《初等几何研究》课程教学资源（书籍文献）初等几何研究 Geometry Transformed（Euclidean Plane Geometry Based on Rigid Motions）.pdf_P21-P25

Sameness Properties of Congruence 5 F' E' D' C' B' A' F D A B C E Figure 5. Quadrilaterals in Congruent Triangles Proof. A student steeped in triangle congruence might set off to find equal angles and lengths by proving congruence of some of the five subtriangles in each figure. After quite a number of steps, this will provide a proof that the corresponding sides and angles of the quadrilateral are congruent. But we are given a rigid motion 𝑇 that maps△𝐴𝐵𝐶 to△𝐴′𝐵 ′𝐶 ′ since the triangles are congruent. Once we show 𝑇 also maps 𝐷𝐶𝐸𝐹 to 𝐷 ′𝐶 ′𝐸 ′𝐹 ′ , this will prove the quadrilaterals are congruent. This follows from preservation of distance, since the distances defining 𝐷 and 𝐸 are the same as the distances defining 𝐷 ′ and 𝐸 ′ . In particular, 𝐷, which is the point on 𝐵𝐶 equidistant from 𝐵 and 𝐶, maps to the point on 𝐵 ′𝐶 ′ equidistant from 𝐵 ′ and 𝐶 ′ . Likewise, the image of 𝐸 is 𝐸 ′ , the point on segment 𝐴 ′𝐶 ′ at distance ‖𝐶𝐸‖ from 𝐶 ′ . Finally, the intersection 𝐹 of segments 𝐴𝐷 and 𝐵𝐸 must map by 𝑇 to the intersection of the segments 𝐴 ′𝐷 ′ and 𝐵 ′𝐸 ′ . □ The key idea is to exploit the explicit presence of a rigid motion 𝑇 whenever two figures are known to be congruent. Notice that we did not have to know anything about the behavior of 𝑇 beyond that it maps △𝐴𝐵𝐶 to △𝐴′𝐵 ′𝐶 ′ . Note. This proof used an important and convenient notational convention about congruence. When we write that △𝐴𝐵𝐶 is congruent to △𝐷𝐸𝐹, this means more than that there is a rigid motion 𝑇 mapping one triangle to the other. It also means that 𝑇(𝐴) = 𝐷, 𝑇(𝐵) = 𝐸, 𝑇(𝐶) = 𝐹. So the statement that △𝐴𝐵𝐶 is congruent to △𝐸𝐹𝐷 is a different statement. If we want merely to say the triangles are congruent without this convention, we can name 𝑡1 = △𝐴𝐵𝐶 and 𝑡2 = △𝐷𝐸𝐹 and then say that 𝑡1 is congruent to 𝑡2 . This convention carries over to segments 𝐴𝐵, quadrilaterals 𝐴𝐵𝐶𝐷, and to any figure that is defined in terms of a list of points. Sameness Properties of Congruence In the first examples in this chapter, we have seen that the word “same” in our ordinary language has such rich and varied meanings that it is not precise enough for mathematical use. However, a concept of sameness or equivalence is an important one that we need in mathematics. The following theorem says that congruence has some key

6 1.Congruence and Rigid Motions properties that we expect from our intuition about sameness.We writeUto de. note that U is congruent to V Theorem 1.4.Congruence has the following three properties: (1)For any set U in the plane.UU. (2)fU兰，then V≥U. (3)fU兰V and V兰w,then U兰w Informally,if we think of congruence in terms of su erp osition,this is easy to see If we the s the otion T to r n the es v to U.so is s etric.Finally,if we motion S.the motion Tand ST of first then S move w This hows that The formal proof is pretty much the same,except there are surprisingly many steps to check off since we will need to prove some basic properties of rigid motions along the way.The main things to prove are that inverses and products of rigid motions are rigid motions(properties that we casually assumed with our informal motions but need to check to see that our formal definition really works as expected). We need to prove these properties of rigid motions that correspond to the three points of Theorem 1.4 Theorem 1.5.These statements about rigid motions are true: (1)The identity mapI is a rigid motion. (2)IfT is a rigid motion of the plane,then T-is also a rigid motion. (3)IfS and T are rigid motions of the plane,then