Congruence. Chapter The Three Points Theorem

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1 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 and, fixes every point on the line Proof. Let and be distinct fixed points of an isometry and let be any point on line distinct from and Note that = since and are distinct and collinear. But 0 = () since = 0 and 0 = () since = 0 Therefore 0 = (see Figure 2.1) 39

2 40 CHAPTER 2. CONGRUENCE Figure 2.1. Theorem 74 An isometry that fixes three non-collinear points is the identity. Proof. Let and be three non-collinear points that are fixed under an isometry By Theorem 73, fixes every point on 4 Let be any point in the plane o of 4 and choose a point 6= interior to 4 (see Figure 2.2). Figure 2.2. Then line intersects 4 in two distinct points, which are fixed by so each point on is fixed by by Theorem 73. In particular, the arbitrarily chosen point is fixed by Therefore = as claimed. Theorem 75 (The Three Points Theorem) Two isometries that agree on three non-collinear points are equal.

3 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 41 Proof. Suppose that and are isometries and and are noncollinear points such that ( )=( ) () =() and () =() (2.1) Apply 1 to both sides of each equation in (2.1) and obtain =( 1 )( ) =( 1 )() and =( 1 )() Thus 1 is an isometry that fixes three non-collinear points and by Theorem 74, 1 = Apply 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 be distinct parallel lines and let be a common perpendicular. Let = and = Then is the translation by vector 2LM i.e., = 2LM Proof. Let 0 = (); then ( )() = ( ()) = () = 0 = LL 0() (2.2) Let be a point on distinct from and let 0 = LL 0() Then by Theorem 29. Therefore KK 0 = LL 0 ( )() = ( ()) = () = 0 = KK 0() = LL 0() (2.3) Let = (); then is the midpoint of and and is the midpoint of and 0 Hence JL = LM = ML 0 so that JM = JL + LM = LM + ML 0 = LL 0 and JM = LL 0 Therefore ( )() = ( ()) = () = = JM () = LL 0() (2.4) (see Figure 2.3).

4 42 CHAPTER 2. CONGRUENCE Figure 2.3. By equations (2.2), (2.3), and (2.4), the isometries and LL 0 agree on three non-collinear points and so that = LL 0 by Theorem 75. Since is the midpoint of and 0 we have LL 0 = 2LM as desired. In the proof above, is the midpoint of 0 so by Theorem 56 we have LL 0 = which proves the next corollary: Corollary 77 Let and be parallel lines and let be a common perpendicular. Let = and = Then = Figure 2.4. The next theorem tells us that the converse is also true:

5 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 43 Theorem 78 An isometry is a translation if and only if is a product of two reflections in parallel lines. Proof. Let be any line. Then = (the identity transformation) if and only if = 2, and the statement holds in this case. So assume that 6= and note that implication was proved in Theorem 76. For the converse, let and be distinct points and consider the non-identity translation LN ;we must show that LN is a product of two reflections in distinct parallel lines. Let be the midpoint of and ; let and be the perpendiculars to at and, respectively (see Figure 2.5). '= l ( ) '' = ( ') Then by Theorem 56, l Figure 2.5. and by Corollary 77, Therefore with k as desired. LN = = LN = Here is a useful trick that transforms a product of reflections in parallel lines into a product of halfturns. Given parallel lines and choose any common perpendicular ; let = and = Then = = ( ) =( ) ( )= Conversely, given distinct points and let = let be the line through perpendicular to and let bethelinethrough perpendicular to 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 and are points such that PQ = RS then PQ = RS Since the product of two reflections in distinct parallel lines is a translation, there

6 44 CHAPTER 2. CONGRUENCE is the following analogous statement for parallels: If lines and are parallel lines passing through collinear points and, respectively, then = Theorem 79 Let and be distinct parallel lines. There exist unique lines and parallel to such that = = Proof. Given distinct parallels and choose any common perpendicular. By Theorem 58, there exist unique points and on such that = and = Thus = = Let and be the lines perpendicular to at and, respectively (see Figure 2.6). Figure 2.6. Then by Corollary 77, = = = and = = = Thus given distinct parallels and, Theorem 79 tells us that determines unique parallels and such that = =. Furthermore, line is the unique line parallel to such that the directed distance from to equals the directed distance from to ; line is the unique line parallel to such that the directed distance from to equals the directed distance from to Thus:

7 2.2. TRANSLATIONS AS PRODUCTS OF REFLECTIONS 45 Corollary 80 Let and be distinct points and let be any line perpendicular to Then there exists a unique line parallel to such that (See Figure 2.7.) PQ = Figure 2.7. l Corollary 81 Let and be distinct parallel lines. There exists a unique line parallel to and such that = 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 parallel to and such that = (see Figure 2.8). l Figure 2.8.

