tact, so that the plane DVP touches the conical surface in the slant side V P. Let the plane PVO cut the plane PQR in the straight line PN, and the conical surface in the second slant side V p'. Then, if PN cuts Vp' in a point P', the section P Q R shall be an ellipse; if PN is parallel to V p', the section shall be a parabola; and if PN cuts Vp produced beyond the vertex in a point P' the section shall be an hyperbola. First, let us suppose that PN is not parallel to V P', and therefore, if produced, cuts it in a point P' below or above the vertex V. Take any two points Q, Q' in the curve P Q R, and through these points draw the planes DVQ, DV Q (IV. 1.) cutting the plane PQR in the straight lines QR, Q'R' and the plane V P P in the straight lines VN, V N' respectively (IV. 2.); also through the point P draw a plane parallel to the base of the cone, and therefore cutting the cone in a circular section P qr (22.), and let the same plane cut the plane PQR in the straight line PH. Then, because the plane DVP touches the conical surface, the straight line PH touches both the circular section Pqr and the curve PQ R, and consequently the diameter POp' of the circle is perpendicular to PH. Let qnr be the projection of QNR on the plane Pqr by straight lines drawn from V, or, which is the same thing, the common section of the plane p q r with the plane DVQ; then, because VD is parallel to the plane of the circle, and likewise to the plane PQR, it is parallel to PH, which is the common section of these two planes (IV. 40. Cor.); but qr is parallel to VD, because it is the common section of a plane passing through VD with the plane pq", to which VD is parallel (IV. 10.): therefore qr is parallel to PH (IV. 6.); and, for the like reasons, QR is likewise parallel to PH, that is (IV. 6.), to qr. Also, because qr is parallel to PH, which touches the circle Pqr in P, it is perpendicular to the diameter P O p' (III. 2. and I. 14.), and is therefore (III.3.) bisected in the point n; wherefore, also, since QR is parallel to q'r, it is bisected in the point N (II. 30.). Through N and N' draw KL and K'L', each of them, parallel to POp. Then, because the triangles V K N, VNL are similar to the triangles Vp'n, VnP respectively (I. 15.) KN: p'n :: VN : Vn (II. 31. and II. 19.), and NL: nPVN Vn, and consequently (II. 37. Cor. 3.), KNX N L is to p' nx n P as V N to V n2, that is, since the triangles V N Q, Vnq are similar, as QN to q n2 (II. 37. Cor. 4.): but p' n x n P is equal to q n2 (III. 20.): therefore, also, K N x N L is equal to Q No (II. 18 Cor.). In the same manner, it may be shown that K'N' x N' L' is equal to Q'N'2. But, because the triangles P' KN, PN Lare similar to the triangles P' K'N', PN'L' respectively (I. 15.) KN: NP'::K'N: N' P', and NL NP :: N'L': N'P (II. 31.), and consequently, KNxN L or Q N2 is to PN× NP' as K'N' × N' L' or Q' N/o is to PN' x N' P (II. 37. Cor. 3.). Therefore, alternando, QN2 is to Q' N/ as PN x NP to PN × N' P'; and consequently, PQR is an ellipse or an hyperbola having the diameter P P' and tangent PH (19.); an ellipse, if PN cuts V P' below the vertex; an hyperbola, if above. And, by a similar construction, if PN be parallel to V p', it may be shown, in the same manner, that K N × NL is equal to Q N2, and K' N' x N' L' to N. But, because K N' is a parallelogram, KN is equal to K'N' (Í. 22.), and, because NL is parallel to N'L', NL is to N' L' as PN to PN' (II. 30. Cor. 2.); therefore, KN NL is to K'N' x N' L' as PN to PN' (II. 35.). Therefore, Q N is to Q'N' as PN to PN; and, consequently, PQR is a parabola having the diameter PN and tangent PH (19.). Therefore, &c. PROP. 25. Every section of a cylinder which is made by a plane parallel to its base, is a circle having its centre in the axis of the cylinder, whether the cylinder be right or oblique. Let A B C, abc be the bases of a cylinder, and Oo its axis, and let PQR be a section made by any plane which is parallel to ABC or abc, and cuts the axis Oo in E. The section P Q R shall be a circle having the centre E. Let P be any point in the curve P QR; join PE; through P draw PA parallel to E O, and, consequently (V. def. 1.), lying in the convex surface of the cylinder, to meet the circumference A B C in A, and join O A. Then, because the parallels PA, EO are intercepted between parallel planes, they are equal to one another (IV. 13.); and, because PA and E O are both equal and parallel, EP is equal to OA (I. 21.), that is, to the radius of the circle A B C. And, in the same manner, it may be shown that the straight line drawn from E to any other point Q of PQR is equal to the same radius. Therefore, the point E is at the same distance from every point of PQR; and, consequently, PQR is a circle having the centre E. Therefore, &c. Cor. The