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With an understanding of corresponding angles and corresponding sides, we can define similar polygons. When pairs of consecutive and corresponding vertices are associated, the corresponding sides are included between the corresponding angles or vertices. AB 4 HJ, BC 4 JK, CD 4 KL, and AD 4 HL 쮿 Geometry in Nature The segments of the chambered nautilus are similar not congruent in shape. © Joao VirissimoShutterstock Two polygons are similar if and only if two conditions are satisfied: 1. All pairs of corresponding angles are congruent.

2. All pairs of corresponding sides are proportional. DEFINITION

The second condition for similarity requires that the following extended proportion exists for the sides of the similar quadrilaterals of Example 1. Note that both conditions for similarity are necessary Although condition 1 is satisfied for square EFGH and rectangle RSTU [see Figures 5.6a and b], the figures are not similar—that is, one is not an enlargement of the other—because the extended propor- tion is not true. On the other hand, condition 2 is satisfied for square EFGH and rhom- bus WXYZ [see Figures 5.6a and 5.6c], but the figures are not similar because the pairs of corresponding angles are not congruent. AB HJ = BC JK = CD KL = AD HL G H a F E Figure 5.6 U b R S Z W Y c EXAMPLE 2 Which figures must be similar? a Any two isosceles triangles c Any two rectangles b Any two regular pentagons d Any two squares Solution a No; pairs need not be , nor do the pairs of sides need to be proportional. b Yes; all angles are congruent measure 108° each, and all pairs of sides are proportional. c No; all angles measure 90°, but the pairs of sides are not necessarily proportional. d Yes; all angles measure 90°, and all pairs of sides are proportional. ⬵ ∠ 쮿 Exs. 1–4 It is common practice to name the corresponding vertices of similar polygons in the same order. For instance, if pentagon ABCDE is similar to pentagon MNPQR, then we know that , , , , , , , , , and . Because of the indicated correspondence of vertices, we also know that AB MN = BC NP = CD PQ = DE QR = EA RM ∠E ⬵ ∠R ∠D ⬵ ∠Q ∠C ⬵ ∠P ∠B ⬵ ∠N ∠A ⬵ ∠M E 4 R D 4 Q C 4 P B 4 N A 4 M EXAMPLE 3 If in Figure 5.7, use the indicated measures to find the measures of the remaining parts of each of the triangles. 䉭ABC ~ 䉭DEF 5 B C A 4 3 37 ° Figure 5.7 E F D 6 Solution Because the sum of the measures of the angles of a triangle is 180°, And because of the similarity and the correspondence of vertices, The proportion that relates the lengths of the sides is From , we see that From , we see that DE = 10 3 DE = 6 5 so that 3 DE = 30 3 6 = 5 DE FE = 8 3 FE = 6 4 so that 3 FE = 24 3 6 = 4 FE AC DF = CB FE = AB DE so 3 6 = 4 FE = 5 DE m ∠D = 53°, m ∠E = 37°, and m ∠F = 90° m ∠A = 180 - 90 + 37 = 53° 쮿 In a proportion, the ratios can all be inverted; thus, Example 3 could have been solved by using the proportion DF AC = FE CB = DE AB In an extended proportion, the ratios must all be equal to the same constant value. By designating this number which is often called the “constant of proportionality” by k, we see that It follows that , , and . In Example 3, this con- stant of proportionality had the value , which means that the length of each side of the larger triangle was twice the length of the corresponding side of the smaller triangle. If , the similarity leads to an enlargement, or stretch. If , the similarity results in a shrink. The constant of proportionality is also used to scale a map, a diagram, or a blue- print. As a consequence, scaling problems can be solved by using proportions. 0 6 k 6 1 k 7 1 k = 2 DE = k AB FE = k CB DF = k AC DF AC = k, FE CB = k, and DE AB = k Exs. 5–10 EXAMPLE 4 On a map, a length of 1 in. represents a distance of 30 mi. On the map, how far apart should two cities appear if they are actually 140 mi apart along a straight line? Solution Where the map distance desired in inches, Then and x = 4 2 3 in. 30x = 140

1 30

= x 140 x = 쮿 D A E C B 16 x x 3 Figure 5.8 EXAMPLE 5 In Figure 5.8, with . If , , and , find the length BC. Solution From the similar triangles, we have . With and representing the lengths of the congruent segments and by x, we have Substituting into the proportion, we have It follows that Now x or BC equals 4 or 12. Each length is acceptable, but the scaled drawings differ, as illustrated in Figure 5.9 on next page. x - 4x - 12 = 0 x 2 - 16x + 48 = 0 16x - x 2 = 48 x16 - x = 3

16 3

x = 16 - x 16 16 = AE + x so AE = 16 - x BC EC AC = AE + EC DE BC = AE AC EC = BC AC = 16 DE = 3 ∠ADE ⬵ ∠B 䉭ABC 䉭ADE