THEOREM LXXIII. In any triangle a straight line parallel to the base has to the base the ratio which the part of either side intercepted between this line and the vertex has to the whole side. Let A B C be the triangle and DE a line parallel to BC; then shall DE: BC=AD:AB=AE: AC. Through E draw E F parallel to A B; Any two parallels are cut proportionally by a series of secants drawn through the same point; and the ratio of any two corresponding parts is equal to the ratio of the parts of any secant contained between the point and the parallels. If A D, a d be two parallels, and OA, OB, OC, OD secants drawn from the same point, O, situated either between or without the two parallels; then ab: AB = bc: BC=cd: CD (I.) For in the triangles OA B, O a b by Theorem LXXVI. and in the triangles O B C, Obc, we have Ob:OB = bc: BC = Oc: OC. с B C D In like manner, from the triangles O CD, Ocd, we obtain and so forth. Oc: OC cd: CD, Any two of these proportions have a common ratio. Consequently all the ratios are equal, or ab: AB = bc: BC = cd: CD. By the same demonstration it may be shown that THEOREM LXXV. If two parallelograms are equal in area, and have an angle of the one equal to an angle of the other, then the sides which contain the angle of the first are the extremes of a proportion of which the sides containing the equal angle of the second are the means. The converse is also true. Let A B and B C be equal parallelograms, having E B Then AB 4DBF = 4GBE; then DB: BE=GB:BF. Let DB and B E be placed in the same straight line; then FB, BG are in one straight line. Complete parallelogram F E. FE=DB:BE, and BC: FE=GB: BF; but since A B = BC, AB: FE =BC: FE; therefore DB: BE = GB: BF. Conversely, let DB: BEC B: B F. Because A B:FE=DB: BE, and BC: FE=GB:BF; therefore A B: FE=BC: FE; hence A B = BC. Cor.-Read "triangle" for "parallelogram" and the proof will be the same. THEOREM LXXVI. If four straight lines be proportionals the rectangle contained by the extremes is equal to the rectangle contained by the means. Conversely, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals. Since rectangles are equiangular parallelograms, the proof in this case is identical with that of the preceding theorem. THEOREM LXXVII. The bisector of the vertical angle of a triangle divides the base into parts proportional to the sides. The converse is also true. Let A B C be the triangle, and let B D be the bisector of the angle B; then shall AD: DC = AB: BC. E Produce A B to E, and make B E = BC. Then the exterior angle ABC = the two interior angles, BCE, BEC. But since BE BC, the angles BCE, BEC are equal. Therefore BCE D Hence, the alternate angles being equal, BD is parallel to C E, and AD: DC ABBE = AB: BC. Conversely, let AB: BC= AD: DC; then shall ▲ ABD = 4 CBD. The same construction being made, because B C = B E, therefore A B: BEAD: DC. Therefore B D is parallel to E C. Hence ABD = LE; DBC = 4 BCE; therefore ▲ A B D = 4 C B D. THEOREM LXXVIII. If the bisector of the angle exterior to the vertical angle of a triangle be continued to meet the base produced, the parts intercepted between the bisector and the extremities of the base are proportional to the sides. The con verse is also true. Let A B C be the triangle and BD the bisector of the angle exterior to ABC; then AD: AC AB: BC. Through but = C draw C E parallel to DB; then 4 CEB = DB F and 4 ECB = 4 CBD; D therefore ▲ BEC = 4 BCE; therefore A B:BC=AD:DC. Conversely, if AB: BC=AD: DC, then CBD = 4 DBF. THEOREM LXXIX. In equal circles, angles at the centres or circumferences have to one another the same ratio as the arcs on which they stand. Let A M B, B M C be angles at the centre, M, on arcs AB, BC; then shall H ZAMB: 4 B M C = arc A B : arc BC. If the arcs are commensurable, and contain their G.C.M. m and ʼn times respectively, by dividing AB and BC into equal parts and joining the points A M B will be divided into m equal angles, BMC into n; therefore of division, and B ▲AMB:4B MC :: m : n, and arc A B : arc B C :: m : n. The theorem may be extended to incommensurables precisely as in Theorem LXXII. or by Theorem LXVIII. SECTION III.-Similar Figures. Definition. When two figures are so related that one of them is an enlarged or reduced copy of the other, they are said to be similar. Figures 162 and 163 represent pairs of figures having this relationship. Although the two figures forming a pair are unequal in area, we at once recognise their similarity in all other respects. On further examination it will be found that all the angles of one are respectively equal to the corresponding angles of the other, and that the sides of one are proportional to the corresponding sides of the other. If, for instance, a line of the larger figure is three times the corresponding line of the smaller, every line in the larger is three times the corresponding line in the smaller. From the fact that the angles of one are respectively equal to those of the other, it will at once be seen that when the figures are placed so that a side of the one is parallel to the corresponding side of the other, and directed the same way from the vertices of the equal angles, then every pair of corresponding lines will be parallel. It is very frequently necessary to draw figures having this similarity. A map, for example, is a figure similar to that of the country which it represents. Two maps of the same country are similar figures. Plans of the same estate or of the same building are similar to one another, and also to the estate or building. To draw a figure similar to a given figure, two things have to be secured. First.-The angles of the new figure must be respectively equal to the angles of the given figure. This may be effected by drawing the sides of the former parallel to those of the latter. Secondly. The corresponding sides of the two figures must have a constant ratio. There are several ways of securing this : 1. We may use the reduction scale or proportion-compasses, as described in the preceding section. If the button of the compasses be screwed so as to divide the whole length of the instrument into two |