the composition ST is also a rigid motion Note.We write composition of transformations as a product,without the little circle that is sometimes used. Before proving this theorem,we first show how it is used to prove Theorem 1.4. Each item in Theorem 1.5 corresponds to the same-numbered item in Theorem 1.4. This proof is a formal version of our informal superposition reasoning above. Proof of Theorem 1.4.In each case,we prove E F by finding a rigid motion that maps E to F. (1)The identity map Iis a rigid motion,and I(U)=U,soUU. (2)If U V,there is a rigid motion T with T(U)=V.Then T-(V)=U.Since T-1 is a rigid motion,VU. (3)If U V,there is a rigid motion T so that T(U)=V.IfV W,there isa rigic motion S so that S(V)=W.Then ST(U)=W,so U W since ST is a rigid motion

6 1. Congruence and Rigid Motions properties that we expect from our intuition about sameness. We write 𝑈 ≅ 𝑉 to denote that 𝑈 is congruent to 𝑉. Theorem 1.4. Congruence has the following three properties: (1) For any set 𝑈 in the plane, 𝑈 ≅ 𝑈. (2) If 𝑈 ≅ 𝑉, then 𝑉 ≅ 𝑈. (3) If 𝑈 ≅ 𝑉 and 𝑉 ≅ 𝑊, then 𝑈 ≅ 𝑊. Informally, if we think of congruence in terms of superposition, this is easy to see. If we draw the figures on paper, then not moving 𝑈 at all shows the first item. If we use a motion 𝑇 to move 𝑈 to 𝑉, then the reverse motion 𝑇 −1 moves 𝑉 to 𝑈, so congruence is symmetric. Finally, if we move 𝑈 to 𝑉 by a motion 𝑇 and then move 𝑉 to 𝑊 by the motion 𝑆, the combined motion 𝑆𝑇 of first 𝑇, then 𝑆 moves 𝑈 to 𝑊. This shows that congruence is transitive. The formal proof is pretty much the same, except there are surprisingly many steps to check off since we will need to prove some basic properties of rigid motions along the way. The main things to prove are that inverses and products of rigid motions are rigid motions (properties that we casually assumed with our informal motions but need to check to see that our formal definition really works as expected). We need to prove these properties of rigid motions that correspond to the three points of Theorem 1.4. Theorem 1.5. These statements about rigid motions are true: (1) The identity map 𝐼 is a rigid motion. (2) If 𝑇 is a rigid motion of the plane, then 𝑇 −1 is also a rigid motion. (3) If 𝑆 and 𝑇 are rigid motions of the plane, then the composition 𝑆𝑇 is also a rigid motion. Note. We write composition of transformations as a product, without the little circle that is sometimes used. Before proving this theorem, we first show how it is used to prove Theorem 1.4. Each item in Theorem 1.5 corresponds to the same-numbered item in Theorem 1.4. This proof is a formal version of our informal superposition reasoning above. Proof of Theorem 1.4. In each case, we prove 𝐸 ≅ 𝐹 by finding a rigid motion that maps 𝐸 to 𝐹. (1) The identity map 𝐼 is a rigid motion, and 𝐼(𝑈) = 𝑈, so 𝑈 ≅ 𝑈. (2) If 𝑈 ≅ 𝑉, there is a rigid motion 𝑇 with 𝑇(𝑈) = 𝑉. Then 𝑇 −1(𝑉 ) = 𝑈. Since 𝑇 −1 is a rigid motion, 𝑉 ≅ 𝑈. (3) If 𝑈 ≅ 𝑉, there is a rigid motion 𝑇 so that 𝑇(𝑈) = 𝑉. If 𝑉 ≅ 𝑊, there is a rigid motion 𝑆 so that 𝑆(𝑉 ) = 𝑊. Then 𝑆𝑇(𝑈) = 𝑊, so 𝑈 ≅ 𝑊 since 𝑆𝑇 is a rigid motion. □

