PROPOSITION I. THEOREM.

Every section of a sphere, made by a plane, is a circle.

Let AMB be a section, made by a plane, in the sphere, whose centre is C. From the point C, draw CO perpendicu- A lar to the plane AMB; and different lines CM, CM, to different points of the curve AMB, which terminates the section.

The oblique lines CM, CM, CA, are equal, being radii of the sphere; hence (Prop. VI. B. I. El. S. Geom.) they are

D

B

equally distant from the perpendicular CO; therefore all the lines OM, MO, OB, are equal. Consequently, the section AMB is a circle, whose centre is O.

Cor. 1. If the section passes through the centre of the sphere, its radius will be the radius of the sphere; hence, all great circles are equal.

Cor. 2. Two great circles always bisect each other; for their common intersection, passing through the centre, is a diameter.

Cor. 3. Every great circle divides the sphere and its surface into two equal parts; for, if the two hemispheres were separated, and afterwards placed on the common base, with their convexities turned the same way, the two surfaces would exactly coincide, no point of the one being nearer the centre than any point of the other.

Cor. 4. The centre of a small circle, and that of the sphere, are in the same straight line, perpendicular to the plane of the small circle.

Cor. 5. Small circles are the less the further they lie from the centre of the sphere; for, the greater CO is, the less is the chord AB, the diameter of the small circle AMB.

Cor. 6. An arc of a great circle may always be made to pass through any two given points of the surface of the sphere; for the two given points, and the centre of the sphere, make three points, which determine the position of a plane. But if the two given points were at the extremities of a diameter, these two points and the centre would then lie in one straight line, and an infinite number of great circles might be made to pass through the two given points.

PROPOSITION II. THEOREM.

In every spherical triangle, any side is less than the sum of the other two.

Let O be the centre of the sphere, and ACB the triangle: draw the radii OA, OB, OC. Imagine the planes AOB, AOC, COB, to be drawn; those planes will form a solid angle at the centre O; and the angles AOB, AOC, COB, will be measured by AB, AC, BC, the sides of the spherical triangle. But each of the three plane angles forming a solid

A

B

angle is less than the sum of the other two (Prop. XXI., B. I. El. S. Geom.): hence any side of the triangle ABC is less than the sum of the other two.

PROPOSITION III. THEOREM.

The shortest distance from one point to another, on the surface of a sphere, is the arc of the great circle which joins the two given points.

Let ADB be the arc of the great circle which joins the points A and B; and without this line, if possible, let M be a point of the shortest path between A and B. Through the point M, draw MA, MB, arcs of great circles; and take BD=MB.

M

Α

B

D

By the last Proposition, the arc ADB is shorter than AM+MB; take BD=BM respectively from both; there will remain AD AM. Now, the distance of B from M, whether it be the same with the arc BM, or with any other line, is equal to the distance of B from D; for, by making the plane of the great circle BM to revolve about the diameter which passes through B, the point M may be brought into the position of the point D; and the shortest line between M and B, whatever it may be, will then be identical with that between D and B: hence the two paths from A to B, one passing through M, the other through D, have an equal part in each, the part from M to B equal to the part from D to B. The first part is the shorter by hypothesis ; hence the distance

from A to M must be shorter than the distance from A to D, which is absurd; the arc AM being proved greater than AD. Hence no point of the shortest line from A to B can lie out of the arc ADB; hence this arc is itself the shortest distance between its two extremities.

PROPOSITION IV. THEOREM.

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

Let ABC be any spherical triangle; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be semicircumferences, since (Prop. I. Cor. 2.) two great circles always bisect each other. But in the triangle BCD, we have (Prop. II.) the side BC<BD+CD; add AB+AC to both; we shall have AB+AC+BC<ABD+ACD:that is to say, less than a circumference.

E

C

D

B

PROPOSITION V. THEOREM.

The sum of all the sides of any spherical polygon is less than the circumference of a great circle.

