Transformations in High-School Geometry: A Workable Interpretation of Common Core UCLA Curtis Center: March 5, 2016 John Sarli, Professor Emeritus of Mathematics, CSUSB MDTP Workgroup Chair Abstract. Previous attempts in schools to interpret congruence and similarity in terms of transformations have largely failed because this fundamental idea was typically presented as an appendix to the geometry syllabus. The Common Core has addressed this situation systemically, but a successful geometry course requires proper organization in order to motivate the importance of geometric transformations. We outline a workable approach at the high-school level, with focus on the synthetic treatment of rigid motions, emphasizing the important distinction between congruence and measurement. I. Key statements in CA Mathematics Framework interpretation of CCSSM II. Synthetics rst III. Why are "rigid motions" rigid? IV. What is a "collineation"? V. Parallelism 1
I. Key statements in CA Mathematics Framework interpretation of CCSSM CCSSM: Conceptual Category: Geometry Understand congruence in terms of rigid motions. G-CO.6 Use geometric descriptions of rigid motions to transform gures and to predict the e ect of a given rigid motion on a given gure; given two gures, use the de nition in terms of rigid motions to decide if they are congruent. G-CO.7 Use the de nition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of lines and corresponding pairs of angles are congruent (sic). G-CO.8 Explain how the criteria for triangle congruence (ASA, SAS, and SSS) follow from the de nition of congruence in terms of rigid motions (sic). G-CO.10 Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180 ; base angles of an isosceles triangle are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point. Mathematics Framework: Key Advances. Because concepts such as rotation, re ection, and translation were treated in the grade-eight standards mostly in the context of hands-on activities and with emphasis on geometric intuition, the Geometry course places equal weight on precise de nitions. Table G-1. Standards for Mathematical Practice: Explanation and Examples for Geometry MP.8 Look for and express regularity in repeated reasoning. Students explore rotations, re ections, and translations, noticing that some attributes of shapes (e.g., parallelism, congruency, orientation (sic)) remain the same. They develop properties of transformations by generalizing these observations. 2
Conceptual Category: Geometry. Two shapes are congruent if there is a sequence of rigid motions (sic) in the plane that takes one shape exactly onto the other. Figure G-2. Congruent Corresponding Parts Imply Triangle Congruence (sic) Above all else, instructors should emphasize the reasoning involved in connecting one step in the logical argument to the next. II. Synthetics First It had for a long time been evident to me that geometry can in no way be viewed, like arithmetic or combination theory, as a branch of mathematics; instead geometry relates to something already given in nature, namely, space. (Hermann Grassmann, Die Ausdehnungslehre, 1844) Grassmann s observation helps us understand the special challenge that Geometry presents in addressing MP.6 Attend to precision. We need to allow students to become familiar with the plane, without coordinates, so that transformations of the plane can be properly understood and then, later, represented with coordinates. Transformations Are Not New: Some History Is Euclid I.4 (SAS) actually the proof of a theorem? Pappus and other geometers didn t think so, nor did they believe Euclid s proof of I.8 (SSS), because both proofs require that one triangle "be applied" to the other. Pappus produced alternative proofs for many of Euclid s propositions by de ning "congruence" to mean "relatable by rigid motion". His critiques were part of an inquiry that lasted for centuries and culminated in David Hilbert s formulation (1899) of a complete and consistent set of axioms for Euclidean geometry, and the formalization of congruence in terms of transformations. 3
