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Chapter 2 Congruence In this chapter we use isometries to study congruence. We shall prove the Fundamental Theorem of Transformational Plane Geometry, which states that two plane figures are congruent if and only if the first is the image of the second under a composition of three or fewer reflections. This is a truly remarkable fact. Our approach will be to study the fixed point sets for various families of isometries. We shall observe that an isometry is completely determined by its set of fixed points. This powerful idea enables us to identify all plane isometries and understand their properties. 2.1 The Three Points Theorem The facts that appear in our next three theorems are quite surprising: Theorem 73 An isometry that fixes two distinct points P and Q, fixes every ←→ point on the line P Q. Proof. Let P and Q be distinct fixed points of an isometry α and let R be ←→ any point on line P Q distinct from P and Q. Note that PR ∩ QR = R since P, Q and R are distinct and collinear. But R0 = α (R) ∈ PR since P R = P R0 and R0 = α (R) ∈ QR since QR = QR0 . Therefore R0 = R (see Figure 2.1). 39 40 CHAPTER 2. CONGRUENCE Figure 2.1. Theorem 74 An isometry that fixes three non-collinear points is the identity. Proof. Let P, Q, and R be three non-collinear points that are fixed under an isometry α. By Theorem 73, α fixes every point on 4P QR. Let Z be any point in the plane off of 4P QR and choose a point M 6= Z interior to 4P QR (see Figure 2.2). Q P M Z R Figure 2.2. ←−→ Then line ZM intersects 4P QR in two distinct points, which are fixed by α, so ←−→ each point on ZM is fixed by α, by Theorem 73. In particular, the arbitrarily chosen point Z is fixed by α. Therefore α = ι, as claimed. Theorem 75 (The Three Points Theorem) Two isometries that agree on three non-collinear points are equal. 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 41 Proof. Suppose that α and β are isometries and P, Q, and R are noncollinear points such that α(P ) = β(P ), α(Q) = β(Q), and α(R) = β(R). (2.1) Apply α−1 to both sides of each equation in (2.1) and obtain P = (α−1 ◦ β)(P ), Q = (α−1 ◦ β)(Q), and R = (α−1 ◦ β)(R). Thus α−1 ◦β is an isometry that fixes three non-collinear points and by Theorem 74, α−1 ◦ β = ι. Apply α to both sides of this last equation gives the desired result. In the next two sections, we use the Three Points Theorem to characterize translations and rotations in a new and very important way. 2.2 Translations as Products of Reflections Theorem 76 Let and m be distinct parallel lines and let n be a common perpendicular. Let L = ∩ n and M = m ∩ n. Then σ m ◦ σ is the translation by vector 2LM, i.e., σ m ◦ σ = τ 2LM . Proof. Let L0 = σ m (L); then (σ m ◦ σ )(L) = σ m (σ (L)) = σ m (L) = L0 = τ LL0 (L). Let K be a point on (2.2) distinct from L, and let K 0 = τ LL0 (K). Then τ KK0 = τ LL0 by Theorem 29. Therefore (σ m ◦ σ )(K) = σm (σ (K)) = σ m (K) = K 0 = τ KK0 (K) = τ LL0 (K). (2.3) Let J = σ (M ); then L is the midpoint of J and M and M is the midpoint of L and L0 . Hence JL = LM = ML0 so that JM = JL + LM = LM + ML0 = LL0 and τ JM = τ LL0 . Therefore (σ m ◦ σ )(J) = σ m (σ (J)) = σ m (M ) = M = τ JM (J) = τ LL0 (J) (see Figure 2.3). (2.4) 42 CHAPTER 2. CONGRUENCE Figure 2.3. By equations (2.2), (2.3), and (2.4), the isometries σ m ◦ σ and τ LL0 agree on three non-collinear points J, K, and L, so that σ m ◦ σ = τ LL0 by Theorem 75. Since M is the midpoint of L and L0 we have τ LL0 = τ 2LM , as desired. In the proof above, M is the midpoint of LL0 so by Theorem 56 we have