Notes — Chapter 10: Similarity
To the Student. A model car is the same shape as the original but at a smaller size. The original and the model are similar geometric shapes. You can transform the original shape so that it can be superimposed on the smaller model by first shrinking it to the size of the model. The shrunken car is then congruent to the model, and the shrunken car can be sent onto the model by an isometry. The end result of this shrinking followed by an isometry is a transformation from the original to the model which preserves angles between corresponding segments and reduces all distances by the same scale factor.
Model cars are made at scales such as 1/20th. At this scale, a car that was 15 feet long and 5 feet wide would have a model that was 15/20 =.75 feet long and 5/20 =.25 feet wide. In inches, the model would be.75 x 12
= 9 inches long and.25 x 12 = 3 inches wide, and all the other dimensions would be in proportion.
Scale models are not just toys, they are a big business that includes maps, which are schematic scale models of cities and countries. On the Web, you will find companies that make scale models for architects and for movie and television productions.
With a reduced scale model, you can hold something like the Empire State Building in your hands and examine it. With an enlarged model, you can see how a living cell works or how atoms are arranged in a crystal. The geometry of similar figures, maps, and scale models is explored in this chapter.
General Remarks about Similarity.
▶ Scale Models and Similarity. Similarity is the basis for making scale models. You can’t make a strictly correct scale model in spherical geometry, because as you scale down the lengths of the sides of a triangle, you change the sum of its angles. The distortions are negligible if you are making scale models following the surface of the earth in your.backyard, but they become important on larger scales. Similarity and scale models are practical tools used by all of us, from casual readers of maps to architects planning cities or modeling new buildings.
▶ Similarity, Dilations, & Congruence. Two shapes are congruent if they are the same size and shape. Using the transformational point of view, this is equivalent to saying that there is an isometry that carries one onto the other. In exactly the same way, two shapes are similar if they have the same shape, although they may differ in size. In terms of transformations, shapes are similar if there is a stmilarity that carries one onto the other. A similarity is a mapping in which the distance between transformed points is a constant multiple of the distance between the points before transformation. In terms of segment lengths, the transformation T is a similarity if T(P)T(Q) = APQ, for all points P and Q. The number A is the scale factor of the similarity. The SSS similarity theorem (page 412) tells us that similarities send angles onto equal angles, and thus, they preserve the shapes of triangles.
The AA similarity theorem is something new. It holds because the shape of a triangle in Euclidean geometry depends only on its angles, and because by the Angle Sum Theorem, any two of its angles determine the third angle. If the Parallel Postulate were false, we would not be able to construct scale models and there would be no AA similarity theorem for triangles.
A dilation with magnitude λ about O is a similarity. A dilation followed or preceded by an isometry multiplies the lengths of segments by the of the dilation, and so such compositions are also similarities. Indeed, every similarity may be written as an isometry followed by
dilation (or just as easily, a dilation followed by an isometry). We could restrict ourselves to dilations about O, the origin of our coordinate system. In terms of coordinates, dilations about the origin have the form T((x,y)) = (λx, λy). This generalizes to three dimen-. sions where T((z, y,z)) = (λx, λy, λz). Dilations allow us to change size leaving the shape the same. _
li” Side-Splitter Theorem is Key. The study of similar triangles follows the pattern we used for studying congruent triangles. We can conclude triangles are congruent by SAS, ASA, or SSS, and after the Parallel Postulate is assumed, we have SAA as well. For △ABC and AA’‘B’C’ to be similar, written △ABC ∼ AA‘B’C’, The corresponding angles must be equal (same shape), and the corresponding sides must be proportional (so pipe NAB B Ce ABC Wand CA) =’ )2 ‘The key to establishing all of the properties of similar triangles is the Side-Splitter Theorem and its Corollary, Lesson 3.
▶ Transformations & Similarity Proofs. The proofs given in this chapter follow the pattern used used in our study of congruent triangles. For example, we might copy △ABC onto AA’‘B/C’. This static picture makes it easy to follow the details. For a more dynamic view, think of using a dilation to shrink or expand one of the triangles being studied so that you then have two triangles the same size, then show that these same-sized triangles are congruent. The composition of the dilation and isometry gives you a similarity between the original triangles. The static and dynamic arguments are essentially the same, but the first emphasizes construction while the second emphasizes transformations.