Sameness Properties of Congruence 1 Theorem 1.5 about rigid motions is very important.But we will not really use it until Chapter 3,so if you would prefer to read and understand the statement of the theorem now but defer going through the formal proof until we have spent some time with distance and angle measure,that is fine.(In fact,once you get the idea of how repetitious the proof of this theorem is.you may be satisfied without reading every step!) Proof of Theorem 1.5.First,we prove that distance is preserved in each case.Then afterwards,we will take up angle measure.We are not assuming any properties of distance except that it is a real-valued function d(A,B)of two points.But instead of the usual function notation.we will write llABll in place of d(A.B).You will see that this reasoning is very general and quite formal,without regard for any special properties of d.By Definition 1.1.whenever we know a transformation T isa rigid motion. we know that for any A and B.T(A)T(B)Il = (1)Theidentity mapisI.Bydefinition,forany pointsAand B,I(A)=AandI(B)=B. Therefore.(A)I(B)I=AB by substitution. (2)We need to prove that for any points A and B.(A)T-(B)Il =ABll.Since T is a rigid motion,we apply it to the image points of T-to get two things: IIT(T-1(A))T(T-1(B))=IT-(A)T-(B)II because T is a rigid motion.But T(T-1(A))=A and T(T-1(B))=B since TT-1=I.so llT(T-1(A))T(T-1(B))I= IABll by substitution. (3)For nts A and B need to prove ST(A)ST(B)I=ABII.By definition. ST(AS(Aand ST()S(T().Since Sisa rigd motion IIS(T(A))S(T(B))II=IT(A)T(B)II and this equals AB because T is a rigid motion. The proof of angle (A, efined f ny th B.C w we h ve s,we will denc teA,B,C)bym∠AE but for no we wi stick with the usual f onal not: seems a t this proc In every c ow,we wil the th ed.For e points and ysuch points,ifTisa rigid motion, then by Definition 1.1.g(T(A).T(B).T(C))=j(A.B.C). IC)=C.Therefore.((A)(B)I(C))A.B.C)by substitution (2)We need to prove that for any pointsA,B,and C, (T-4),T-1(B),T-(C)=44,B,C) We apply T to the image points of T-1 to get u(T(T-(A),T(T-(B),T(T-1(C)). Since T is a rigid motion this equals u(T-(A).T-1(B).T-1(C)).But TT-1(A)= A,TT-1(B)=B,and TT-1(C)=C so the expression equals u(A,B,C)by substi- tution

Sameness Properties of Congruence 7 Theorem 1.5 about rigid motions is very important. But we will not really use it until Chapter 3, so if you would prefer to read and understand the statement of the theorem now but defer going through the formal proof until we have spent some time with distance and angle measure, that is fine. (In fact, once you get the idea of how repetitious the proof of this theorem is, you may be satisfied without reading every step!) Proof of Theorem 1.5. First, we prove that distance is preserved in each case. Then afterwards, we will take up angle measure. We are not assuming any properties of distance except that it is a real-valued function 𝑑(𝐴, 𝐵) of two points. But instead of the usual function notation, we will write ‖𝐴𝐵‖ in place of 𝑑(𝐴, 𝐵). You will see that this reasoning is very general and quite formal, without regard for any special properties of 𝑑. By Definition 1.1, whenever we know a transformation 𝑇 is a rigid motion, we know that for any 𝐴 and 𝐵, ‖𝑇(𝐴)𝑇(𝐵)‖ = ‖𝐴𝐵‖. (1) The identity map is𝐼. By definition, for any points𝐴and 𝐵, 𝐼(𝐴) = 𝐴and 𝐼(𝐵) = 𝐵. Therefore, ‖𝐼(𝐴)𝐼(𝐵)‖ = ‖𝐴𝐵‖ by substitution. (2) We need to prove that for any points 𝐴 and 𝐵, ‖𝑇−1(𝐴)𝑇−1(𝐵)‖ = ‖𝐴𝐵‖. Since 𝑇 is a rigid motion, we apply it to the image points of 𝑇 −1 to get two things: ‖𝑇(𝑇−1(𝐴))𝑇(𝑇−1(𝐵))‖ = ‖𝑇−1(𝐴)𝑇−1(𝐵)‖ because 𝑇 is a rigid motion. But 𝑇(𝑇−1(𝐴)) = 𝐴 and 𝑇(𝑇−1(𝐵)) = 𝐵 since 𝑇𝑇−1 = 𝐼, so ‖𝑇(𝑇−1(𝐴))𝑇(𝑇−1(𝐵))‖ = ‖𝐴𝐵‖ by substitution. (3) For any points 𝐴 and 𝐵, we need to prove ‖𝑆𝑇(𝐴)𝑆𝑇(𝐵)‖ = ‖𝐴𝐵‖. By definition, 𝑆𝑇(𝐴) = 𝑆(𝑇(𝐴)) and 𝑆𝑇(𝐵) = 𝑆(𝑇(𝐵)). Since 𝑆 is a rigid motion, ‖𝑆(𝑇(𝐴))𝑆(𝑇(𝐵))‖ = ‖𝑇(𝐴)𝑇(𝐵)‖ and this equals ‖𝐴𝐵‖ because 𝑇 is a rigid motion. The proof of angle preservation follows the identical pattern, except that we have a function of three points, 𝜇(𝐴, 𝐵, 𝐶), that is defined for any three points 𝐴, 𝐵, 𝐶 with 𝐵 ≠ 𝐴 and 𝐶 ≠ 𝐴. Once we have defined angles, we will denote 𝜇(𝐴, 𝐵, 