τ LL0 = ϕM ◦ ϕL , which proves the next corollary: Corollary 77 Let and m be parallel lines and let n be a common perpendicular. Let L = ∩ n and M = m ∩ n. Then σm ◦ σ = ϕM ◦ ϕL . Figure 2.4. The next theorem tells us that the converse is also true: 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 43 Theorem 78 An isometry α is a translation if and only if α is a product of two reflections in parallel lines. Proof. Let be any line. Then α = ι (the identity transformation) if and only if α = σ 2 , and the statement holds in this case. So assume that α 6= ι and note that implication ⇐ was proved in Theorem 76. For the converse, let L and N be distinct points and consider the non-identity translation τ LN ; we must show that τ LN is a product of two reflections in distinct parallel lines. Let ←→ M be the midpoint of L and N ; let and m be the perpendiculars to LM at L and M , respectively (see Figure 2.5). P ' = σl (P) P P '' = σm (P ') L M l N m Figure 2.5. Then by Theorem 56, τ LN = ϕM ◦ ϕL and by Corollary 77, ϕM ◦ ϕL = σ m ◦ σ . Therefore τ LN = σ m ◦ σ with km as desired. Here is a useful trick that transforms a product of reflections in parallel lines into a product of halfturns. Given parallel lines and m, choose any common perpendicular n; let L = ∩ n and M = m ∩ n. Then σ m ◦ σ = σ m ◦ ι ◦ σ = σ m ◦ (σ n ◦ σ n ) ◦ σ = (σ m ◦ σn ) ◦ (σ n ◦ σ ) = ϕM ◦ ϕL . ←→ Conversely, given distinct points L and M, let n = LM , let be the line through L perpendicular to n, and let m be the line through M perpendicular to n. Then reading the calculation above from right to left we see how to get from a product of halfturns to a product of reflections in parallel lines. If P, Q, R and S are points such that PQ = RS, then τ PQ = τ RS . Since the product of two reflections in distinct parallel lines is a translation, there 44 CHAPTER 2. CONGRUENCE is the following analogous statement for parallels: If lines p, q, r and s are parallel lines passing through collinear points P, Q, R and S, respectively, then σq ◦ σp = σs ◦ σr . Theorem 79 Let , m and n be distinct parallel lines. There exist unique lines p and q parallel to such that σm ◦ σ = σn ◦ σp = σq ◦ σn . Proof. Given distinct parallels , m and n, choose any common perpendicular b. By Theorem 58, there exist unique points P and Q on b such that ϕP = ϕN ◦ ϕM ◦ ϕL and ϕQ = ϕM ◦ ϕL ◦ ϕN . Thus ϕM ◦ ϕL = ϕN ◦ ϕP = ϕQ ◦ ϕN . Let p and q be the lines perpendicular to b at P and Q, respectively (see Figure 2.6). Figure 2.6. Then by Corollary 77, σ m ◦ σ = ϕM ◦ ϕL = ϕN ◦ ϕP = σ n ◦ σ p and σ m ◦ σ = ϕM ◦ ϕL = ϕQ ◦ ϕN = σ q ◦ σ n . Thus given distinct parallels , m and n, Theorem 79 tells us that n determines unique parallels p and q such that σm ◦ σ = σ n ◦ σp = σ q ◦ σ n . Furthermore, line p is the unique line parallel to n such that the directed distance from p to n equals the directed distance from to m; line q is the unique line parallel to n such that the directed distance from n to q equals the directed distance from to m. Thus: 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 45 Corollary 80 Let P and Q be distinct points and let be any line perpendicular ←→ to P Q. Then there exists a unique line m parallel to such that τ PQ = σ m ◦ σ . (See Figure 2.7.) Q P M L m l Figure 2.7. Corollary 81 Let , m, and n be distinct parallel lines. There exists a unique line p parallel to , m, and n such that σp = σn ◦ σm ◦ σ , i.e., the product of three reflections in parallel lines is a reflection in some unique line parallel to them. Proof. By Theorem 79, there exists a unique line p parallel to , m, and n such that σm ◦ σ = σn ◦ σp (see Figure 2.8). l m p Figure 2.8. n 46 CHAPTER 2. CONGRUENCE Apply σ n to both sides and obtain σp = σn ◦ σm ◦ σ . Exercises 1. Lines and m have respective equations Y = 3 and Y = 5. Find the equations of the translation σ m ◦ σ . 