Plane Geometry - Bruce E. Shapiro

Foundations of GeometryLecture Notes for Math 370California State University, NorthridgeRevised for Spring 2013

ContentsBUGGY DRAFT – MAY CONTAIN ERRORS1 Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 NCTM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 CA Standard . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Common Core . . . . . . . . . . . . . . . . . . . . . . . . . . 175 Logic and Proof . . . . . . . . . . . . . . . . . . . . . . . . . 196 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Euclid’s Elements . . . . . . . . . . . . . . . . . . . . . . . . 298 Hilbert’s Axioms . . . . . . . . . . . . . . . . . . . . . . . . 339 Birkhoff/MacLane Axioms . . . . . . . . . . . . . . . . . . 3710 The SMSG Axioms . . . . . . . . . . . . . . . . . . . . . . . 4111 The UCSMP Axioms . . . . . . . . . . . . . . . . . . . . . . 4512 Venema’s Axioms . . . . . . . . . . . . . . . . . . . . . . . . 4913 Incidence Geometry . . . . . . . . . . . . . . . . . . . . . . 5314 Betweenness . . . . . . . . . . . . . . . . . . . . . . . . . . . 6115 The Plane Separation Postulate . . . . . . . . . . . . . . . 7516 Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7917 The Crossbar Theorem . . . . . . . . . . . . . . . . . . . . 8718 Linear Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . 9119 Angle Bisectors . . . . . . . . . . . . . . . . . . . . . . . . . 9520 The Continuity Axiom . . . . . . . . . . . . . . . . . . . . . 9921 Side-Angle-Side . . . . . . . . . . . . . . . . . . . . . . . . . 10322 Neutral Geometry . . . . . . . . . . . . . . . . . . . . . . . 10723 Angle-Side-Angle . . . . . . . . . . . . . . . . . . . . . . . . 11124 Exterior Angles . . . . . . . . . . . . . . . . . . . . . . . . . 11525 Angle-Angle-Side . . . . . . . . . . . . . . . . . . . . . . . . 11926 Side-Side-Side . . . . . . . . . . . . . . . . . . . . . . . . . . 12127 Scalene and Triangle Inequality . . . . . . . . . . . . . . . 12528 Characterization of Bisectors . . . . . . . . . . . . . . . . . 12929 Transversals . . . . . . . . . . . . . . . . . . . . . . . . . . . 133iii

ivCONTENTS30 Triangles in Neutral Geometry . . . . . . . . . . . . . . . . 13731 Quadrilaterals in Neutral Geometry . . . . . . . . . . . . . 14532 The Euclidean Parallel Postulate . . . . . . . . . . . . . . 15533 Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16734 The Parallel Projection Theorem . . . . . . . . . . . . . . 17335 Similar Triangles . . . . . . . . . . . . . . . . . . . . . . . . 17736 Triangle Centers . . . . . . . . . . . . . . . . . . . . . . . . 18137 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18738 The Pythagorean Theorem . . . . . . . . . . . . . . . . . . 19139 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20140 Circles and Triangles . . . . . . . . . . . . . . . . . . . . . . 21341 Euclidean Circles . . . . . . . . . . . . . . . . . . . . . . . . 21942 Area and Circumference of Circles . . . . . . . . . . . . . 22943 Indiana Bill 246 . . . . . . . . . . . . . . . . . . . . . . . . . 23744 Estimating π . . . . . . . . . . . . . . . . . . . . . . . . . . . 23945 Euclidean Constructions . . . . . . . . . . . . . . . . . . . . 25146 Hyperbolic Geometry . . . . . . . . . . . . . . . . . . . . . 26347 Perpendicular Lines in Hyperbolic Geometry . . . . . . . 27148 Parallel Lines in Hyperbolic Geometry . . . . . . . . . . . 27549 Triangles in Hyperbolic Geometry . . . . . . . . . . . . . . 28350 Area in Hyperbolic Geometry . . . . . . . . . . . . . . . . 28951 The Poincare Disk Model . . . . . . . . . . . . . . . . . . 29752 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30553 Spherical Geometry . . . . . . . . . . . . . . . . . . . . . . 313A Symbols Used . . . . . . . . . . . . . . . . . . . . . . . . . . 319Bibliography . . . . . . . . . . . . . . 323« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 1PrefaceReflectionsFoundations of Geometry is a content class for future high school teachers.In most college math classes, the focus is on a logical developmentof mathematical tools and techniques; their underlying theory, derivationsand proofs; and application to the real world or other mathematical subdisciplines.As a content class we will do all this, but we will also makean (albeit limited) attempt to integrate the material in a way that it ispertinent to your future as a an educator.On Pedagogy. As teachers you will need to know the most effective wayto present the material to others. To be effective, you should be clear, concise,and interesting. To be successful your presentation needs to promoteunderstanding and retention.Most of this cannot be learned in the classroom. It requires hard workand lots of practice. I can tell you what I find effective but that might nothelp very much. My students are more mathematically mature than yourswill be - after all, they are college mathematics majors. My students arepeople who like math, want to learn more about it, and (in the your case)are anxious to share that excitement with others. Middle school and highschool is different. So most of what works for me probably won’t work verywell in your classrooms.As you progress towards your California teaching credential you will takea lot of classes in pedagogy, the art or science of teaching. Some of thesewill focus on pedagogy in mathematics and others will not. A few of you1

2 SECTION 1. PREFACEare enrolled in integrated programs that teach content simultaneously withpedagogy (but in different classes), or already have teaching experience andwant a refresher in the content area. To these students what I have to sayhere is probably nothing new.Integrating Content with Pedagogy. Our focus is on mathematicalcontent. But can content be separated from pedagogy? Should it be? Thisis not an easy question. Some would say that as mathematicians we should“leave sociology to the sociologists.” This has a grain of truth to it - wecan best teach what we know best - and instructors might do more harmthan good by straying outside their own specialized disciplines. It seemsnearly everybody has an opinion on the matter. But to leave pedagogy outof the equation completely would be you, as a future teacher, a disservice.This is not a class in pedagogy. I am not going to tell you how to teach.I couldn’t even if I wanted to. But what I want you to do as you progressthrough the semester is to think about how you can communicate yourknowledge to others. Start with your colleagues in this class. If you reallyunderstand something you should be able to explain it to people with backgroundssimilar to yours. Ask for their criticism and listen to what theysay. Take their suggestions into account the next time you work together.Of course being able to explain your proof of the Pythagorean theorem tothe student sitting next to you does not mean you’ll be able to explain itto your ninth grade students. But it will get you thinking about explainingthings in your own words. Listen to what you say. Explain things the wayyou would want them explained to you. Think about what questions youmight have, and answer them to yourself.As a teacher in training, you should continually be asking yourself “howam I ever going to teach this?” whenever you are presented with newmathematical content. Yes, you should certainly master the material. Yes,you should learn the “big picture” - how does this new content mesh withwhat you already know about math? Where else can we go from here?What other theorems, results, and methods can we obtain? How can thiscontent be applied to other applications outside pure math?Topics and Presentation. This class will give you the big picture.You’ll get “down and dirty” in other classes that are more focused on specificgrade level requirements. How to present this big picture has a longand well storied history.You could say it all started with Euclid – but all he really did was write« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 1. PREFACE 3down in a formal manner the knowledge of his day. He started by layingout lists of assumptions and basic axioms, and then asking what couldbe derived, using the logical methods of his day, from those assumptions.The result has come to be known as Euclidean geometry. Euclid boiledthe basics down to five axioms (after writing a whole mess of somewhatconfusing definitions). And followed it with a zillion pages of theorems.In the past century or so other mathematicians have tried to re-formulatedEuclid’s axioms so that they are easier to learn. These other presentations,starting with Hilbert and continuing up to the common core standardsused today, are all completely equivalent, in the sense that the same massof mathematical knowledge is derivable from them. This process of redefiningEuclid to make things easier to teach – and learn – has not beenwithout controversy.On Standards. The Common Core State Standards Initiative has nowbeen adopted by 45 states, the District of Columbia, Guam, AmericanSamoa, and the U.S. Virgin islands. 1 These standards give teachers (orschools, or boards of education, depending upon where you end up teaching)a framework within which to begin your presentation. These recommendationsserve as a resource for educators, and help provide a guide for the developmentof curriculum, instruction, and assessment. Since the the choiceof textbook will, in all likelihood, be determined by factors outside yourcontrol, the formal method of presentation may also be pre-determined.The common core standards adopted in California for a high school geometryclass give a list of 48 specific topics that should be covered divided intocategories that include congruence, similarity, right triangles, circles, arclengths, measurement and dimension (among others). As you go throughthis course I want you to think about how what we are studying fits intothis material. Ask yourself questions like this: how can I put this togetherinto a coherent, logical sequence? How can I make it interesting and fun,challenging and not boring? How can I promote comprehension? How canI make my students more autonomous? How can I communicate high expectationsto all my students without bias? How can I treat all my studentsequally? How can I assess my students’ retention and understanding? Howcan technology help? What sort of examples would make this easier to understand?What sort of activities would promote retention? What shouldgo into my lesson plan?These are not easy question and there are no easy answers. Thinking aboutthem now, while you are learning geometry for the first time (or perhaps1 Holdouts are Alaska, Texas, Nebraska, Minnesota, Virginia, and Puerto Rico.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

4 SECTION 1. PREFACEre-learning it from a different perspective) should help make you betterteachers in the end. There is a lot of work to be done, and a long andbumpy road ahead of you.However, the more traditional approach in ourstate is to learn the content material first, then study pedagogy.Spring 2012 ReflectionsThe notes have been only marginallymodified this semester, primarilyto reflect an easier to read pagesize 2 . I have also run thingsthrough the spell-checker and fixedprobably half of the spelling errors.Unfortunately the grammarand style is still atrocious. I didhave fun discovering such interestingspelling corrections that weresuggested such as circumciser forcirumcenter; orthodontic for orthocenter;verticals for vertices; Nebraskafor Bhaksara; saccharine forSaccheri; Stouffer for Plouffe; geographyfor geogebra; confiscatefor Chudnuvsky; and YardmasterKandahar for Yasumasa Kandahar.The Mohr-Moscheroni Theorembecame the More MaraschinosTheorem; Clairut’s Axiom the Clairol Axiom; Boylai’s Theorem the BonsaiTheorem; and our venerable textbook by Venema, was reduced to a pitifulenema. I also discovered that non-mathematicians spell pointwise astwo words or with a dash, as in point-wise, and that I can’t spell isoscelesworth beans. I guess, like the scarecrow, I deserve my Th.D. (Doctorate ofThinkology), for that.Spring 2013. I’ve fixed all of the bugs that were discovered last spring,added some stuff on the common core, and shortened the original preface,though I think its still too wordy.2 That is, for me, with low vision, standing in front of the class, lecturing.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 2NCTM MathRecommendationsThe National Council of Teachers ofMathematics (http://www.nctm.org)is the world’s largest mathematicseducation organization, with almost100,000 members. They have publishedseveral highly influential – andat times, controversial – standards overthe past several years. The first setof curriculum standards was released in1989; professional standards for teachersin 1991; and assessment standardsin 1995. A more focused set of contentstandards was released in their 2000 report.1 Although California was one ofthe first states to adopt much of theoriginal 1989 standards, they remaincontroversial, even in our own university.The 2000 report listed four goals of its study:1. To set forth a comprehensive set of goals for students at all levels(PK-12);2. To serve as a resource for educators and policy makers;1 Principles and Standards for School Mathematics, NCTM, 2000, ISBN 0873534808.5

6 SECTION 2. NCTM3. To guide in the development of curriculum frameworks, assessments,and instruction; and4. To stimulate discussion at all levels on how best to help students learnmath.It recommendations included sixprinciples for primary and secondarymathematics education anda set of content standards for eachsubject area. Recommended contentis discussed in the next section.The six principles of schoolmathematics recommended by theNCTM were Equity, Curriculum,Teaching, Learning, Assessment,and TechnologyEquity. The report challengesthe stereotype that only somestudents are capable of learningmathematics, and compares it tothe pervasive belief that everyoneshould learn to read and write inEnglish. This bias leads to low expectations for many students, especiallyamong students who live in poverty, are non-native English speakers, havedisabilities, or are female, among other groups. The Equity Principle demandsthat high expectations be communicated in words and deeds to allstudents with instructional programs that are interesting to students andhelp them see the importance of mathematical study.Curriculum. A curriculum is more than a collection of activities: it mustbe coherent, focused on important mathematics, and well articulated acrossthe grades. Lessons should be planned so that fundamental ideas form anintegrated whole.Teaching. Effective mathematics teaching requires understanding whatstudents know and need to learn and then challenging and supporting themto learn it well. Effective teaching requires knowing and understandingmathematics, students as learners, and pedagogical strategies. Teachersneed to understand the “big ideas” in math and present them as a coherent« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 2. NCTM 7and connected subject. They must know where commen misunderstandingsarise, and be able to present important concepts like “a fraction ispart of the whole” with different or multiple representations of each ideaor concept. Effective teaching also requires a challenging and supportiveclassroom environment. Well chosen tasks can encourage students’ curiosity.Teachers must decide which aspect of a task to highlight, what questionsto ask, and how to support students without doing the thinking forthem. Finally, effective teaching requires continually seeking improvement.You will learn much by observing your students, listening carefully to theirquestions, ideas, and explanations, and analyzing what your students aredoing and how your actions affect your students’ learning.Learning. Students must learn mathematics with understanding, activelybuilding new knowledge from experience and prior knowledge. Factualknowledge, procedural proficiency, and conceptual understanding are thethree components to this. Memorization of facts without understandingleaves students unable to apply these facts or procedures to knew domains,while conceptual understanding together with factual and proceduralknowledge makes subsequent learning easier. Autonomous learningthen becomes possible, and students can eventually take control of theirown learning.Assessment. Assessment shouldsupport the learning of importantmathematics and furnish useful informationto both teachers and students.In the modern age of standardizedtests and numerical metricsfor corporate quality and productionthis seems to have been forgotten.Good assessment practicesprovide information to both thestudent and the teacher. Scoringguides or rubrics can help teachersunderstand students’ proficiencies.And poor student performance is just as often an indication that the teacheris not providing sufficient grounding in the content being assessed. Finally,since students have different learning methods different types of assessmentsmay be necessary to make an effective decision regarding student success.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

8 SECTION 2. NCTMTechnology. Technology is essential in teaching and learning mathematics;it influences the mathematics that is taught and enhances students’learning. Used properly, technology can help students learn math. This isparticularly so where a number of programs allow students to perform thetraditional compass-and-straightedge constructions with infinite accuracyand neatness. But these constructions can do something that constructionson paper cannot: students can drag points and move lines and shapesaround and see the results immediately. This is a type of feedback neverbefore possible and we are still grappling with its implications and possibilities.The Standards. Four basic standards are defined for all grades; at eachgrade level certain expectations are to be met. The four basic geometrystandards deal with (1) geometric properties; (2) spatial relationships; (3)transformations and symmetry; and (4) problem solving. They are listedin more detail below, along with the specific expectations for grades for9 through 12. Portions of the standards are published on-line at http://standards.nctm.org/document/ but beyond this list specific details arelacking from the report; the entire geometry standard is only 10 pages. 2Details are given in the box on the following page.Figure 2.1: Part of an example analyzing properties of two- and threedimensionalgeometric shapes to develop mathematical arguments aboutgeometric relationships from the NCTM Standards for high school geometry.[figure from NCTM Standards (2000), Chapter 7]2 For whatever reason NCTM has chosen not to make these details open-source butinstead require membership ($53/year) or document purchase ($54.95) for completeaccess. Thus many teachers never see this document. There is a copy in the CSUNlibrary, though.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 2. NCTM 9California Framework1. Analyze characteristics and properties of two- and three-dimensional geometric shapesand develop mathematical arguments about geometric relationships to:(a) analyze properties and determine attributes of two- and three-dimensional objects;(b) explore relationships (including congruence and similarity) among classes of twoandthree-dimensional geometric objects (figure 2.1), make and test conjecturesabout them, and solve problems involving them;(c) establish the validity of geometric conjectures using deduction, prove theorems,and critique arguments made by others; and to(d) use trigonometric relationships to determine lengths and angle measures.2. Specify locations and describe spatial relationships using coordinate geometry andother representational systems:(a) use Cartesian coordinates and other coordinate systems, such as navigational,polar, or spherical systems, to analyze geometric situations;(b) investigate conjectures and solve problems involving two- and three-dimensionalobjects represented with Cartesian coordinates.3. Apply transformations and use symmetry to analyze mathematical situations:(a) understand and represent translations, reflections, rotations, and dilations ofobjects in the plane by using sketches, coordinates, vectors, function notation,and matrices;(b) use various representations to help understand the effects of simple transformationsand their compositions.4. Use visualization, spatial reasoning, and geometric modelling to solve problems:(a) draw and construct representations of two- and three-dimensional geometricobjects using a variety of tools;(b) visualize three-dimensional objects and spaces from different perspectives andanalyze their cross sections;(c) use vertex-edge graphs to model and solve problems;(d) use geometric models to gain insights into, and answer questions in, other areasof mathematics;(e) use geometric ideas to solve problems in, and gain insights into, other disciplinesand other areas of interest such as art and architecture.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

10 SECTION 2. NCTM« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 3California Framework andStandardsIn 1997 the California State Boardof Education adopted MathematicsContent Standards for Public Schools.These standards list specific contentmaterial that should be covered at eachlevel of public education in Mathematics.This material has since been supersededby the Common Core Standards,which were adopted by Californiain 2010.“by meeting the goals ofstandards-based mathematics,students will achieve greaterproficiency in the practicaluses of mathematics in everydaylife, such as balancing acheck book, purchasing a car,and understanding the dailynews.” Furthermore, “when students delve deeply into mathematics,they gain not only conceptual understanding of mathematicalprinciples but also knowledge of and experience with pure reasoning.One of the most important goals of mathematics is to teachstudents logical reasoning 1 .”1 “If you don’t know how to add fractions, you don’t know how to add.”[Attributed to11