One of Hilbert s axioms (IV-6) is equivalent to SAS: If triangles ABC and DEF are such that AB = DE, BC = EF and ]B = ]E, then CA = F D, ]A = ]D and ]C = ]F. This statement is in terms of measurements - no mention of "congruence" yet. In a high-school course we pare down Hilbert s work to a few essentials. For example, we accept without deeper justi cation some of Euclid s "common notions" such as: If C is on segment AB but is neither A nor B then AB is greater than AC. In triangle ABC, if D is on segment AC but is neither A nor C then ]ABC is greater than ]ABD. Euclid uses these as givens, together with SAS, to prove proposition I.16: The exterior angle of a triangle is greater in measure than either of its opposite interior angles. The rst twenty-eight propositions in Book I of Euclid s Elements do not require his parallel postulate but many, such as I.16 and I.26 (ASA), use SAS. Look at Euclid s proof of I.26 (ASA). It uses only SAS and the "common notions" above. (He also uses I.16 in the proof but does not really need it.) The proof of I.27 (If a transversal to two given lines makes alternate angles equal in measure then the given lines are parallel) requires I.16. Later we will see that, using transformations, we can get SSS from SAS. So, we add SAS to Euclid s " ve axioms": 4
Two distinct points determine a segment. Any segment extends uniquely and continuously to a straight line. The is a unique circle with given center and radius. All right angles are equal. SAS The unique parallel postulate. Geometers would later unpack the assumptions hidden in these axioms. What, for example, did Euclid mean by the fourth postulate? He was at once asserting that angles can be measured unambiguously, and that this measurement does not change with location. (Pappus noted that Euclid never de nes congruence but refers to superposition of gures on occasion. Thus any right angle can be superimposed on any other right angle.) From this postulate we obtain the notion of complementary and supplementary angles, and nd that vertical angles have equal measures. a g b + = 180 = + ) = 5
Again, the emphasis is on measurement. We are not ready to talk about congruence yet. Hilbert s work was motivated by the discovery of non-euclidean geometry, which results when we postulate that there is more than one parallel through a point not on a given line. Hilbert s contemporary, Felix Klein, was motivated by the Pappus interpretation of congruence that uses rigid motion, so he de ned geometry by the transformations that preserve important properties. This de nition can help us avoid a contemporary problem in education: Many students perceive Geometry as a random collection of unorganized claims and individual facts. As a result they will have di culty applying Mathematical Practice Standards such as: MP.8 Look for and express regularity in repeated reasoning. III. Why are "rigid motions" rigid? Re ection is a Rigid Motion - Why? Pappus did not de ne transformations as functions but understood that re ection in a line is the most fundamental step in producing superposition. We still use his de nition: The re ection in line l is the mapping P 7 P 0 such that l is the perpendicular bisector of segment P P 0. Note that if P is on l then P 0 = P, whereas if P is not on l then P 0 and P are in opposite half-planes. Any re ection is a transformation (invertible mapping). Further, if C is on line AB then there is a unique point D on A 0 B 0 such that l is the perpendicular bisector of segment CD. Thus D = C 0 and so a re ection takes lines to lines: It is a collineation. 6
The following results can be taught as classroom activities. Allow students to draw their own diagrams in order to reinforce the important idea that no one gure represents every possible interpretation of a geometric statement. Di erent diagrams can be compared for discussion purposes. 1. Let A 0 and B 0 be the re ections of points A and B, respectively, in a line l. Then AB = A 0 B 0. Proof. Let P be the midpoint of AA 0 and Q be the midpoint of BB 0. (So P and Q are on l.) Then triangles AP Q and A 0 P Q correspond by SAS because: P Q is a common side, A 0 P = AP since l bisects AA 0, and both angles at P are right angles. Thus AQ = A 0 Q, ]P AQ = ]P A 0 Q, and ]AQP = ]A 0 QP. Then, triangles ABQ and A 0 B 0 Q correspond by SAS because AQ = A 0 Q, BQ = B 0 Q, and ]AQB = 90 ]AQP = 90 ]AQP = ]A 0 QB 0. Thus AB = A 0 B 0 (and ]ABQ = ]A 0 B 0 Q, ]BAQ = ]B 0 A 0 Q). Note. This proof has also shown that if the vertex of an angle is on l then