In the triangle CID right-angled, suppose at 1, when ID is greater and Ic less than a quadrant, the other side, or hypothenuse, CD is greater than a quadrant (322): and as the angle DIC increases, viz. when it becomes equal to DIG, the side DC increases till it becomes equal to DG; hence the first part of the proposition is evident.

In the second part, the supplemental triangle (304), will have exactly the same properties as the first part of this proposition.

PROPOSITION XIX.

(341) In any right-angled spherical triangle ABC, if the sides be produced, viz. AC to 1, AB to H, and BC to D, so that CI, CD, and AH be quadrants, or each 90°; and if from the point ▲, as a pole, the great circle HGFE be described,

and from the point c, as a pole, IDE be drawn to intersect HGFE in E; then shall the triangles CGF and EDF have their respective sides and angles either equal to those of the triangle ABC, or they will be complements of each other; and eight right-angled spherical triangles will be formed, having (every two of them) equal angles at their bases.

E

D

Since A is the pole of the circle HGFE, AG and AH are quadrants and perpendiculars to FGH; and the arcs FGH, FCB, being perpendicular to ABH, are quadrants, and F is the pole of ABH. Therefore the triangle CGF is right-angled at G; CG is the complement of AC, CF of BC; BH, the complement of AB, is the measure of the angle F; and GH, the complement of FG, is the measure of the angle A.

The arcs EFG, EDI, being each perpendicular to ACGI, are quadrants, and E is the pole of ACGI; and the arcs CFD and CGI being quadrants, for c is the pole of IDE, are perpendicular to IDE. Therefore the triangle EDF is right-angled at D; ED is the complement of ID, and ID measures the angle ACB equal to DCI (295); DF is equal to BC; the hypothenuse EF is equal to the angle A, for it is equal to GH the measure of A, and for the same reason FED is equal to AC, and EFD is equal to the complement of AB.

The right-angled triangles ABC, AHG, have the angle a common; FGC, FHB have the angle F common; CGF, CID have the angle c common; and EDF, EIG have the angle E common.

CHAP. II.

THE STEREOGRAPHIC PROJECTION OF THE SPHERE.

(342) The stereographic projection of the sphere, is such a representation of the various parts of its surface, on the plane of one of its great circles, as would be formed by lines drawn from the pole of that great circle, to every point of the circle, to be projected, viz.

Conceive a point E (Plate IV. Fig. 8.) situated any where on the surface of a globe, and a plane CD to stand at right angles to an imaginary line EO, connecting the centre of the globe and the point E. Then, if an indefinite number of lines be supposed to be drawn from the point E, to every point of the circumference of any circle described upon the spherical surface, they will trace out upon the cutting plane CD a stereographic projection of the sphere.

(343) The plane on which the points, lines, and circles of the sphere are represented, is called the plane of projection; and the point from which all the lines are drawn through the several parts of the circles of the sphere, to this plane, is called the projecting point.

pro

(344) The primitive circle is situated in the plane of jection; and the projecting point, on the sphere, is one of the poles of this circle; but on the plane of projection, the projected poles of the primitive are in its centre.

(345) A circle, the plane of which is parallel to the plane of the primitive, is called a parallel circle, and is represented on the plane of projection by a circle parallel to, and comprehended within, the primitive.

(346) A circle, whose plane is perpendicular to the plane of the primitive, is called a right circle; because, passing through the projecting point, its circumference becomes a straight line on the plane of projection.

(347) A circle whose plane is oblique to the plane of the primitive, is called an oblique circle.

(348) Lines drawn from the projecting point to every part of the circumference of a circle to be projected, will form the convex surface of a cone, whose vertex is the projecting point. Thus if AB be a diameter of a circle to be projected, and E the projecting point, then AEB will be a cone, of which E is the

vertex.

(349) All writers on conic sections have demonstrated, that if a cone be cut by a plane ab parallel to the base AB, the

section will be a circle; hence it follows, that all small circles parallel to the primitive will be projected into circles. And the radii of all projected small circles, parallel to the primitive, will be equal to the semi-tangent of their distances from the remoter pole; thus ao=ob is the tangent of the angle aEO, or the semi-tangent of the arc ae where e is the remoter pole.

(350) It is also shown by writers on conic sections, that if a scalene, or oblique cone, be cut by a plane not parallel to the base, but in such a manner that the cone cut from the base towards the vertex be equiangular with the original cone, the section shall be a circle.

Thus (Plate IV. Fig. 9.) if the cone ABE be so cut by a plane CD, that the angle CDE be equal to the angle BAE, and DCE equal to ABE, then the section will be a circle, having CD for its diameter. And this is what is termed cutting a cone in a subcontrary position.

Emerson's Conic Sections, Book I. Prop. 89th.

(351) Every circle of the sphere which does not pass through the poles of the primitive, is projected into a circle.

Let DC be the diameter of a circle to be projected on the plane of the primitive FB, from the point E. Lines from the point E to the circumference of that circle form a cone, whose triangular section is CED.

