Prop. 11. To set off any number of degrees upon a right circle, as DE. See the figure to proposition 10. Draw CA perpendicular to DE, and make the angle HCK equal the degrees given; make CF equal to the radius AE ; take FG, the nearest distance, and set it off from A to B, then AB will be equal to E the degrees in the angle HCK as proposed. A Another Method. On ED describe the semicircle END, then, by Prop. 11, set off NP equal to the degrees given, draw PL perpendicular to ED, then AL will contain the degrees required. Prop. 12. To measure an arc of an ellipsis, or to set off any number of degrees upon it. About AR, the transverse axis of the B elipsis, describe a circle ABR; erect the R perpendiculars BED, KFI, on AR; then BK is the measure of EF, or EF is the representation of the arc BK, which may be measured, or any degrees set off upon it, as in Prop. 9. These problems are all evident from the three first propositions, and need no other demonstration. If the sphere be projected on any plane, parallel to the primitive circle, the projection will be the very same, for being effected by parallel lines, which are always at the same distance, the same figure o: representation will be invariably formed. K SECTION II. The Stereographic Projection of the Sphere Prop. I. Any circle passing through the projecting point, is projected into a right line. For, all lines, drawn from the projecting point to this c'rele, pass through its intersection with the plane of projection, which is a night line. Cor. 1. A great circle passing through the poles of the primitive circle, is projected into a right line passing through the centre. Cor. 2. Any circle passing throngh the projecting point, is projected into a right line perpendicular to the line of measures, and distant from the celltre, the semi-tangent of its nearest distance from the pole opposite to the pro. jecting point. Prop. 2. Every circle that does not pass through the projecting point, is projected into a circle. Case 1. Let the circle EF be parallel to the E priinitive BD, lines drawn to all points of it, from the projecting point A, will form a conic surface, which, being cut parallel to the base, B H by the plane BD, the section GH, into which EF is projected, will be a circle ; by the conic sections. Case 2. Let BH be the line of mea. EM sures to the circle EF ; draw FK parallel to BD, then will the arc AK equal AF, and therefore the angle AFK or AHG B is equal to AEF; wherefore, in the triangles AEF and AGH, the angles at E and H are equal, and the angle A common ; therefore, the angles at F and G are equal. Whence the cone of rays AEF, the base of which EF is a circle, is cut by a sub-contrary section, by the plane of projection BD; and, therefore, by the conic sections, the section GH, which is the projection of the circle EF, will also be a circle. Cor. When AF is equal to AG, the circle EF is projected into a circle equal t, itself. For then the similar triangles AHG and AEF will also be equal, and GH equal to EF. Prop. 3. Any point on the surface of the sphere is projected into a point, distant from the centre, the semi-tangent of its distance from the pole opposite to the projecting point. Thus the point E, in the last figure, is projected into G, and Finto H; and CG is the semi-tangent of EM, and CH of MF. Cor. 1. A great circular perpendicular to the primitive, is projected into a line of semi-tangents passing throagh the centre, and produced ad infinitum. For MF is projected into CH its semi-tangent, and FM into the semi. tangent CG. Cor. 2. Any arc EM of a great circle, perpendicular to the primitive, is projected into its semi-tangent. Thus, EM is projected into GC. Cor. 3. Any arc EMF of a great circle is projected into the sum of its semitangents, of its greatest and Icast distances from the opposite pole M, if it lie on both sides of M; or the difference of the semi-tangents when all on one side. Prop. 4. The angle made by two circles on the surface of the sphere, is equal to that made by their representations upon the plane of projection. Let the angle BPK be projected; through the angular point P and the B H centre C draw the plane of a great circle PED, perpendicular to the plane of projection EFG. Let a plane D PHG touch the sphere in P; then, since the circle EPD is perpendicular G both to this plane and to the plane of projection, it is perpendicular to their A. intersection GH. But the angles made by circles are the same as those made by their tangents; therefore, in the plane PGH, draw the tangents PH, PF, PG, to the arcs PB, PD, PK, and they will be projected into the lines pH, pF, PK; then the 2 CPA will = 2 HpG. For the angle CPF is a right angle = CPA + CAP; therefore, taking away the equal angles CPA and CAP, and Z PPE = CpA or PpF; consequently, pF = PF. Therefore, in the right-angled triangles PFG and PFG there are two sides equal, and the included angle is a right angle; therefore, the hypothenuse PG=PG; and, for the same reason, in the right-angled triangles PFH and PFH, PH=pH. Lastly, in the triangles PHG and pHG, all the sides are respectively equal ; and, therefore ZP=lp. Q. E. D. Cor. The angle made by two circles on the sphere, is equal to the angle made by the radii of their projection at the point of