(B. VI., P. IV.): hence, the point D is the pole of the arc AM.

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. For, by turning the arc DF about 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 an arc of a great circle.

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

Let the angle BAC be formed by the two arcs AB, AC: then is it equal to the angle FAG formed by the tangents AF, AG, and is measured by the arc DE of a great circle, 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 ABDH, ACEH (B. VI., D. 4); which is that of the arcs AB, AC. Now, if the arcs AD and AE are both quadrants, the

lines OD, OE, are perpendicular to OA, and the angle DOE is equal to the angle of the planes ABDH, ACEH: hence, the arc DE is the measure of the angle contained by these planes, or of the angle CAB; which was to be proved.

Cor. 1. The angles of spherical triangles may be compared by means of the arcs of great circles described from their vertices as poles, and included between their sides.

A spherical angle can always be constructed equal to a given spherical angle.

Cor. 2. Vertical angles, such as ACP and BCN, are equal; for either of them is the angle formed by the two planes ACB, PCN. When two arcs ACB, PCN, intersect, the sum of two adjacent angles, as ACP, PCB, is equal to two right angles.

P

N

B

PROPOSITION V. THEOREM.

If from the vertices of the angles of a spherical triangle, as poles, arcs be described forming a second spherical triangle, the vertices of the angles of this second triangle are respectively poles of the sides of the first.

From the vertices A, B, C, as poles, let the arcs EF, FD, DE, be described, forming the triangle DFE : then are the vertices D, E, and F, respectively poles of the sides BC, AC, AB.

M/

Ο

K

For, the point A being the

pole of the arc EF, the distance AE is a quadrant; the point C being the pole of the arc DE, the distance CE is likewise a quadrant: hence, the point E is at a quadrant's distance from the points A and C: hence, it is the pole of the arc AC (P. III., C. 4). It may be shown, in like manner, that D is the pole of the arc BC, and F that of the arc AB; which was to be proved.

Cor. The triangle ABC, may be described by means of DEF, as DEF is described by means of ABC. Triangles so related that any vertex of either is the pole of the side opposite it in the other, are called polar triangles.

PROPOSITION VI. THEOREM.

Any angle, in one of two polar triangles, is measured by a semi-circumference, minus the side lying opposite to it in the other triangle.

Let ABC, and EFD, be any two polar triangles on a sphere whose centre is O: then is any angle in either triangle measured by a semi-circumference, minus the side lying opposite to it in the other triangle.

For, produce the sides AB,

AC, if necessary, till they meet

EF in G and H. The point A

being the pole of the angle A is that arc (P. IV.).

the arc GH, measured by But, since E

is the pole of AH, the arc EH

is a quadrant; and since F is

the pole of AG, FG is a quad

M

rant: hence, the sum of the arcs EH and GF is equal to

a semi-circumference.

But, the sum of the arcs EH and

GF is equal to the sum of the arcs EF and GH: hence, the arc GH, which measures the angle A, is equal to a semi-circumference minus the arc EF. In like manner, it may be shown, that any other angle, in either triangle, is measured by a semi-circumference minus the side lying opposite to it in the other triangle; which was to be proved

Cor. 1. Beside the triangle DEF, three other triangles, polar to ABC, may be formed by the intersection of the arcs DE, EF, DF, prolonged. But the proposition is applicable only to the central triangle, ABC, which is distinguished from the three others by the circumstance, that the vertices A

F

and D lie on the same side of BC; B and E, on the same side of AC; C and F, on the same side of AB. The polar triangles ABC and DEF are called supplemental triangles, any part of either being the supplement of the part opposite it in the other.

Cor. 2. Arcs of great circles, drawn from corresponding vertices of two supplemental polar triangles perpendicular to the respective sides opposite, are supplements of each other. For, from A draw the arc of a great circle, AN, perpendicular to BC; it must, when prolonged, pass through D, the pole of BC, and

B

N

E

P

must also, when prolonged to P, be perpendicular to EF (P. III., C. 3): DN and AP being quadrants (P. III. C. 1), DP and AN are supplements of each other.

If from the vertices of any two angles of a spherical triangle, as poles, arcs of circles are described passing through the vertex of the third angle; and if from the second point in which these arcs intersect, arcs of great circles are drawn to the vertices, used as poles, the parts of the triangle thus formed are equal to those of the given triangle, each to each.

Let ABC be a spherical triangle situated on a sphere whose centre is O, CED and CFD arcs of circles described about B and A as poles, and let DA and DB be arcs of great circles: then are the parts of the triangle ABD equal to those of the given triangle ABC, each to each. For, by construction, the side

B

D

AD is equal to AC, the side BD is equal to BC, and the side AB is common: hence, the sides are equal, each to each. Draw the radii OA, OB, OC, and OD. The radii OA, OB, and OC, form the edges of a triedral angle whose vertex is 0; and the radii OA, OB, and OD, form the edges of a second triedral angle whose vertex is also at O; and the plane angles formed by these edges are equal, each to each: hence, the planes of the equal angles are equally inclined to each other (B. VI., P. XXI.). But, the angles made by these planes are equal to the corresponding spherical angles; consequently, the angle BAD is equal to BAC, the angle ABD to ABC, and the angle ADB to ACB: hence, the parts of the triangle ABD are equal to the parts of the triangle ACB, each to each; which was to be proved.