𝐶) by 𝑚∠𝐴𝐵𝐶, but for now, we will stick with the usual functional notation because it seems a bit more readable for this proof. In every case below, we will assume the three points satisfy 𝐵 ≠ 𝐴 and 𝐶 ≠ 𝐴 so that 𝜇 is defined. For any such points, if 𝑇 is a rigid motion, then by Definition 1.1, 𝜇(𝑇(𝐴), 𝑇(𝐵), 𝑇(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). (1) The identity map is𝐼. By definition, for any points𝐴, 𝐵, and 𝐶, 𝐼(𝐴) = 𝐴, 𝐼(𝐵) = 𝐵, 𝐼(𝐶) = 𝐶. Therefore, 𝜇(𝐼(𝐴), 𝐼(𝐵), 𝐼(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶) by substitution. (2) We need to prove that for any points 𝐴, 𝐵, and 𝐶, 𝜇(𝑇−1(𝐴), 𝑇−1(𝐵), 𝑇−1(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). We apply 𝑇 to the image points of 𝑇 −1 to get 𝜇(𝑇(𝑇−1(𝐴)), 𝑇(𝑇−1(𝐵)), 𝑇(𝑇−1(𝐶))). Since 𝑇 is a rigid motion this equals 𝜇(𝑇−1(𝐴), 𝑇−1(𝐵), 𝑇−1(𝐶)). But 𝑇𝑇−1(𝐴) = 𝐴, 𝑇𝑇−1(𝐵) = 𝐵, and 𝑇𝑇−1(𝐶) = 𝐶 so the expression equals 𝜇(𝐴, 𝐵, 𝐶) by substitution

8 1. Congruence and Rigid Motions (3) For any points 𝐴 and 𝐵, we need to prove 𝜇(𝑆𝑇(𝐴), 𝑆𝑇(𝐵), 𝑆𝑇(𝐶)) = 𝜇(𝐴, 𝐵, 𝐶). By definition, 𝑆𝑇(𝐴) = 𝑆(𝑇(𝐴)), 𝑆𝑇(𝐵) = 𝑆(𝑇(𝐵)), and 𝑆𝑇(𝐶) = 𝑆(𝑇(𝐶)). Since 𝑆 is a rigid motion, 𝜇(𝑆(𝑇(𝐴)), 𝑆(𝑇(𝐵)), 𝑆(𝑇(𝐶))) = 𝜇(𝑇(𝐴), 𝑇(𝐵), 𝑇(𝐶)). and this equals 𝜇(𝐴, 𝐵, 𝐶) because 𝑇 is a rigid motion. □ T S A'' C'' B'' C' B' A' A B C Figure 6. Arrows and Primed Points Showing Transformations A Few Words about Functional Notation. You may have noticed how extensively functional notation was used in the proofs above. From one point of view, this seems appropriate since transformations are functions. But despite its precision, functional notation can seem dense and hard to follow on the page, not to mention on a blackboard or a whiteboard. An alternative notation might be to write something like this, with the reasoning a bit less explicit: In Figure 6, let the 𝑇-images of𝐴, 𝐵, 𝐶 be𝐴 ′ , 𝐵 ′ , 𝐶 ′ , and let the 𝑆-images of𝐴 ′ , 𝐵 ′ , 𝐶 ′ be 𝐴 ″ , 𝐵 ″ , 𝐶 ″ . Then if 𝑇 and 𝑆 are rigid motions, 𝑚∠𝐴𝐵𝐶 = 𝑚∠𝐴′𝐵 ′𝐶 ′ = 𝑚∠𝐴″𝐵 ″𝐶 ″ . This can be a more readable form of notation in the right context. But it does require an introductory explanation of what the primed images are, and it can be confusing when there are complicated things going on. If one is writing on a board or a document camera, drawing a diagram with a few well-placed arrows to go with 𝑆 and 𝑇 can be very helpful, as in Figure 6. In the coming chapters, we will sometimes use functional notation and sometimes we will use other ways to express what is going on, always aiming for clarity and readability. Circles. At this stage, with almost zero information about the plane, we cannot say much about any objects or shapes. But we can define one geometrical object just by having a distance function on the plane. Definition 1.6 (Circle). If 𝑂 is a point and 𝑟 is a positive real number, the circle with center point 𝑂 and radius 𝑟 is the set of points 𝑃 with distance ‖𝑂𝑃‖ = 𝑟. This is a familiar definition, but we will see in the exercises that, with an unfamiliar distance, the shape of the circle is not what we think of as circular

Exercises and Explorations Exercises and Explorations 1.(Taxicab Circles).This problem calls on your informal background knowledge and not anything we have proved so far.On a sheet of graph paper with(x.y)- coordinates,define the distance d(P.Q)to bex-x+ly1-yal for P=(x1.y1) and Q=(x2,y2).This is called the taxicab distance because it is the distance to go from P to Q by traveling on streets,if the graph paper is viewed as a grid of city streets parallel to the axes. Picka pointAon theg aph paper and draw a circle of radius 5 using the taxicab distance in the circle definition.Draw some other circles and see how they are related.What transformations will map such circles to each other? 