12 SECTION 3. CA STANDARDIn 2005 California also adopted a Mathematics Framework for CaliforniaPublic Schools. The purpose of this framework was to provide for “instructionalprograms and strategies, instructional materials, professionaldevelopment, and assessments that are aligned with the standards” withthe intent of providing a “context for continuing a coordinated effort to enableall California students to achieve rigorous, high levels of mathematicsproficiency.” These are largely derived from the NCTM standards, and aresummarized in the boxes on this and the following pages.An important part of the framework is balance. A balanced program shouldgive the student (a) proficiency in basic skills; (b) conceptual understanding;and (c) problem solving ability.1. Become proficient in basic computational and procedural skills.These are the basic skills that all students should learn to use routinely andautomatically. They should be practiced sufficiently and used frequentlyenough to commit them to memory and ensure that these skills are retainedand maintained over the years.2. Develop conceptual understanding. Students who do not have a deepunderstanding of mathematics suspect that it is just a jumble of unrelatedprocedures and incomprehensible formulas. In seeing the larger pictureand in understanding the underlying concepts, students are in a strongerposition to apply their knowledge to new situations and problems and torecognize when they have made procedural errors.3. Become adept at problem solving. Problem solving in mathematicsis a goal-related activity that involves applying skills, understandings, andexperiences to resolve new, challenging, or perplexing mathematical situations.Problem solving involves a sequence of activities directed toward aspecific mathematical goal, such as solving a word problem, a task that ofteninvolves the use of a series of mathematical procedures and a conceptualrepresentation of the problem to be solved.All three components are important; none is to be neglected or underemphasized.Balance, however, does not imply allocating set amounts oftime for each of the three components.professor Barry Simon, Caltech, in a freshman mathematics class.]« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 3. CA STANDARD 13Goals for Teachers (2005 CA. Framework)1. Increase teachers’ knowledge of mathematics content through professional developmentfocusing on standards-based mathematics.2. Provide an instructional program that preserves the balance of computational andprocedural skills, conceptual understanding, and problem solving.3. Assess student progress frequently toward the achievement of the mathematics standardsand adjust instruction accordingly.4. Provide the learning in each instructional year that lays the necessary groundwork forsuccess in subsequent grades or subsequent mathematics courses.5. Create and maintain a classroom environment that fosters a genuine understandingand confidence in all students that through hard work and sustained effort, they canachieve or exceed the mathematics standards.6. Offer all students a challenging learning experience that will help to maximize theirindividual achievement and provide opportunities for students to exceed the standards.7. Offer alternative instructional suggestions and strategies that address the specificneeds of California’s diverse student population.8. Identify the most successful and efficient approaches within a particular classroom sothat learning is maximized.Goals for Students (2005 CA. Framework)1. Develop fluency in basic computational and procedural skills, an understanding ofmathematical concepts, and the ability to use mathematical reasoning to solve mathematicalproblems, including recognizing and solving routine problems readily andfinding ways to reach a solution or goal when no routine path is apparent.2. Communicate precisely about quantities, logical relationships, and unknown valuesthrough the use of signs, symbols, models, graphs, and mathematical terms.3. Develop logical thinking in order to analyze evidence and build arguments to supportor refute hypotheses.4. Make connections among mathematical ideas and between mathematics and otherdisciplines.5. Apply mathematics to everyday life and develop an interest in pursuing advancedstudies in mathematics and in a wide array of mathematically related career choices.6. Develop an appreciation for the beauty and power of mathematics.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

14 SECTION 3. CA STANDARDThe California Standards (1997)The California Board of Education sets forth 22 specific requirements forgeometry in secondary education. 2 These specifics are listed on pages 42-43 of Mathematics Content Standards for California Public Schools. TheMathematics Framework for California Public Schools expands on these 22requirements with specific examples (pages 85-89 of the Framework) anddiscuss specific considerations for high school geometry on pages 184-189of the Framework.California Framework1. Introduce students to the basic nature of logical reasoning.2. Use inductive reasoning and geometric constructions to build up a breadth ofknowledge from from a few basic axioms.3. Become proficient in proofs and learn the basic principles of plane geometry.4. Study the basic properties of triangles, quadrilaterals, circles, and parallel lines.5. Use the system built up to prove the Pythagorean theorem and from theredevelop and understanding of areas of differently shaped objects.6. Use coordinates and shapes as a natural jumping off into trigonometry.7. Show that the proofs of geometry can be repeated analytically using trigonometry.8. Develop the connection between geometry and algebra, introducing the conceptsof analytic geometry.According to the framework document, the main purpose of the geometrycurriculum is to develop geometric skills and concepts and the abilityto construct formal logical arguments and proofs in a geometric setting.“The geometry skills and concepts developed in this discipline are usefulto all students. Aside from learning these skills and concepts, studentswill develop their ability to construct formal, logical arguments and proofsin geometric settings and problems.” The curriculum is weighed towardsplane Euclidean geometry but allows (and encourages) some use on coordinatesystems and transformations. The “considerations” section (pages184-189 of the Framework) walk us through the standards and give us aperspective on how a geometry course might be structured.2 This material has been largely superseded by the adoption of the common core standardsby California in 2010.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 3. CA STANDARD 15California Geometry Content Standards (1 of 2)1. Students demonstrate understanding by identifying and giving examples ofundefined terms, axioms, theorems, and inductive and deductive reasoning.2. Students write geometric proofs, including proofs by contradiction.3. Students construct and judge the validity of a logical argument and give counterexamplesto disprove a statement.4. Students prove basic theorems involving congruence and similarity.5. Students prove that triangles are congruent or similar, and they are able to usethe concept of corresponding parts of congruent triangles.6. Students know and are able to use the triangle inequality theorem.7. Students prove and use theorems involving the properties of parallel lines cutby a transversal, the properties of quadrilaterals, and the properties of circles.8. Students know, derive, and solve problems involving the perimeter, circumference,area, volume, lateral area, and surface area of common geometricfigures.9. Students compute the volumes and surface areas of prisms, pyramids, cylinders,cones, and spheres; and students commit to memory the formulas for prisms,pyramids, and cylinders.10. Students compute areas of polygons, including rectangles, scalene triangles,equilateral triangles, rhombi, parallelograms, and trapezoids.11. Students determine how changes in dimensions affect the perimeter, area, andvolume of common geometric figures and solids.12. Students find and use measures of sides and of interior and exterior angles oftriangles and polygons to classify figures and solve problems.13. Students prove relationships between angles in polygons by using properties ofcomplementary, supplementary, vertical, and exterior angles.14. Students prove the Pythagorean theorem.15. Students use the Pythagorean theorem to determine distance and find missinglengths of sides of right triangles.16. Students perform basic constructions with a straight edge and compass, suchas angle bisectors, perpendicular bisectors, and the line parallel to a given linethrough a point off the line.17. Students prove theorems by using coordinate geometry, including the midpointof a line segment, the distance formula, and various forms of equations of linesand circles.18. Students know the definitions of the basic trigonometric functions defined bythe angles of a right triangle. They also know and are able to use elementaryrelationships between them. For example, tan(x) = sin(x)/cos(x), (sin(x)) 2 +(cos(x)) 2 = 1.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

16 SECTION 3. CA STANDARDCalifornia Geometry Content Standards (Continued)19. Students use trigonometric functions to solve for an unknown length of a sideof a right triangle, given an angle and a length of a side.20. Students know and are able to use angle and side relationships in problemswith special right triangles, such as 30 ◦ , 60 ◦ , and 90 ◦ triangles and 45 ◦ , 45 ◦ ,and 90 ◦ triangles.21. Students prove and solve problems regarding relationships among chords, secants,tangents, inscribed angles, and inscribed and circumscribed polygons ofcircles.22. Students know the effect of rigid motions on figures in the coordinateThe Dragon of Proof« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 4Common CoreThe Common Core Standards Initiative has been an attempt to unify thevariously confusing and conflicting state curricula and to bring them intoalignment with one another. It is coordinated by the National GovernorsAssociation for Best Practices (NGA) and the Council of Chief StateSchools Officers (CCSSO). These standards 1 attempt to defined the knowledgeand skills that students should acquire during their K-12 education,in order to survive basic college curricula once they graduate.Common Core Standards: Goals1. Are aligned with college and work expectations;2. Are clear, understandable and consistent;3. Include rigorous content and application of knowledge through high-order skills;4. Build upon strengths and lessons of current state standards;5. Are informed by other top performing countries, so that all students are prepared tosucceed in our global economy and society; and6. Are evidence-based.The mathematics standards are progressive, with each level keyed to thenext, so that upon successful completion at any given level, the student isprepared for the subsequent level. They provide for a strong foundation innumbers and basic operations (addition, subtraction, multiplication, division)and fractions at the elementary school levels; geometry, algebra, andprobability and statistics in middle school; and and am emphasis on mathematicalmodeling (applying math to practical problems) in high school.1 see http://www.corestandards.org/about-the-standards17

18 SECTION 4. COMMON COREAt all levels the standards stress both procedural skill and conceptual understanding.The state of California has rewritten its content standards interms of the common core. 2California Common Core Standards for High School Geometryˆ Congruence– Experiment with transformations in the plane– Understand congruence in terms of rigid motions– Prove geometric theorems– Make geometric constructionsˆ Similarity, Right Triangles, and Trigonometry– Understand similarity in terms of similarity transformations– Prove theorems involving similarity– Define trigonometric ratios and solve problems involving right triangles– Apply trigonometry to general trianglesˆ Circles– Understand and apply theorems about circles– Find arc lengths and areas of sectors of circlesˆ Expressing Geometric Properties with Equations– Translate between the geometric description and the equation for a conic section– Use coordinates to prove simple geometric theorems algebraicallyˆ Geometric Measurement and Dimension– Explain volume formulas and use them to solve problems– Visualize relationships between two-dimensional and three-dimensional objectsˆ Modeling with Geometry– Apply geometric concepts in modeling situationsWhat the standards – NCTM, California, or Common Core – do NOTspecify is how to develop these concepts. Whether one should follow anaxiomatic Euclidean development, or the logical framework established byHilbert, or start with the smaller sets of axioms developed by groups such asthe School Mathematics Study Group (SMSG), the University of ChicagoSchool Mathematics Group (UCSMG) is left up to the individual school(teacher, district, etc). As a teacher learning the advanced mathematicalcontent behind these standards, it is important for you to understand eachof these perspectives so that you will understand where the frameworkthat is actually in use in your eventual school placement fits into the “bigpicture.”2 See http://www.cde.ca.gov/be/st/ss/. More detail is given in the standard; the boxgiven here is just from the summary page.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 5Logic and Proof inMathematicsThis section is intentionally concise as it should be a review of Math 320.The language of mathematics is formal. Statements can be written down ina form that separates their content from their meaning in order to establishconsistency and validity. We start with a set of undefined terms that areaccepted as given without further explanation, such as point, line, plane.Usually these terms can be defined in some further reduced terms but, likea dictionary, we will eventually run into circular definitions if we attemptto continue to refine the definitions, or we end up with a statement thatdoesn’t really make much sense:A point is that of which there is no part. [Euclid, definition 1]Does this really clarify what a point is? Thus it is best to choose ourundefined terms as something that is more-or-less agreed upon.Following the undefined terms, we can define additional objects in terms ofthe undefined.We then need to state our assumptions. These are called postulates oraxioms; in Euclid’s system there are five.Next, we have a system of rules for obtaining new true statements fromour postulates. This is our logical system. We will sometimes define newsymbols as shorthands to represent parts of our logical system.The true statements are called theorems, and the sequence of steps thatjustifies the validity of the theorem is called a proof. We will write all of19

20 SECTION 5. LOGIC AND PROOFour theorems in the following manner:If [hypothesis] then [conclusion]We can give our statements names, like A and B, in which case we write:A ⇒ Bwhich we read as “If A then B” or “A implies B.”For a theorem to be accepted as a true, it must have a proof. A is alist of statements that justifies a theorem. Each step must be justified (orexplained) by one of the following methods:ˆ By hypothesis ... (assume that ...)ˆ By axiom X ... (or theorem, definition, postulate, ...)ˆ By step Y ... (an earlier step in the proof)ˆ By a rule of symbolic logicWe will discuss some of the rules of symbolic (formal) logic shortly.There are two special types of theorems: a lemma, and a corollary. Logicallythere is no difference between a theorem, a lemma, and a corollary. Alemma is a theorem which is not really interesting (according to the author)in itself, or is a result that is not pertinent to the subject at hand, but isonly stated only because it makes the proof of some other theorem moreinteresting. A corollary is a theorem that follows almost immediately as aresult of another theorem with very little proof.A statement A in our logical system will only be allowed to have two values:True and False (we may denote these values by T and F).Just writing a statement does not make it true:If △ABC is any triangle then it is equilateralwould have a truth value of False.The value of any implication (A ⇒ B) is given by the following truthtable:A B A ⇒ Btrue true truetrue false falsefalse true truefalse false trueThe negation operation (∼ A) turns true to false and false to true. We usenegation to prove theorems according to method of RAA (Reductio ad« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 5. LOGIC AND PROOF 21absurdum): To prove that A ⇒ B, assume ∼ B and deduce somethingthat is not true.We define the operations of and (A∧B) and or (A∨B) as meaning follows:A B A ∧ B A ∨ BT T T TT F F TF T F TF F F FWe can remember this truth table by the following rule: A∨B means eitherA is true or B is true, or both are true; and A ∧ B means both A and Bare true.The Law of Excluded Middles says that for any statement A, either Ais true, or ̸ A is true:∀P, P ∨ ∼ PThe converse of the theorem A ⇒ B is B ⇒ A. The converse is a completelydifferent statement, and may or may not be true.If both a theorem and its converse are true, we call the theorem a logicalequivalence. We write this as A ⇐⇒ B or “A iff B” is read as “A if andonly if B” and means(A ⇒ B) ∧ (B ⇒ A)The contrapositive of a theorem A ⇒ B is ∼ B ⇒∼ A. The contrapositiveis equivalent to the original theorem. This is demonstrated by thefollowing truth table.A B A ⇒ B ∼ A ∼ B ∼ A ⇒∼ BT T T F F TT F F F T FF T T T F TF F T T T TThe universal quantifier ∀x is read as “for all x.” The statement (∀x)(S(x))means “for all x, the statement S(x) is true.”The existential quantifier ∃y is read as “there exists y.”Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

22 SECTION 5. LOGIC AND PROOFRene Descartes explaining math to Queen Christina of Sweden (Pierre Louis Dumesnil,1698-1781); copy by Nils Forsberg (1884).We can then derive the following additional rules of logic using truth tables.∼ (∼ A) ⇐⇒ A (5.1)∼ (A ⇒ B) ⇐⇒ A∧ ∼ B (5.2)∼ A ⇒∼ B ⇐⇒ A ⇒ B (5.3)∼ (A ∧ B) ⇐⇒ (∼ A) ∨ (∼ B) (5.4)∼ [∀xF (x)] ⇐⇒ ∃x ∼ F (x) (5.5)∼ [∃xF (x)] ⇐⇒ ∀x ∼ F (x) (5.6)The following rule is known as the rule of detachment or modus ponens,[A ∧ (A ⇒ B)] ⇒ B (5.7)In other words, if A is true, and we know that A ⇒ B, then we also know« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 5. LOGIC AND PROOF 23that B is true. We also have the following rules of deduction:[(A ⇒ B) ∧ (B ⇒ C)] ⇒ [A ⇒ C] (5.8)[A ∧ B] ⇒ A (5.9)For the following example, recall that a rational number is any number xthat can be expressed as the ratio of two integers, e.g., x = p/q, where bothp and q are integers. An irrational number is a number that is not rational.Example 5.1 Prove that √ 2 is irrational.Proof.1. Assume that √ 2 is rational. (RAA hypothesis)2. There exists integers p and q such that √ 2 = p/q (Definition of arational number)3. Assume that p and q have no common factors (from our knowledgeof numbers we know that we can cancel out all common factors).4. Therefore at least one of p and q is odd. (otherwise there would be acommon factor of 2, which contradicts step 35. Since p 2 = 2q 2 , p is divisible by 2, hence it is even.6. Write p = 2s where s is an integer (definition of even).7. Then 4s 2 = p 2 = 2q 2 ⇒ q 2 = 2s 28. Then q 2 is divisible by 2, hence it is even (definition of even).9. Hence q is even because the square of any odd number must be odd.10. Since p and q are even this contradicts step 4. Hence our RAA assumptionis false.We numbered the steps in this proof for clarity. In general we don’t do this,because it is somewhat tedious and takes up a lot of space, and insteadusually write them as a paragraph. However, until you are comfortablewriting proofs, you should write them out in a step-by-step manner.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

24 SECTION 5. LOGIC AND PROOF“I ♥ Geometry!”« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 6The Real NumbersWe will take the terms set and element as undefined terms. We write aset as the list of elements surrounded by curly brackets:or by a rule{A, B, C, ...}{x|S(x)}where S(x) is some rule such as “x is even.”We use x ∈ S to represent “x is an element of the set S.”We denote the natural numbers bythe integers byand the positive integers byN = {1, 2, 3, . . . },Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . . },Z + = {0, 1, 2, ..}The union of two sets S and T is given byS ∪ T = {x|x ∈ S ∧ x ∈ T }0 This section is intended primarily as a review and hence is necessarily concise.25