8 46 CHAPTER 2. CONGRUENCE Apply to both sides and obtain = Exercises 1. Lines and have respective equations =3and =5 Find the equations of the translation 2. Lines and have respective equations = and = +4 Find the equations of the translation 3. The vector of the translation is 4 3 Find the equations of lines and such that = 4. The translation has equations 0 = +6 and 0 = 3 Find equations of lines and such that = 5. Lines and have respective equations =3 =5and =9 a. Find the equation of line such that = b. Find the equation of line such that = 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 are distinct intersecting lines, there is one directed angle from and with measure 1 in the range and another with measure 2 in the range In fact, 2 = Theorem 82 Let and be distinct lines intersecting at a point and let be the measure of the angle from to Then = 2 for every, i.e., is the rotation about through twice the directed angle from to.

9 2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS 47 Proof. First observe that ( )() = ( ()) = () = = 2 () (2.5) Let let be a point on distinct from and let be the point in such that the directed angle measure from to is Let 0 = 2 (); then 0 lies on and is the perpendicular bisector of 0 (see Figure 2.9). Therefore 0 = () and we have ( )() = ( ()) = () = 0 = 2 () (2.6) Let = (); then is the perpendicular bisector of in which case lies on and the directed angle measure from to is 2. '= ( ), l = ( ) l Figure 2.9. Hence = 2 () so that ( )() = ( ()) = () = = 2 () (2.7) By equations (2.5), (2.6), and (2.7), the two isometries and 2 agree on non-collinear points and Therefore by Theorem 75, as claimed. = 2 As with translations, the converse is also true:

10 48 CHAPTER 2. CONGRUENCE Theorem 83 A non-identity isometry is a rotation if and only if is the product of two reflections in distinct intersecting lines. Proof. The implication was proved in Theorem 82. For the converse, given 2 let be any line through and let betheuniquelinethrough such that the directed angle measure from to is (see Figure 2.10). By Theorem 82, 2 =. Figure There is the following analogue to Theorem 79: Theorem 84 Let and be distinct lines concurrent at point. There exist unique lines and concurrent at such that = = Proof. Given distinct lines and concurrent at point choose points on and on distinct from such that ; let By Theorem 82, 2 = (2.8) Choose a point on line distinct from and consider ray There exist unique rays and such that the directed angle measure from to and from to equals So let = and = (see Figure 2.11).

11 2.3. ROTATIONS AS PRODUCTS OF REFLECTIONS 49 l C Figure Then by Theorem 82 we have 2 = and 2 = (2.9) The result now follows from the equations in (2.8) and (2.9). Atranslation is a product of two reflections in distinct parallels lines and i.e., = Similarly, a rotation 2 is a product of two reflections in distinct lines and intersecting at i.e., 2 = Furthermore, if and are concurrent at Theorem 84 tells us that uniquely determines lines and also concurrent with at such that = = where is the unique line through such that the measure of the directed angle from to is ; istheuniquelinethrough such that the measure of the directed angle from to is Thus: Corollary 85 Let and let be an arbitrarily chosen line passing through point If there exist unique lines and passing through such that 2 = = Finally, if we multiply both sides of the equation = on the left by we obtain = Hence: Corollary 86 Let and be distinct lines concurrent at point. There exists a unique line passing through such that =

12 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 Ahalfturn is the product (in either order) of two reflections in perpendicular lines intersecting at Note that the identity = PP = 0 = for all, 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 have respective equations =3and = a. Find the equations of the rotation = b. Find the center and angle of rotation and 2. Lines and have respective equations = and = +4 a. Find the equations of the rotation = b. Find the center and angle of rotation and 3. Find the equations of lines and such that 90 = 4. Let = 3 4 Find equations of lines and such that 60 = 5. Lines and have respective equations =0 =2 and =0 a. Find the equation of line such that = b. Find the equation of line such that = 6. Let be a point on line Prove that = 7. If = prove that lines and are either concurrent or mutually parallel. 8. If lines and are either concurrent or mutually parallel, prove that =

13 2.4. THE FUNDAMENTAL THEOREM Construct the following in the figure below: l a. Line such that = b. Line such that = c. The fixed point of 10. Given distinct points and construct the point such that PQ 45 = Let = be a pair of congruent rectangles. Describe how to find a rotation such that ( ) = 12. Given distinct points and construct the point such that QR 120 = If and are the respective perpendicular bisectors of sides and of 4 find the line such that = 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 and be distinct points and let = We know that the identity and the reflection are isometries that fix both and Are there any other isometries with this property? Let be a non-identity isometry that fixes both and and let be any point not fixed by By Theorem 73,