2.(Rigid Motions and Circles).Prove that a rigid motion of the plane maps a circle to a circle.(Actually an isometry also maps a circle to a circle.) 3.(Informal Exploration).Draw by tracing around ajar lid or other method to obtain a pair of identical shapes like the one on the left and another pair like the one on the right in Figure 7.Experiment with each pair.Superimpose one of the shapes onto the other as many ways as you can.What number of rigid motions did you find?How would you describe those rigid motions?How do the numbers and nature of the rigid motions for the two pairs of shapes compare? Figure 7.Two Figures for Experiments 4.(Geometry of the Real Line).The real numbers R have a geometry,with the dis tance between a and b defined as b-al.A transformation G of the real numbers is an isometry if it preserves distance;in other words,G(a)-G(b)=la-bl for any a and b. (a)For G,an isometry of R,suppose G(1)=8 and G(3)=10.What is G(2)? What is G(9)?What is G(t)for any t? (b)For H,an isometry ofR,suppose H(1)=8 and H(3)=6.What is H(9)?What is H(9/2)?What is H(t)for any t? (c)Show that any transformation F of R that preserves distance must have one of these two formulas:F(t)=t+k or F(t)=-t+k,for some constant k. (d)One of the forms of F above always has a fixed point e so that F(c)=c.For which form is this true,and what number is c?

Exercises and Explorations 9 Exercises and Explorations 1. (Taxicab Circles). This problem calls on your informal background knowledge and not anything we have proved so far. On a sheet of graph paper with (𝑥, 𝑦)- coordinates, define the distance 𝑑(𝑃, 𝑄) to be |𝑥1 − 𝑥2 | + |𝑦1 − 𝑦2 | for 𝑃 = (𝑥1 , 𝑦1 ) and 𝑄 = (𝑥2 , 𝑦2 ). This is called the taxicab distance because it is the distance to go from 𝑃 to 𝑄 by traveling on streets, if the graph paper is viewed as a grid of city streets parallel to the axes. Pick a point𝐴on the graph paper and draw a circle of radius 5 using the taxicab distance in the circle definition. Draw some other circles and see how they are related. What transformations will map such circles to each other? 2. (Rigid Motions and Circles). Prove that a rigid motion of the plane maps a circle to a circle. (Actually an isometry also maps a circle to a circle.) 3. (Informal Exploration). Draw by tracing around a jar lid or other method to obtain a pair of identical shapes like the one on the left and another pair like the one on the right in Figure 7. Experiment with each pair. Superimpose one of the shapes onto the other as many ways as you can. What number of rigid motions did you find? How would you describe those rigid motions? How do the numbers and nature of the rigid motions for the two pairs of shapes compare? Figure 7. Two Figures for Experiments 4. (Geometry of the Real Line). The real numbers ℝ have a geometry, with the distance between 𝑎 and 𝑏 defined as |𝑏 − 𝑎|. A transformation 𝐺 of the real numbers is an isometry if it preserves distance; in other words, |𝐺(𝑎) − 𝐺(𝑏)| = |𝑎 − 𝑏| for any 𝑎 and 𝑏. (a) For 𝐺, an isometry of ℝ, suppose 𝐺(1) = 8 and 𝐺(3) = 10. What is 𝐺(2)? What is 𝐺(9)? What is 𝐺(𝑡) for any 𝑡? (b) For 𝐻, an isometry of ℝ, suppose 𝐻(1) = 8 and 𝐻(3) = 6. What is 𝐻(9)? What is 𝐻(9/2)? What is 𝐻(𝑡) for any 𝑡? (c) Show that any transformation 𝐹 of ℝ that preserves distance must have one of these two formulas: 𝐹(𝑡) = 𝑡 + 𝑘 or 𝐹(𝑡) = −𝑡 + 𝑘, for some constant 𝑘. (d) One of the forms of 𝐹 above always has a fixed point 𝑐 so that 𝐹(𝑐) = 𝑐. For which form is this true, and what number is 𝑐?