20 2.Axioms for the Plane The Other Axioms The last two axioms are stated here.What follows from assuming these axioms will be investigated in coming chapters. The Reflection Axiom is an existence axiom for a very particular kind of rigid mo tion,not described in much detail in the axiom.The geometry of this rigid motion wil be described in Chapter 3.Then by composing the transformations for different lines. in coming chapters we will prove many theorems of geometry and will generate famil iar transformations such as rotations and translations.Already in Chapter 3 we will begin to see consequences for the geometry of triangles and other figures. Axiom5(Reflection).Forany line m,thereexistsa rigid motion distinct from the identity transformation that fixes the points of m. We note by Theorem 2.12 that it is actually sufficient that this rigid motion fixes two points. The last of the six axioms concerns the properties of dilations.These transforma- tions are not rigid motions but are transformations that will be defined later as simili- tudes. Definition 2.13.For point A and a real number k 0,the dilation centered at A with dilation ratio k.denoted is a transformation of the plane defined as follows: (a)DA.k(A)=A:(b)ifXA.X )is the point on X with IlAX'll =kllAXll Axiom 6(Dilation).For every point A and every k>0,the dilation DAk preserves angle measure and scales every distance by k:for all P'=DAk(P)and Q'=Dak(Q). lIP'Q'll kllPQll. Note the difference in the nature of these two axioms.The Reflection Axiom is an existence axiom while the Dilation Axiom asserts properties of an existing transforma- tion that is already defined. We will see in Chapter 7 that the Dilation Axiom implies the famous Euclidea axioms is true not only in Euclidean geometry,but in hyperbolic non-Euclidean geom- etry as well.For many readers,this will not be important,but anyone thinking abou proving theorems that are true in both Euclidean and non-Euclidean geometry will note that,except for a very few well-identified and isolated exceptions,only the first five axioms are used in proofs until the end of Chapter 6. In Chapter 8,dilations will be central in the study of similarity.At that time,the definition of dilation will be e extended to negative
20 2. Axioms for the Plane The Other Axioms The last two axioms are stated here. What follows from assuming these axioms will be investigated in coming chapters. The Reflection Axiom is an existence axiom for a very particular kind of rigid motion, not described in much detail in the axiom. The geometry of this rigid motion will be described in Chapter 3. Then by composing the transformations for different lines, in coming chapters we will prove many theorems of geometry and will generate familiar transformations such as rotations and translations. Already in Chapter 3 we will begin to see consequences for the geometry of triangles and other figures. Axiom 5 (Reflection). For any line 𝑚, there exists a rigid motion distinct from the identity transformation that fixes the points of 𝑚. We note by Theorem 2.12 that it is actually sufficient that this rigid motion fixes two points. The last of the six axioms concerns the properties of dilations. These transformations are not rigid motions but are transformations that will be defined later as similitudes. Definition 2.13. For point 𝐴 and a real number 𝑘 > 0, the dilation centered at 𝐴 with dilation ratio 𝑘, denoted 𝒟𝐴,𝑘, is a transformation of the plane defined as follows: (a) 𝒟𝐴,𝑘(𝐴) = 𝐴; (b) if 𝑋 ≠ 𝐴, 𝑋 ′ = 𝒟𝐴,𝑘(𝑋) is the point on 𝐴𝑋⃗ with ‖𝐴𝑋′‖ = 𝑘‖𝐴𝑋‖. Axiom 6 (Dilation). For every point 𝐴 and every 𝑘 > 0, the dilation 