Now the extremity c, of the diameter CD, will be projected into the point A, and the other extremity D into the point B in the plane of the primitive. Through D draw od parallel to FB, that is, perpendicular to Ee, then will the arc Ed be equal to the arc ED; and since an angle at the circumference of a circle is measured by half the arc on which it stands (20 Euclid III), the angle ECD will be equal to the angle dDE, or, which is the same thing, equal to the angle ABE, because AB is parallel to do.

Hence the cone ECD is equiangular with the cone EBA, and it is cut by the plane of projection in a subcontrary position, therefore (350) the section is a circle.

(352) COROLLARY. The centre, and consequently the poles of every circle oblique to the primitive, have their projections in the line of common section of the primitive, and a great circle perpendicular both to the primitive and to the oblique circle.

For the centre and poles of the circle whose diameter is CD, have their projections in the line FB, the common section

of the primitive, and the great circle FCDE which is perpendicular to the primitive and to the circle whose diameter is CD.

PROPOSITION II. (Plate IV. Fig. 11.)

(353) The distance of the centre of any projected oblique great circle from the centre of the primitive, is equal to the tangent, and the radius of the projected circle is equal to the secant of the inclination of the oblique circle and primitive.

Let E be the projecting point, FG the diameter of the primitive EFeG, and CD the diameter of a circle to be projected. The point c will appear at A, and D at B; therefore AB is the projected diameter. Bisect AB in p, and p will be the centre of the projected circle eAEB.

Now CED CEB, being an angle in a semicircle, is a right angle; and EPA and EPB are likewise right angles; hence the triangles AEB, APE, and EPB are equiangular and similar. The angle epc is double of the angle eEc, the one being at the centre and the other at the circumference of the circle, and for the same reason ape is double of ABE; but the angle eEc has been shown to be equal to ABE, therefore ePC PPE, and CPF-PEP: and it is plain that pp is the tangent of PEP to the radius of the sphere, therefore it is equal to the tangent of the angle CPF, the angle of inclination of the oblique circle and primitive.

It is also obvious that Ap=Ep, the radius of the projected circle, is the secant of the angle PEP or CPF.

(354) COROLLARY. The distance (AP) of the extremity of the diameter of any projected circle from the centre of the primitive, is the semi-tangent of the complement of (CF) the circle's inclination to the primitive.

For AP is the tangent of the angle AEP, which is the half of cre, the complement of FPC.

PROPOSITION III. (Plate IV. Fig. 11.)

(355) The distances of the projected poles of any oblique great circle from the centre of the primitive, are equal to the tangent and cotangent of half the inclination of the oblique circle and primitive.

Let E be the projecting point, FG the diameter of the primitive, and CD the diameter of the circle to be projected.

Make eм=CF, then CM will be a quadrant; draw MPN, which will be the axis of CD; and M, N, its poles. If EM be joined, the pole м will appear at m in the primitive; but Pm is the tangent of the angle PEM, which is the half of epm, or of its equal CPF, the angle of inclination.

Likewise a line drawn from E through N, the other pole, will cut PF, produced beyond F, in the exterior projected pole. Now PF, produced to meet EN, is evidently the cotangent of the angle PEM, or half of (CPF) the angle of inclination.

PROPOSITION IV. (Plate IV. Fig. 12.)

(356) The distance of the centre of a projected small circle (perpendicular to the primitive) from the centre of the primitive, is equal to the secant of its distance from its own pole; and the radius thereof is equal to the tangent of that distance.

Let CD be the diameter of a small circle cutting the primitive at right angles, or, which amounts to the same thing, whose poles G and F lie in the circumference of the plane of projection. Now AB is the diameter of its representative on the plane of projection FG; bisect AB in p, and draw pc, cp, and CB; then the radius pc of the projected small circle is the tangent of the arc GC, the distance of the small circle from its pole G, and Pp is equal to the secant of the same arc.

For since the triangles Ece, BPe, are right-angled at c and P, and have the angle e common, the remaining angle PEA will be equal to the remaining angle CBP-PCB, for pc=pB; but the angle PCA is equal to PEA, because PC is equal to PE, hence PCA PCB, therefore the angle ACB is equal to the angle Pcp, but ACB is a right angle, therefore Pcp is a right angle; and consequently pc is the tangent of the arc GC, and Pp its secant.

PROPOSITION V. (Plate IV. Fig. 13.)

(357) The angle comprehended between two great circles, passing through the same point, on the surface of the sphere, is equal to the angle comprehended between their representatives on the plane of projection.

Let BGK be any spherical angle. Through the angular point G and the centre P, draw the plane of a great circle GDE, perpendicular to the plane PHL of projection. Let a plane GHL touch the sphere in G; then since the plane of the circle GDE is perpendicular to this plane and to the plane of projection, it is perpendicular to their intersection HL. The angles made by great circles at any point on the surface of a sphere are equal to the angles made by their tangents (299); therefore in the plane GHL draw the tangents GH, GF, GL to the arcs GB, GD, GK, and their projections are the lines gн, gF, gL: the angle HGL is equal to HgL. For the angle PGF= a right angle PgE+PEg; therefore taking away the equal angles PGE and PEG, we have the angle gGF=PgE=GgF, and consequently gr=GF. Hence in the right-angled triangles