26 SECTION 6. REAL NUMBERSThe intersection of two sets S and T is given byS ∩ T = {x|x ∈ S ∨ x ∈ T }We will use the notation A − B to indicate set difference, which we reada “A minus B”A − B = {x|(x ∈ A) ∧ (x ∉ B)}Venema (and some other texts) use the notation A B for this set.The symbol ∅ represents the empty set. The symbol Q represents the setof all rational numbers.A rational number r is a quotient of two integers p and q, wherer = p/qThe symbol R represents the set of all real numbers. We will not give adefinition of real numbers, but example 5.1 shows that there are numbersthat are not rational. Any real number that cannot be expressed as arational number is called an irrational number. We will see later thatthere is a one-to-one correspondence between the points on a line and thereal numbers, so in a sense, the real numbers give us anything we canmeasure.Axiom 6.1 (Trichotomy of the Real Numbers) Let x, y ∈ R. Thenexactly one of the following is true:x < y, x = y, or x > yAxiom 6.2 (Density) Let x < y ∈ R. Then both of the following aretrue:1. There exists a rational number q such that x < q < y2. There exists an irrational number z such that x < z < yCorollary 6.3 There is an irrational number between any two rationalnumbers.Corollary 6.4 There is a rational number between any two irrational numbers.Theorem 6.5 (Comparison Theorem) Suppose that x, y ∈ R satisfy1. For every rational number q < x, q < y2. For every rational number q < y, q < xthen x = y.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 6. REAL NUMBERS 27Definition 6.6 (Upper Bound) A number M is called an upper boundfor a set A if ∀x ∈ A, x ≤ M.Definition 6.7 (Least Upper Bound) A number m is called a leastupper bound for a set A if for all upper bounds M of A, m ≤ M, and wewrite m = lubA.Axiom 6.8 (Least Upper Bound Axiom) Every bounded non-emptysubset of the real numbers has a least upper bound.The following property expresses the notionthat you can fill up a bucket with spoonfuls ofwater. We will accept it as an axiom althoughit fact it can be derived from the Least UpperBound Axiom.Axiom 6.9 (Archimedian Property) IfM > 0, ɛ > 0 are both real numbers than thereexists a postive integer n such that nɛ > M.Definition 6.10 (Function) A function f isa rule that assigns to each element a ∈ A anelement b = f(a) ∈ B. We call A the domainof f, and we call the subset of B to whichelements of A are mapped by f the range off. We write f : A ↦→ B.Definition 6.11 A function f : A ↦→ B isone-to-one (sometimes 1-1 or (1:1)) if a 1 ≠a 2 ⇒ f(a 1 ) ≠ f(a 2 )Babylonian clay tablet YBC7289 (c 1800-1600 BC) showing√2 ≈ 1 +2160 + 5160 2 + 1060 3(figure Bill Casselmanhttp://www.math.ubc.ca/~cass/Euclid/ybc/ybc.html.)Definition 6.12 A function f : A ↦→ B is onto if (∀b ∈ B)(∃a ∈ A) suchthat b = f(a).Definition 6.13 A function f : A ↦→ B that is 1-1 and onto is called aone-to-one-correspondence.Definition 6.14 A function f(x) is continuous on an interval (a, b) if forevery ɛ > 0 there exists a δ > 0 such that whenever |x − y| < δ, x, y ∈ (a, b),then |f(x) − f(y)| < ɛ.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

28 SECTION 6. REAL NUMBERS“The Sum of the Squares ...”« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 7Euclid’s ElementsHere and in the following sections we willpresent some of the axiomatic systems thathave been used to develop geometry in thewest since Euclid’s time. We will not be usingany of thse specific systems in our own developmentbut they are worth being familiarwith.Euclid’s Elements was written around 300 BC.It consists of 13 Books that are a compilationof geometric knowledge that had beed acquiredover the previous several centuries. Wedon’t know what (if any) parts of it are originalto Euclid. The great contribution of thisdocument is that it sets forth a set of basic assumptions(five postulates) from which all ofthe remainder of the treatise is logically derived.Euclid at the Oxford HistoryMuseum.Any logical system requires one to first define some basic concepts whichare accepted on faith; for geometry these are thinks like points and lines.In fact, Eculid states 23 definitions before his postulates, and follows themwith five additional common notions, which are statements that he expectsare so obvious they can be accepted by anyone with reason.It is worth looking at these basic statements to get a feeling for whereEuclid’s geometry starts, so they are extracted below[Euclid].29

30 SECTION 7. EUCLID’S ELEMENTSEuclid’s Axioms1 Let it have been postulated to draw a straight-line from any point to any point.2 And to produce a finite straight-line continuously in a straight-line.3 And to draw a circle with any center and radius.4 And that all right-angles are equal to one another.5 Euclid’s Parallel postulate. And that if a straight-line falling across two(other) straight-lines makes internal angles on the same side (of itself whosesum is) less than two right-angles, then the two (other) straight-lines, beingproduced to infinity, meet on that side (of the original straight-line) that the(sum of the internal angles) is less than two right-angles (and do not meet onthe other side) (see figure 7.1.)Historically the 5thpostulate has beenconsidered separatefrom the first fourand was followed byover 2000 years of attemptsto derive itfrom them. It wasnot until the 19thcentury (by EugenioBeltrami in 1868)that it was shownthat postulate 5 couldnot be derived fromthe other four. In the process, the existence of non-Euclidean geometrieswas proven, as well as neutral geometry, the study of the consequencesof the first four postulates.Euclid presumably stated his common notions to make clear what assumptionshe was making that he thought were obvious. Today we would probablystate them using algebra, but such expressions had not been inventedyet.0 Euclid’s line is what we call a plane curve.0 Euclid’s straight-line is what we would call a line segment. The modern concept of aline that extends infinitely in each direction was unknown to Euclid.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 7. EUCLID’S ELEMENTS 31Euclid’s Definitions1 A point is that of which there is no part.2 And a line 1 is a length without breadth.3 And the extremities of a line are points.4 A straight-line 2 is (any) one which lies evenly with points on itself.5 And a surface is that which has length and breadth only.6 And the extremities of a surface are lines.7 A plane surface is (any) one which lies evenly with the straight-lines on itself.8 And a plane angle is the inclination of the lines to one another, when two lines in aplane meet one another, and are not lying in a straight-line.9 And when the lines containing the angle are straight then the angle is called rectilinear.10 And when a straight-line stood upon (another) straight-line makes adjacent angles(which are) equal to one another, each of the equal angles is a right-angle, and theformer straight-line is called a perpendicular to that upon which it stands.11 An obtuse angle is one greater than a right-angle.12 And an acute angle (is) one less than a right-angle.13 A boundary is that which is the extremity of something.14 A figure is that which is contained by some boundary or boundaries.15 A circle is a plane figure contained by a single line [which is called a circumference],(such that) all of the straight-lines radiating towards [the circumference] from onepoint amongst those lying inside the figure are equal to one another.16 And the point is called the center of the circle.17 And a diameter of the circle is any straight-line, being drawn through the center,and terminated in each direction by the circumference of the circle. (And) any such(straight-line) also cuts the circle in half. †18 And a semi-circle is the figure contained by the diameter and the circumference cutsoff by it. And the center of the semi-circle is the same (point) as (the center of) thecircle.19 Rectilinear figures are those (figures) contained by straight-lines: trilateral figuresbeing those contained by three straight-lines, quadrilateral by four, and multilateralby more than four.20 And of the trilateral figures: an equilateral triangle is that having three equal sides,an isosceles (triangle) that having only two equal sides, and a scalene (triangle) thathaving three unequal sides.21 And further of the trilateral figures: a right-angled triangle is that having a rightangle,an obtuse-angled (triangle) that having an obtuse angle, and an acute-angled(triangle) that having three acute angles.22 And of the quadrilateral figures: a square is that which is right-angled and equilateral,a rectangle that which is right-angled but not equilateral, a rhombus that which isequilateral but not right-angled, and a rhomboid that having opposite sides and anglesequal to one another which is neither right-angled nor equilateral. And let quadrilateralfigures besides these be called trapezia.23 Parallel lines are straight-lines which, being in the same plane, and being produced toinfinity in each direction, meet with one another in neither (of these directions).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

32 SECTION 7. EUCLID’S ELEMENTSFigure 7.1: Illustration of Euclid’s Parallel postulate. The two angles αand β add to les than 180 degrees, hence the lines h and k meet at a pointS on the same same side of g as α and β.Euclid’s Common Notions1 Things equal to the same thing are also equal to one anotherWe might write this today as: If a = b and b = c then a = c.2 And if equal things are added to equal things then the wholes are equal.We might write this as: if a = c then a + b = c + b.3 And if equal things are subtracted from equal things then the remainders areequal.Which we might write this as: if a = c then a − b = c − b.4 And things coinciding with one another are equal to one another.By this Euclid meant he could imagine picking up the “picture” of a triangle(or some other object) and lay it on top of another; if they were the same thenthey were considered “equal” (or maybe congruent).5 And the whole [is] greater than the partWhich we might write as: if a > 0 and b > 0 then a + b > a and a + b > b.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 8Hilbert’s AxiomsDavid Hilbert (1862-1943) boiled geometry down to 20 axioms which heclassified into seven axioms of connection (we now use the term incidence);five axioms of order; one axiom of parallels; six axioms of congruence; andone axiom of continuity. He did this because it had been discovered overthe centuries that Euclid had left out parts of his arguments and Hilbertwas attempting to fill in all the blanks. The axioms below are taken fromthe lecture notes of his course in geometry given in 1898 and translated byE.J. Townsend in 1902.Hilbert’s system begins with the followingundefined terms: point, line, plane,lie on, between, congruent. His axiomsare divided up into different sub-areasof geometry: connection, order, parallels,congruence, continuity, and completeness.The axioms of connection definethings like how points form linesand planes. The axioms of order expressthe concept of betweenness of points.They are classified into four linear axiomsof order and one plane axiom of order.The axiom of parallels is a equivalentto Euclid’s parallel postulate. Theaxioms of congruence formalize our intuitivenotions of equivalences among linesegments, angles, and triangles. The ax-33

34 SECTION 8. HILBERT’S AXIOMSiom of continuity introduces the continuity of real numbers to geometry,and completeness tells us that everything we can possibly know we canlearn from these axioms.Hilbert’s Axioms of Connection1. Two distinct points A and B always completely determine a straight line a. We writeAB = a or BA = a.2. Any two distinct points of a straight line completely determine that line; that is, ifAB = a and AC = a, where B ≠ C, then also BC = a.3. Three points A, B, C not situated in the same straight line always completely determineaplane α. We write ABC = α.4. Any three points A, B, C of a plane α, which do not lie in the same straightline,completely determine that plane.5. If two points A, B of a straight line a lie in a plane α, then every point of a lies in α.6. If two planes α, β have a point A in common, then they have at least a second pointB in common.7. Upon every straight line there exist at least two points, in every plane at least threepoints not lying in the same straight line, and in space there exist at least four pointsnot lying in a plane.Hilbert’s Axioms of Order1. If A, B, C are points of a straight line and B lies between A and C, then B lies alsobetween C and A.2. If A and C are two points of a straight line, then there exists at least one point Blying between A and C and at least one point D so situated that C lies between Aand D.3. Of any three points situated on a straight line, there is always one and only one whichlies between the other two.4. Any four points A, B, C, D of a straight line can always be so arranged that B shalllie between A and C and also between A and D, and, furthermore, that C shallliebetween A and D and also between B and D.5. Let A, B, C be three points not lying in the same straight line and let a be a straightline lying in the plane ABC and not passing through any of the points A, B, C.Then, if the straight line a passes through a point of the segment AB, it will alsopass through either a point of the segment BC or a point of the segment AC.Hilbert’s Axiom of ParallelsIn a plane α there can be drawn through any point A, lying outside of a straight line a, oneand only one straight line which does not intersect the line a. This straight line is called theparallel to a through the given point A.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 8. HILBERT’S AXIOMS 35Hilbert’s Axioms of Congruence1. If A, B are two points on a straight line a, and if A ′ is a point upon the same oranother straight line a1, then, upon a given side of A ′ on the straight line a ′ , we canalways find one and only one point B ′ so that the segment AB (or BA) is congruentto the segment A ′ B ′ . We indicate this relation by writing AB ∼ = A ′ B ′ Every segmentis congruent to itself; that is, we always have AB ∼ = AB.2. If AB ∼ = A ′ B ′ and also AB ∼ = A ′′ B ′′ , then A ′ B ′ ∼ = A ′′ B ′′3. Let AB and BC be two segments of a straight line a which have no points in commonaside from the point B, and, furthermore, let A ′ B ′ and B ′ C ′ be two segments of thesame or of another straight line a having, likewise, no point other than B ′ in common.Then, if AB ≡ A ′ B ′ and BC ∼ = B ′ C ′ , we have AC ∼ = A ′ C ′ .4. Let an angle (h, k) be given in the plane α and let a straight line a ′ be given in aplane α ′ . Suppose also that, in the plane α ′ , a definite side of the straight line a ′be assigned. Denote by h ′ a half-ray of the straight line a ′ emanating from a pointO ′ of this line. Then in the plane α ′ there is one and only one half-ray k ′ such thatthe angle (h, k), or (k, h), is congruent to the angle (h ′ , k ′ ) and at the same timeall interior points of the angle (h ′ , k ′ ) lie upon the given side of a ′ . We express thisrelation by means of the notation ∠(h, k) ∼ = ∠(h ′ , k ′ ) Every angle is congruent toitself; that is, ∠(h, k) ∼ = ∠(h, k).5. If the ∠(h, k) ∼ = ∠(h ′ , k ′ ) and ∠(h, k) ∼ = ∠(h ′′ , k ′′ ) then ∠(h ′ , k ′ ) ∼ = ∠(h ′′ , k ′′ )6. If, in the two triangles ABC and A ′ B ′ C ′ the congruences ABV A ′ B, AC ∼ = A ′ C ′and ∠BAC ∼ = ∠B ′ A ′ C ′ then ∠ABC ∼ = ∠A ′ B ′ C ′ and ∠ACB ∼ = ∠A ′ C ′ B ′ .Hilbert’s Axiom of Continuity (Archimedian Axiom)Let A 1 A be any point on a straight line AB. Choose A 2 , A 3 , .. so that A 1 is between Aand A 2 , A 2 is between A 1 and A 3 , etc, such thatAA 1 = A 1 A 2 = A 2 A 3 = · · ·Then there always exists a certain point A n such that B lies between A and A n.In more modern terms, ∀ɛ > 0 and ∀x > 0, then ∃m ∈ N such that mɛ > x.Hilbert’s Axioms of CompletenessTo a system of points, straight lines, and planes, it is impossible to add other elements insuch a manner that the system thus generalized shall form a new geometry obeying all ofthe five groups of axioms. In other words, the elements of geometry form a system which isnot susceptible of extension, if we regard the five groups of axioms as valid.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

36 SECTION 8. HILBERT’S AXIOMS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 9Birkhoff/MacLaneAxiomsGeroge Birkhoff (1884-1944) is bestknown for his works on differential equations,taught at Univ. of Wisconsin,Princeton, and Harvard. In 1932 he proposeda very compact set of axioms whichallow you to use a ruler (a straight edgewith marks on it) and a protractor. Hispurpose was to make geometry more understandableto high school students. 1Birkhoff (left); MacLane (right)Saunders MacLane (1909-2005) was afriend (and professional collaborator) of George Birkhoff’s son Garrett, andworked primarily at Harvard (where he met the Birkhoffs) and the Univ.of Chicago. 2 In 1959 [MacLane, 1959] proposed an extension of Birkhoff’sAxioms that included a distance measure thereby making the system somewhatmore intuitive than Hilbert’s. MacLane introduced the concept ofdistance metrics into the axioms, and added an axiom of continuity.1 The system was published in the paper ”A Set of Postulate for Plane Geometry basedon a Scale and Protractor,” in Annals of Mathematics, 33:329-345 (1932).2 he photo is by Konrad Jacobs (CCASA license) and from Wikimedia).37

38 SECTION 9. BIRKHOFF/MACLANE AXIOMSBirkhoff’s AxiomsUndefined Terms: point, line, distance, angle.1. Axiom of Line Measure. The points A, B, .. of any line l can be put into (1, 1)correspondence with the real numbers x so that |x B − x A | = d(A, B) for all pointsA, B. Here d(A, B) denotes the distance between the points A and B. In otherwords, you are allowed to use a rule to measure the length of a line.2. The point-line postulate. One and only one straight line l contains two given pointsP, Q (P ≠ Q).3. The Axiom of angle measure. The half lines l, m, .. through any point O can be putinto (1,1) correspondence with teh real numbers a mod 2π so that, if A ≠ 0 andB ≠ O are points of l and m respectively, the difference a m − a l mod 2π is ∠A0B.4. The Postulate of triangle similarity. If in two triangles, △ABC and △A ′ B ′ C ′ ,of for some constant k > 0, d(A ′ , B ′ ) = kd(A, B), d(A ′ , C ′ ) = kd(A, C), andalso ∠B ′ A ′ C ′ = ±∠BAC then also d(B ′ , C ′ ) = kd(B, C), ∠C ′ B ′ A ′ = ±∠CBA,∠A ′ C ′ B ′ = ±∠ACB.1. There are at least two points.MacLane’s Axioms on Distance2. If A and B are points, d(AB) is a nonnegative number (that gives the distancebetween the points).3. For points A and B, d(AB) = 0 if and only if A = B.4. If A and B are points then d(AB) = d(BA).MacLane’s Axioms on Lines1. A l ine is a set of points containing more than one points.2. Through two distinct points there is one and only one l ine.3. Three distinct points on a line if and only if one of them is between the other two.4. On each ray from a point O and to each positive real number b there is a point Bwith d(OB) = bMacLane’s Axioms on Angles1. If r and s are rays from the same point, then ∠rs is a real number (mod 360).2. If r, s, t are three rays from the same point, the ∠rs + ∠st = ∠rt.3. If r is a ray from O and c is a real number, then there is a ray s from O such that∠rs = c ◦ .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 9. BIRKHOFF/MACLANE AXIOMS 39McLane’s Axiom on SimilarityIf two triangles △ABC and △A ′ B ′ C ′ have ∠ABC = ±∠A ′ B ′ C ′ , d(AB) = kd(A ′ B ′ ), andd(BC) = kd(B ′ C ′ ) for some postive number k then they are similar.MacLane’s Axiom of ContinuityLet ∠AOB be proper. If D is between A adn B then 0 < ∠AOD < ∠AOB.Here a proper angle is an angle x such that 0 < x < 180.The 37th View of Mt Fuji by Katsushika Hokusai (1760-1849) never madeit to wood block.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