its re ection in l has the same measure. Thus: 2. Re ection preserves the measure of angles. Proof. Let \A 0 B 0 C 0 be the re ection of \ABC in l and let P = l \ CB (let P = l \ AB if lk CB). Let D be the intersection of AB with the perpendicular to l at P. Then, by our note, ]BP D = ]B 0 P D 0 and so, by SAS, ]P BD = ]P B 0 D 0. But then ]ABC = 180 ]P BD = 180 ]P B 0 D 0 = ]A 0 B 0 C 0. We now know that re ection is a rigid motion, so we do not need to postulate (as the Framework seems to suggest) that re ections preserve length and angle measure. Pappus s Proofs of Euclid I.5 and I.6 G-CO.10 recommends proving Euclid I.5 but, using re ections, it is easy to get proofs of both I.5 and its converse I.6. 7
I.5 The base angles of an isosceles triangle have equal measure. Proof. Re ect in the bisector of vertex angle A. Then A 7 A, and B 7 B 0 with B 0 on ray AC because re ection preserves angle measure. But it is given that AB = AC, so B 0 = C because re ection preserves length. Similarly, C 0 = B, so \ABC re ects to \ACB. Therefore these base angles have equal measure. I.6 If two angles in a triangle have equal measure then their opposite sides have equal length. Proof. Given that ]ABC = ]ACB, re ect in the perpendicular bisector of segment BC. Then ABC A 0 CB and so ]A 0 CB = ]ABC = ]ACB. But then A and A 0 are both on ray CA. Similarly, ]A 0 BC = ]ACB = ]ABC and so A and A 0 are both on ray BA. Thus A = A 0 and so AB re ects to AC. Therefore AB = AC. As a corollary we get the following special case () of SSS: () For distinct points C and D, let triangles ABC and ABD be such that AC = AD and BC = BD. Then the three pairs of corresponding angles are equal in measure. Proof. If we construct segment CD we nd, by I.5, that ]ACD = ]ADC and that ]BCD = ]BDC. Then ]ACB = ]ADB, and so, by SAS the remaining pairs of corresponding angles are also equal in measure. Corollary to the Proof. If two circles with centers A and B intersect at the two points C and D, then line AB is the?-bisector of CD, so the re ection in AB interchanges C and D. Translations and Rotations Using SAS we have shown that re ection is a rigid motion: Re ection preserves length and, as a consequence, also preserves measure of angle. 8
We can now make new rigid motions by composing re ections. A translation is a mapping that moves every point a given distance in a given direction. (Thus, the translation is determined by P 7 P 0, for any given point P. Any translation is a transformation.) The rotation about point P through angle is the mapping A 7 A 0 such that (1) if point A is di erent from P, then P A 0 = P A and ]AP A 0 = ; and (2) if point A is the same as point P, then A 0 = A. (Thus, A 0 is on the circle with center P and radius P A. Any rotation is a transformation.) 3. The composition of two re ections in parallel lines is a translation perpendicular to the lines through twice the distance between the lines. The composition of two re ections in lines that intersect at point O is a rotation about O through twice the angle between the lines. The proof of this theorem requires an understanding of signed distances and signed (oriented) angle measures. At the high-school level it is probably best to take an exploratory approach to this important result. In the process of "discovering the kaleidoscope" students may become curious as to what other rigid motions, if any, can be produced from re ections. Composing three re ections will lead them to a fourth possibility - something other than a re ection, translation, or rotation - called a glide (a re ection followed by a translation parallel to the line of re ection). In a college geometry course it is shown that no other type of rigid motion can result from a composition of re ections; and conversely, that every rigid motion can be produced from three or fewer re ections. In a high-school course we are more concerned with using the four rigid motions to establish congruence of gures: We say two gures are congruent if there is a rigid motion of the plane that moves one gure to the other (Pappus superposition). 9