Take the pentagon ABCDE, for example. Produce the sides AB, DC, till they meet in F; then, since BC is less than BF÷CF, the peri- G meter of the pentagon ABCDE will be less than that of the quadrilateral AEDF. Again produce the sides AE, FD, till they meet in G; we

C

F

D

B

shall have ED<EG+DG; hence the perimeter of the quadrilateral AEDF is less than that of the triangle AFG; which last is itself less than the circumference of a great circle: hence, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference.

Scholium. This proposition is fundamentally the same as Prop. XXII. B. I. El. S. Geom. ; for O being the centre of the

sphere, a solid angle may be conceived as formed at O, by the plane angles AOB, BOC, COD, &c.; and the sum of these angles must be less than four right angles; which is exactly the proposition we have been engaged with. The demonstration here given is different from that of Prop. XXII. B. I. El. S. Geom.; both, however, suppose that the polygon ABCDE is convex, or that no side produced will cut the figure.

PROPOSITION VI. THEOREM.

The poles of a great circle of the sphere, are the extremities of that diameter of the sphere which is perpendicular to this circle; and these extremities are also the poles of all small circles parallel to it.

wise are all the arcs EA, EM, EB, &c. ; hence the points D and E are each equally distant from all the points of the circumference AMB; hence (Def. 5.) they are the poles of that circumference.

Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence (Prop. I. Cor. 4.) it passes through O the centre of the circle FNG; hence, if the oblique lines DF, DN, DG be drawn, these oblique lines will diverge equally from the perpendicular DO, and will themselves be equal. But, the chords being equal, the arcs are equal; hence the point D is the pole of the small circle FNG; and, for like reasons, the point E is the other pole.

Cor. 1. Every arc DM, drawn from a point in the arc of a great circle AMB to its pole, is a quarter of the circumfer

1

ence, which, for the sake of brevity, is usually named a quadrans, or quadrant: and this quadrant at the same time makes a right angle with the arc AM. For (Prop. XVII. B. I. El. S. Geom.) the line DC being perpendicular to the plane AMC, every plane DMC passing through the line DC, is perpendicular to the plane AMC; hence the angle of these planes, or (Def. 6.) the angle AMD, is a right angle.

Cor. 2. To find the pole of a given arc AM, draw the indefinite arc MD perpendicular to AM; take MD equal to a quadrant; the point D will be one of the poles of the arc AM: or thus, at the two points A and M, draw the arcs AD and MD perpendicular to AM; their point of intersection, D, will be the pole required.

Cor. 3. Conversely, if the distance of the point D from each of the points A and M, is equal to a quadrant, the point D will be the pole of the arc AM, and also the angles DAM, AMD, will be right.

For, let C be the centre of the sphere, and draw the radii CA, CD, CM. Since the angles ACD, MCD are right, the line CD is perpendicular to the two straight lines CA, CM; hence it is perpendicular to their plane (Prop. V. B. I. El. S. Geom.); hence the point D is the pole of the arc AM; and consequently the angles DAM, AMD are right.

Scholium. The properties of these poles enable us to describe arcs of a circle on the surface of a sphere, with the same facility as on a plane surface. It is evident, for instance, that by turning the arc DF, or any other line extending to the same distance, round the point D, the extremity F will describe the small circle FNG; and, by turning the quadrant DFA round the point D, its extremity A will describe the arc of the great circle AM.

If the arc AM were required to be produced, and nothing were given but the points A and M through which it was to pass, we should first have to determine the pole D, by the intersection of two arcs described from the points A and M as centres, with a distance equal to a quadrant. The pole D being found, we might describe the arc AM and its prolonga. tion, from D as a centre, and with the same distance as before.

In fine, if it is required from a given point P to let fall a perpendicular on the given arc AM, produce this arc to S, till the distance PS be equal to a quadrant; then, from the pole S, and with the same distance, describe the arc PM, which will be the perpendicular required.