40 SECTION 9. BIRKHOFF/MACLANE AXIOMSA Geometric Mind« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 10The SMSG AxiomsThe School Mathematics Study Group (SMSG) at Yale University wasfunded by the US National Science Foundation to reform mathematics educationin the 1950’s and developed mathematical curricula that came beknow as the “new math” during the 1960’s. A set of 22 s that were intendedto make geometry more intuitive and understandable were produced by thisgroup, as was a mimeographed textbook that was later used as the basis ofa geometry textbook by E.E. Moise and F.L.Downs (1964) that is still incirculation. Some of the axioms are redundant in the sense that they canbe derived from the others.The undefined terms are point, line, plane, lie on, distance, angle measure,area, volume, and there are 22 axioms.Axiom 1 Given any two distinct points there is exactly one line that containsthem.Axiom 2 Distance Postulate. To every pair of distinct points there correspondsa unique positive number. This number is called the distancebetween the two points.Axiom 3 Ruler Postulate. The points of a line can be placed in a correspondencewith the real numbers such that: (1) To every point of the linethere corresponds exactly one real number; (2) To every real number therecorresponds exactly one point of the line. (3) The distance between twodistinct points is the absolute value of the difference of the correspondingreal numbers.Axiom 4 Ruler Placement Postulate. Given two points P and Q of a line,the coordinate system can be chosen in such a way that the coordinate of41

42 SECTION 10. THE SMSG AXIOMSP is zero and the coordinate of Q is positive.Axiom 5 Every plane contains at least three non-collinear points, andspace contains at least four non-coplanar points.Axiom 6 If two points lie in a plane, then the line containing these pointslies in the same plane.Axiom 7 Any three points lie in at least one plane, and any three noncollinearpoints lie in exactly one plane.Axiom 8 If two planes intersect, then that intersection is a line.Edward Griffith Begle (1941-1978), director of SMSG for ten years. Photograph by PaulHalmos, Archives of American Mathematics, Dolph Briscoe Center for American History,University of Texas at Austin.Axiom 9 Plane Separation Postulate. Given a line and a plane containingit, the points of the plane that do not lie on the line form two sets suchthat: (1) each of the sets is convex; and (2) if P is in one set and Q is inthe other, then segment PQ intersects the line.Axiom 10 Space Separation Postulate. The points of space that do not liein a given plane form two sets such that: (1) Each of the sets is convex;and (2) If P is in one set and Q is in the other, then segment PQ intersectsthe plane.Axiom 11 Angle Measurement Postulate. To every angle ∠x there correspondsa real number between 0 ◦ and 180 ◦ .The real number is called the measure of the angle and denoted by m(∠x).Axiom 12 Angle Construction postulate. Let AB be a ray on the edge ofthe half-plane H. For every r between 0 and 180 there is exactly one ray« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 10. THE SMSG AXIOMS 43AP , with P in H such that m(∠P AB) = r.Axiom 13 Angle Addition postulate. If D is a point in the interior of∠BAC, then m∠BAC) = m(∠BAD) + m(∠DAC).Axiom 14 Supplement postulate. If two angles form a linear pair, thenthey are supplementary.Axiom 15 SAS postulate. Given a one-to-one correspondence between twotriangles (or between a triangle and itself). If two sides nd the includedangle of the first triangle are congruent to the corresponding parts of thesecond triangle, then the correspondence is a congruence.Axiom 16 Parallel postulate. Through a given external point there is atmost one line parallel to a given line.Axiom 17 To every polygonal region there corresponds a unique positivereal number called its area.Axiom 18 If two triangles are congruent, then the triangular regions havethe same area.Axiom 19 Suppose that the region R is the union of two regions R 1 andR 2 . If R 1 and R 2 intersect at most in a finite number of segments andpoints, then the area of R is the sum of the areas of R 1 and R 2 .Axiom 20 The area of a rectangle is the product of the length of its andthe length of its altitude.Axiom 21 The volume of a rectangle parallelepiped is equal to the productof the length of its altitude and the area of its base.Axiom 22 Cavalieri’s Principle. Given two solids and a plane. If for everyplane that intersects the solids and is parallel to the given plane the twointersections determine regions that have the same area, then the two solidshave the same volume.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

44 SECTION 10. THE SMSG AXIOMS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 11The UCSMP AxiomsThe University of Chicago School MathematicsProject was founded in 1983 with the aimof upgrading mathematics education in elementaryand secondary schools throughoutthe United States. They have developed a setof axioms that are in wide use today, and arealso redundant in the sense that some axiomscan be proved from others. The purpose of theredundancy was to make the learning of geometrymore intuitive. These axioms used incorporateda transformational approach. Detailsof the projects history is given on itsweb page at (http://ucsmp.uchicago.edu/history.html). Many of today’s elementaryand secondary textbooks are based on these standards, which encompassall of mathematics, not just geometry.The only undefined terms are point, line, and plane.Point-Line-Plane AxiomsAxiom 1 Through any two points there is exactly one line.Axiom 2 Every line is a set of points that can be put into a one-to-onecorrespondence with the real numbers, with any point corresponding tozero and any other point corresponding to the number 1.Axiom 3 Given a line in a plane, there is at least one point in the planethat is not on the line. Given a plane in space, there is at least one point45

46 SECTION 11. THE UCSMP AXIOMSin space that is not on the plane.Axiom 4 If two points lie in a plane, the line containing them lies in theplane.Axiom 5 Through three non-collinear points, there is exactly one plane.Axiom 6 If two different planes have a point in common, then their intersectionis a line.Distance AxiomsAxiom 7 On a line, there is a unique distance between two points.Axiom 8 If two points on a line have coordinates x and y the distancebetween them is |x − y|.Axiom 9 If point B is on the line segment AC then AB + BC = AC,where AB, BC, AC denote the distances between the points.Triangle InequalityAxiom 10 The sum of the lengths of two sides of a triangle is greater thanthe length of the third side.Angle MeasureAxiom 11 Every angle has a unique measure from 0 ◦ to 180 ◦ .Axiom 12 Given any ray −→ V A and a real number r between 0 and 180 thereis a unique angle ∠BV A in each half-plane of ←→ V A such that ∠BV A = r.Axiom 13 If −→ V A and −→ V B are the same ray, then ∠BV A = 0.Axiom 14 If −→ V A and −→ V B are opposite rates, then ∠BV A = 180.Axiom 15 If −→ V C (except for the point V ) is in the interior of angle ∠AV Bthen ∠AV C + ∠CV B = ∠AV B.Corresponding Angle AxiomAxiom 16 Suppose two coplanar lines are cut by a transversal. If twocorresponding angles have the same measure, then the lines are parallel. Ifthe lines are parallel, then the corresponding angles have the same measure.Reflection AxiomsAxiom 17 There is a one to one correspondence between points and theirimages in a reflection.Axiom 18 Collinearity is preserved by reflection.Axiom 19 Betweenness is preserved by reflection.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 11. THE UCSMP AXIOMS 47Axiom 20 Distance is preserved by reflection.Axiom 21 Angle measure is preserved by reflection.Axiom 22 Orientation is reversed by reflection.Area AxiomsAxiom 23 Given a unit region, every polygonal region has a unique area.Axiom 24 The area of a rectangle with dimensions l and w is lw.Axiom 25 Congruent figures have the same area.Axiom 26 The areas of the union of two non-overlapping regions is thesum of the areas of the regions.Volume AxiomsAxiom 27 Given a unit cube,every solid region has a unique volume.Axiom 28 The volume of box with dimensions l, w, and h is lwh.Axiom 29 Congruent solids have the same volume.Axiom 30 The volume of the union of two non-overlapping solids is thesum of their volumes.Axiom 31 Given two solids and a plane. If for every plane which intersectsthe solids and is parallel to the given plane the intersections have equalareas, then the two solids have the same volume.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

48 SECTION 11. THE UCSMP AXIOMS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 12Venema’s AxiomsVenema [Venema, 2006] introduces a set of axioms that meld together partsof the axiom systems of Birkhoff, MacLane, the SMSG, and the UCSMP.We present them here for reference, since we will be using this system inthe remainder of the class.Venema’s Undefined TermsThe undefined terms are point, line, distance, half-plane, angle-measure,areaVenema’s Axiom’s of Neutral GeometryExistence says that at least some points exist, and incidents says that everypair of distinct points defines a line.1. Existence Postulate. The collection of all points forms a nonemptyset with more than one (i.e., at least two) points. The set of all points inthe plane is called P.2. Incidence Postulate. Every line is a set of points. For every pair ofdistinct points A, B there is exactly one line l = ←→ AB such that A, B ∈ l.The ruler postulate allows us to associate real numbers – and hence measurements– with distances and line segments.49

50 SECTION 12. VENEMA’S AXIOMS3. Ruler Postulate. For every pair of points P, Q there is a numberP Q called the distance from P to Q. For each line l there is a one-to-onemapping f : l ↦→ R such that if x = f(P ) and y = f(Q) then P Q = |x − y|.The Plane Separation Postulate says that a line has two sides; it is used todefine the concept of a half-plane.4. Plane Separation Postulate. For every line l the points that donot lie on l form two disjoint, convex non-empty sets H 1 and H 2 , calledhalf-planes, bounded by l such that if P ∈ H 1 and Q ∈ H 2 then P Qintersects l.The protractor postulate encapsulates our “common notions” (to use aEuclidean term) about angles: they can be measure, added, and ordered.5. Protractor Postulate. For every angle ∠BAC there is a realnumber µ(∠BAC) called the measure of ∠BAC such that1. 0 ≤ µ(∠BAC) < 1802. µ(∠BAC) = 0 ⇐⇒ −→ −→ AB = BC3. Angle Construction Postulate. ∀r ∈ R such that 0 < r < 180,and for every half-plane H bounded by ←→ AB, there exists a uniqueray −→ AE such that E ∈ H and µ(∠BAE) = r.4. Angle Addition Postulate. If the ray −→ AD is between the rays−→AB and −→ AC thenµ(∠BAD) + µ(∠DAC) = µ(∠BAC)As we shall see later in the discussion, the first five postulates are notsufficient to ensure that our common notions about triangle congruencewill hold (for example, the following postulate fails in taxi-cab geometry).6. SAS (Side Angle Side Postulate) If △ABC and △DEF are twotriangles such that AB ∼ = DE,BC ∼ = EF , and ∠ABC ∼ = ∠DEF then△ABC ∼ = △DEF .Parallel PostulatesThe combination of the first six postulates, when taken together, are knownas neutral geometry. They can be extended with one of three possible parallelpostulates. It turns out that the second of these – the elliptic parallelpostulate, is inconsistent with the plane separation postulate and the conceptof betweenness – but the other two postulates are each consistent withneutral geometry. In fact, it can be proven that they are mutually exclusive– if you accept either one of the Hyperbolic or Euclidean parallel postulates« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 12. VENEMA’S AXIOMS 51under neutral geometry, the other is provably false; and if either is takenas false, the other is provably true.Euclidean Parallel Postulate For every line l and for every externalpoint P , there is exactly one line m such that P lies on m and m ‖ lElliptic Parallel Postulate For every line l and for every external pointP , there is no line m such that P lies on m and m ‖ lHyperbolic Parallel Postulate For every line l and for every externalpoint P , there are at least two lines m and n such that P lies on bothm and m and m ‖ l and n ‖ l.Area PostulatesNeutral Area Postulate Associated with each polygonal region R thereis a nonnegative number α(R), called the area of R, such that:1. (Congruence) If two triangles are congruent, then their associate regionshave equal area; and2. (Additivity) If R = R 1 ∪ R 2 and R 1 and R 2 do not overlap, thenα(R) = α(R 1 ) + α(R 2 )Euclidean Area Postulate (Venema 9.2.2)α(R) = length(R) × width(R).If R is a rectangle, theReflectionThe Reflection Postulate (Venema 12.6.1) For every line l thereexists a transformation ρ l : P ↦→ P such that:1. If P ∈ l then ρ l (P ) = P2. If P ∉ l, then P and ρ l (P ) lie on opposite half planes of l.3. ρ l preserves distance, collinearity, and angle measure.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

52 SECTION 12. VENEMA’S AXIOMS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 13Incidence GeometryWe will use the expression “a geometry” to refer to the consequences ofa particular set of axioms. For example by Hilbert Geometry we meanthe geometry that is the consequence of Hilbert’s axioms; by EuclideanGeometry we mean the consequences of Euclid’s Axioms, etc. Here we willdescribe a particular type of finite geometry, that is, a geometry witha finite number of points.Incidence Geometry is a term we will use for the geometry that we canderive from the following three axioms.Axiom 13.1 (Incidence Axiom 1) For every pair of distinct points P andQ there exists exactly one line l such that both P and Q lie on l.Axiom 13.2 (Incidence Axiom 2) For every line l there exists at least twodistinct points P and Q such both P and Q lie on l.Axiom 13.3 (Incidence Axiom 3) There exist three points that do not alllie on the same line, i.e., there exists three non-collinear points.Definition 13.4 Three points A, B, C are collinear if there exists at leastone line l such that all three points line on l. They are said to be noncollinearif no such line l exists.Example 13.1 3-Point Plane Geometry. In this example of a finitegeometry that obeys all the axioms of incidence geometry, we define:ˆ A point is an element of the set {A, B, C}ˆ A line is a pair of points such as l = {A, B}ˆ A point P lies on a line l if P ∈ l.53

54 SECTION 13. INCIDENCE GEOMETRYFigure 13.1: Graph diagram illustrating three-point geometry. Parallel linesdo not exist in this geometry.There are three possible lines in this geometry:{A, B}, {B, C}, {A, C}We will describe finite geometries with graph-diagrams consisting of nodesand line segments (figure 13.1). The nodes represent the points, and theline segments connecting the nodes represent the sets that represent lines.This is a representation of a finite geometry, not a picture of the points oflines in the usual sense. In other words, the 3-point geometry does not looklike a triangle; we just represent it by a graph that looks like a triangle.Example 13.2 Four-point Geometry. Here we define (see figure 13.2):ˆ A point is an element of the set {A, B, C, D}ˆ A line is a pair of points such as l = {A, B}ˆ A point P lies on a line l if P ∈ l.There are six lines in this example: {A, B}, {A, C}, {A, D}, {B, C}, {B, D},and {C, D}.Example 13.3 Fano’s Geometry. Here we have seven points given bythe set {A, B, C, D, E, F, G} and we define lines as any of the followingseven specific subsets:as illustrated in figure 13.3{A, B, C}, {C, D, E}, {E, F, A}, {A, G, D},{C, G, F }, {E, G, B}, {B, D, F }« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 13. INCIDENCE GEOMETRY 55Figure 13.2: Graph diagram illustrating four-point geometry. Every linehas precisely one other line that is parallel to it, and and there is preciselyone parallel line through each point that is not on a given line, and hencefour point geometry obeys the Euclidean parallel postulate.Figure 13.3: Graph diagram illustrating Fano’s geometry. Each line segmentand the circle represents a line in this geometry.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

56 SECTION 13. INCIDENCE GEOMETRYExample 13.4 The Cartesian Plane. This is the traditional example weuse in geometry and it obeys all the rules of incidence geometry. Define apoint as an ordered pair of real numbers {x, y}. Then a line is the collectionof all points ax + by + c = 0 for some choice of real numbers a, b, c. Theusual notation for this set is R 2 .Example 13.5 Surface of a Sphere. Consider a unit sphere centered atthe origin in normal 3-space. The surface of this sphere is given by the setof all points {x, y, z} such thatx 2 + y 2 + z 2 = 1Define a point as any point on the surface of the sphere, and define aline as any great circle on the plane (a great circle is the intersection ofthe unit sphere with any plane that goes through the center of sphere; orequivalently, it is any circle on the sphere whose radius is equal to the radiusof the sphere, which is 1). This geometry does not obey incidence geometrybecause any two antipodal points (points at opposite poles of the sphere)are on an infinite number of common lines. This violates Incidence Axiom1. Furthermore, there are no parallel lines in this geometry because anytwo great circles meet.Example 13.6 The Klein Disk. Consider the interior of the unit diskcentered at the origin of R 2 . This is the set of all points such thatx 2 + y 2 < 1Define a point as any ordered pair of numbers (x, y) such that x 2 + y 2 < 1,and define a line as any the part of any Euclidean line that lies inside thiscircle. See figure 13.4. The Klein Disk obeys incidence geometry but doesnot obey Euclid’s fifth postulate.Definition 13.5 (Parallel Lines) Two lines l and m are said to be parallelif there is no point P such that P lies on both l and m. We denote this byl ‖ m.Euclid’s fifth axiom is equivalent to the following statement:Axiom 13.6 (Euclidian Parallel Postulate) For every line l and forevery point P ∉ l there is exactly one line m such that P lies on m andm ‖ l.Four point geometry and geometry of the Cartesian plane each satisfy theEuclidean Parallel Postulate.There are two other possible parallel postulates that are incompatible withthe Euclidean Parallel Postulate but which lead to consistent geometries.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 13. INCIDENCE GEOMETRY 57Figure 13.4: A Klein Disk. Lines l and n pass through point P , and areboth parallel to line m. The Klein Disk is a model of Hyperbolic Geometry.Axiom 13.7 (Elliptic Parallel Postulate) For every line l and for everypoint P ∉ l there is no line m such that P lies on m and m ‖ l.Three-point geometry and geometry on the sphere satisfy the Elliptic parallelpostulate.Axiom 13.8 (Hyperbolic Parallel Postulate) For every line l and forevery point P ∉ l there are at least two lines m such that P lies on mand m ‖ l.The Klein Disk and Five Point geometry (figure 13.5) satisfy the hyperbolicparallel postulate.Theorem 13.9 If l and m are distinct, nonparallel lines, then there existsa unique point P such that P lines on both l and m.Proof. By hypothesis, l ≠ m and l ̸‖ m. Then by the negation of thedefinition of parallel lines, there is a point P that lines on both l and m.To proove that P is unique, we assume that there is a second point Q ≠ Pthat also lies on both lines as our RAA hypothesis.By incidence axiom 1, there is exactly one line n that contains both P andQ.Since P is on l, then since n is unique, l = n.But since Q is on m, then since n is unique, m = n.Hence l = m. This contradicts the hypothesis that l ≠ m. Hence our RAARevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