Lesson 1. Ratio and Proportion. We recognize many things by their shape without regard to their size. As the Lichtenstein mural thatthe opens this lesson shows, reference objects found in a picture let us estimate the true sizes of all the objects shown. This is why geologists include rulers, rock hammers, or their colleagues in their photographs of rock formations. The algebraic treatment of ratio and proportion in this lesson puts our intuitions about size and shape on a. firm mathematical basis. These algebraic tools are used throughout this chapter. They are what let us work things out using scale drawings and scale models.
Exercises and Calculators. Calculators allow us to work out proportions even when the numbers are awkward. Most of the exercises involve ratios that can be handled easily by hand, but some students will want to use their calculators for all the exercises. Set IT, Sinker or Floater (53 and 54) involves densities and a calculator is needed for it.
Applications include art, architecture, and music. A project for students interested in music might be to compare the Pythagorean Tuning (41-47) to the welltempered scale. The underlying assumption in these
exercises is that the strings and string tensions are constant, so that Frequency x String Length is a constant. A good source on musical notes and frequencies is The Science of Musical Sound by John R. Pierce, SCIENTIFIC AMERICAN Books, New York, 1983.
Lesson 2. Similar Figures. Similar figures have the same shape, but what is shape? Everyone who has watched a movie on both small and large screens knows that things are missing when you see the movie on a TV screen. Here we learn that the reason is because these two types of screens are rectangles of different shapes.
Two shapes are similar if there is a correspondence between their points such that corresponding segments are proportional. Such a correspondence is a similarity.
Note that the definition of similarity for triangles includes the fact that corresponding angles are equal. This allows us to freely use the properties of similar triangles without relying on the SSS similarity theorem (page 412). That theorem tells us that if the corresponding sides of two triangles are proportional then their corresponding angles are equal.
Dilation about a point O with magnitude r is a similarity. The illustration on page 387 shows how dilations act on triangles and circles, showing both magnification and reduction. Students will be familiar with the similarities you get by using the enlarging or reducing features of a copying machine. (As an aside, a curious fact is that copy machines set at 100% usually enlarge slightly to avoid any dark lines at the edges when a document is not precisely centered. After five generations of copies on my machine, a line that was 8.5 inches long had grown longer by a sixteenth of an inch. The magnitude of a single dilation in this series is (8.5625/8.5)5 ∼ 1.0015. Knowing the enlargement at each stage, a detective can estimate how many generations of copying were needed to get to the degree of enlargement seen in a crime document from a given original.)
Set I. Plane Table (3-80) shows how accurate maps were made without calculations or numerical measurements. This uses Angle Side Angle similarity. We may think of this as AA similarity, but it is useful to keep the S in mind because the ratio of the baseline side of the map to the baseline on the ground determines the scale for the map. There is a nice section in The Ring of Truth by Philip and Phylis Morrison, Random House, New York, 1987, on making a plane table map of Thomas Jefferson’s estate, Monticello. This book is based on a PBS series which is available on video. In Exercises 9 — 11 you see Kempe’s Linkage that trisects angles. The black circles with white centers are pinned joints. Those at H and J bisect the rods they lie on. The solid black dots are sliders that can move along the rods they are on. The sliders keep the configuration in one plane. The linkage is built out of units like that made up of points OAHC in the diagram. These Kempe
called “contra-parallelograms” in his pamphlet How to Draw a Straight Line. This pamphlet was republished in Squaring the Circle and Other Monographs, Hobson et al., Chelsea Publishing Company, 1953, which is now available from Chelsea at the American Mathematical Society. There was also an NCTM reprint of Kempe’s booklet.
A. B. Kempe (pronounced Kemp) is also famous for his contributions to the proof of the Four Color Map Conjecture. (Any map of the globe can be colored with four colors so that no two adjacent countries have the same color.) See page 6 or Four Colors Suffice by Robin Wilson, Penguin Books and Princeton University Press, 2002. In 1879 Kempe published his results on this problem, giving a proof that was seen as convincing until P. J. Heawood published an example in 1889 that showed that one of Kempe’s methods failed for certain sorts of maps. Later, Kempe’s techniques were used in work that culminated in a proof of the Four Color Map Conjecture by Kenneth Appel and Wolfgang Haken in 1976. Their proof required more than 1200 hours of computer time to check its technical details. See, for instance, Appel and Haken’s October 1977 SctENTIFIC AMERICAN article.