2. Lines and m have respective equations Y = X and Y = X + 4. Find the equations of the translation σ m ◦ σ . 3. The vector of the translation τ is m such that τ = σ m ◦ σ . £ 4 −3 ¤ . Find the equations of lines and 4. The translation τ has equations x0 = x + 6 and y 0 = y − 3. Find equations of lines and m such that τ = σ m ◦ σ . 5. Lines , m and n have respective equations Y = 3, Y = 5 and Y = 9. a. Find the equation of line p such that σ m ◦ σ = σ p ◦ σ n . b. Find the equation of line q such that σ m ◦ σ = σ n ◦ σ q . 2.3 Rotations as Products of reflections In the previous section we observed that every non-identity translation is a product of two reflections in distinct parallel lines. In this section we prove the analogous statement for rotations: Every non-identity rotation is the product of two reflections in distinct intersecting lines. Thus translations and rotations are remarkably similar and have many analogous properties. Note that if and m are distinct intersecting lines, there is one directed angle from and m with measure Θ1 in the range 0 < Θ1 < 180 and another with measure Θ2 in the range 180 < Θ2 < 360. In fact, Θ2 = Θ1 + 180. Theorem 82 Let and m be distinct lines intersecting at a point C and let 0◦ < Θ◦ < 180◦ be the measure of the angle from to m. Then σ m ◦ σ = ρC,2Θ , for every Θ ∈ Θ◦ , i.e., σ m ◦ σ is the rotation about C through twice the directed angle from to m. 2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS 47 Proof. First observe that (σ m ◦ σ ) (C) = σ m (σ (C)) = σ m (C) = C = ρC,2Θ (C). (2.5) ◦ Let Θ ∈ Θ , let L be a point on distinct from C and let M be the point in −→ −−→ m ∩ CL such that the directed angle measure from CL to CM is Θ. Let L0 = ρC,2Θ (L); then L0 lies on CL and m is the perpendicular bisector of LL0 (see Figure 2.9). Therefore L0 = σ m (L) and we have (σm ◦ σ ) (L) = σ m (σ (L)) = σ m (L) = L0 = ρC,2Θ (L). (2.6) Let J = σ (M ); then is the perpendicular bisector of JM , in which case J lies −→ −−→ on CL and the directed angle measure from CJ to CM is 2Θ. L ' = ρC,2Θ (L) CL m M Θ C Θ Θ L l J = σl (M) Figure 2.9. Hence M = ρC,2Θ (J) so that (σ m ◦ σ ) (J) = σ m (σ (J)) = σ m (M ) = M = ρC,2Θ (J). (2.7) By equations (2.5), (2.6), and (2.7), the two isometries σ m ◦ σ and ρC,2Θ agree on non-collinear points C, J and L. Therefore σ m ◦ σ = ρC,2Θ by Theorem 75, as claimed. As with translations, the converse is also true: 48 CHAPTER 2. CONGRUENCE Theorem 83 A non-identity isometry α is a rotation if and only if α is the product of two reflections in distinct intersecting lines. Proof. The implication ⇐ was proved in Theorem 82. For the converse, given ρC,2Θ , let be any line through C and let m be the unique line through C such that the directed angle measure from to m is Θ (see Figure 2.10). Figure 2.10. By Theorem 82, ρC,2Θ = σ m ◦ σ . There is the following analogue to Theorem 79: Theorem 84 Let , m, and n be distinct lines concurrent at point C. There exist unique lines p and q concurrent at C such that σm ◦ σ = σn ◦ σp = σq ◦ σn . Proof. Given distinct lines , m, and n concurrent at point C, choose points L on and M on m distinct from C such that 0◦ < m∠LCM < 180◦ ; let Θ ∈ m∠LCM. By Theorem 82, ρC,2Θ = σ m ◦ σ . (2.8) −−→ Choose a point N on line n distinct from C and consider ray CN . There exist −−→ −−→ −−→ −−→ unique rays CP and CQ such that the directed angle measure from CP to CN −−→ −−→ ←→ ←→ and from CN to CQ equals Θ. So let p = CP and q = CQ (see Figure 2.11). 