58 SECTION 13. INCIDENCE GEOMETRYFigure 13.5: Illustration of five point geometry: the points are representedby the symbols A, B, C, D.E and the lines are represented by any subset ofprecisely two points. Since lines {B, D} and {B, C} are both parallel to line{A, E} and both pass through point B, this model satisfies the hyperbolicparallel postulatehypothesis must be wrong, i.e., P = Q. Hence the point P is unique.Here are some other useful results that hold in incidence geometry.Theorem 13.10 Let l be a line. Then there exists at least one point Pthat does not lie on l.Theorem 13.11 Let P be a point. Then there exist at least two distinctlines that contain P .Proof. By Incidence Axiom 3 there exist at least three points R, S, T thatare non-collinear.Either P ∈ {R, S, T } or P ∉ {R, S, T }.If P ∈ {R, S, T } then without loss of generality we can relabel the points sothat P = R. Define l = ←→ P S and m = ←→ P T . Since P, S, T are non-collinear,then l and m are the desired lines.If P ∉ {R, S, T } then either P ∈ ←→ RS or P ∉ ←→ RS.If P ∈ ←→ RS then let l = ←→ RS and m = ←→ P T . By construction, T ∉ RS, elseR, S, T would be collinear. Since T ∈ m and T ∉ l, l ≠ m. Hence l and mare the two distinct lines that contain P .If P ∉ ←→ RS then let l = ←→ P R and m = ←→ P S. By construction R ∈ l, but« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 13. INCIDENCE GEOMETRY 59R ∉ m, else P, R, S would be collinear, and we have assumed otherwise.Hence l and m are distinct lines that contain P .Theorem 13.12 Let l be a line. Then there exist two distinct lines m andn that intersect l.Proof. Let l be a line.By incidence axiom 2 there are two points P, Q thatlie on l.By theorem 13.10 there exists a third point R that does lie on l.By incidence axiom 1, there exist lines m = ←→ P R and n = ←→ QR.Since P ∈ l and P ∈ m, l intersects m (definition of intersection).Since Q ∈ l and Q ∈ n, l intersects n (definition of intersection).Since R ∉ l then any line that contains R is different from l. The two linesm and n contain R, hence m ≠ l and n ≠ l.Suppose m = n. By definition of m, P ∈ m. By defintion of n, Q ∈ n.Hence if m = n, Q ∈ m, i.e., both P and Q line on m. Hence m = l bythe uniqueness part of incidence axiom 1. This contradicts the previousparagraph, m ≠ l. Hence the assumption m = n must be wrong.Hence m ≠ n, which means there are two distinc lines that intersect l.Theorem 13.13 Let P be a point. Then there exists at least one line thatdoes not contain P .Theorem 13.14 There exist three distinct lines such that no point lies onall three of them.Theorem 13.15 Let P be a point. Then there exist points Q and R suchthat P, Q, R are non-collinear.Proof. Let P and Q be points such that Q ≠ P . By incidence axiom 1,there is a unique line l that contains P and Q.By theorem 13.10 there exists at least one point R that does lie on l. Thepoints P, Q, R are non-collinear.Theorem 13.16 Let P ≠ Q be points. Then there exists a point Q suchthat P, Q, R are non-collinear.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

60 SECTION 13. INCIDENCE GEOMETRY« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 14BetweennessIn this section we will begin our formulation of plane geometry based onVenema’s axiomatic system. Our undefined terms are: point, line, distance,half-plane, and angle measure.Axiom 14.1 (Existence Postulate) The collection of all points forms anonempty set with more than one (i.e., at least two) points.Definition 14.2 The set of all points in the plane is called P.Definition 14.3 We denote the line through the points A and B by ←→ AB.Definition 14.4 We will write A ∈ l to indicate that the point A lies onthe line l, and A is said to be incident with l.Axiom 14.5 (Incidence Postulate) Every line is a set of points. Forevery pair of distinct points A, B there is exactly one line l = ←→ AB suchthat A, B ∈ l.Definition 14.6 A point Q is external to a line l if Q ∉ l.Two lines are said to be parallel if the do not intersect, i.e., l ∩ m = ∅.Definition 14.7 (Parallel) Two lines l and m are said to be parallel ifthere is no point P such that P ∈ l and P ∈ m , and we write l ‖ m.Theorem 14.8 If l and m are two distinct, nonparallel lines, then thereexists exactly one point P that lines on both l and m.We proved this as theorem 13.9.61

62 SECTION 14. BETWEENNESSFigure 14.1: The Ruler Postulate tells us that there is a 1-1 correspondencebetween the points on a line and the real numbersCorollary 14.9 If l and m are lines, the either (1) l = m; (2)l ‖ m; or (3)l ∩ m contains precisely one point.Definition 14.10 (coordinate) Let l be a line. A 1-1 correspondence 1f : l ↦→ R such that P Q = |f(P )−f(Q)| for every point P, Q ∈ l is called acoordinate function for l and the number f(P ) is called the coordinateof P .We will assume that a coordinate function exists for every line, and use itto define the distance P Q.Axiom 14.11 (Ruler Postulate) For every pair of points P, Q there is anumber P Q called the distance from P to Q. For each line l there is aone-to-one mapping f : l ↦→ R such that if x = f(P ) and y = f(Q) thenP Q = |x − y| is the value of the distance.Definition 14.12 (Betweenness) Let A,B,C are distinct points. We saythat B is between A and C, and write A∗B∗C, if C ∈ ←→ AB and AC+CB =AB.Definition 14.13 The (line) segment (joining A and B) isAB = {A, B} ∪ {P |A ∗ P ∗ B}Definition 14.14 The ray (from A in the direction of B) is−→AB = AB ∪ {P |A ∗ B ∗ P }Definition 14.15 The length of segment AB, denoted by AB, is the distancefrom A to B.Definition 14.16 We call the points A and B the endpoints of AB.Definition 14.17 Two segments AB and BC are said to be congruent ifthey have the same length, and we say AB ∼ = BC, i.e.,AB = BC ⇐⇒ AB ∼ = BC1 Recall from 6.13 that a 1-1 correspondence is a function that is 1-1 and onto.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 63Figure 14.2: Betweenness on a line segment (top), ray (middle), and line(bottom).Definition 14.18 A metric is a function d : P × P ↦→ R that satisfies:1. (∀P, Q ∈ P) d(P, Q) = d(Q, P )2. (∀P, Q ∈ P) d(P, Q) ≥ 03. d(P, Q) = 0 ⇐⇒ P = QThe following theorem is usually also taken as a fourth requirement of ametric in most analysis books.Theorem 14.19 (Triangle Inequality)d(P, Q) ≤ d(P, R) + d(R, Q)We will not be able to prove this theorem until we learn more about planegeometry (see theorem 27.2).Lemma 14.20 Given any two points P, Q ∈ P, there exists a line containingboth P and Q.Proof. Either P = Q or P ≠ Q.If P ≠ Q then there is exactly one line l = ←→ P Q such that P and Q bothline on l (incidence postulate).If P = Q then by the existence postulate there must be a second pointR ≠ P and by the incidence postulate there is a unique line l that containsboth P and R; since P = Q then Q ∈ l.Hence there is a line l that contains both P and Q.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

64 SECTION 14. BETWEENNESSTheorem 14.21 Distance is a metric.Proof. Let P and Q be points. Then we need to show that each of thefollowing hold:(a) P Q = QP(b) P Q ≥ 0(c) P Q = 0 ⇐⇒ P = QBy lemma 14.20 there is a line l that contains both P and Q.By the ruler postulate (axiom 14.11), there is a one to one function f : l ↦→R, with let x = f(P ) and y = f(Q) such that the distance is given byTo see (a),To see (b),P Q = |f(P ) − f(Q)| = |x − y|P Q = |x − y| = |y − x| = QPP Q = |x − y| ≥ 0To see (c), first suppose that P Q = 0. Then0 = P Q = |x − y|⇒ x = y⇒ P = Qwhere the last line follows because f is one-to-one. To verify the converseof (c) suppose that P = Q. Then x = f(P ) = f(Q) = y so thatwhich verifies the converse of (c).P Q = |x − y| = 0Example 14.1 (Euclidean Metric) Let P = (x 1 , y 1 ), Q = (x 2 , y 2 ). Then»d(P, Q) = (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2To show that this is a metric, calculate»d(P, Q) = (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2 ]»= (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 ]= d(Q, P )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 65which verifies property (1).To get property (2), observe that the argument of the square root is anon-negative number, hence the square root is defined and positive or zero.For property (3), first assume P = Q. Thend(P, Q) = d(P, P )»= (x 1 − x 1 ) 2 − (y 1 − y 1 ) 2= 0This shows that (P = Q) ⇒ (d(P, Q) = 0), To get the converse, assumethat d(P, Q) = 0. Then»0 = (x 1 − x 2 ) 2 + (y 1 − y 2 ) 20 = (x 1 − x 2 ) 2 + (y 1 − y 2 ) 2 (14.1)If either x 1 − x 2 ≠ 0 or y 1 − y 2 ≠ 0 then the right hand side of equation14.1 is non-zero. Hence x 1 = x 2 and y 1 = y 2 , which means P = Q. Thus(d(P, Q) = 0) ⇒ (P = Q).Example 14.2 (Cartesian Coordinates in the Euclidean Metric)In the Cartesian Plane any (non-vertical) line l can be described by someequationy = mx + band any vertical line byLetif l is non-vertical, and setif l is vertical.y = af(x, y) = x √ 1 + m 2 (14.2)f(x, y) = f(a, y) = yTo see that f is a coordinate function and that this works, we need toconsider each case (vertical and non-vertical) separately, and to show thatf is 1-1, onto, and satisfiesin each case.P Q = |f(P ) − f(Q)| (14.3)Suppose first that l is non-vertical, and define f as given by equation 14.2.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

66 SECTION 14. BETWEENNESS(a) To show that f is 1-1, letP = (x 1 , y 1 ), Q = (x 2 , y 2 )and suppose that f(P ) = f(Q) (see definition 6.11). Thenx 1√1 + m2 = f(P ) = f(Q) = x 2√1 + m2Since √ 1 + m 2 ≠ 0 it can be cancelled out, giving x 1 = x 2 . Thusy 1 = mx 1 + b= mx 2 + b= y 2Hence f is 1-1 (P = Q ⇒ f(P ) = f(Q)).(b) To show that f is onto, pick any z ∈ R and definex =z√1 + m2andy = mx + b (14.4)Then P = (x, y) ∈ l because of 14.4 andf(P ) = f(x, y) = x √ 1 + m 2 = zThus (definition 6.12) f is onto R.(c) To verify the distance formula (equation 14.3), let P = (x 1 , y 1 ), Q =(x 2 , y 2 ) ∈ l. Theny 1 = mx 1 + by 2 = mx 2 + b« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 67henceP Q = d(P, Q)»= (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2»= (x 2 − x 1 ) 2 + (mx 2 + b − mx 1 − b) 2»= (x 2 − x 1 ) 2 + (m(x 2 − x 1 )) 2»= (1 + m 2 )(x 2 − x 1 ) 2= √ 1 + m 2 |x 2 − x 1 |∣√ √ = ∣x 2 1 + m2 − x 1 1 + m2∣= |y 2 − y 1 |= |f(P ) − f(Q)|Thus if l is not a vertical line, f is a coordinate function.Now suppose that l is a vertical line with equation x = a and define f :l ↦→ R by f(a, y) = y.(a) Let P = (a, y 1 ), Q = (a, y 2 ) ∈ l where P ≠ Q, hence y 1 ≠ y 2 andf(P ) = y 1 ≠ y 2 = f(Q)which shows that f is 1-1 (P ≠ Q ⇒ f(P ) ≠ f(Q)).(b) Let y ∈ R be any number. Then P = (a, y) ∈ l and f(P ) = y. Hence lis onto.(c) Let P = (a, y 1 ) and Q = (a, y 2 ). ThenP Q =»(a − a) 2 + (y 2 − y 1 ) 2= |y 2 − y 1 |= |f(Q) − f(P )|Example 14.3 (The “Taxicab” Metric) Let P = (x 1 , y 1 ), Q = (x 2 , y 2 ).Thend(P, Q) = |x 1 − x 2 | + |y 1 − y 2 |A coordinate function for the taxicab metric is given byf(x, y) =®x(1 + |m|), if l is not vertical;y, if l is verticalRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

68 SECTION 14. BETWEENNESSThe following theorem tells us that we can place the origin of the ruler atany place we want, and orient the ruler in any direction we want.Theorem 14.22 (Ruler Placement Postulate) For every pair of distinctpoints P, Q there is a coordinate function f : ←→ P Q ↦→ R such that f(P ) = 0and f(Q) > 0.Lemma 14.23 Let f : l ↦→ R be a coordinate function for l and let c ∈ R.Then g : l ↦→ R given by g(P ) = f(P ) + c is also a coordinate function forl.Proof. (Lemma 14.23) We need to show three things: that g is 1-1, onto,and P Q = |g(Q) − g(P )|.(a) Suppose g(P ) = g(Q). Thenf(P ) + c = f(Q) + cso f(P ) = f(Q). Since f is 1-1, P = Q. Thus (g(P ) = g(Q)) ⇒ (P = Q)so g is 1-1.(b) Let x ∈ R. Since f is onto, ∃P ∈ l such thatso thatf(P ) = x − cg(P ) = f(P ) + c = xHence ∀x ∈ R, ∃P ∈ l such that g(P ) = x. Thus g is onto.(c) P Q = |f(P ) − f(Q)| = |f(P ) + c − f(Q) − c| = |g(P ) − g(Q)|Lemma 14.24 Let f : l ↦→ R be a coordinate function. Then g(x) = −f(x)is a coordinate function.Proof. (Lemma 14.24) We need to show three things: that g is 1-1, onto,and P Q = |g(Q) − g(P ).(a)Let g(P ) = −f(P ). Suppose that g(P ) = g(Q). Then −f(P ) = −f(Q)hence P = Q. Hence g is 1-1.(b) Let x ∈ R. Since f is onto, there is some point P ∈ l such thatf(P ) = −x. Hence ∃P ∈ l such that −g(P ) = −x, i.e., there is some P ∈ lsuch that g(P ) = x. Hence g is onto.(c) P Q = |f(Q) − f(P )| = | − g(Q) + g(P )| = |g(Q) − g(P )|Proof. (Theorem 14.22) Pick any two distinct points P ≠ Q.incidence postulate (Axiom 14.5) there is a line l = ←→ P Q.Buy the« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 69By the Ruler postulate (axiom 14.11) there exists a coordinate functiong : l ↦→ R. Definec = −g(P )and define h : l ↦→ R byh(x) = g(x) + cThen h is a coordinate function by Lemma 14.23.Since h(P ) = 0, it must be the case that h(Q) ≠ 0 because h is 1-1. Wehave two cases to consider: h(Q) > 0 or h(Q) < 0.If h(Q) > 0 then h has the desired properties and the theorem is proven.If h(Q) < 0, define a new function j : l → R by j(x) = −h(x). Then bylemma 14.24, j is a coordinate function. Since it has the required properties,the theorem is proven.Figure 14.3: Circles that intersect in the real plane do not necessarily intersectin the rational plane.Distances Must Be Real. The following examples illustrate why rulers(hence distances) require real numbers and not rational numbers.Example 14.4 The distance between the points (1, 0) and (0, 1) in P is√2.Example 14.5 Find the intersection of the line y = x and the unit circleusing whatever knowledge you may already have of circles and triangles.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

70 SECTION 14. BETWEENNESSExample 14.6 Let P = (0, 0), Q = (2, 0). The circles of radius 2 centeredat P and Q do not interset in Q 2 . Their intersection in R 2 is (1, ± √ 3)(figure 14.3).Theorem 14.25 (Betweenness Theorem for Points) Let A, B, C bedistinct points on the line l, and let f : l ↦→ R be a coordinate function forl. Then the A ∗ C ∗ B if and only if eitherf(A) < f(C) < f(B) (14.5)orf(A) > f(C) > f(B) (14.6)Proof. Suppose that f(A) < f(C) < f(B). ThenAC + CB = |f(C) − f(A)| + |f(B) − f(C)|= f(C) − f(A) + f(B) − f(C)= f(B) − f(A)= |f(B) − f(A)|= ABso that A ∗ C ∗ B by definition 14.12.A similar argument holds if f(A) > f(C) > f(B).Now consider the converse. Assume that A ∗ C ∗ B so thati.e.,AC + CB = AB|f(C) − f(A)| + |f(B) − f(C)| = |f(B) − f(A)|But by algebra, we also havehencef(C) − f(A) + f(B) − f(C) = f(B) − f(A)|f(C) − f(A) + f(B) − f(C)| = |f(B) − f(A)|Let u = f(C) − f(A) and v = f(B) − f(C). Then|u| + |v| = |u + v|From algebra we know that this means either u and v are both positive orboth negative. [Assume the converse. If u > 0 and v < 0 then this gives« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 71u − v = u + v which implies v = 0 or f(B) = f(C); but C and B aredistinct points so f(B) ≠ f(C). If u < 0 and v > 0 then −u + v = u + vwhich implies u = 0 or f(C) = f(A) which is impossible because A and Care distinct points.]Since u and v have the same sign, then both f(C) − f(A) and f(B) − f(C)have the same sign.If both f(C) − f(A) > 0 and f(B) − f(C) > 0 then f(C) > f(A) andf(B) > f(C) so that f(B) > f(C) > f(A).If both f(C) − f(A) < 0 and f(B) − f(C) < 0 then f(C) < f(A) andf(B) < f(C) so that f(B) < f(C) < f(A).Corollary 14.26 If A, B, C are distinct collinear points then exactly oneof them lies between the other two.Proof. Since A, B, C are distinct then they correspond to real numbersx, y, z. Then this is a property of the real numbers, exactly one of x,y, andz lies between the other two.Corollary 14.27 Let A, B, C be points such that B ∈ −→ AC. ThenA ∗ B ∗ C ⇐⇒ AB < ACProof. By theorem 14.25 one the following must hold:If 14.7 holds thenIf 14.8 holds thenf(A) < f(B) < f(C) (14.7)f(A) > f(B) > f(C) (14.8)AB = f(B) − f(A) < f(C) − f(A) = ACAB = f(A) − f(B) < f(A) − f(C) = ACTo prove the converse, suppose that AB < AC.By the corollary one of A,B, C lies between the other two. We have threepossibilities: A ∗ B ∗ C, B ∗ A ∗ C or A ∗ C ∗ B.But B ∗ A ∗ C is not possible B ∈ −→ AC and B is distinct from A.So suppose A ∗ C ∗ B. Then eitherf(A) < f(C) < f(B)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