Dilation Problem (15-19) shows dilation at work. Practical measurements are used here.
Calculator Answers. In Billboards (12-14), we start by observing a ≅ oe Then in Exercise 13 we solve the exact relationship - = u for x. On my calculator if I find x by x = 1/[5 + (11 x 11)], I get x = 24.20000 00001. But the more sensible x = (11 x 11) +5 gives x = 24.20000 00000. This second calculator result correctly suggests that the exact answer xr = 245. Students should understand that when it comes to practical measurements, answers that differ by 1 in the twelfth significant figure may be considered the same. However, if they were asked for an exact algebraic solution, then the 1 in the tenth decimal place of my first answer should be counted as an error.
In connection with rounding for convenience, students might make a project of finding the International Standards Organization (ISO) paper sizes on the Web and looking at how these were worked out and comparing this with the numbers used in Set II Paper Sizes (41- 47). This project shows how rigid mathematical requirements must be relaxed to give numbers that are easily dealt with in practice.
Violin Family (35-40) is about the scaling of similar instruments. Carleen Maley Hutchins worked this out for the violin family. Hutchins was a master of theory and practice, making many fine instruments herself. Her contributions are well documented on the Web. See www.newviolinfamily.org, her November 1962 SCIENTIFIC AMERICAN article, “The Physics of Violins,” or her book The Physics of Music, W. H. Freeman and Company, San Francisco, 1978.
Similar-Triangles Proof (24-34) gives a special case of the SAS Similarity Theorem. See page 412 and its figures. Given △ABC, if D lies on segment AB and E lies on segment AC so that AD/AB = AE/AC =r = 3, then △ABC ∼ △ADE. In this figure the angles that are between the two sets of sides that are proportional are equal angles — because they are both angle A. The conclusion is △ABC ∼ △ADE. The first steps of the proof on page 412 show how we pass between △ADE and a general AA’B’C’ & △ADE. See comments above on Transformations and Similarity Proofs.
▶ Digression: Midsegments and SAS Similarity. If ∠A’ = ∠A and A’‘B’/AB = A’/C’/AC =r we have SAS conditions for AA’B’C’ ∼ △ABC with a similarity ratio of r. Exercises 24 — 34 show that the Midsegment Theorem proves that these conditions imply similarity when the ratio r = 4 = 271. Iterating this argument shows that AA’B’/C’ ∼ △ABC for similarity ratio r = k27”, where n is a positive integer and1<k<n.
To see how this works you need to draw a diagram as follows: Use the midpoints of the sides to split △ABC into (21)? = 4 congruent subtriangles all similar to △ABC. See the related figure for Exercises 1-9 in Lesson 5 of this chapter. Iterating this subdividing process through n stages gives (2”)? = 4” congruent subtriangles, all similar to △ABC at ratio r = 27”. Start from vertex A and move down side AB k of these small triangles to locate point X. Along side AC move down k small triangles from A to locate point Y. Then AX/AB = AY/AC = k²-” so △AXY satisfies the SAS condition for similarity with △ABC. Segment XY is made up of k segments that are sides of the smallest subtriangles in our subdivision, each parallel to BC and each 2~” times the length of BC, the sides of △AXY are all proportional to the corresponding sides of △ABC. Because XY is made up of segments all parallel to BC, it is parallel to BC, and the corresponding angles △AXY are equal to those of △ABC. So △AXY ∼ △ABC
Suppose that AA’B’C’ satisfies SAS conditions for similarity to △ABC with similarity ratio r of the form considered above. Then AA’B’/C’ & △AXY, where △AXY is constructed as above. Then AA’/B/C’ © △AXY ∼ △ABC, so AA’B/C’ ∼ △ABC. Numbers of the form k2~“, where k and n are integers are dyadic rationals. Every number can be approximated as closely as you please by dyadic rationals, so the fact that SAS conditions imply similarity for dyadic rational ratios of similarity plus the assumption that things behave continuously implies that SAS conditions imply similarity for all ratios.
The arguments above shows the power of the Midsegment Theorem. The proof of the Side-Splitter Theorem in the next lesson uses area, and it is much easier than the proof based on the Midsegment Theorem outlined above. Harold Jacobs tells me that he has seen proofs as described in geometry books that were around in his school days. _ “A
Set III. In the Villa Foscari plan notice that if a smaller similar rectangle is snuggled in the corner of a large rectangle and if these have the same orientation, then the diagonal of the larger rectangle lies on the diagonal of the smaller one. This observation is used by graphic. designers in scaling drawings. By drawing the diagonal of an existing rectangle, they can use this diagonal to draw other similar rectangles at any given size.