radius of every circular section of a cylinder, which is made by a plane parallel to its base, is equal to the radius of the base. PROP. 26. In an oblique cylinder, if Aa and A' a' are the parallel straight lines in which the surface of the cylinder is cut by a plane passing through the axis Oo perpendicular to the base, and if the cylinder be cut by a plane PQR which is perpendicular to the plane A a a' A', and is inclined to either of the parallel straight lines, A a, at the same angle at which the base is inclined to the other A' a', the section made by the plane PQR shall be a circle having its centre in the axis of the cylinder; or, in other words, every subcontrary section of an oblique cylinder is a circle having its centre in the axis of the cylinder. Let the plane PQR cut the plane Aaa' A' in the straight line P P' (IV. 2.), take any point Q in the curve P QR, and through Q draw the plane DQ D'R parallel to the base of the cylinder (IV. 43.), and let this plane cut the plane Aaa' A' in the straight line DE D' and the plane PQR in the straight line QR; then DQD' is a circle having the centre E (25.). And, because the Flanes D. Q D', PQR are each of them perpendicular to the plane A a a' A', their common section QR is perpendicular to the same plane (IV. 18.), and consequently to the straight lines DD', PP, which meet it in that plane (IV. def. 1.). Now, because, by the supposition, the planes PQR, DQD' are equally inclined to the straight lines A a, A'a' respectively, the angles of inclination N P D, N D'P' are equal to one another; but N D'P' is equal to N D P, because A'a' is parallel to A a (I. 15.); therefore, the angle NPD is likewise equal to N D P, and consequently (I. 6.) the side N P is equal to the side ND. And. for the like reasons, N P' is equal to N D'. Therefore, the rectangle P N x N P' is equal to the rectangle DN x ND'. But, because D E D' is the diameter of the circle D Q D', and is perpendicular to the chord QR at the point N, QR is bisected in N (III. 3.), and Q N2 is equal to DN × ND' (III. 20.). Therefore, Q N2 is equal to PNxN P'. Therefore, because PP' bisects every straight line QR which is drawn perpendicular to it from a point Q of the section PQR, and that the square QN of the half of such straight line is equal to the rectangle PN×N P under the segments of PP, the section PQR is a circle having P P' for its diameter (III. 3. and III. 20.). Also, the middle point of PP' is the centre of the circle. But, because Oo, A a and A'a OA, P'F is equal to FP (II. 29.), that are parallel, and that A'O is equal to is, F is the middle point of PP. Therefore, F is the centre of the circle P Q R. Therefore, &c. 2 Cor. The radius of every subcontrary section of an oblique cylinder is equal to the radius of the base of the cylinder. PROP. 27. Every plane section of a cylinder which is neither parallel to the axis* nor parallel to the base, nor subcontrary, is an ellipse having its centre in the A plane which is parallel to the axis of a cylin der, cuts the convex surface in two straight lines which are parallel to the axis. ABA', nor subcontrary: PQR shall be an ellipse having for its centre the point C in which its plane cuts the axis of the cylinder. Through O draw a plane parallel to the plane PQR (IV. 43.), and let it cut the plane ABA' in the straight line OB; draw the diameter AA' perpendicular to OB, and let the plane AOo cut the convex surface of the cylinder in the parallel straight lines A a, A' a', and the plane P Q R in the straight line PCP': in the curve PQR take any two points Q, Q', and through these points draw the planes KQLR, K'Q'L'R' parallel to the base A B A' (IV. 43.) and cutting the plane PQR in the straight lines QNR, Q'N'R' and the plane Aaa' A' in the straight lines K N L, K'N' L' respectively (IV. 2.). Then, because QR is the common section of two planes which are parallel respectively to the two passing through OB, QR is parallel to OB (IV. 12. Cor.); and, for the like reason, Q'R' is parallel to O B or QR. Also, because K L and A' A are the common sections of parallel planes by the same plane A ad A', K L is parallel to A' A (IV. 12.); and, for the like reason, K' L' is parallel to A'A or K L. But A'OA is at right angles to OB. Therefore, QR is at right angles to KL. and Q'R' is at right angles to K'L' (IV. 15.). And, because the diameter KL of the circle KQLR is at right angles to the chord QR, it bisects Q R in the point N (III. 3.); and, for the like reason, Q'R' is bisected in N'. Therefore (III. 20.) Q N2 is equal to K N x NL, and Q'N' to K'N'x N'L'. But, because the triangles P' KN, PN L are similar to the triangles P' K'N', PN'L' respectively (I. 15.) KN NP':: K'N': N'P', and NL: PN:: N' L: PN' (II. 31.), and, consequently, KNxN L or QN PNXN