72 SECTION 14. BETWEENNESSorf(A) > f(C) > f(B).If f(A) > f(C) > f(B) then−AC = f(C) − f(A) > f(B) − f(A) = −ABso AB > AC which is a contradiction.If f(A) < f(C) < f(B) thenAC = f(C) − f(A) < f(B) − f(A) = ABwhich is also a contradiction. Hence A ∗ C ∗ B is also not possible. All thatis left is A ∗ B ∗ C.Definition 14.28 The point M is the midpoint of the segment AB if A ∗M ∗ B and AM = MB.Theorem 14.29 (Existence and Uniqueness of Midpoints) If A andB are distinct points then there exists a unique point M that is the midpointof AB.Proof. To prove existence, let f be a coordinate function for the line ←→ AB,and definex =f(A) + f(B)2Since f is onto, there exists some point M ∈ ←→ AB such that f(M) = x.Hence2f(M) = f(A) + f(B)orf(M) − f(B) = f(A) − f(M)Thus AM = MB. To see that A ∗ M ∗ B, leta = min(f(A), f(B))b = max(f(A), f(B))Since A and B are distinct then a ≠ b and we havewith a < b. Hencex = a + b2x < 2b2 = b« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 14. BETWEENNESS 73andx > 2a 2 = agivinga < x < bHence eitherf(A) < f(M) < f(B)orf(A) > f(M) > f(B).By theorem 14.25 A ∗ M ∗ B.To verify uniqueness, let M ′ ∈ AB, where M ′ ≠ M and AM ′ = M ′ B.Suppose that f(A) < f(B). Then both the following hold:f(A) < f(M) < f(B)f(A) < f(M ′ ) < f(B)Furthermore, since M and M ′ are midpoints|f(A) − f(M)| = AM= 1 2 AB= AM ′= |f(A) − f(M ′ )|Since f(A) < f(M) and f(A) < f(M ′ ), this givesf(M) − f(A) = f(M ′ ) − f(A)f(M) = f(M ′ )Since f is 1-1 then M = M ′ , which proves uniqueness when f(A) < f(B).A similar argument holds to prove uniqueness when f(A) > f(B).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

74 SECTION 14. BETWEENNESS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 15The Plane SeparationPostulateIntuitively, we know that a line divides a plane into two halves. These twohalves are called half-planes. We will take this observation as an axiom.Definition 15.1 A set of points S is convex if for every P, Q ∈ S, theentire segment P Q ∈ S.Axiom 15.2 (Plane Separation Postulate) For every line l the pointsthat do not lie on l form two disjoint, convex non-empty sets H 1 and H 2 ,called half-planes, bounded by l such that if P ∈ H 1 and Q ∈ H 2 thenP Q intersects l.More specifically, the plane separation postulate tells us the following (seefigure 15.1).H 1 ∪ H 2 = P − l (15.1)(E, F ∈H 1 ) ⇒H 1 ∩ H 2 = ∅ (15.2)(EF ⊆ H 1 ) ∧ (EF ∩ l = ∅) (15.3)(G, H ∈H 2 ) ⇒(GH ⊆ H 2 ) ∧ (GH ∩ l = ∅) (15.4)P ∈ H 1 ∧ Q ∈ H 2 ⇒ P Q ∩ l ≠ ∅ (15.5)75

76 SECTION 15. THE PLANE SEPARATION POSTULATEFigure 15.1: Plane Separation Postulate (axiom 15.2). P Q intersects theline because P and Q are in different half planes.Definition 15.3 Let l be a line and A a point not on l. Then we use H A,lto denote the half-plane of l that contains A. When the line is clear fromthe context we will just the notation H A .Definition 15.4 Two points A, B are said to be on the same side of theline l if they are both in the same half-plane. They are said to be onopposite sides of the line if they are in different half planes.In figure 15.1 points P and Q are on opposite sides of l, while points P andF are on the same side of l. In terms of this notation, we can restate theplane separation postulate as follows.Axiom 15.5 (Plane Separation Postulate, Second Form) Let l be aline and let A, B be points not on l. Then A and B are on the same sideof l if and only ifAB ∩ l = ∅,and are on opposite sides of l if and only if.AB ∩ l ≠ ∅Definition 15.6 Two rays −→ −→BA and BC having the same endpoint are oppositerays if BA ≠ BC and−→ −→←→ −→ −→AB = BA ∪ BCor equivalently, A ∗ B ∗ CDefinition 15.7 An angle is the union of two non-opposite rays −→ AB and−→AC having the same endpoint, and is denoted by ∠BAC or ∠CAB. Thepoint A is called the vertex of the angle and the two rays are called thesides of the angle.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 15. THE PLANE SEPARATION POSTULATE 77Figure 15.2: Definition of angle ∠CAB. The interior of the angle is theintersection of the two half-planes, and is shaded darker..BACDefinition 15.8 Let A, B, C be points such that the rays −→ −→ AB ≠ AC arenot opposite. The interior of ∠BAC isH B,←→ AC∩ H C,←→ AB,i.e., the intersection of the half plane H B determined by B and ←→ AC andthe half plane H C determined by C and AB (figure 15.2).Definition 15.9 Three points A, B, C are collinear if there exists a singleline l such that A, B, and C all lie on l. If no such line exists, then thepoints are non-collinear.Corollary 15.10 If A, B, and C are non-collinear, then the rays −→ AB and−→AC are neither opposite nor equal.Definition 15.11 Let A, B, and C be non-collinear points.triangle △ABC is the union of the three segments,Then the△ABC = AB ∪ BC ∪ CAThe points A, B, and C are called the vertices of the triangle, and thesegments AB, BC and CA are called the sides of the triangle.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

78 SECTION 15. THE PLANE SEPARATION POSTULATETheorem 15.12 (Pasch’s Theorem) Let △ABC be a triangle and supposethat l is a line that does not include A, B, or C. Then if l intersectsAB then it also intersects either BC or CA.Proof. Suppose l intersects AB and does not include any of the verticesA,B, or C.Figure 15.3: Pasch’s Theorem. Any line that intersects AB must alsointersect either AC or BC.BACLet H 1 and H 2 be the two half planes determined by l. Then the points Aand B are in opposite half-planes by the plane separation postulate and thehypothesis. Suppose A ∈ H 1 and B ∈ H 2 (this is just notation; we couldhave made the alternative assignment without any loss of generality).Then either C ∈ H 1 or C ∈ H 2 .If C ∈ H 1 , then B and C are in opposite half planes so BC intersects l, bythe plane-separation postulate.Alternatively, if C ∈ H 2 , then A and C are in opposite half-planes, so ACintersects l, also by the plane separation postulate.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 16AnglesDefinition 16.1 Ray −→−→ −→AD is between rays AB and AC if D is in theinterior of angle ∠BAC (figure 16.2), and we write 1−→AB ∗ −→ −→ AD ∗ ACAxiom 16.2 (Protractor Postulate) For every angle ∠BAC there is areal number µ(∠BAC) called the measure of ∠BAC such that1. 0 ≤ µ(∠BAC) < 1802. µ(∠BAC) = 0 ⇐⇒ −→ −→ AB = AC3. Angle Construction Postulate: For every number r ∈ (0, 180)and for each half plane H bounded by ←→ AB there exists a unique ray−→AE such that E ∈ H and µ(∠BAE) = r.4. Angle Addition Postulate If −→−→ −→AD is between rays AB and AC thenµ(∠BAD) + µ(∠DAC) = µ(∠BAC)When it is clear from the context that we are referring to the measure ofthe angle rather than the angle itself we will sometimes drop the µ andinstead of writing, say µ(∠BAC) = 45 we will just say ∠BAC = 45.If µ(∠BAC) < 90 we call angle ∠BAC an acute angle.If µ(∠BAC) > 90 we call angle ∠BAC an obtuse angle.1 The asterisk (∗) notation for rays is not used in the text but we will find it a convenientshorthand.79

80 SECTION 16. ANGLESFigure 16.1: The protractor postulate allows to associate a number between0 and 180 with every angle.BACIf µ(∠BAC) = 90 we call angle ∠BAC a right angle.Figure 16.2: Since point D is in the interior of angle ∠BAC, we say thatray −→−→ −→AD is between ray AB and AC. Furthermore, by the angle additionpostulate (axiom 16.2), γ = α + β, where α = µ(∠BAD), β = µ(∠DAC),and γ = µ(BAC).Definition 16.3 Two angles ∠BAC and ∠EDF are called congruent ifµ(∠BAC) = µ(∠EDF ), and we write ∠BAC ∼ = ∠EDFTheorem 16.4 Let l be a line, A ∈ l, B ∉ l, and A ∗ C ∗ B. Then B andC are on the same side of l (figure 16.3.)Proof. Let m = ←→ AB. Lines m and l are not parallel since they share pointA, and since B ∉ l then m ≠ l. Hence by theorem 13.9 they interesect at« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 16. ANGLES 81Figure 16.3: Points B and C are on the same side of the line l (theorem16.4).Figure 16.4: Illustration of the Z-theorem. Points B and E are on oppositesides of the line, hence rays −→ −−→AB and DE can not intersect.precisely one point, A.A ∗ C ∗ B ⇒∼ B ∗ A ∗ C (by corollary 14.26)⇒ A ∉ BC⇒ BC ∩ l = ∅By Axiom 15.5 (The Plane Separation Postulate), B and C are on the sameside of l.Corollary 16.5 Let l be a line, A ∈ l, B ∉ l, and C ∈ −→ AB where C ≠ A.The B and C are on the same side of l.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

82 SECTION 16. ANGLESCorollary 16.6 (The Z-Theorem) Let l be a line and let A, D ∈ l,A ≠ D. If B, E ∉ l are points on opposite sides of l then −→ −−→ AB ∩ DE = ∅.Proof. By corrollary 16.5,∀P ≠ A ∈ −→ AB, P ∈ HB,l∀Q ≠ D ∈ −−→ DE, Q ∈ H E,lSince B and E are on opposite sides of l,H B,l ∩ H E,l = ∅by the plane separation postulate (Axiom 15.2).Thus the only place the rays could intersect is the endpoints (A and D) butthese points are distincy by hypothesis. Hence the rays do not intersect.Corollary 16.7 (Betweenness for Rays is Well Defined) Let D bea point in the interior of angle ∠BAC. Then every point on the ray −→ AD(except for A) is in the interior of ∠BAC.Proof. Let A, B, C be noncollinear and let D be a point in the interior ofangle ∠BAC (figure 16.2).D and C are on the same side of ←→ AB by the definition of interior of anangle.Let P be any point on −→ AD except A.By Corollary 16.5 every point on −→−→AD is on the same side AB as C, i.e.,P ∈ H ←→ C, AB.By the same argument D and B are on the same side of ←→ AB and henceevery point on −→−→AD is on the same side of AC as B, i.e., P ∈ HB, ←→Hence P ∈ H ←→ C, AB∩ H ←→ B, ACHence P is in the interior of ∠BAC (defintion 15.8). Hence every point onthe ray −→ AD is in the interior of the angle.Theorem 16.8 Let A, B, C be distinct, noncollinear points. Let D ∈ ←→ BC.ThenB ∗ D ∗ C ⇐⇒ −→ −→ −→AB ∗ AD ∗ ACAC.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 16. ANGLES 83Figure 16.5: Illustration of theorem 16.8, illustrating B ∗ D ∗ C ⇐⇒ −→ AB ∗−→AD ∗ −→ ACProof. Suppose that B ∗ D ∗ C (hypothesis).Then by theorem 16.4, C and D are on the same side of ←→ AB, and B and Dare on the same of line ←→ ACHence by the definnition of interior of an angle (definition 15.8), D is inthe interior of angle ∠BAC.Hence by the definition of betweenness for rays (definition 16.1),To prove the converse, assume equation 16.1.−→AB ∗ −→ −→ AD ∗ AC (16.1)Then by corollary 16.7 point D is in the interior of angle ∠BAC.Hence B and D are on the same side of ←→ AC by the defition of interior ofan angle.Therefore C ∉ BD and hence B ∗ C ∗ D is not possible.By a similar argument, C and D are on the same side of ←→ AB, B ∉ CD, andC ∗ B ∗ D is not possible.Since D ∈ ←→ BC then one of the following must hold:(B ∗ C ∗ D) ∨ (C ∗ B ∗ D) ∨ (B ∗ D ∗ C)Since the first two possibilities have been eliminated then we must haveB ∗ D ∗ C by corollary 14.26.Theorem 16.9 Let A, B, C, D be four distinct points such that C and Dare on the same side of ←→ AB and D ∉ ←→ AC. Then either C is in the interiorof ∠BAD or D is in the interior of ∠BAC.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

84 SECTION 16. ANGLESFigure 16.6: Illustration of Theorem 16.9.Proof. Assume that D is not in the interior of ∠BAC ( figure 16.6) anduse this to prove that C is in the interior of ∠BAD. The proof of thealternative case is analogous.C and D are on the same side of ←→ AB by hypothesis. HenceD ∈ H C,←→ AB(16.2)Since D is not in the interior of ∠BAC by assumption, by the definition ofinterior,D ∉ H ←→ C, AB∩ H ←→ B, AChence either D ∉ H C,←→ ABor D ∉ H B,←→ AC.But by equation 16.2, D ∈ H C,←→ ABso we concludeD ∉ H B,←→ ACHence by the plane separation postulate, B and D must be on oppositesides of ←→ AC, andBD ∩ ←→ AC ≠ ∅−→ −→ −→Let E be the unique point of intersection. Since B ∗ E ∗ D, AB ∗ AE ∗ AD(theorem 16.8) and therefore E is in the interior of ∠BAD (definition ofbetweenness for rays), andE ∈ H ←→ D, ABHence E and D lie on the same side of ←→ AB. By hypothesis we alreadyassumed that C, D are on the same side of ←→ AB. Hence C, D, and E are onthe same side of line ←→ AB.Therefore −→ AC and −→ AE can not be opposite rays.Therefore −→ AC = −→ AE.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 16. ANGLES 85Figure 16.7:16.10).Illustration of the betweenness theorem for rays (theoremSince ray −→ AE is in the interior of angle ∠BAD, that means −→ AC must alsobe in the interior of ∠BAD.Hence C is in the interior of angle ∠BAD.Theorem 16.10 (Betweenness Theorem for Rays) Let A, B, C, andD be four distinct points such that C and D lie on the same side of ←→ AB.Thenµ(∠BAD) < µ(∠BAC) ⇐⇒ −→ −→ −→AB ∗ AD ∗ ACProof. See figure 16.7.To prove (⇐), assume −→ AB ∗−→ AD ∗−→ AC.D is in the interior of ∠BAC by Corollary 16.7.By the angle addition postulateµ(∠BAC) = µ(∠BAD) + µ(∠DAC)and by the protractor postulate ∠DAC > 0 because AB ≠ AC. Henceµ(∠BAD) < µ(∠BAC)To prove the (⇒) we will prove the contrapositive, which is∼ ( −→ AB ∗−→ AD ∗−→ AC) ⇒ [µ(∠BAC) ≤ µ(∠BAD)]So we start by assuming ∼ ( −→ −→ −→AB ∗ AD ∗ AC).Since D and C are on the same side of ←→ AB we must rule out the possibilitythat −→ −→ −→AD ∗ AB ∗ ACRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

86 SECTION 16. ANGLESD cannot be in the interior of ∠BAC because then we would have −→ AB ∗−→AD ∗ −→ AC.Hence eitheror−→AD ∗ −→ AC ∗ −→ AB−→AC = −→ −→AD, i.e., D ∈ ACIf D ∈ −→ AC then∠BAD = ∠BAC (16.3)If D ∉ −→ AC then −→ −→ −→AD∗AC ∗AB. By the proof of the first part of the theorem(the (⇐)),∠BAC < ∠BAD (16.4)Combining equations 16.3 and 16.4 gives the desired result.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 17The Crossbar TheoremTheorem 17.1 (The Crossbar Theorem) Let △ABC be a triangle andD a point in the interior of angle ∠BAC. Then there is a point E that lieson both −→ AD and BC.Figure 17.1: Illustration of the crossbar theorem (theorem 17.1). If D is inthe interior of ∠CAB then it must cross side BC of △ABCProof. The proof will apply the z-theorem three times.Choose points P and Q such that P ∗ A ∗ D and Q ∗ A ∗ C as shown infigure 17.2. Define l = ←→ P D.Since D is in the interior of ∠BAC neither B nor C lie on l.We can apply Pasch’s Theorem (theorem 15.12) to △BCQ. Since ADcrosses CQ of △BCQ it must cross either BQ or BC. This gives us four87