2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS m p l Θ n q 49 Θ C Θ Figure 2.11. Then by Theorem 82 we have ρC,2Θ = σ n ◦ σ p and ρC,2Θ = σ q ◦ σ n . (2.9) The result now follows from the equations in (2.8) and (2.9). A translation τ is a product of two reflections in distinct parallels lines and m, i.e., τ = σ m ◦ σ . Similarly, a rotation ρC,2Θ is a product of two reflections in distinct lines and m intersecting at C, i.e., ρC,2Θ = σm ◦ σ . Furthermore, if n, and m are concurrent at C, Theorem 84 tells us that n uniquely determines lines p and q also concurrent with n at C such that σ m ◦ σ = σ n ◦ σ p = σ q ◦ σ n , where p is the unique line through C such that the measure of the directed angle from p to n is Θ; q is the unique line through C such that the measure of the directed angle from n to q is Θ. Thus: Corollary 85 Let 0◦ < Θ◦ < 180◦ and let n be an arbitrarily chosen line passing through point C. If Θ ∈ Θ◦ , there exist unique lines p and q passing through C such that ρC,2Θ = σ n ◦ σ p = σ q ◦ σ n . Finally, if we multiply both sides of the equation σ m ◦ σ = σ n ◦ σ p on the left by σ n we obtain σp = σn ◦ σm ◦ σ . Hence: Corollary 86 Let , m, and n be distinct lines concurrent at point C. There exists a unique line p passing through C such that σp = σn ◦ σm ◦ σ , 50 CHAPTER 2. CONGRUENCE i.e., the product of three reflections in concurrent lines a reflection in some unique line concurrent with them. Corollary 87 A halfturn ϕP is the product (in either order) of two reflections in perpendicular lines intersecting at P. Note that the identity ι = τ PP = ρC,0 = σ ◦σ for all P, C, and . Therefore Theorem 88 A product of two reflections is either a translation or a rotation; only the identity is both a translation and a rotation. Exercises 1. Lines and m have respective equations X = 3 and Y = X. a. Find the equations of the rotation ρC,Θ = σ m ◦ σ . b. Find the center and angle of rotation C and Θ◦ . 2. Lines and m have respective equations Y = X and Y = −X + 4. a. Find the equations of the rotation ρC,Θ = σ m ◦ σ . b. Find the center and angle of rotation C and Θ◦ . 3. Find the equations of lines 4. Let C = and m such that ρO,90 = σ m ◦ σ . £3¤ 4 . Find equations of lines and m such that ρC,60 = σ m ◦ σ . 5. Lines , m and n have respective equations X = 0, Y = 2X and Y = 0. a. Find the equation of line p such that σ m ◦ σ = σ p ◦ σ n . b. Find the equation of line q such that σ m ◦ σ = σ n ◦ σ q . 6. Let C be a point on line . Prove that σ ◦ ρC,Θ ◦ σ = ρC,−Θ . 7. If σp = σ n ◦ σ m ◦ σ , prove that lines , m and n are either concurrent or mutually parallel. 8. If lines , m and n are either concurrent or mutually parallel, prove that σn ◦ σm ◦ σ = σ ◦ σm ◦ σn. 2.4. THE FUNDAMENTAL THEOREM 51 9. Construct the following in the figure below: b a l n m c a. Line s such that σs = σ n ◦ σ m ◦ σ . b. Line t such that σ t = σ c ◦ σ b ◦ σ a . c. The fixed point of σ t ◦ σ s . 10. Given distinct points P and Q, construct the point R such that τ PQ ◦ ρP,45 = ρR,45 . 11. Let ¤ABCD ∼ = ¤EF GH be a pair of congruent rectangles. Describe how to find a rotation ρP,Θ such that ρP,Θ (¤ABCD) = ¤EF GH. 12. Given distinct points P, Q and R, construct the point S such that τ QR ◦ ρP,120 = ρS,120 . 13. If , m and n are the respective perpendicular bisectors of sides AB, BC and CA of 4ABC, find the line p such that σ p = σ n ◦ σ m ◦ σ . 