292. THEOREM [AC.7].—In equal circles the arcs of sectors are proportional to their angles. In Fig. 427, 0 and P are the centres of the two equal circles ABC, DEF: Cons.-Assuming that arcs AB, DE are commensurable there is some arc L which is contained an exact number of times in each of them. Suppose AB and DE divided respectively into m and n parts of length L. Join 0 to the points of division of AB, and P to the points of division of DE. Then AOB is divided into m angles, and DPE into n angles. Proof. In equal circles ABC, DEF, which AB and DE are divided are all equal, .. the angles into which ▲ AOB and divided are all equal. Let each of these angles be 0. the arcs into me, and ▲ DPE = n0, m n mL, and arc DE m n LAOB LDPE LDPE are AC. 6c. EXERCISES CXII (Riders). 1. If the vertical angle of a triangle is bisected both internally and externally, the bisectors divide the base harmonically. 2. Two intersecting chords in a circle are such that the segments of one are in the same ratio as the segments of the other. Prove that the line bisecting the angle formed by corresponding segments passes through the centre of the circle. 3. AB is a diameter of a circle and CD is a chord at right angles to it. E is any point on CD and AE, BE meet the circle in F, G. Prove that any two adjacent sides of the quadrilateral CFDG are in the same ratio as the other two. 4. ABCD is a quadrilateral and P is any point. Prove that the centroids of the triangles PAB, PBC, PCD, PDA form a parallelogram. 5. OA, OB are any two straight lines intersecting in O, and R and S are any two points on OB, OA respectively; BN parallel to AR meets OA in N, and AM parallel to BS meets OB in M. Prove that MN is parallel to RS. (Pascal.) 6. P is a point outside a circle, and chords PAB, PCD are drawn to meet the circle in A, B, C, D. PEF bisects APC, and meets the chord AC in E, and the chord BD in F; prove that BF: FD = CE : EA. 7. Trisect a given straight line by means of Theorem T.12. 8. Draw through a point between two given convergent straight lines a straight line terminated by these lines and divided at the point in a given ratio. 9. Find a point on a given line (or circle) whose distances from two given points are in a given ratio. 10. Determine a point whose distances from three given points are in given ratios. 11. Construct a triangle given one side, an angle adjacent to it, and the ratio of the other two sides. 12. Construct a triangle given one side, the angle opposite to it, and the ratio of the other two sides. 13. Construct a triangle given one side, the ratio of the other two sides and the area. 14. Construct a triangle given the base, the altitude and the ratio of the other two sides. CHAPTER XX. SIMILAR FIGURES. 293. The student is already familiar with the idea of diagrams "drawn to scale," such as a map of a town, a plan of a garden, or a "reduced" copy of a geometrical figure. The logical treatment of the properties of such diagrams is known as the theory of similar figures. EXERCISES CXIII (Practical). = 1.4", BC 1. Draw two triangles abc, ABC, such that be ba find e such that be = BA. On Make the necessary measurements, and evaluate carefully each of the following ratios (as a decimal): : What inferences are suggested by your results? 2. Draw a convex polygon HKLMN in which HN = KL = 4 cms., LM = MN = 6 cms., HK = 8 cms., LH = LK = 90°. Take any point O [a convenient figure will be obtained if OM is, say, nearly perpendicular to, and nearly bisected by, HK]. On OH, OK, OL, OM, ON take points h, k, l, m, n respectively, such that Oh: OH = Ok: OK 01: OL= = = 1:3. What do you notice about the polygon hklmn? Take a few suitable measurements to verify your statement. 3. In the figure of question 2, produce HO to h', KO to k', LO to l', making Oh' Oh, Ok' = Ók, Ol' = Ol, = What do you notice about the polygon h'k'l'm'n'? 294. DEFINITION.-Two triangles (or two polygons) are said to be similar if their corresponding angles are equal and if also the sides joining corresponding vertices are in a constant ratio. This constant ratio is called the ratio of similitude of the two triangles (or polygons). We shall use the symbol to denote that two figures are similar. Thus 66 A ABCDEF" means A ABC is similar to ▲ DEF." 66 295. THEOREM [S.1].—If two triangles are equiangular they are similar. In Fig. 428: Given that LA = La, LB = b, and .:. ▲ C Cons. FIG. 428. From BA mark off BH ba, and from BC mark off BK = bc. Join HK. Proof.-In As HBK, abc, ·· BH = ba, BK = Similarly by marking off lengths along CA and CB equal to ca and cb respectively it can be proved that COROLLARY 1.-If two parallels are cut by two intersecting transversals the two triangles so formed are similar. Thus in Figs. 434, 435, PQ, pq are the parallels and AOa, BOb are the intersecting transversals. Then it is easily proved that as OAB, Oab are equiangular and therefore similar. COROLLARY 2.-If the three sides of one triangle are respectively parallel (or respectively perpendicular) to the three sides of another then the triangles are similar. Suppose that the sides AB, BC, CA of ABC are respectively parallel (or respectively perpendicular) to the sides ab, bc, ca of ▲ abc. It is easy to prove that, since the arms of LA are respectively parallel (or respectively perpendicular) to the arms of a; therefore SA and a are either equal or supplementary. Similarly for Ls B and b, and for s C and c. Now LA Lα + LB + Lb + LC + c = 4R. T.1. (i) It is not possible for ▲ s A, B, C to be respectively supplementary to Ls a, b, c; for this would give LA+La+ LB + Lb + LC + Lc = 6 R. (ii) It is not possible for two of the angles A, B, C to be respectively supplementary to two of the angles a, b, c, and the remaining angles to be equal, for then LA + La + LB + Lb + LC + Lc would necessarily be greater than 4R. (iii) Hence at least two of the angles A, B, C must be respectively equal to two of the angles a, b, c. But in this case (by T.1) the third angles are also equal. Thus the two triangles are equiangular, and therefore similar. NOTE.-Theorem S.1 may obviously be stated as follows: If two triangles have two angles in the one triangle respectively equal to two angles in the other triangle, then the two triangles shall be similar. Compare this statement with Theorem C.2, § 83. DEFINITION.-Two triangles which are equiangular are said to be of the same species. Thus a triangle is given "in species" if we know the magnitude of its angles (or of two of its angles). SIM. GEO. EE |