88 SECTION 17. THE CROSSBAR THEOREMFigure 17.2: Illustration for proof of the crossbar theorem (theorem 17.1).possibilities:−→AD ∩ BC ≠ ∅ (17.1)−→AD ∩ BQ ≠ ∅ (17.2)−→AP ∩ BC ≠ ∅ (17.3)−→AP ∩ BQ ≠ ∅ (17.4)Since P ∗ A ∗ D, P and D lie on opposite sides of ←→ AC (Plane Separation).But B and D are on the same side ←→ AC (because D is in the interior of∠CAB), so B and P are on opposite sides of ←→ AC.−→ −→By the Z-theorem (see figure 17.3), QB ∩ AP = ∅.Hence QB ∩ −→ AP = ∅. This eliminates equation 17.4.Figure 17.3: Elimination of equation 17.4 by the Z-theorem (heavy lines).Since B and P are on opposite sides of ←→−→AC, as we have just argued, CB ∩−→AP = ∅ (which which also follows from the Z-theorem).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 17. THE CROSSBAR THEOREM 89Hence BC ∩ −→ AP = ∅. This eliminates equation 17.3 (see figure 17.4).Figure 17.4: Elimination of equation 17.3 by the Z-theorem (heavy lines).Since Q ∗ A ∗ C, Q and C are on opposite sides of ←→ AB.Since C and D are on the same side of ←→ AB (because D is in the interiorof ∠CAB), then Q and D are on opposite sides of ←→ AB (plane separationpostulate).By the Z-theorem −→ −→−→BQ ∩ AD = ∅. Hence BQ ∩ AD = ∅. This elimninatesequation 17.2 (see figure 17.5).Figure 17.5: Elimination of equation 17.2 by the Z-theorem (heavy lines).Since equations 17.2, 17.3, and 17.4 have been eliminated, then equation17.1 must be true.Since ←→ AD intersects BC they have a unique point of intersection. Call thispoint E. This is the point that the theorem says exists.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

90 SECTION 17. THE CROSSBAR THEOREMTheorem 17.2 (MacLane’s Continuity Axiom a ) A point D is in theinterior of angle ∠BAC if and only if the ray −→ AD intersects the interior ofthe segment BC.a MacLane [MacLane, 1959] takes this result as an axiom and uses this name because itimplies Birkhoff’s [Birkhoff, 1932] continuity axiom (theorem 20.2).Figure 17.6: Illustration of MacLane’s Continuity Axiom (theorem 17.2.)Proof. (⇒) Suppose that D is in the interior of ∠BAC.−→crossbar theorem, AD intersects BC.Then by theSince the intersection point lies on a ray that is interior to ∠BAC, it doesnot lie on an endpoint of BC (definition of interior; else it would lie on one ofthe two rays that define the angle, not the intersection of their half-planes);hence the intersection point is interior to BC.(⇐) Suppose that −→ AD intersects the interior of BC. Call the point ofintersection E.−→ −→ −→ −→ −→Hence B ∗ E ∗ C. By theorem 16.8, AB ∗ AE ∗ AC. Hence AE = AD is inthe interior of angle ∠BAC. Hence D is in the interior of ∠BAC.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 18Linear PairsDefinition 18.1 Two angles ∠BAD and ∠DAC are called a linear pair−→ −→if AB and AC are opposite rays.Definition 18.2 Two angles ∠BAD and ∠DAC are called supplementaryangles if ∠BAD + ∠DAC = 180.Figure 18.1: Angles angles ∠BAD and ∠DAC form a linear pairLemma 18.3 (Linear Pair Lemma) If C ∗ A ∗ B and D is in the interiorof ∠BAE then E is in the interior of ∠DAC.Proof. Since D is in the interior of ∠BAE, E and D are on the same sideof line ←→ AB by the definition of interior of an angle.Since ←→ AB = ←→ AC, this means that E and D are on the same side of line ←→ AC,i.e.,E ∈ H ←→ D, ACBy the crossbar theorem, ray −→ AD intersects segment BE (see figure 18.2).91

92 SECTION 18. LINEAR PAIRSFigure 18.2: Illustration of proof of Linear Pair Lemma (Lemma 18.3).Therefore E and B are on different sides of ←→ AD by the plane separationpostulate.Since C ∗ A ∗ B then B and C are on opposite sides of ←→ AD.Therefore C and E are on the same side of ←→ AD, i.e.HenceE ∈ H C,←→ ADE ∈ H ←→ D, AC∩ H ←→ C, ADThen by definition E is in the interior of angle ∠CAD.Theorem 18.4 (Linear Pair Theorem) If ∠BAD and angle ∠DAC fora linear pair then∠BAD + ∠DAC = 180 (18.1)Proof. Suppose that ∠BAD and ∠DAC form a linear pair. Then −→ AB and−→AC are opposite.Let α = ∠BAD and let β = ∠DAC (figure 18.3).α + β = 180.We must show thatThere are three possible cases: α + β > 180, α + β < 180, and α + β = 180.Suppose that α + β < 180. By the angle construction postulate there is apoint E on the same side of ←→ AB as D such that ∠BAE = α + β < 180.Define γ = ∠DAE.By the betweenness theorem for rays, D is in the interior of angle ∠BAE.Hence∠BAE = ∠BAD + ∠DAE« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 18. LINEAR PAIRS 93Figure 18.3: Illustration of symbols in the proof of the linear pair theorem(theorem 18.4).Figure 18.4: Illustration of symbols in the second part of the proof of thelinear pair theorem (theorem 18.4).by the angle addition postulate. Henceα + β = α + γγ = βBut E is in the interior of ∠DAC by lemma 18.3. Hence by the angleaddition postulate,γ + ∠EAC = βHence ∠EAC = 0. By the protractor postulate, this means that −→ AE =−→AC,which contradicts the fact that E is in the interior of ∠DAC. Hencewe must conclude that α + β ≥ 180.Now suppose that α + β > 180. Choose a point F on the same side of ←→ ABas D such that γ = ∠BAF = α + β − 180 (figure 18.4).Since β < 180,α + β − 180 < αBy the betweenness theorem for rays this means that F is in the interior ofangle ∠BAD. By the angle addition postulate∠BAF + ∠F AD = ∠BADRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

94 SECTION 18. LINEAR PAIRSγ + ∠F AD = αα + β − 180 + ∠F AD = α∠F AD = 180 − βAlso, by lemma 18.3, D is in the interior of angle ∠F AC, and by the angleaddition postulate∠F AC = ∠F AD + ∠DAC= 180 − β + β= 180which contradicts the protractor postulate. Hence α + β > 180 is false.Hence α + β = 180« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 19Angle BisectorsDefinition 19.1 (Angle Bisector) Let A, B, C be distinct non-collinearpoints. The ray −→ AD is the angle bisector of ∠BAC if D is in the interiorof ∠BAC and ∠BAD = ∠DACFigure 19.1:−→ AD is the angle bisector of ∠BAC if α = β.Theorem 19.2 (Angle Bisector Existence and Uniqueness) IfA, B, C are distinct and non-collinear then there exists a unique angle bisectorfor ∠BAC.Proof. To prove existence, define the measures of the various angles as illustratedin figure 19.1. We observe that by the angle construction postulatethere exists a unique ray −→ AD such thatβ = ∠BAD = 1 2 ∠BAC95

96 SECTION 19. ANGLE BISECTORSwith D on the same side of ←→ AB as C.By the Betweenness theorem for rays,By the angle addition postulateThis proves existence.−→AB ∗ −→ −→ AD ∗ AC∠BAC = ∠BAD + ∠DAC= β + α= 1 2 ∠BAC + α⇒ α = 1 2 ∠BAC = β−→ −→Figure 19.2: Uniqueness of Angle bisectors. AD and AE are both bisectorsof ∠BAC. To prove uniqueness, they must be the same ray.To prove uniqueness, let −→ AE be another angle bisector of ∠BAC (see figure19.2.)If we do the same construction as before we find that ∠BAE = ∠BAD;by the uniqueness part of the angle construction postulate, this means that−→AE = −→ AD.Definition 19.3 Two lines l and m are perpendicular (l ⊥ m) if thereexists a point A on both l and m, a point B ∈ l, and a point C ∈ m suchthat ∠BAC is a right angle.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 19. ANGLE BISECTORS 97Definition 19.4 Let A and B be distinct points. A perpendicular bisectorof AB is a line l such that the midpoint of AB lies on l, and l ⊥ ←→ AB.Theorem 19.5 (Existence and Uniqueness of Perpendicular Bisectors)If A and B are distinct points then there exists a unique perpendicularbisector of AB.Proof. By the midpoint existence theorem, the segment AB has a uniquemidpoint M.By the protractor postulate there exists a point C ∉ ←→ AB such that ∠AMC =90.Let l = ←−→ MC.Since l ⊥ ←→ AB and M ∈ l, l is the perpendicular bisector. This provesexistence.The midpoint is unique by the midpoint existence/uniqueness theorem. Bythe protractor postulate there is only one ray on each side of ←→ AB that isperpendicular to AB and passes through M. These two rays are oppositeand on the same line and hence the bisector is unique.Definition 19.6 Angles ∠BAC and ∠DAE form a vertical pair (are verticalangles) if −→ −→−→ −→AB and AE are opposite and rays AC and AD are oppositeor rays −→ −→−→ −→AB and AD are opposite and AC and AE are opposite.Figure 19.3: Angles α and β are vertical anglesTheorem 19.7 (Vertical Angle Theorem) Vertical angles are congruent.Proof. Assume the geometry shown in figure 19.3. Then −→ −→AB and AE areopposite; and −→ AC and −→ AD are opposite. We need to show that α = β.Angles β = ∠EAD and ∠EAC are a linear pair. Henceβ + ∠EAC = 180Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

98 SECTION 19. ANGLE BISECTORSAngles ∠EAC and ∠CAB = α are also a linear pair, so∠EAC + α = 180Using algebra,180 − β = ∠EAC = 180 − αα = β« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 20The Continuity AxiomRecall that a function f is continuous if ∀ɛ > 0, ∃δ > 0 such that|x − y| < δ ⇒ |f(x) − f(y)| < ɛTheorem 20.1 Let [a, b] and [c, d] be closed intervals of real numbers andlet f : [a, b] ↦→ [c, d] be a function. If f is strictly increasing and onto, thenf is continuous.Proof. Let ɛ > 0 be given (and sufficiently small) and let x ∈ (a, b).Since f is strictly increasing, c < f(x) < d for all x ∈ (a, b). We takec < f(x) − ɛf(x) + ɛ < dIf we did not start with an ɛ that was sufficiently small than reduce thevalue of ɛ to, say, (1/2) min[|f(x) − c|, |f(x) − d|]Since f is onto, there exist numbers x 1 , x 2 ∈ [a, b] such thatf(x 1 ) = f(x) − ɛ and f(x 2 ) = f(x) + ɛChoose δ = 1 2 min [|x − x 1|, |x − x 2 |]Then x 1 < x − δ and x 2 > x + δSuppose that |x − y| < δ. Thenx − δ < y < x + δ99

100 SECTION 20. THE CONTINUITY AXIOMSince f is increasing,f(x) − ɛ = f(x 1 )< f(x − δ)< f(y)< f(x + δ)< f(x 2 ) = f(x) + ɛ−ɛ < f(y) − f(x) < ɛHence |f(x) − f(y)| < ɛ.Figure 20.1: If a function is strictly increasing and onto it is continuousdy 2yεεy 1δδcax 1x x 2bMotivation for the Continuity Axiom. Let A, B, C be non-collinearpoints, and d = BC, the distance from B to C (see figure 20.2).From the ruler placement postulate there is a one-to-one correspondencewith r(B) = 0 and r(C) = d.r : BC ↦→ [0, d]Since r is a one-to-one correspondence, then for any point D ∈ BC thereexists an x ∈ [0, d], x = r(D) or D = r −1 (x).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 20. THE CONTINUITY AXIOM 101Figure 20.2: The continuity axiom says that the function mapping thedistance x = BD to the measure of the angle α is continuous.Theorem 20.2 (The Continuity Axiom) The functiondefined byf : [0, d] ↦→ [0, µ(∠BAC)]f(x) = µ(∠BAD)is continuous, invertible, and its inverse is continuous.Proof. Let D, E be points such thatBy the definition of betweenness, this is true iffB ∗ D ∗ E ∗ C (20.1)BD < BE < BCFurthermore, equation 20.1 is possible iffby theorem 16.8.−→AB ∗ −→ −→ −→AD ∗ AE ∗ AC (20.2)By the betweenness theorem for rays (theorem 16.10), equation 20.2 holdsiffµ(∠BAD) < µ(∠BAE) < µ(∠BAC)Thus f(x) is strictly increasing.To show that the function is onto, pick any α ∈ (0, µ(∠BAC)) and constructa ray r with angle α.Since α < µ(∠BAC), by theorem 16.10 (betweenness for rays),−→AB ∗ −→ r ∗ −→ ACRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

102 SECTION 20. THE CONTINUITY AXIOMhence r is in the interior of ∠(BAC).Therefore it intersects BC at some point D such that B ∗ D ∗ C. Since0 < BD < BC, there is some number x such that 0 < x < d and α = f(x).Hence the function is onto.Since the function f(x) is strictly increasing and onto, by theorem 20.1,f(x) is continuous.Now suppose there exist x and y in (0, d) such that f(x) = f(y).By the ruler postulate there must be points D and E on BC such thatx = BD and y = BE.Since f(x) = f(y), we know that µ(∠BAD) = µ(∠BAE). This is onlypossible of −→ AE = −→ AD. Since both rays intersect BC, and since they arethe same ray, their intersection points must be identical, giving D = E ⇒BD = BE ⇒ x = y.Thus f(x) = f(y) ⇒ x = y, which means f is one-to-one. We have alreadyshown that f is onto. Since f is one-to-one and onto, it is invertible.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 21Side-Angle-SideDefinition 21.1 Two triangles △ABC and △DEF are congruent(△ABC ∼ = △DEF ) if there is a correspondence between the vertices ofthe first triangle and the vertices of the second triangle such that the correspondingangles are congruent and corresponding sides are congruent.Figure 21.1: SAS postulate. Distances AB = DE, BC = EF and anglesα = β hence the two triangles are congruent.The Side-Angle-Side Congruence Condition (SAS) is Euclid’s Proposition4; it was taken as an axiom by Hilbert as well as the SMSG (Axiom 15).103

104 SECTION 21. SIDE-ANGLE-SIDEBirkhoff and MacLane instead chose instead to start with axioms of trianglesimilarity rather than congruence. The UCSMP uses a set of reflectionpostulates instead of SAS.Axiom 21.2 (Side-Angle-Side (SAS) Postulate) If △ABC and△DEF are two triangles such that AB ∼ = DE, ∠ABC ∼ = ∠DEF , andBC ∼ = EF then △ABC ∼ = △DEF .Why can’t we just derive SAS from the other axioms, e.g., like Euclid? Forone thing, Euclid proved his proposition 4 using the concept of placing onefigure on top of another to show congruence (common notion 4). In fact,this should probably have been a separate axiom, and we have not statedany axiom of correspondence.The following example shows that the axioms we have so far are not sufficientto prove SAS.Figure 21.2: SAS failure in taxicab geometry.Example 21.1 Failure of SAS in Taxicab GeometryConsider triangles △ABC and △DEF as shown in figure 21.2.taxicab metric,AB = DE = 2In theAC = DF = 2∠EDF = ∠BAC = 90If SAS held in this model then the two triangles would be congruent. However,EF = 2 ≠ 4 = BCso the two triangles are not congruent.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 21. SIDE-ANGLE-SIDE 105This SAS is not a logical conclusion of the other axioms we have produced!By adding the addtional axiom, we rule out models that use a metric suchas the taxicab distance.Definition 21.3 A triangle is called an isosceles triangle if it has twocongruent sides.Theorem 21.4 (Isosceles Triangle Theorem) If △ABC is a trianglewith AB ∼ = BC then ∠BAC ∼ = ∠BCA.Proof. See figure 21.3. Let D be a point in the interior of ∠ABC such that−−→BD bisects ∠ABC (by the theorem on the existence of angle bisectors).Figure 21.3: Proof the isocelese triangle theorem (theorem 21.4).By the crossbar theorem, there is a point E at which the ray −−→ BD intersectsthe edge AC.Since AB = BC, BE = BE and ∠ABE ∼ = ∠CBE, the conditions for SASare met.Then △BAE ∼ = △BCE by SAS.Hence ∠BAC ∼ = ∠BCE.Theorem 21.5 Existence of Perpendiculars. For every line l thereexists another line m and a point P such that (P ∈ m) ∧ (l ⊥ m).Proof. Let l be a line and let C ∉ l be a point.By the ruler postulate there exist distinct points A, B ∈ l.By the protractor postulate there exists a point D on the opposite of l fromC such that α = β, where α = ∠BAC and β = ∠BAD (see figure 21.4).Define a point E on −→ AD such that AE = AC (ruler postulate). Since Cand D are on opposite sides of ←→ AB; and D and E are on the same side ofRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

106 SECTION 21. SIDE-ANGLE-SIDEFigure 21.4: Proof of the existence of perpendiculars (see theorem 21.5).the line, then C and E are on opposite sides. Hence segment CE intersectsline ←→ AB at some point F .Then △F AE ∼ = △F AC (SAS). Hence δ = µ(∠AF E) = µ(∠AF C) = γ.Since γ and δ form a linear pair, γ + δ = 180.γ = δ = 90.Hence m ⊥ l, proving existence of the perpendicular line.Hence (since γ = δ),« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 22Neutral GeometryNeutral Geometry is the study of geometry without any parallel postulate.Traditionally, it is the geometry based on Euclid’s first four axioms whichare summarized here.Euclid’s Axiom 1 A line may be drawn between any two points.Euclid’s Axiom 2 A line segment may be extended indefinitely.Euclid’s Axiom 3 A circle may be drawn with any give point as centerand any given radius.Euclid’s Axiom 4 All right angles are congruent.Neutral geometry is based entirely on these four postulates, or some equivalentset.To extend neutral geometry we can attempt to add one of the followingparallel postulates giving us either Euclidean Geometry or Hyperbolic Geometry.Euclidean Parallel Postulate. For every line l and every point P ∉ lthere is exactly one line m such that P ∈ m and m ‖ l.Hyperbolic Parallel Postulate. For every line l and every point P ∉ lthere are at least two distinct lines m and n (m ≠ n) such that P ∈ m andP ∈ n and m ‖ l and n ‖ l. (Note that m ̸‖ n because they intersect at P !)If one can imagine a single unique parallel line through a given point ormultiple lines through a given point, one might also ask the question ofwhether there are geometries without any parallel lines. This leads to thefollowing postulate:107