SSS. If the lengths of the sides of two triangles are correspondingly equal in pairs then the pairs of corresponding angles are equal in measure. Proof. Let ABC and DEF be the two triangles with corresponding sides equal in length. Apply the translation A 7 A 0, where A 0 = D. Since translation is a rigid motion, we have DB 0 = AB = DE, so we next apply the rotation about D that takes B 0 7 B 00, where B 00 = E. Where is C 00, the image of C after these two rigid motions? On the one hand, it must be on the circle with center D that passes through F, but it must also be on the circle with center E that passes through F. So either C 00 = F (in which case A 7 D, B 7 E, C 7 F ), or we are in the special case () of SSS, above. In either case the corresponding angles are equal in pairs. Corollary to the Proof. Triangles ABC and DEF are congruent. The Corollary to the Proof is what Pappus was getting at in his interpretation of Euclidean geometry. It took centuries before it became the accepted way of looking at congruence. Small wonder that introduction into classrooms meets with resistance and misinterpretation. IV. What is a Collineation? Functions on the Plane Mappings - assign an image point to a given point Transformations - mappings that are invertible (one-to-one and onto): 1) Collineations - transformations that take lines to lines 2) Similarities - collineations that preserve absolute angle measurement 3) Isometries (Rigid Motions) - similarities that preserve length (every isometry is the composition of three or fewer re ections) Collineations are Determined by Triangles 10
1) A correspondence from a given triangle to any other given triangle determines a collineation. 2) A correspondence from a given triangle to another with proportional sides determines a similarity. 3) A correspondence from a given triangle to another with equal sides determines a rigid motion. Remember: Two gures are congruent if there is a rigid motion of the plane that moves one gure to the other. For example, for two triangles ABC and DEF we only write ABC ' DEF to mean there is a rigid motion T such that T (A) = D T (B) = E T (C) = F Every point on gure ABC is mapped to some point on gure DEF, because T is a collineation, and every point on DEF is mapped back to some point on ABC by T 1. Given two gures that we "suspect" are congruent, how do we nd a rigid motion T that maps one to the other? Why would we "suspect" that two gures are congruent? When we compare two gures we may notice that corresponding "parts" may have equal measure. If the gures appear to have no obvious "parts" we may use our intuition about "size and shape". In either case, we look for a rigid motion. Example. Figure G-2 (page 7, CA Mathematics Framework) Revisited 11
G-CO.7 Use the de nition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of lines and corresponding pairs of angles are congruent (sic). G-CO.7 (restated): Use the de nition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are equal in measure. As the Framework states, one direction is easier than the other. In fact, if ABC ' DEF then, by de nition, there is a rigid motion such that A 7 D, B 7 E, C 7 F, and this transformation preserves the measures of the parts. Conversely, suppose corresponding sides and angles are equal in measure. Let s reword the justi cation on page 7 of the Framework Geometry Chapter that exhibits the desired rigid motion: First, apply the translation T de ned by A 7 D. We know that T (B) is on the circle centered at D with radius AB = DE, so we can apply a rotation R about D that moves T (B) to E (if necessary). The rigid motion R T has sent A to D and B to E. If it has also sent C to C 0 = F then R T is the rigid motion that allows us to claim ABC ' DEF. But if C 0 is not F then triangles DEC 0 and DEF are related as in the special case () of SSS, so we can apply the re ection S in the line DE; in this case, S R T is the rigid motion that allows us to claim ABC ' DEF. Orientation When worded properly, the above argument that triangles with corresponding parts equal in measure are actually congruent is more than a justi cation: It is a valid proof, in fact, one that typi es the level of argument we expect in a Common Core high-school Geometry course. The most important proofs provide opportunities to explain simple, modern ideas. For example, the above proof is constructive and gives students an opportunity to understand that two triangles either have the same or opposite orientation: If R T was the rigid motion that worked then the triangles had the same orientation. If the rigid motion was S R T then the triangles had opposite orientations. 12