The Villa Foscari exercise yields only to a systematic search of all the rectangles in the figure. Systematic search is an important problem-solving technique. Students should be encouraged to think about how they managed to check every possibility or how were missed. This exercise would be good for group work.
Lesson 3. The Side-Splitter Theorem. The dam pictured has a vertical upstream face to simplify the discussion. Most dams have slanted faces and are held in place largely by gravity. The trick is to make the dam so that it will be stable whether it is full or empty. Students may be interested in looking up how dams are constructed. J. E. Gordon’s book Structures: Or Why things Don’t Fall Down, a Penguin Pelican Book, 1978, reprinted by Da Capo Press 1988, pages 184-187 is a good source and has many examples that can be understood using simple geometric or algebraic reasoning.
The Side-Splitter Theorem and its Corollary (page 394) are more important and more intuitive that the proof of these results. Given △ABC we split its sides AB and AC with points D and E so that DE ∥ BC. By what we know about parallels, the corresponding angles of △ABC and △ADE are equal. The corollary tells us that AD/AB = AE/AC. We would like to know that △ABC ∼ △ADE. For this we need to show that AE/AC = DE/BC also. This follows by applying the Corollary to a congruent copy of △ADE along the lines used to prove the AA Similarity Theorem in the next lesson.
Set I. Turkish Theorem (1-3) and Errors of Omission (4-6) are review. Proportion is related to products in Picturing Products (7-12). Supply and Demand (13- 17) relates geometric proportions and linear relations between two variables. Side-Splitter Practice (18-21) is presented visually. Set II. Note the algebraic strand in Corollary to Theorem 44 (29-33). Other applications are to the geometry of bridges and, in Set III, to the pantograph. Building a pantograph would be a good project.
Lesson 4. The AA Similarity Theorem. Indirect measurement is a great triumph of geometry. Thales lived from 624-546 B.C., according to Isaac Astmov’s Biographical Encyclopedia of Science and Technology. It is reported that Thales used similar triangles to determine the height of the Great Pyramid. Records of this feat can be traced to Hieronymus (4th century BC) as reported by Diogenes (2nd century AD).
You might think that after some 2500 years nothing new could be said about this, but Lothar Redlin, Ngo Viet, and Saleem Watson’s article, “Thales’ Shadow,” in the December 2000 issue of Mathematics Magazine has a number of interesting points to make, backed by computer-drawn pictures. The problem is that at many times and on many days the shadow of the apex of the Great Pyramid falls at inconvenient places, such as on the sides of the pyramid. So when could Thales have made his measurements? Thinking about this could be a big project for students.
Redlin, Ngo, and Watson conclude, “Did Thales actually measure the height of the pyramid at all? It is impossible to say for sure, but the zdea of measuring the height of such a tall object using only its shadow is so beautiful and striking that it overshadows any of its practical applications….”
Theorem 45, AA Similarity, is a key result for practical purposes, and it allows us to prove that similarity is a transitive relation for triangles. Intuitively we know that this must be true if similar figures have the same shape.
Set I. The exercises begin with Thales’ Method (1- 4). Triangle Average (5 and 6) explores a fascinating fact about directly similar triangles. The more general result can be established by understanding the algebraic nature of similarity transformations. The details are outlined below.
▶ Digression: Similarities and Complex Numbers. Similarities in the plane can be handled algebraically using complex numbers. This digression draws on the lesson Proportion and Dilations, as well as on ideas about complex numbers, including de Moivre’s Theorem. Students will see this result later in their secondlevel algebra course.
We start with a Cartesian coordinate system. The point A will have coordinates (a1, a²). To use complex numbers, we identify the point A with a complex number A = a; + ia, where i² = −1. Given any complex numbers w # O and v, the function T(z) = wz+ vu of z defines a mapping of the complex plane to itself. We will see that T is a similarity.
We represent w in polar form using the de Moivre formula. Thus, w = re’?, where r is positive and 0 <6 < 2x. Think of carrying z onto T(z) in three steps:
Ze Oz 3, r(ez) = wz & wz +0 = T(z).