𝒟𝐴,𝑘 preserves angle measure and scales every distance by 𝑘: for all 𝑃 ′ = 𝒟𝐴,𝑘(𝑃) and 𝑄 ′ = 𝒟𝐴,𝑘(𝑄), ‖𝑃′𝑄 ′‖ = 𝑘‖𝑃𝑄‖. Note the difference in the nature of these two axioms. The Reflection Axiom is an existence axiom while the Dilation Axiom asserts properties of an existing transformation that is already defined. We will see in Chapter 7 that the Dilation Axiom implies the famous Euclidean Parallel Postulate as a theorem. In contrast, everything proved using only the first five axioms is true not only in Euclidean geometry, but in hyperbolic non-Euclidean geometry as well. For many readers, this will not be important, but anyone thinking about proving theorems that are true in both Euclidean and non-Euclidean geometry will note that, except for a very few well-identified and isolated exceptions, only the first five axioms are used in proofs until the end of Chapter 6. In Chapter 8, dilations will be central in the study of similarity. At that time, the definition of dilation will be extended to negative 𝑘
Exercises and Explorations Exercises and Explorations 1.(Rulers and Protractors).Letp(x)be a ruler for a line m.It was claimed in Remark 2.1 that for a constant real number k,both p(x+k)and p(-x+k)are also rulers. Later it was similarly noted that if (t)is a protractor function at A.so are (t+k) and (-t+k). (a)Show that p(x+k)satisfies the definition of a ruler.Next show that p(-x)is a ruler.Combine the two results to prove the statement in the remark.You )hmcne bout po tors. 2.(Midpoints).Let points Aand Bcorrespond by a ruler on line ABtoaand b. (a)What real number m corresponds to the midpoint M of AB? (b)If D is a point corresponding to real number d and if B is the midpoint of AD, what number is d? 3.(Angle Bisectors).From Definition 2.9,an angle bisector is a ray that divides an ang mea ctor at A.let the pola r angles of B,C,and D be b,c,d.Assumin lb-cl<180 and D is interior to BAC,for what value of d will AD bisect this angle?In other words,for what D is mBAD =mDAC? (b)Two lines AB and AC form four angles at A.Explain how the four angle bi- sectors are related. 4.(Relationships on a Line).Let points Aand B correspond by a ruler on line AB to =3and b= 5.what number c corresponds to the point C on the line with =7a nd BCl =9? (b)If points A.B on line m correspond by a ruler to real numbers 3 and 5,wha numb er c corresponds to the point C=DA.s(B) (c)What is a general formula for c in terms of a and b if C=DA.k(B)? 5.(Fixed points on R).Suppose that f is a map from the real numbers R toR that (a)If f(3)=3.what are the possible values of f(7)? (b)If f(3)=3 what are the (c)If point a is a fixed what are the possible values of ny other (d),if fhastwo fixed points,what can you say about f(? 6.(Plane Separation and Triangles).The plane separation axiom allows one to prove some facts that are basic and visually very believable but rather technical.One may choose to believe them without going through the proofs.However,they do provide good practice in reasoning with the Plane Separation Axiom. (a)In AABC,let Dbe an interior point of BC and let Ebe an interior point of CA. Prove that AD and BE intersect at a point G,an interior point of the triangle