2.4 The Fundamental Theorem In this section we observe that every isometry is a product of three or fewer reflections. This important result will be obtained by carefully analyzing the set of points fixed by an isometry. Theorem 89 An isometry that fixes two distinct points is either a reflection or the identity. ←→ Proof. Let P and Q be distinct points and let m = P Q. We know that the identity ι and the reflection σ m are isometries that fix both P and Q. Are there any other isometries with this property? Let α be a non-identity isometry that fixes both P and Q, and let R be any point not fixed by α. By Theorem 73, 52 CHAPTER 2. CONGRUENCE R is off line m and P, Q, and R are non-collinear. Let R0 = α(R); since α is an isometry, P R = P R0 and QR = QR0 . Consider the two circles PR and QR , which intersect at points R and R0 (see Figure 2.12). R '=α(R) P m Q PR R QR Figure 2.12. Since P and Q are equidistant from R and R0 , line m is the perpendicular bisector of chord RR0 . Hence α(R) = R0 = σm (R), α(P ) = P = σ m (P ), and α(Q) = Q = σ m (Q) and by Theorem 75, α = σm . Theorem 90 An isometry that fixes exactly one point is a non-identity rotation. Proof. Let α be an isometry with exactly one fixed point C, let P be a point distinct from C, and let P 0 = α(P ). Let m be the perpendicular bisector of P P 0 (see Figure 2.13). Since α is an isometry, CP = CP 0 . Hence C is on m and σm (C) = C. By the definition of a reflection, σ m (P 0 ) = P. Therefore (σ m ◦ α)(C) = σ m (α(C)) = σ m (C) = C and (σ m ◦ α)(P ) = σ m (α(P )) = σ m (P 0 ) = P so that σ m ◦α is an isometry that fixes two distinct points C and P . By Theorem ←→ 89, either σ m ◦ α = ι or σm ◦ α = σ where = CP . 2.4. THE FUNDAMENTAL THEOREM 53 P m C P' l Figure 2.13. Lines and m are distinct since P 6= P 0 . However, if σ m ◦ α = ι, then σ m = α, which is impossible because α fixes exactly one point while σm fixes infinitely many points. Therefore we are left with σm ◦ α = σ and hence α = σm ◦ σ . The fact that α is a non-identity rotation follows from Theorem 83. In summary we have: Theorem 91 Every isometry with a fixed point is either the identity, a reflection or a rotation. An isometry with exactly one fixed point is a non-identity rotation. Proof. By Theorem 89, an isometry with two (or more) distinct fixed points is the identity or a reflection. By Theorem 90 and Theorem 83, an isometry with exactly one fixed point is a rotation. We are ready for our first of three important results in this course: Theorem 92 (The Fundamental Theorem of Transformational Plane Geometry): A transformation α : R2 → R2 is an isometry if and only if α factors as a product of three or fewer reflections. Proof. By Exercise 1.1.3, the composition of isometries is an isometry. Since reflections are isometries, every product of reflections is an isometry. Conversely, let α be an isometry. If α = ι, choose any line and write ι = σ ◦ σ , in which case the identity factors as a product of two reflections. So assume that α 6= ι and choose a point P such that P 0 = α(P ) 6= P. Let m be the perpendicular bisector of P P 0 and observe that (σ m ◦ α)(P ) = σ m (α(P )) = σ m (P 0 ) = P, 54 CHAPTER 2. CONGRUENCE i.e., β = σ m ◦α fixes the point P. By Theorems 89 and 90, β factors as a product of two or fewer reflections, which means that α = σm ◦ β factors as a product of three or fewer reflections. Theorem 92 tells us that a product of eight reflections, say, can be simplified to a product of three or fewer reflections. But how would one actually perform such a simplification process? One successful strategy