Step 1 is a counterclockwise rotation of magnitude 0 about the origin, O, and it preserves lengths and angles. Step 2 is a dilation about O of magnitude r, and it preserves angles and multiplies the lengths of all segments by r. Step 3 is a translation, and it preserves lengths and angles. The combined result is T, which is a similarity that preserves angles (because each step preserves them), rotates lines through a clockwise angle of @, and multiplies the lengths of segments by r.
Suppose that △ABC ∼ △DEF, where the similarity is direct, meaning that when if go around △ABC in clockwise order, then the corresponding vertices of △DEF are also traversed in clockwise order. There is a positive ratio of similarity r such that rAB = DE (equality of lengths) and an angle 0, 0 < @ < 2m such that after counterclockwise rotation through 6 radians AB becomes parallel to DE. Recalling that we have identified points with complex numbers, we may let w = re”? and define v = D—wA. Then the transformation T defined above, with these values of w and v, sends △ABC onto a triangle that is the same size as △DEF and whose sides are parallel to those of △DEF, and whose clockwise sequence of vertices matches that of the target triangle, △DEF. Since T(A) = D, the similarity T must send △ABC onto △DEF. Informally, T(/△ABC) = △DEF. This shows how we can use operations defined on complex numbers to construct a similarity mapping of the sort described above that sends △ABC onto △DEF, if these triangles are directly similar.
If △ABC ∼ △DEF have opposite orientations of their vertices in this similarity, then we need an operation to switch counterclockwise to clockwise ordering as we go from △ABC to △DEF. A reflection will do this. The complex conjugate of z = z, + izg is denoted z² or Z and is defined by Z = z; — iz. The mapping z > Z is the reflection of the complex plane in the real axis. It reverses the clockwise orientation of the vertices of a triangle. Sending △ABC — △ABC by conjugation and then △ABC — △DEF by T, we get a similarity S(z) = wz+v that sends △ABC to △DEF, as required.
triangle Average (5 and 6) concerns △GHI which is “midway” between △ABC ∼ △DEF, where the similarity preserves the clockwise orientation of the triangles. Students will see that all these triangles are similar.
More generally, we could look at intermediate triangles A of the way from △ABC to △DEF. The vertices of such a △GHI are:
Cat Ae D
H =(1—)B +2,
Il=(1—-A)C+AF,
with 0 < A < 1 and with G, H, and I distinct. If T(△ABC) = △DEF, where T(z) = wz + v, then D = wA+v, E=wB+v0, and F = wC+v. Using these expressions you can find G, H, and I by algebra. The formulas are all of the same form.
Example: G = (1—A+ Aw)A+F Av.
The similarity S(z) = [(1—A)+Aw]z+ [dv] with = 1/2 sends △ABC to △GHI, so these triangles are similar, which is what was to be shown.
This is what you might expect. The mapping z > 1z+ Ov, with a multiplier of 1 sends △ABC onto itself. The
mapping z — wz +1v, with a multiplier of w rotates and shrinks △ABC so that it matches △DEF in size and orientation. To get » of the way between these extremes, we would need a multiplier of (1 — A)1+ Aw. The corresponding translation needed is A of the way from 0 to v, or (1 — A)O+ Az = Xz, which is what we have.
There is one thing more to check. If the multiplier of z were zero, the points G, H, and I would have to be identical and G, H, and I would not be vertices of a triangle that is similar to △ABC. Since G, H, and I are distinct, [A+(1—A)w] 4 0, and S is a genuine similarity.
There are many relationships that students can discover by experiment, either with paper and pencil or using Geometer’s Sketchpad™. There must be some way to prove that these relationships must hold. However, the proofs may require methods beyond the range of introductory geometry. This seems to be such a result. The difficulty in this case is that we need some way to describe all similarities of the plane, so that we can smoothly interpolate between the identity map that sends △ABC onto itself and the similarity between △ABC and △DEF.
Matrices could be used in place of complex numbers, but this would require an understanding of rotation matrices, a matter no less technical than the use of complex numbers. We could work with rotations and dilations about the origin, but this is equivalent to using complex numbers in polar form, as above. If you or your students should find a royal road to this result, I would be delighted to see it.