Exercises and Explorations 21 Exercises and Explorations 1. (Rulers and Protractors). Let 𝜌(𝑥) be a ruler for a line 𝑚. It was claimed in Remark 2.1 that for a constant real number 𝑘, both 𝜌(𝑥 + 𝑘) and 𝜌(−𝑥 + 𝑘) are also rulers. Later it was similarly noted that if 𝛿(𝑡) is a protractor function at 𝐴, so are 𝛿(𝑡 + 𝑘) and 𝛿(−𝑡 + 𝑘). (a) Show that 𝜌(𝑥 + 𝑘) satisfies the definition of a ruler. Next show that 𝜌(−𝑥) is a ruler. Combine the two results to prove the statement in the remark. You may wish to supplement words and symbols with drawings. (b) Show how almost identical reasoning justifies the statement about protractors. 2. (Midpoints). Let points 𝐴 and 𝐵 correspond by a ruler on line 𝐴𝐵 to 𝑎 and 𝑏. (a) What real number 𝑚 corresponds to the midpoint 𝑀 of 𝐴𝐵? (b) If 𝐷 is a point corresponding to real number 𝑑 and if 𝐵 is the midpoint of 𝐴𝐷, what number is 𝑑? 3. (Angle Bisectors). From Definition 2.9, an angle bisector is a ray that divides an angle into two angles of equal measure. (a) For a protractor at 𝐴, let the polar angles of 𝐵, 𝐶, and 𝐷 be 𝑏, 𝑐, 𝑑. Assuming |𝑏 − 𝑐| < 180 and 𝐷 is interior to ∠𝐵𝐴𝐶, for what value of 𝑑 will 𝐴𝐷⃗ bisect this angle? In other words, for what 𝐷 is 𝑚∠𝐵𝐴𝐷 = 𝑚∠𝐷𝐴𝐶? (b) Two lines 𝐴𝐵 and 𝐴𝐶 form four angles at 𝐴. Explain how the four angle bisectors are related. 4. (Relationships on a Line). Let points 𝐴 and 𝐵 correspond by a ruler on line 𝐴𝐵 to 𝑎 and 𝑏. (a) If 𝑎 = 3 and 𝑏 = 5, what number 𝑐 corresponds to the point 𝐶 on the line with ‖𝐴𝐶‖ = 7 and ‖𝐵𝐶‖ = 9? (b) If points 𝐴, 𝐵 on line 𝑚 correspond by a ruler to real numbers 3 and 5, what number 𝑐 corresponds to the point 𝐶 = 𝒟𝐴,5(𝐵)? (c) What is a general formula for 𝑐 in terms of 𝑎 and 𝑏 if 𝐶 = 𝒟𝐴,𝑘(𝐵)? 5. (Fixed points on ℝ). Suppose that 𝑓 is a map from the real numbers ℝ to ℝ that preserves distance. (a) If 𝑓(3) = 3, what are the possible values of 𝑓(7)? (b) If 𝑓(3) = 3, what are the possible values of 𝑓(𝑡), for any 𝑡? (c) If a point 𝑎 is a fixed point of 𝑓 (i.e., 𝑎 = 𝑓(𝑎)), what are the possible values of 𝑓(𝑡) for any other number 𝑡? (d) Based on your answers, if 𝑓 has two fixed points, what can you say about 𝑓(𝑡)? 6. (Plane Separation and Triangles). The plane separation axiom allows one to prove some facts that are basic and visually very believable but rather technical. One may choose to believe them without going through the proofs. However, they do provide good practice in reasoning with the Plane Separation Axiom. (a) In△𝐴𝐵𝐶, let 𝐷 be an interior point of 𝐵𝐶 and let 𝐸 be an interior point of 𝐶𝐴. Prove that 𝐴𝐷 and 𝐵𝐸 intersect at a point 𝐺, an interior point of the triangle
22 2.Axioms for the Plane (b)(Pasch Axiom).There are other axioms of plane separation that are used ir of Axiom4.One is called the Pasch Axiom:ifa line m intersects side AB fBCat one pointD,aineror point of side AB,then malso intersect either BC or CA.Prove this statement as a theorem using the axioms in this chapter. (c)(Quadrilateral Shapes).Sketch a few quadrilaterals,convex and nonconvex. In your examples,is there always at least one diagonal that separates the other two vertices into opposite half-planes of the diagonal?If this is always true, how can one prove it from the axioms?If there are two such diagonals,what does this imply about the convexity or nonconvexity of the quadrilateral? Suggestion:For each statement,begin by drawing possible cases and explain with informal reasoning what is possible before undertaking a more formal explanation
22 2. Axioms for the Plane (b) (Pasch Axiom). There are other axioms of plane separation that are used in place of Axiom 4. One is called the Pasch Axiom: if a line 𝑚 intersects side 𝐴𝐵 of △𝐴𝐵𝐶 at one point 𝐷, an interior point of side 𝐴𝐵, then 𝑚 also intersects either 𝐵𝐶 or 𝐶𝐴. Prove this statement as a theorem using the axioms in this chapter. (c) (Quadrilateral Shapes). Sketch a few quadrilaterals, convex and nonconvex. In your examples, is there always at least one diagonal that separates the other two vertices into opposite half-planes of the diagonal? If this is always true, how can one prove it from the axioms? If there are two such diagonals, what does this imply about the convexity or nonconvexity of the quadrilateral? Suggestion: For each statement, begin by drawing possible cases and explain with informal reasoning what is possible before undertaking a more formal explanation
