PROPOSITION I. THEOREM.

In every spherical triangle, any side is less than the snm 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; these 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 angle is less than the sum of the other two (Book VI. Prop. XIX.); hence any side of the triangle ABC is less than the sum of the other two.

PROPOSITION II. THEOREM.

The shortest path 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 ANB be the arc of a great circle which joins the points A and B; then will it be the shortest path between them.

1st. If two points N and B, be taken on the arc of a great circle, at unequal distances from the point A, the shortest distance from B to A will be greater than the shortest distance from N to A.

NP

D+

B

For, about A as a pole describe a circumference CNP. Now, the line of shortest distance from B to A must cross this circumference at some point as P. But the shortest distance from P to A whether it be the arc of a great circle or any other line, is equal to the shortest distance from N to A; for, by passing the arc of a great circle through P and A, and revolving it about the diameter passing through A, the point P may be made to coincide with N, when the shortest distance from P to A will coincide with the shortest distance from N to A: hence, the shortest distance from B to A, will be greater than the shortest distance from N to A, by the shortest distance from B to P.

If the point B be taken without the arc AN, still making AB greater than AN, it may be proved in a manner entirely similar to the above, that the shortest distance from B to A will be greater than the shortest distance from N to A.

If now, there be a shorter path between the points B and A, than the arc BDA of a great circle, let M be a point of the short

est distance possible; then through M draw MA, MB, arcs of great circles, and take BD equal to BM. By the last theorem, BDA BM+MA; take BD-BM from each, and there will remain ADAM. Now, since BM=BD, the shortest path from B to M is equal to the shortest path from B to D: hence if we suppose two paths from B to A, one passing through M and the other through D, they will have an equal part in each; viz. the part from B to M equal to the part from B to D.

But by hypothesis, the path through M is the shortest path from B to A: hence the shortest path from M to A must be less than the shortest path from D to A, whereas it is greater since the arc MA is greater than DA: hence, no point of the shortest distance between B and A can lie out of the arc of the great circle BDA.

PROPOSITION III. 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 two great circles always bisect each other (Book VIII. Prop. VII. A Cor. 2.). But in the triangle BCD, we have the side BC<BD+CD (Prop I.); add AB+ AC to both; we shall have AB+ AC+BC<ABD+ACD, that is to say,less than a circumference.

PROPOSITION IV. THEOREM

C

B

D

E

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 perimeter of the pentagon ABCDE will be less than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till

C

F

D

B

E

A

they meet in G; we 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 (Book VI.. Prop. XX.); 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 here proved. The

demonstration here given is different from that of Book VI. Prop. XX.; both, however, suppose that the polygon ABCDE is convex, or that no side produced will cut the figure.

PROPOSITION V. THEOREM.

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

Let ED be perpendicular to the great circle AMB; then will E and D be its poles; as also the poles of the parallel small circles HPİ, FNG.

For, DC being perpendicular to the plane A AMB, is perpendicular to all the straight lines CA, CM, CB, &c. drawn through its foot in this plane; hence all the arcs DA, DM, DB, &c. are quarters of the circumference. So likewise 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, they are the poles of that circumference (Def. 7.). Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FNG; hence, it passes through O the centre of the circle FNG (Book VIII. Prop. VII. Cor. 4.); 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 circumference, which for the sake of brevity, is usually named a quadrant: and this quadrant at the same time makes a right angle with the arc AM. For, the line DC being perpendicular to the plane AMC, every plane DME, passing through the line DC is perpendicular to

the plane AMC (Book VI. Prop. XVI.); hence, the angle of these planes, or 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 angles.

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

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 AMB.

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 prolongation, from D as a centre, and with the same distance as before.

In fine, if it be required from a given point P, to let fall a perpendicular on the given arc AM; find a point on the arc AM at a quadrant's distance from the point P, which is done by describing an arc with the point P as a pole, intersecting AM in S: S will be the point required, and is the pole with which a perpendicular to AM may be described passing through the point P.

PROPOSITION VI. THEOREM.

The angle formed by two arcs of great circles, is equal to the angle formed by the tangents of these arcs at their point of intersection, and is measured by the arc described from this point of intersection, as a pole, and limited by the sides, produced if necessary.

Let the angle BAC be formed by the two arcs AB, AC; then will it be equal to the angle FAG formed by the tangents AF, AG, and be measured by the arc DE, described about A as a pole.

For the tangent AF, drawn in the plane of the arc AB, is perpendicular to the radius AO; and the tangent AG, drawn in the plane of the arc AC, is perpendicular to the same radius AO. Hence the angle FAG is equal to the angle contained by the planes ABO, OAC (Book VI. Def. 4.); which is that of H the arcs AB, AC, and is called the angle BAC.

In like manner, if the arcs AD and AE are both quadrants, the lines OD, OE, will be perpendicular to OA, and the angle DOE will still be equal to the angle of the planes AOD, AOE: hence the arc DE is the measure of the angle contained by these planes, or of the angle CAB.

Cor. The angles of spherical triangles may be compared together, by means of the arcs of great circles described from their vertices as poles and included between their sides: hence it is easy to make an angle of this kind equal to a given angle