Seeing and Calculating. The first thing students must do in Set I, Newton’s Figure (7 and 8), Nine Triangles (9-11), and Chess Puzzle (12-18) is to see what is going on in the figures. Then they go on to description, logical analysis, and calculation. Chessboard Puzzle is unmasked as a trick by simple calculations. Some students will want to see exactly where the fraud is. Don’t give the show away and spoil their fun, but one approach uses similar triangles to calculate how far the point B lies above the segment AE. The answer is 15 /8 units, just short of the 2 units needed to make the picture correct, and the method of the fraud becomes clear.
Art, Electricity, and Algebra. Set II. Exercises 19- 24 are tied to the Italian painter Piero della Francesca’s work on proportions and perspective. In Electrician’s Formula, (25-30), we see that when two resistors of values R; and Reg work in parallel, they are equivalent to a resistance of Rory This expression gives the harmonic mean of R; and Ry. The harmonic mean also appears in the formula for the length of time it takes two workers to complete a job if one can complete the Job in time R; and the other can complete it in time Ra. We see similar algebra in Deja Vu (43-47) used in a geometric context.
Original Proofs (37-42) and Mascheroni Constructions (48-50), which are constructions done with compass alone, are pure geometry.
The Set III puzzle on areas has a solution within reach _of students who follow the line suggested here.
Lesson 5. Proportions and Dilations. The Side- Splitter theorem and the AA Similarity Theorem establish much of what we need for applications of similarity. In this lesson we see how what we know can be used to show the proportionality of corresponding segments in similar figures. See Theorem 46 on the altitudes of similar triangles. The figure on page 408 correctly suggests that we should think of the similarity AABC ~ ADEF as establishing a correspondence not just between the vertices of these triangles but also between the feet of their altitudes and between all corresponding pairs of points on or within the triangle as well. The lengths of the segments between corresponding pairs of points under a similarity transformation are always proportional with the same proportionality ratio as the sides of the triangles.
Set I. Fish Story (1-18) drives home more general ideas about similar figures visually, and shows how these ideas are used in computations. Grid Exercises (25-30) relate similarity transformations to coordinates.
Set II. Congruence and Similarity (35-61) links these ideas. Geometrically similar shapes are congruent, if they are the same size. We would expect that the conditions that force congruence of triangles would also force similarity of triangles in a weakened form. This is true. For instance, ASA conditions on two triangles force congruence, while the weaker AA condition forces the triangles to be similar. Likewise, the equality of the corresponding sides of two triangles, SSS, forces congruence. Requiring only that that the sides of the one triangle be proportional to the corresponding sides of the other triangle forces the triangles to be similar (SSS similarity). No student would doubt that a triangle with sides 6, 8, and 12 is similar to one with sides 3, 4, and 6, and we hope that students will freely use such facts, whether or not they are asked to work through the proofs of the SAS and SSS Similarity Theorems (Exercises 40-61).
A practical understanding of similarity can be based on Exercises 35-39 and class discussion. At this stage, interested students could pursue the proofs outlined in Exercises 40-61, if they wish. These proofs are routine applications of methods we have already seen.
Set III lets students explore the camera theme of this lesson. This is good project.
Lesson 6. Perimeters and Areas of Similar Figures. If AB and A’B’ are corresponding segments of similar figures and if A’B’/AB = r, then the figures’ perimeters, P and P’, have ratio P’/P = r, and their areas, A and A’, have ratio A’/A = r?. These are rules
of scaling. More generally, the 3-dimensional volumes of similar figures scale as r°, where r is the ratio of the lengths of corresponding segments in these figures. If a quantity is n-dimensional it will scale as r” in similar figures, where lengths of corresponding segments in these figures scale as r!. See Lesson 8, Similar Solids in Chapter 15.
The exercises show scaling ideas at work in sports, in reading maps, in quilting designs, and in comparing the sizes of circles in a cranberry harvest. Notice that the areas and perimeters of circles scale the same way as the areas and perimeters of polygonal figures. Scaling of perimeters and areas of similar figures has many applications. Set III drives home this point with a surprising version of the Pythagorean Theorem. (Compare Set III with Euclid VI.31.)
Summary and Review. The key idea in this chapter is that similar figures have the same shape, so that. the lengths of corresponding segments are in the same ratio, r, which determines the proportionality of the similarity, and corresponding angles in similar figures are equal. The number r that gives the proportionality between the two figures lets us work out the numbers in all of the applications in the Summary and Review exercises. The Set I exercises cover all of the ideas in the chapter. The Trick Card and other examples in Set II are appealing and should be considered.