Chapter 3 Existence and Properties of Reflections In the previous chapters,we have been describing properties of rigid motions,but with- out the Reflection Axiom.there is nothing that tells us that any rigid motions actually exist(except for the identity)! states the exist en e of some rigid motionsthat will turnou tobe linere l geometri in the axiom.A bare n imum of existence is promi is not even asserte eclared But using the strong restrictions on behavior mandated by preservation of distance and angle measure,we will see that these vaguely described rigid motions must behave in exactly one way. Recall the statement of the axiom: Axiom5(Reflection).Foranylinem.thereexistsarigidmotion distinct from theidentity transformation that fixes the points of m Deducing the Properties of Reflections At thipoiwedo not havegpmetrical deco of thim that this ie the in a lin sCiptioawlco with the e from the examination of a key figure Lemma 3.1(Y Figure).Given aline AB,take any point C not on the line.Then there is exactly one other point D in the plane so that AC=AD and m/BAC m/BAD The points C and D are in opposite half-planes of AB. Proof.Choose the protractorforrayswithendpointAso that the polar angle(AB)= 0.In the following.we choose polar angles in the interval-180<<+180. Letm∠BAC=t.Then the ony rays for whichm∠BAE=have polarangle +tor-t.There are only two such rays.One is AC itself.Call the other one AF. 2石
Chapter 3 Existence and Properties of Reflections In the previous chapters, we have been describing properties of rigid motions, but without the Reflection Axiom, there is nothing that tells us that any rigid motions actually exist (except for the identity)! This axiom states the existence of some rigid motions that will turn out to be line reflections, with the usual geometric description. However, this description is not stated in the axiom. A bare minimum of existence is promised; in fact it is not even asserted that there is only one rigid motion whose existence is declared. But using the strong restrictions on behavior mandated by preservation of distance and angle measure, we will see that these vaguely described rigid motions must behave in exactly one way. Recall the statement of the axiom: Axiom 5 (Reflection). For any line 𝑚, there exists a rigid motion distinct from the identity transformation that fixes the points of 𝑚. Deducing the Properties of Reflections At this point, we do not have a geometrical description of this rigid motion. The proof that this rigid motion is the usual reflection in a line with the usual geometrical description will come from the examination of a key figure. Lemma 3.1 (Y Figure). Given a line 𝐴𝐵, take any point 𝐶 not on the line. Then there is exactly one other point 𝐷 in the plane so that ‖𝐴𝐶‖ = ‖𝐴𝐷‖ and 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷. The points 𝐶 and 𝐷 are in opposite half-planes of 𝐴𝐵. Proof. Choose the protractor 𝛿 for rays with endpoint𝐴so that the polar angle 𝜃(𝐴𝐵) = ⃗ 0. In the following, we choose polar angles in the interval −180 < 𝜃 < +180. Let 𝑚∠𝐵𝐴𝐶 = 𝑡. Then the only rays 𝐴𝐸⃗ for which 𝑚∠𝐵𝐴𝐸 = 𝑡 have polar angle +𝑡 or −𝑡. There are only two such rays. One is 𝐴𝐶⃗ itself. Call the other one 𝐴𝐹⃗ . 23