intersection. For the angle made by two circles on a plave, is the same with that made by their radii drawn to the point of intersection. Prop. 5. The centre of a lesser circle projected, perpendicular to the primitive, is in the line of measures, distant from the centre of the primitive, the secant of the lesser circle's distance from its own pole, and its radius is the tangent of that distance. Let A be the projecting point, EF the circle to be projected, GH the E projected diameter. From the centre C and D draw CF, DF; and the tri K BD JH angles CFI, DFI, are right-angled at I, then ZIFC= _ FCA=2Z FEA, or 2 Z FEG = 2 Z FHG = 2 FDG. Therefore, iFC+IFD=FDG +IFD equal a right-augle; that is, CFD is a right-angle, and the line CD is the secant of BF, and the radius FD is the tangent of it. Q. E, D. Cor. If these circles be actually described, it is plain the radius FD is a tangent to the primitive at F, where the lesser circle cuts it. Prop. 6. The centre of projection of a great circle is in the line of measures, distant from the centre of the primitive the tangent of its inclination to the primitive ; and its radius is the secant of its inclination. C Let A be the projecting point, B EF the great circle, GH the projected diameter, D the centre, Iraw DA. The angle EAF being in a semi-circle is a right-angle. In the right-angled triangle GAH, T G AC is perpendicular to GH; therefore, żGAC = AHC and their F double, ECB = ADC, and their complements ECI=CAD. There. fore, CD is the tangent of ECI, and the radius AD is its secant. Q. E. D. Cor. 1. If the great obliqne circle AGBHbe actually described upon the primitive AB, all great circles passing through (), will liave the centre of their projection in the line RS, drawn through the centre D perpendicular to the line of measures IH. For, since all great 1 circles cut one another at the distance of a semicircle, all circles passing through G must cnt at the opposite point H; and therefore their centres must be in the line RDS. Cor. 2. Hence, also, if any oblique circle GLH be required to make any IS given angle with another circle BGAH, it will be projected the same way, with regard to GAH considered as a primitive, and KS its line of measnres, as the circle BGA is on the primitive BIA, and line of measures ID; and, therefore, the tangent of the angle AGL, to the radius GD, set off from D to N, gives the centre of GL. For the <NGD will then be equal to AGL, by Cor. ?, Prop. 4; and, therefore, GLH is truly projecteil. N.B. Of all great circles in the projection the primitive is the least; for the radins of any obliqne great circle being the secant of the inclination, is greater than the radius of the primitive; as the secant is always greater than the primitive. Therefore, every obliqne greicine in the projection is greater than the primitive. Prop. 7. The projected extremities of the diameter of any circle (see the first figure of Prop. 6), are in the line of measures, distant from the centre of the primitive circle the semi-tangent of its nearest and greatest distances from the pole of projection opposite to the projocting point. For the diameter of the circle EF is projected into GH from the projecting point E IC IC circle cuts the line of measnres, within and without the primitive, is distant from the centre of the primitive the tangent and co-tangeot of half the complement of the inclination of the circle to the primitive. For CG is egnal to the tangent of half EB, or of half the complement of IE the inclioation, and because the < EAF is a right aogle, CH is the co-langent of GAC, or half EB. Prop. 8. The projected poles of any circle are in the line of measures within and without the primitive, and distant from its centre the tangent and co-tangent of half the inclination to the primitive. For the poles Pp of the circle EF, are E G Cor. 1. The pole of the primitive is its cen. Cor. 2. The projected centre of any circle is always between the projected pole nearest to it on the sphere and the centre of the primitive ; and the projected ceptres of all circles lie between the projected poles. For the middle point of EF, or its centre, is projected into S; and all the points in Pp, in which are all the centres, are projected into Dd. Cor. 3 If P be the projected centre of any circle EFG, any right line EG, FH passing through P will intercept equal arcs EF, GH. For in any circle of the sphere, any two TA lines passing through the centre intercept G equal arcs, and these are projected into ECP 19 right lines passing through the projected centre P, and therefore EF, GH, represent equal arcs. Prop. 9. If EFGH, efgh (see the last figure), represent two equal circles, whereof EFG is as far distant from its pole P, as efg is from the projecting point, any two right lines, eEP and fFP being drawn ihrough P, will intercept equal arcs in the representations of these circles ; if P falls within the circles on the same side ; but on the contrary side if without; that is, EF=ef, and GH=gh. For, by the nature of the section of a sphere, any two circles passing H through two given points, or poles, h on the surface of the sphere, will inter Pa G cept equal arcs of two other circles 91 EC cqui-castant from these poles. Therefore, the circles EFG and efg on the |