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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

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the plane AMC (Book VI. Prop. XVI.); hence, the angle of these planes, or the angle AMD, is a right angle.

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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 A 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.

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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.

Scholium. Vertical angles, such as ACO and BCN are equal; for either of them is still the angle formed by the two planes ACB, OCN.

It is farther evident, that, in the intersection of two arcs ACB, OCN, the two adjacent angles ACO, OCB, taken together, are equal to two right angles.

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PROPOSITION VII. THEOREM.

From the vertices A, B, C, as poles, let the arcs EF, FD, ED, be described, forming on the surface of the sphere, the triangle DFE; then will the points D, E, and F, be respectively poles of the sides BC, E

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AC, AB.

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

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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 removed the length of a quadrant from each of the points A and C; hence, it is the pole of the arc AC (Prop. V. Cor. 3.). It might be shown, by the same method, that D is the pole of the arc BC, and F that of the arc AB.

Cor. Hence the triangle ABC may be described by means of DEF, as DEF is described by means of ABC. Triangles so described are called polar triangles, or supplemental tri angles.

PROPOSITION VIII. THEOREM.

The same supposition continuing as in the last Proposition, euch angle in one of the triangles, will be measured by a semicir cumference, minus the side lying opposite to it in the other triangle.

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For, produce the sides AB, AC, if necessary, till they meet EF, in G and H. The point A being the pole of the arc GH, the angle A will be measured by that arc (Prop. VI.). But the arc EH is a quadrant, and likewise GF, E being the pole of AH, and F of AG; hence EH+GF is equal to a semicircumference. Now, EH+ GF is the same as EF+GH; hence the arc GH, which measures the angle A, is equal to a semicircumference minus the side EF. In like manner, the angle B will be measured by circ.-DF: the angle C, by circ.-DE.

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And this property must be reciprocal in the two triangles, since each of them is described in a similar manner by means of the other. Thus we shall find the angles D, E, F, of the triangle DEF to be measured respectively by circ.-BC, circ.—AC,

circ.-AB. Thus the angle D, for example, is measured by the arc MI; but MI+BC=MC+BI={ circ.; hence the arc MI, the measure of D, is equal to circ.-BC: and so of all the rest.

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Scholium. It must further be observed, that besides the triangle DEF, three others might be formed by the intersection of the three arcs DE, EF, DF. But the proposition immediately before us is applicable only to the central triangle, which is distinguished from the other three by the circumstance (see the last figure) that the two ar gles A and D lie on the same side of B the two B and E on the same side of AC, and the two C F on the same side of AB.

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PROPOSITION IX. THEOREM

If around the vertices of the two angles of a given spherical triangle, as poles, the circumferences of two circles be described which shall pass through the third angle of the triangle; if then, through the other point in which these circumferences intersect and the two first angles of the triangle, the arcs of great circles be drawn, the triangle thus formed will have all its parts equal to those of the given triangle.

Let ABC be the given triangle, CED, DFC, the arcs described about A and B as poles; then will the triangle ADB have all its parts equal to those of ABC.

For, by construction, the side AD= AC, DB=BC, and AB is common; hence these two triangles have their sides equal, each to each. We are now to show, that the angles opposite these equal sides are also equal.

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If the centre of the sphere is supposed to be at O, a solid angle may be conceived as formed at O by the three plane angles AOB, AOC, BOC; likewise another solid angle may be conceived as formed by the three plane angles AOB, AOD, BOD. And because the sides of the triangle ABC are equal to those of the triangle ADB, the plane angles forming the one of these solid angles, must be equal to the plane angles forming the other, each to each. But in that case we have shown that the planes, in which the equal angles lie, are equally inclined to each other (Book VI. Prop. XXI.); hence all the angles of the spherical triangle DAB are respectively equal to those of the triangle CAB, namely, DAB=BAC, DBA=ABC, and ADB ACB; hence the sides and the angles of the triangle ADB are equal to the sides and the angles of the triangle AC3.

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Scholium. The equality of these triangles is not, however, an absolute equality, or one of superposition; for it would be impossible to apply them to each other exactly, unless they were isosceles. The equality meant here is what we have already named an equality by symmetry; therefore we shall call the triangles ACB, ADB, symmet. al triangles.

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