is to consider three noncollinear points and their images and follow the procedure that appears in the proof of the next important theorem: Theorem 93 4P QR ∼ = 4ABC if and only if there is a unique isometry α such that α(P ) = A, α(Q) = B and α(R) = C. ∼ 4ABC, Theorem 75 tells us that if an isometry Proof. Given 4P QR = α with the required properties exists, it is unique. Our task, therefore, is to show that such an isometry α does indeed exist; we’ll do this by constructing α explicitly as a product of three isometries α3 ◦ α2 ◦ α1 each of which is either the identity or a reflection. Q R P B A C Figure 2.14. Begin by noting that AB = P Q, AC = P R, and BC = QR, (2.10) by CP CT C (see Figure 2.14). The isometry α1 : If P = A, let α1 = ι. Otherwise, let α1 = σ where is the 2.4. THE FUNDAMENTAL THEOREM 55 perpendicular bisector of P A. In either case, α1 (P ) = A. Let Q1 = α1 (Q) and R1 = α1 (R) and note that P Q = AQ1 , P R = AR1 , and QR = Q1 R1 (2.11) (see Figure 2.15). Q R P R1 l B α1(P)=A C Q1 Figure 2.15. The isometry α2 : If Q1 = B, let α2 = ι. Otherwise, let α2 = σ m where m is the perpendicular bisector of Q1 B. By (5.2) and (5.3) we have AB = P Q = AQ1 so the point A is equidistant from points B and Q1 . Therefore A lies on m and in either case we have α2 (A) = A and α2 (Q1 ) = B. Let R2 = α2 (R1 ) and note that AR1 = AR2 (see Figure 2.16). and Q1 R1 = BR2 (2.12) 56 CHAPTER 2. CONGRUENCE Q R P R1 l α2(Q1)=B m α2(A)=A C Q1 Figure 2.16. The isometry α3 : If R2 = C, let α3 = ι. Otherwise, let α3 = σ n where n is the perpendicular bisector of R2 C. By (5.2), (5.3), and (2.12) we have AC = P R = AR1 = AR2 so the point A is equidistant from points C and R2 and lies on n. On the other hand, (5.2), (5.3), and (2.12) also give BC = QR = Q1 R1 = BR2 so the point B is equidistant from points C and R2 and also lies on n. In either case we have α3 (A) = A, α3 (B) = B, and α3 (R2 ) = C. Let α = α3 ◦ α2 ◦ α1 and observe that α(P ) = α3 (α2 (α1 (P ))) = α3 (α2 (A)) = α3 (A) = A α(Q) = α3 (α2 (α1 (Q))) = α3 (α2 (Q1 )) = α3 (B) = B α(R) = α3 (α2 (α1 (R)) = α3 (α2 (R1 )) = α3 (R2 ) = C. Therefore α is indeed a product of three or fewer reflections. The converse follows from the fact that isometries preserve lengths and angles. The following remarkable characterization of congruent triangles is an immediate consequence of Theorem 93: Corollary 94 4P QR ∼ = 4ABC if and only if 4ABC is the image of 4P QR under three or fewer reflections. 2.4. THE FUNDAMENTAL THEOREM 57 Corollary 95 Two segments or two angles are congruent if and only if there exists an isometry mapping one onto the other. Proof. Two congruent segments or angles are contained in a pair of congruent triangles so such an isometry exists by Theorem 93. Since isometries preserve length and angle the converse also follows. Now we can define a general notion of congruence for arbitrary plane figures. Definition 96 Two plane figures s1 and s2 are congruent if and only if there is an isometry α such that s2 = α (s1 ). Exercises £¤ £¤ £0¤ £¤ £1¤ £ 12 ¤ 1. Let A = 00 ; B = 50 ; C = 10 ; D = 42 ; E = −2 ; F = −4 . Given that 4ABC ∼ = 4DEF, find three or fewer lines such that the image of 4ABC under reflections in these lines is 4DEF. £¤ £3¤ £8¤ 2. Let A = 67 ; B = 14 ; C = 15 . In each of the following, 4DEF ∼ = 4ABC. Find three or fewer lines such that the image of 4ABC under reflections in these lines is 4DEF. a. b. c. d. e. D E F £12¤ £9¤ £14¤ 1 £−1¤ 10 £ ¤ 4 −9 £ −4 ¤ −15 £−5¤ −4 8 9 £−8¤ 7 £ ¤ 7 −16 £−11¤ £−9¤ 12 £ 2 −17 ¤ −12 £−12¤ −1 −6 £−12¤ −17 £−13¤ 58 CHAPTER 2. CONGRUENCE