E

and let the sections which will be great circles, because the line GH passes through the centre of the sphere-meet ABCD in the lines of common section FA, FB, FC. A Then GA, GB, GC, being joined, the right-angled triangles GFA, GFB, GFC, have the sides FA, FB, FC, equal, because they are radii of the same circle, and one side GF common to all; and the angles at F are right angles (Sp. Ge. I. Cor. 4); therefore (Pl. Ge. I. 4) the hypothenuses GA, GB, GC, and consequently the arcs which they subtend, are likewise equal. COR. 1.-The pole of a great circle is at the distance of a quadrant from its circumference.

H

COR. 2.-Hence any plane passing through the centre of the sphere, divides it into two equal parts, which are therefore called hemispheres.

COR. 3.-If a point in the surface of the sphere be at the distance of a quadrant from other two points not diametrically opposite, it will be the pole of the great circle passing through them.

For only one great circle can pass through these two points, and its pole is distant from them by a quadrant. (Sp. Ge. I. Cor. 2, and II. Cor. 1.)

COR. 4.-The radius of a small circle is the sine of its distance from either pole to the radius of the sphere, or the cosine of its distance from the parallel great circle. COR. 5.-Hence, those small circles, whose planes are equally distant from the centre, are equal; and conversely; and of two circles unequally distant, that which is nearer the centre is the greater; and conversely. COR. 6.-Parallel circles intercept equal arcs on those great circles which pass through their poles.

PROPOSITION III.

The intercepted arc of a great circle, whose pole is the angular point, is the measure of a spherical angle.

Let ABC be a spherical angle, of which the angular point

B is the pole of the great circle ACD. Then is the intercepted arc AC the measure of ABC.

M

N

B

For, let the tangents MB, NB, and the radii of the sphere EA, EB, EC, be drawn. The angle MBN is the same with the spherical angle ABC, for the tangents are perpendicular to BE (So. Ge. I. Def. 4, and Sp. Ge. Def. 7); but MBN is equal to AEC; A because, since AB, BC, are quadrants, and BEA, BEC, right angles, MB, BN, are parallel to AE, EC. Wherefore the spherical angle ABC

E

F

P

Q

is equal to AEC, the measure of which is the arc AC to the radius of the sphere.

COR. 1.-The circumferences of two great circles cut each other at right angles, when their planes are perpendicular; and conversely.

COR. 2.-At the point of intersection of two great circles, the opposite angles are equal, the two adjacent angles are together equal to two right angles, and each angle is equal to its opposite one, at the other point of intersection (Sp. Ge. Def. 7).

COR. 3.-The distance of the adjacent poles of two great circles is the measure of their inclination, or of the spherical angle.

For, since AB, BC, pass through the poles of ACD, ACD passes through the poles of AB, BC (Sp. Ge. I. Cor. 7); let P be the pole of AB, and Q the adjacent pole of BC; then AP, CQ, are quadrants, and ACPQ.

COR. 4.-The intercepted arc of any circle, whose pole is the angular point, is the measure of the spherical angle to the radius of that circle.

COR. 5.-Two great circles, which pass through the poles of parallel circles, intercept similar arcs.

COR. 6. A spherical angle is equal to the inclination of two tangents, to the arcs containing it, drawn from the angular point.

PROPOSITION IV.

If a plane be perpendicular to the diameter of a sphere, at one of its extremities, it touches the sphere.

A

Let the plane CD be perpendicular to AB, the diameter of the sphere ABG, at its extremity B; then CD touches the sphere in that point.

For, let F be any other point in CD, and EF, FB, be drawn ; the angle EBF is, by hypothesis, a right angle. Hence EF is greater than EB, and consequently F a point without the sphere. Thus the plane CD

C

A

G

B

F

D

meets the sphere only in the point B, and therefore touches it. COR. 1.-A sphere and a plane can touch one another only in one point.

COR. 2.-If a plane touch a sphere, the radius at the point of contact is perpendicular to it.

COR. 3.-If a plane touch a sphere, a perpendicular to it, at the point of contact, passes through the centre. COR. 4. If a plane touch a sphere, its line of common section, with the plane of any circle of the sphere passing through the point of contact, is a tangent to that circle.

For this line of common section is in the plane of the circle, and it touches the circle.

COR. 5.-A tangent, to any circle of the sphere, is the common tangent of all the circles in whose plane it is.

PROPOSITION 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 perpen

1

diculars 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.

A

F

Because DE is at right angles to each of the straight lines AE, EG, it is at right angles to the plane AEG, which passes through AE, EG (So. Ge. I. 4); and, therefore, every plane that passes through DE is at right angles to the plane AEG (So. Ge. I. 17); 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 (So. Ge. I. 18) to the plane DBC, and the angles AGE, AGF, are consequently right angles.

D

E

G

But, since the arc AB is equal to the arc 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 (Pl. Ge. I. 26), 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 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 (Pl. Ge. I. 8). But the angle AFG is the angle which the plane ADC makes with the plane DBC (So. Ge. I. Def. 4), 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 (Sp. Ge. Def. 7) 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