More generally, the re ected image of any gure has the opposite orientation of the original gure. A second re ection in any line restores the original orientation. Since we are now asking students to construct sequences of rigid motions it is not too much to ask that they understand why translations and rotations preserve orientation, whereas re ections and glides reverse orientation (Table G-1, MP.8). Recommended reading: Chapter 2 of Brannan/Esplen/Gray Geometry (2nd edition), Cambridge, 2012. 13
A Construction Exercise: Determining a Rigid Motion Given: ABC and A 0 B 0 C 0 have equal corresponding parts. 14
Where is the image P 0 of arbitrary point P? 15
A P B' A' B P 0 is now in one of two places. 16
A P C B' P' A' B C' The rigid motion is determined. 17
Orientation Matters P B C A' C' A B' Isosceles triangles with same orientation. Find image of point P. 18
P C B A' C' A B' Isosceles triangles with opposite orientation. Find image of point P. V. Parallelism So far, we have outlined a treatment of synthetic, "neutral" geometry: We have not used Euclid s parallel postulate. Although there are many types of geometry, school mathematics is devoted primarily to plane Euclidean geometry,... Geometry Chapter of the CA Mathematics Framework (page 2) Non-Euclidean geometry was developed under the assumption that that Euclid s Parallel Postulate (EPP) is independent of his other axioms - it was easier to postulate alternatives to EPP by stating it in the Proclus/Playfair/Hilbert (PPH) form. Both EPP and PPH are equivalent to Euclid s proposition I.32 ( ): 19
EPP. If the interior angles on one side of a transversal to two given lines sum to less than two right angles, then the two lines intersect on that side of the transversal. PPH. Through a given point not on a given line there is a unique line that does not intersect the given line.. The interior angles of any triangle sum to two right angles. In a college geometry course the equivalence of these statements is usually proved in the order: EPP =) PPH =) =) EPP The last step actually requires Mathematical Induction and limit arguments, which helps explain why the independence of EPP was disputed for so long. The other two steps are elementary. In high school we focus on PPH =) but, to appreciate why the uniqueness of the parallel line is required, it is also helpful to present EPP =) PPH. Note. Teachers should be aware that I.27 (equal alternate angles implies parallel) does not require the parallel postulate but its converse I.29 (parallel implies equal alternate angles) does require it; Euclid uses I.29 to prove. This crucial distinction between Euclidean and non-euclidean geometry provides an excellent opportunity to exercise: MP.3 Construct viable arguments and critique the reasoning of others. Typically, when you prove that the sum of the interior angles of a triangle is 180 you begin by constructing the parallel to a side through the third vertex. Where do you use the uniqueness of this parallel line in your proof? 20
EPP =) PPH. Proof. Let P be any point not on the given line CD. Let Q be any point, other than C or D, on segment CD and let AB be a line through P such that ]DQP = ]AP Q. By I.27, AB k CD. Let A 0 B 0 be any line through P distinct from AB, with A 0 in the same half-plane relative to P Q as A, and B 0 in the same half-plane as B. If ]A 0 P Q < ]AP Q, then ]A 0 P Q + ]CQP < 180 because ]DQP = 180 ]CQP ; if ]B 0 P Q < ]BP Q, then ]B 0 P Q + ]DQP < 180 because ]CQP = 180 ]DQP. In either case, EPP implies that AB must intersect CD. Thus AB is the unique line through P that is parallel to CD. PPH =). Proof. Given triangle ABC, let DE be the unique parallel to BC through A, with D and E chosen so that \BAD is alternate to \ABC, and \EAC is alternate to \BCA. By I.27, if D 0 is in the same half-plane as D relative to AB such that ]BAD 0 = ]ABC, then AD 0 k BC, so by uniqueness, AD 0 = DE. Thus ]BAD = ]ABC and, similarly, ]EAC = ]BCA. It follows that ]ABC + ]BCA = ]BAD + ]EAC. Since ]BAD + ]EAC + ]CAB = 180 it follows that the sum of the interior angles of triangle ABC also equals 180. These are good examples of theorems whose proofs do not require the use of transformations. 21