P:: K'N'x N' L' or Q'N': PN'x N'P' (II. 37. Cor. 3.). Therefore, alternando, QN: Q'N'2 :: PN X N P PN' x N'P'; and consequently (19.), P Q R is an ellipse having the diameter PP', and the tangent at P parallel to O B. Also, because A A' is bisected in O, and that A a, Oo and A' a' are parallel to one another, P P' is bisected in C (II. 29.). Therefore, C is the centre of the ellipse PQR. Therefore, &c. 24 1 23 39. Cor. 2 read 29. Cor. 2 *143 2 25 1 26 26 1 52 28 2 * 46 1 *46 1 * 46 1 * 46 2 33 equal to read is equal to the points D E read the points D, E triangle ABC read triangle abc 3 from bottom, for H C read AC 16 for [10]. Cor. read [11]. Cor. 2 5 and 2 from bottom, for A': B':: A: B read A': B:: A: B' 4 and 3 from bottom, for A read A', and for A' read A 1 from bottom, for B read B', and for B' read B 1 for B' read B, omitting because it is supposed to be greater than B, which is greater than Q 13 of note, for and d read c and d 45 for def. 7 read def. [7] A, B, A', B read A, B, A', B' 10. Cor. read 11. Cor. 2 homologous, and read homolo 33 for altitude GH read altitude CH 16 ABCD read ABCDE 27 omit the reference (II. 12) *150 1 in the figure the line AE is dotted by mistake 30 for 1. 12 read I. 12. Cor. 1. 16-G H, being read G H being 152 2 154 2 157 1 160 1 *160 1 4 - АБ, АC read OB, ОС 4 from bottom, for FL read EL 160 1 in the lower figure, for u read U 167 2 19 for right angle read right-angled 21 - 1 Cor. 2 read 2 Cor. 2 171 1 172 2 12 from bottom, for pyramid read cone 13 for 5. Cor. 1 read 7. Cor. 1 126- pyramid read cone 182 1 181 1 AC read Ac #182 1 -1.5 read I. 6 75 1 19 D. (I. 48 read D. (I. 48) 56 read 56. Cor. read Gʻ 7 for 1 read 1. Cor. 1 78 1 2 78 2 in the figure of the Scholium, for C' 1 in the figure of the Scholium, for the C nearest to A read c 82 1 83 99 1 96 7 for A B C D E F read A, B, C, D, E, F 39 -31 read 31. Cor. 1 is greater read is, greater 1 from bottom, for impracticable read impracticable in the way of calcu lation Page 93, col. 2, after Book III. prop. 28, add, Cor. In an isosceles triangle, which has each of the equal angles double of the vertical angle, the sides and base are in extreme and mean proportion; and conversely. In such a triangle, each of the equal angles is four-fifths of a right angle (I. 19.) As we are enabled, therefore, to describe such a triangle by II. 59, we can thus divide a right angle into five equal parts, as was observed in the scholium of p. 26. And generally, if a regular polygon of n sides can be inscribed in a circle, (as, in the present instance, the regular polygon of five sides,) a right angle may be divided into n equal parts, by taking for one of those parts (I. 46. Cor.) a fourth of the angle which the side of the polygon subtends at the centre of the circle. (I. 3. Cor.) INDE X. The theorems and problems of Plane Geometry will be found under the heads Straight Line, Angle, Triangle, Square, Rhombus, Rectangle, Parallelogram, Quadrilateral, Rectilinea! Figure, Circle; those of Solid Geometry under Plane, Dihedral Angle, Solid Angle, Tetrahedron, Cube, Rectangular Parallelopiped, Parallelopiped, Prism, Pyramid, Polyhedron, Regular Polyhedron, Cylinder, Cone, Sphere, Solid of Revolution; of Spherical Geometry under that head; of Ratios and Proportion under those respective heads; and so of Projection and the Conic Sections. The parts included in unciæ are additions; having been made, either with the view of supplying such connecting links as seemed wanting in the present digest of the whole work, as in "Circle" (E) and (G); or of completing what had been left imperfect, as in the notes on "Proportion " and "Rectangle;" or of extending and generalising where only partial views had been given, as under the heads "similar," symmetrical," "touch;" or of adding whatever of use or interest had been inadvertently omitted, as in" Annulus," "Lunes," and the note on "Centrolinead.” 20 Algebra, its signs +, -, X, &c., borrowed Alternate, certain angles said to be, which Analysis, (Gr., undoing, or taking to pieces,) in geometry, is that mode of demonstrating a theorem, or solving a problem, which searches into the thing proposed, and takes it (as it were) to pieces, in order to discover the more simple truths and constructions upon which it is built: the reverse process is called Synthesis, (Gr, putting together,) and proceeds in a didactic form, by the putting together of truths and constructions already established, to do, or establish the certainty of the thing proposed 107 Angle, dihedral. See Dihedral Angle." Angle, rectilineal def. 1 When said to be right, oblique, acute, def.2 obtuse (f) To bisect a given angle See "Straight Line." 26 Angle, solid. (See "Solid Angle.") def. 125 sch. 97, 98 (See "Circle.") Are (Lat., a bow) of a circle. |