24 3.Existence and Properties of Reflections Figure 1.Only One Match to the Angle and Distance ofC Let ACk.There is exactly one point on each ray at distancek from A.Oneof these points is C.Let the point on AF be D. Then lACl =ADII k,and mBAC=mzBAD =t.From the Plane Separation Axiom,these pointslie in opposite half-planes ofAB,so segment CDintersectsAB. ◇ Lemma 3.2(Reflection and Perpendicular Bisectors).For any AB,let T be arigid mo tion that fixes the points ofAB.IfC is not a fixed point ofT,then C is not on AB and T(C) is the point D constructed in the Y Figure Lemma.Moreover,the segment CD intersects AB at M.the midpoint of CD,and AB is the perpendicular bisector of CD. Proof.LetD=T(C),a point distinct from C.Tisa rigid motion that preserves angle measure and distance.Therefore,mBAC mBAD and lACll =ADII. So from the Y Figure Lemma,D is uniquely determined and on the opposite side of from C. Let M be the intersection of CD and AB.(If M=A,interchange the roles of A and B in the proof.) Since M is on AB,T(M)=M.Since T is a rigid motion,IICMI =IDMII.Therefore Mis the midpoint of CD. In addition mCMA=mDMA becauseTmapsone angle to the other.Since the gle n means Theorem 3.3(Three Fixed Points).IfT is a rigid motion with three noncollinear fixed points,then T is the identify transformation.Furthermore,if S and T are two rigid motions that agree at three noncollinear points,then the transformations are the same: S=T Proof.Let A,B,Cbe noncollinear fixed points of T.Suppose there isa point Gthat is not a fixed point,so that T(G)=H is distinct from G. each of the three lines emma 3.2, AB is the perpendicula bisector of GH and AC is also the perpendicular bisector of GH,as shown in Figure 2
24 3. Existence and Properties of Reflections M D C A B F E Figure 1. Only One Match to the Angle and Distance of 𝐶 Let ‖𝐴𝐶‖ = 𝑘. There is exactly one point on each ray at distance 𝑘 from 𝐴. One of these points is 𝐶. Let the point on 𝐴𝐹⃗ be 𝐷. Then ‖𝐴𝐶‖ = ‖𝐴𝐷‖ = 𝑘, and 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 = 𝑡. From the Plane Separation Axiom, these points lie in opposite half-planes of 𝐴𝐵, so segment 𝐶𝐷 intersects 𝐴𝐵. □ Lemma 3.2 (Reflection and Perpendicular Bisectors). For any 𝐴𝐵, let 𝑇 be a rigid motion that fixes the points of 𝐴𝐵. If 𝐶 is not a fixed point of 𝑇, then 𝐶 is not on 𝐴𝐵 and 𝑇(𝐶) is the point 𝐷 constructed in the Y Figure Lemma. Moreover, the segment 𝐶𝐷 intersects 𝐴𝐵 at 𝑀, the midpoint of 𝐶𝐷, and 𝐴𝐵 is the perpendicular bisector of 𝐶𝐷. Proof. Let 𝐷 = 𝑇(𝐶), a point distinct from 𝐶. 𝑇 is a rigid motion that preserves angle measure and distance. Therefore, 𝑚∠𝐵𝐴𝐶 = 𝑚∠𝐵𝐴𝐷 and ‖𝐴𝐶|‖ = ‖𝐴𝐷‖. So from the Y Figure Lemma, 𝐷 is uniquely determined and on the opposite side of 𝐴𝐵 from 𝐶. Let 𝑀 be the intersection of 𝐶𝐷 and 𝐴𝐵. (If 𝑀 = 𝐴, interchange the roles of 𝐴 and 𝐵 in the proof.) Since 𝑀 is on 𝐴𝐵, 𝑇(𝑀) = 𝑀. Since 𝑇 is a rigid motion, ‖𝐶𝑀‖ = ‖𝐷𝑀‖. Therefore 𝑀 is the midpoint of 𝐶𝐷. In addition 𝑚∠𝐶𝑀𝐴 = 𝑚∠𝐷𝑀𝐴 because 𝑇 maps one angle to the other. Since the sum of these angle measures is 180, each of these angles is a right angle. This means that 𝐴𝐵 is the perpendicular bisector of 𝐶𝐷. □ Theorem 3.3 (Three Fixed Points). If 𝑇 is a rigid motion with three noncollinear fixed points, then 𝑇 is the identify transformation. Furthermore, if 𝑆 and 𝑇 are two rigid motions that agree at three noncollinear points, then the transformations are the same: 𝑆 = 𝑇. Proof. Let 𝐴, 𝐵, 𝐶 be noncollinear fixed points of 𝑇. Suppose there is a point 𝐺 that is not a fixed point, so that 𝑇(𝐺) = 𝐻 is distinct from 𝐺. Then by the Two Fixed Points Theorem, Theorem 2.12, for each of the three lines 𝐴𝐵, 𝐵𝐶, 𝐶𝐴, all the points are fixed. Then by Lemma 3.2, 𝐴𝐵 is the perpendicular bisector of 𝐺𝐻 and 𝐴𝐶 is also the perpendicular bisector of 𝐺𝐻, as shown in Figure 2