about that angle proportional, are similar. Cor. 1. Since the triangles ABC and DBE are similar C 203°. Theorem.-Two transversals to a system of parallels are divided similarly by the parallels. AA' is to BB' is || to CC', etc. Then AD and A'D' are divided similarly. P B' Proof. Consider three of the ||s, AA', .. (81°, 1) But A'QC' is a triangle and PB' is || to QC'. (194°, 3). A'P A'B' Cor. 2. Applying Cor. 1 of 202°, Def.-A set of three or more lines meeting in a point is a pencil and the lines are rays. The point is the vertex or centre of the pencil. (199°) q.e.d. BC CD Cor. 1. Let the transversals meet in O, and let L denote any other transversal through O. Then AD, A'D', and L are all divided similarly by the parallels. But the parallels are transversals to the pencil. .. parallel transversals divide the rays of a pencil similarly. = = =etc. B 204°. Theorem.-The rectangle on any two sides of a triangle is equal to twice the rectangle on the circumradius (97°, Def.) and the altitude to the third side. BD is to AC and BE is a diameter. LA=LE, Proof and LADB=2ECB= (106°, Cor. 1) E Proof-Draw AE making LAED =LABC. Then, since LBCA=LBDA, the triangles EDA and BCA are similar. BC. AD AC.DE. D Cor. Denoting BD by p and the circumradius by R, ac=2pR, and multiplying by b, and remembering that pb=2▲ (153°, 2), we obtain Rabc 4A' which (with 175°, Ex. 1) gives the means of calculating the circumradius of a triangle when its three sides are given. 205°. Theorem.-In a concyclic quadrangle the rectangle on the diagonals is equal to the sum of the rectangles on the sides taken in opposite pairs. AC.BD=AB. CD + BC. AD. q.e.d. D E B A Again, since LAEB is supp. to LAED, and CDA is supp. to LABC, therefore triangles BEA and CDA are similar, and AB. CD=AC.EB. Adding these results, AB. CD+BC. AD=AC. BD. 206°. Def.-Two rectilinear figures are similar when they can be divided into the same number of triangles similar in pairs and similarly placed. Thus the pentagons X and Y can be divided into c' the same number of triangles. (Y) If then AP ≈ ▲P', ΔΩΣ ΔΩʹ, ΔR = ΔR', and the triangles are similarly placed, the pentagons are similar. E D (X) R B R The triangles are similarly placed if LEAD corresponds to E'A'D', LAED to A'E'D', ¿DAC to D'A'C', etc. This requires that the angles A, B, C, etc., of one figure shall be respectively equal to the angles A', B', C', etc., of the other figure. Hence when two rectilinear figures are similar, their angles taken in the same order are respectively equal, and the sides about equal angles taken in the same order are proportional. Line-segments, such as AD and A'D', which hold similar relations to the two figures are similar or homologous lines of the figures.. 207°. Theorem.--Two similar rectilinear figures have any two line-segments from the one proportional to the homologous segments from the other. Proof. By definition APP', and they are similarly placed, .. AE: A'E' AD: A'D'. For like reasons, AD: A'D' AC: A'C' = AB : A'B'. AE AD AC AB A'E' A'D' A'C' A'B' and the same can be shown for any other sets of homologous line-segments. = etc., = = = Cor. 1. All regular polygons of the same species are similar figures. ...9 ···9 Now, let a, b, c, d', b', ć, be homologous sides of two similar regular polygons, and let r and r' be their circumradii. 'Then and r' are homologous, r = ہے ... = a+b+c+... a+b+c+... perimeter of P perimeter of P’ But at the limit (148°) the polygon becomes its circumcircle. .. the circumferences of any two circles are proportional to their radii, Cor. 2. If c, d denote the circumferences of two circles and ☛ and ' their radii, r Denote this constant by 27, then C=2πr. с == = constant. (195°, 3) L It is shown by processes beyond the scope of this work that stands for an incommensurable numerical quantity, the approximate value of which is 3.1415926... Cor. 3. Since equal arcs subtend equal angles at the centre (102°, Cor. 2), if s denotes the length of any arc of a circle whose radius is, the tensor varies directly as s varies, and also varies directly as the angle at the centre varies. S r Hence is taken as the measure of the angle, subtended r by the arc, at the centre. Denote this angle by 0. Then 0 ==/ and when s=r, 0 becomes the unit angle. .. the unit angle is the angle subtended at the centre by an arc equal in length to the radius. This unit is called a radian, and the measure of an angle in radians is called its radian measure. Radian measure will be indicated by the mark ^. с Cor. 4. When s= =a semicircle, 0=π. 2 π But a semicircle subtends a straight angle at the centre. .. ☛ is the radian measure of a straight angle and of a Now a straight angle contains 180°, π"= 180°. 2 (41°) Hence I^=57°.29578..., and and these multipliers serve to change the expression of a given angle from radians to degrees or from degrees to radians. Cor. 5. Since the area of a circle is equal to one-half that of the rectangle on its radius and a segment equal in length to its circumference, (149°) (Cor. 2) O=cr=1.2πr.r .. the area of a O is π times that of the square on its radius. 208°. Theorem.-The bisectors of the vertical angle of a triangle each divides the base into parts which are proportional to the conterminous sides. BD and BD' are bisectors of LB. Then AD_AD'_AB E or B C EBE' =(45°), and LE=LABD=2DBC. BC=EC=CE'. (88°, 3) But ABD and ABD' are triangles having EE' || to the common base AB. AB AD (203°, Cor.) AD AD' AB g.e.d. DC CD' BC Cor. D and D' divide the base internally and externally in = Proof-Through C draw EE' to AB. Then = AB AD' CE' CD" = |