14 52 CHAPTER 2. CONGRUENCE is o line and and are non-collinear. Let 0 = (); since is an isometry, = 0 and = 0 Consider the two circles and which intersect at points and 0 (see Figure 2.12). ' Figure Since and are equidistant from and 0 line is the perpendicular bisector of chord 0 Hence () = 0 = () ( )= = ( ) and () = = () and by Theorem 75, = Theorem 90 An isometry that fixes exactly one point is a non-identity rotation. Proof. Let be an isometry with exactly one fixed point let be a point distinct from and let 0 = ( ) Let be the perpendicular bisector of 0 (see Figure 2.13). Since is an isometry, = 0 Hence is on and () = By the definition of a reflection, ( 0 )= Therefore ( )() = (()) = () = and ( )( )= (( )) = ( 0 )= so that is an isometry that fixes two distinct points and.bytheorem 89, either = or = where =

15 2.4. THE FUNDAMENTAL THEOREM 53 l Figure ' Lines and are distinct since 6= 0. However, if = then =, which is impossible because fixes exactly one point while fixes infinitely many points. Therefore we are left with = and hence = The fact that 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 : R 2 R 2 is an isometry if and only if 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 be an isometry. If = choose any line and write = in which case the identity factors as a product of two reflections. So assume that 6= and choose a point such that 0 = ( ) 6= Let be the perpendicular bisector of 0 and observe that ( )( )= (( )) = ( 0 )=

16 54 CHAPTER 2. CONGRUENCE i.e., = fixes the point By Theorems 89 and 90, factors as a product of two or fewer reflections, which means that = 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 4 = 4 if and only if there is a unique isometry such that ( )= () = and () = Proof. Given 4 = 4 Theorem 75 tells us that if an isometry with the required properties exists, it is unique. Our task, therefore, is to show that such an isometry does indeed exist; we ll do this by constructing explicitly as a product of three isometries each of which is either the identity or a reflection. Figure Begin by noting that = = and = (2.10) by (see Figure 2.14). The isometry 1 : If = let 1 = Otherwise, let 1 = where is the

17 2.4. THE FUNDAMENTAL THEOREM 55 perpendicular bisector of In either case, 1 ( )= Let and note that 1 = 1 () and 1 = 1 () (see Figure 2.15). = 1 = 1 and = 1 1 (2.11) l 1( )= Figure The isometry 2 : If 1 = let 2 = Otherwise, let 2 = where is the perpendicular bisector of 1 By (6.1) and (6.2) we have = = 1 so the point is equidistant from points and 1 Therefore lies on and in either case we have 2 () = and 2 ( 1 )= Let and note that (see Figure 2.16). 2 = 2 ( 1 ) 1 = 2 and 1 1 = 2 (2.12)

18 56 CHAPTER 2. CONGRUENCE ( )= 2 1 l 2( )= Figure The isometry 3 : If 2 = let 3 = Otherwise, let 3 = where is the perpendicular bisector of 2 By (6.1), (6.2), and (2.12) we have = = 1 = 2 so the point is equidistant from points and 2 and lies on On the other hand, (6.1), (6.2), and (2.12) also give = = 1 1 = 2 so the point is equidistant from points and 2 and also lies on In either case we have 3 () = 3 () = and 3 ( 2 )= Let = and observe that ( )= 3 ( 2 ( 1 ( ))) = 3 ( 2 ()) = 3 () = () = 3 ( 2 ( 1 ())) = 3 ( 2 ( 1 )) = 3 () = () = 3 ( 2 ( 1 ()) = 3 ( 2 ( 1 )) = 3 ( 2 )= Therefore 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 4 = 4 if and only if 4 is the image of 4 under three or fewer reflections.

19 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 1 and 2 are congruent if and only if there is an isometry such that 2 = ( 1 ). Exercises 1. Let = 0 0 ; = 5 0 ; = 0 10 ; = 4 2 ; = 1 2 ; = 12 4 Given that 4 = 4 find three or fewer lines such that the image of 4 under reflections in these lines is 4 2. Let = 6 7 ; = 3 14 ; = 8 15 In each of the following, 4 = 4 Find three or fewer lines such that the image of 4 under reflections in these lines is 4 a. b. c. d. e

20 58 CHAPTER 2. CONGRUENCE

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