108 SECTION 22. NEUTRAL GEOMETRYElliptic Parallel Postulate. For every line l and every point P ∉ l thereis no line m such that P ∈ m and m ‖ l.We will see that this postulate is incompatible with Neutral Geometry.In fact, if we accept Neutral geometry then we may extend it with eitherthe Euclidean axiom or the Hyperbolic axiom, but that these two axiomsare the only possibilities. If we reject one, than the other one immediatelybecomes fact!Theorem 22.1 Under the conditions of neutral geometry, either the EuclideanParallel Postulate or the Hyperbolic Parallel Postulate must hold.The proof will be given later.This does not mean that the Elliptic postulate is without use. In factthe subject of Elliptic, and in particular, spherical geometry has importantpractical applications in navigation and measurement on a planetary surface.However, accepting elliptic measurement means throwing out one fothe other postulates, usually replacing the concept of betweenness with oneof separation. A more general formulation of geometry that was formalizedby Riemann, known as differential geometry or Riemannian geometry,describes geometry on curved surfaces (or manifolds).Following Venema’s development we have defined a different set of axiomsthat give the same results as Euclid’s first four axioms. We will call thesethe Axioms of Neutral Geometry:Neutral Geometry Axiom 1 (Existence Postulate) (Axiom 14.1) Thecollection of all points forms a nonempty set with more than one (i.e., atleast two) points.Neutral Geometry Axiom 2 (Incidence Postulate) (Axiom 14.5) Everyline is a set of points. For every pair of distinct points A, B there isexactly one line l = ←→ AB such that A, B ∈ l.Neutral Geometry Axiom 3 (Ruler Postulate) (Axiom 14.11) Forevery pair of points P, Q there is a number P Q called the distance fromP to Q. For each line l there is a one-to-one mapping f : l ↦→ R such thatif x = f(P ) and y = f(Q) then P Q = |x − y| is the value of the distance.Neutral Geometry Axiom 4 (Plane Separation Postulate) (Axiom15.2) For every line l the points that do not lie on l form two disjoint,convex non-empty sets H 1 and H 2 , called half-planes, bounded by l suchthat if P ∈ H 1 and Q ∈ H 2 then P Q intersects l.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 22. NEUTRAL GEOMETRY 109Neutral Geometry Axiom 5 (Protractor Postulate) (Axiom 16.2)For every angle ∠BAC there is a real number µ(∠BAC) called the measureof ∠BAC such that(a) 0 ≤ µ(∠BAC) < 180(b) µ(∠BAC) = 0 ⇐⇒ −→ −→ AB = BC.(c) Angle Construction Postulate: For every number r ∈ (0, 180)and for each half plane H bounded by ←→ AB there exists a unique ray−→AE such that E ∈ H and µ(∠BAE) = r.−→−→ −→(d) Angle Addition Postulate: If AD is between rays AB and ACthenµ(∠BAD) + µ(∠DAC) = µ(∠BAC) (22.1)Neutral Geometry Axiom 6 (Side Angle Side (SAS) Postulate)(Axiom 21.2) If △ABC and △DEF are two triangles such that AB ∼ = DE,∠ABC ∼ = ∠DEF , and BC ∼ = EF then △ABC ∼ = △DEF .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

110 SECTION 22. NEUTRAL GEOMETRY« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 23Angle-Side-AngleTheorem 23.1 (ASA (Angle Side Angle)) If triangles △ABC and△DEF satisfy (see figure 23.1)Then △ABC ∼ = △DEF .∠CAB ∼ = ∠F DEAB ∼ = DE∠ABC ∼ = ∠DEFFigure 23.1: Angle-Side-Angle.Proof. By the ruler postulate there exists a point G ∈ −→ AC such that AG ∼ =DF (see figure 23.2).Then by SAS△ABG ∼ = △ABC (23.1)111

112 SECTION 23. ANGLE-SIDE-ANGLEFigure 23.2: Illustration of proof of ASA (theorem 23.1).Since ∠ABC ∼ = ∠DEF (given) and ∠DEF ∼ = ∠ABG (by (23.1)), weconclude that ∠ABC ∼ = ∠ABG.By the angle construction postulate the rays −→ −→BG = BC (because theycorrespond to the same angle measured from −→ BA).Hence point G ∈ −→−→←→BC, and we see that ray BC intersects line AC at bothC and G. Hence line ←→ BC intersects ←→ AC at these same two points.But distinct non-parallel lines can only intersect in one point, so C = G.Hence △ABC ∼ = △ABG ∼ = △DEF .Corollary 23.2 (Converse of Isosceles Triangle Theorem) If △ABCis a triangle that satisfies ∠ACB ∼ = ∠ABC then AB = AC.Proof. Since BC ∼ = CB then by SAS △ABC ∼ = △BAC. Hence AB ∼ =AC.Theorem 23.3 Congruent Triangle Construction Theorem. It ispossible to construct a congruent copy of any triangle on a base, i.e., forany triangle △ABC, if DE ∼ = AB and H is either halfplane of ←→ DE thenthere exists a unique point F ∈ H such that △ABC ∼ = △DEF .Proof. Existence. Suppose that DE ∼ = AB, and choose a side H of AB.Let G ∈ H be chosen so that ∠ABC = ∠DEG (angle construction postulate).Choose F ∈ −→ EG such that EF = BC (ruler postulate).Hence by SAS, △ABC ∼ = △DEF« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 23. ANGLE-SIDE-ANGLE 113Figure 23.3: Illustration of proof of existence of construction of triangle ona base, theorem 23.3.Figure 23.4: Illustration of proof of uniqueness of construction of a triangleon a base, theorem 23.3.Uniqueness. To show that F is unique, suppose that there is a second pointH such that △DEF ∼ = △DEH.Since ∠DEF = ∠DEH, H ∈ −→ EG (uniqueness of ray constructed by angleconstruction postulate).Since △DEF ∼ = △DEH, then EF = EH. Hence H = G, because of theuniqueness part of the ruler postulate.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

114 SECTION 23. ANGLE-SIDE-ANGLE« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 24Exterior AnglesDefinition 24.1 Let △ABC be a triangle. The angles ∠ABC, ∠BCA, ∠CABare called interior angles of △ABC.Definition 24.2 An exterior angle of a triangle is any angle that formsa linear pair with an interior angle of the same triangle.Definition 24.3 If an exterior angle forms a linear pair with the interiorangle at one vertex, the other two interior angles are called remote interiorangles.Figure 24.1: Angle α, β, and γ are interior angles of this triangle. ɛ andδ are exterior angles because the each form a linear pair with the interiorangle α. Angles β and γ are remote interior angles of ɛ and δ. By theexterior angle theorem, ɛ > β, ɛ > γ, δ > β, δ > γ.115

116 SECTION 24. EXTERIOR ANGLESTheorem 24.4 (Exterior Angle Theorem a ) The measure of any exteriorangle of a triangle is strictly greater than the measure of either of itsremote interior angles.a Euclid’s Proposition 16In figure 24.1 this means that ɛ > β, ɛ > γ, δ > β, δ > γ.Figure 24.2: Definitions for proof of theorem 24.4Proof. See figure 24.2. We will show that α > β. The proof that α > γ isanalogous.Let E ∈ AC be its midpoint (Every segment has a midpoint).Choose F ∈ −→ BE such that BE = EF . This is possible by the ruler postulate.By the vertical angle theorem δ = ɛ.By SAS △BEA ∼ = △F EC (because BE = EF, δ = ɛ, and AE = EC), andso ∠ζ = ∠β.Since F and B are on opposite sides of ←→ AC (because F ∗ E ∗ B) and B andD are also on opposite sides of line ←→ AC (since B ∗ C ∗ D), then D and F areon the same side of ←→ AC by the plane separation postulate, i.e, F ∈ H ←→ D, ACSimilarly, since A and E are on the same side ←→ CD (since A ∗ E ∗ C), andE and F are on the same side of ←→ CD (since B ∗ E ∗ F ), then A and F areon the same side of the line ←→ CD, also by the plane separation postulate.Hence F ∈ H A,←→ CD.Hence F ∈ H D,←→ AC∪ H A,←→ CD.By the definition of the interior of an angle, this means F is in the interiorof ∠α. Hence ζ < α by the betweenness theorem for rays. Since ζ = β wehave β < α which is what the theorem states.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 24. EXTERIOR ANGLES 117Theorem 24.5 (Uniqueness of Perpendiculars) For every line l andfor every point P ∉ l, there exists exactly one line m ⊥ l such that P ∈ m.Proof. See figure 24.3. Suppose P is a point with P ∉ l and let m ⊥ lthrough P .Suppose n ⊥ l is a second line, distinct from m, through P . (RAA assumption)Let R be the point at which n intersects l, Then R ≠ Q. (Otherwise,R ∈ m and there would be two distinct points, R and P , at which m andn intersect. This would mean that m = n which would violate the RAAassumption.)Triangle △P RQ has an exterior angle β = 90 and an interior angle α = 90at R by the RAA assumption. This contradicts the exterior angle theorem(theorem 24.4) which says that β > α. Hence we reject the RAA hypothesis.Figure 24.3: Since there can be only one perpendicular line to l throughthe point P , β ≠ 90.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

118 SECTION 24. EXTERIOR ANGLES« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 25Angle-Angle-SideTheorem 25.1 (Angle-Angle-Side(AAS) a ) If △ABC and △DEF have∠ABC = ∠DEF , ∠BCA = ∠EF D and AC = DF then △ABC ∼ =△DEF .a Along with Angle-Side-Angle this is Euclid’s Proposition 26.Proof. See figure 25.1. Let P ∈ −→ CB such that CP = F E. Such a pointexists by the ruler postulate.Then by Side Angle Side, △ACP ∼ = △DF E (since AC = DF , β = δ, andCP = F E).There are three possible cases: P = B, C ∗ P ∗ B, and C ∗ B ∗ P .Suppose first that C ∗ P ∗ B. Then ɛ is an exterior angle for △ABP . Henceɛ > α and ɛ > ζ by the Exterior angle theorem (theorem 24.4).Figure 25.1: Angle-Angle-Side with α = γ, β = δ, and AC = DF withC ∗ P ∗ B119

120 SECTION 25. ANGLE-ANGLE-SIDEBy hypothesis α = γ and by construction of CP we have ɛ = γ. Hence α =ɛ, which contradicts the conclusion of the exterior angle theorem (theorem24.4). Hence we must rule out C ∗ P ∗ B.The proof that C ∗ B ∗ P is impossible is completely analogous.P = B, which means △ACB ∼ = △ACP ∼ = △DF EHenceFigure 25.2: Proof of the hypotenuse leg theorem.Definition 25.2 A triangle is called a right triangle if one interior angleis a right angle.Definition 25.3 The side opposite the right angle in a right triangle iscalled the hypotenuse. 1Theorem 25.4 (Hypotenuse-Leg Theorem (ASS for right △)) If△ABC and △DEF are right triangles with right angles at C and F , AB =DE, and BC = EF then △ABC ∼ = △DEF .Proof. Let P be a point on −→ AC such that A ∗ C ∗ P and and CP = F D(ruler postulate) (see figure 25.2).Then by the linear pair theorem (theorem 18.4) ∠BCP = 90.Hence by SAS, △BCP ∼ = △EF D.Hence by congruence P B = ED, and by hypothesis ED = BA henceP B = BA.Thus △ABP is isosceles, and therefore by the isosceles triangle theorem(theorem 21.4) ∠BP C = ∠BACSince △BCP ∼ = △EF D, by congruence ∠BP C = ∠EDF .Hence ∠BAC = ∠EDF . By AAS, this gives △ABC = △DEF .1 From the ancient Greek word hypoteínousa (υπoτɛiνoυσηζ in Euclid, e.g., Proposition47, Book 1).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 26Side-Side-SideTheorem 26.1 (Side-Side-Side (SSS)) If △ABC and △DEF are trianglessuch that AB = DE, BC = EF , and CA = F D then △ABC ∼ =△DEF .Proof. Given AB = DE, BC = EF and AC = F D for triangles △ABCand △DEF .Figure 26.1: Proof of SSS, case H = A.By the triangle construction theorem there is a point G on the oppositeside of ←→ AB from C such that △DEF ∼ = △ABG.Let H = CG ∩ ←→ AB, which exists by the plane separation postulate becauseC and G are on opposite sides of ←→ AB.There are five possible cases: (1) H = A, (2) H = B, (3) A ∗ H ∗ B, (4)H ∗ A ∗ B, and (5) A ∗ B ∗ H.121

122 SECTION 26. SIDE-SIDE-SIDECase 1 (H = A) (see figure 26.1).By construction, △ABG ∼ = △DEF hence BG = EF ; by assumption EF =BC. Hence BC = BG.Thus triangle CGB is isosceles, and by the isosceles triangle theorem (theorem21.4), ∠BCA ∼ = ∠BGA.By congruence of triangles, ∠BGA ∼ = ∠EF D.Hence ∠BCA ∼ = ∠EF D.Since AG = DF = AC, and BG = EF = BC, by side angle side, △ABC ∼ =△ABG ∼ = △DEF .Case 2 (H = B). The proof is completely analogous to Case 1.Case 3 (A ∗ H ∗ B) (see figure 26.2).Figure 26.2: Proof of SSS when A ∗ H ∗ B.Since A ∗ H ∗ B, −→ CA ∗ −−→ CH ∗ −→ CB =⇒ that H is in the interior of ∠ACB(labeled as ɛ in fig. 26.2) (Betweenness for Rays, see 16.8).Similarly, −→ GA ∗ −−→ GH ∗ −→ GB =⇒ that H is in the interior of ∠BGA (labeledas ζ in fig. 26.2).By the protractor postulatem(∠ACB) = ɛ = α + γ = m(∠ACH) + m(∠BCH)m(∠(BGA) = ζ = β + δ = m(∠AGH) + m(∠BGH)By construction AG = DF ; by assumption DF = AC, hence AC = AG.Therefore △AGC is isosceles, and by the isosceles triangle theorem (theorem21.4)m(∠AGH) = β = α = m(∠ACH)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 26. SIDE-SIDE-SIDE 123By construction BG = EF ; by assumption EF = BC, hence BG = BC.Therefore △BCG is isosceles, and by the isosceles triangle theorem (theorem21.4)m(∠BCH) = γ = δ = m(∠BGH)Hencem(∠BCA) = ɛ = ζ = m(∠BGA) = m(∠EF D)so by SAS, △ACB ∼ = △AGB. Since △AGB ∼ = △DEF by construction,we have △ABC ∼ = △DEF .Case 4 (H ∗ A ∗ B) (see figure 26.3).Figure 26.3: Proof of SSS when H ∗ A ∗ B.H ∗ A ∗ B =⇒ −−→ CH ∗ −→ CA ∗ −→ CB =⇒ A is interior to ∠BCH.Similarly, H ∗ A ∗ B =⇒ −−→ GH ∗ −→ GA ∗ −→ GB =⇒ A is interior to ∠BGH.By the angle addition postulate,m(∠BCA) = m(∠BCH) − m(∠ACH)m(∠BGA) = m(∠BGH) − m(∠AGH)By construction AG = DF ; by assumption DF = AC; hence AC = AG.This means that △ACG is isosceles and hence by the isosceles triangletheorem (theorem 21.4), m(∠ACH) = m(∠AGH).By construction BG = EF ; by assumption EF = BC; hence BC = BGand △BCG is also isosceles. By the isosceles triangle theorem (theorem21.4), m(∠BCH) = m(∠BGH)Hencem(∠BCA) = m(∠BCH) − m(∠ACH)= m(∠BGH) − m(∠AGH)= m(∠BGA)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

124 SECTION 26. SIDE-SIDE-SIDEBy side-angle-side (AG = AC, α = β, BC = BG), △ABC ∼ = △ABG. Byconstruction △ABG ∼ = △DEF . Hence △ABC ∼ = △DEF .Case 5 (A ∗ B ∗ H). The proof is analogous to Case 4.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 27Scalene and TriangleInequalityTheorem 27.1 (The Scalene Inequality) Let A, B, C be distinct noncollinearpoints. ThenAB > BC ⇐⇒ m(∠ACB) > m(∠BAC) (27.1)Proof. ( =⇒ ) See figure 27.1 (left hand side); the theorem says that sideA is longer than side C if and only if angle c > a. We will use the notationillustrated on the right hand side of the figure. The if we assume AB > BC,we need to show that γ > δ.Figure 27.1: Proof of the Scalene inequality (theorem 27.1).By the ruler postulate there exists a point D ∈ AB such that DB = BC.By the protractor postulate, γ > α.125

126 SECTION 27. SCALENE AND TRIANGLE INEQUALITYBy the isosceles triangle theorem α = β.Angle β is an exterior angle of triangle △ADC. Hence β > δ.Therefore γ > α = β > δ.( ⇐ ) Assume γ > δ.By trichotomy one of the following must hold: AB > BC, AB = BC, orAB < BC.First, suppose that AB = BC. Then α = δ, which contradicts the hypothesis.Hence AB ≠ BC.Next, suppose that AB < BC. Then γ = ∠ACB < ∠BAC = δ by the firstpart of the theorem. But this contradicts the hypothesis, so AB > BC.Theorem 27.2 (The Triangle Inequality) If A, B, C are non-collinearpoints thenAC < AB + BC (27.2)Proof. Let A, B and C be non-collinear (see figure 27.2).Figure 27.2: Proof of the Triangle Inequality (theorem 27.2.)Let D ∈ ←→ AD such that BD = BC and A ∗ B ∗ D (ruler postulate).Since A ∗ B ∗ D, β > α (protractor postulate).By the isosceles triangle theorem, α = γ (because BC = BD).Hence β > γ. By the scalene inequality,AD > ACSince A ∗ B ∗ D, AD = AB + BD,AB + BD > AC« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 27. SCALENE AND TRIANGLE INEQUALITY 127Figure 27.3: Proof of hinge theorem (theorem 27.3.)Theorem 27.3 (The Hinge Theorem) Suppose △ABC and △DEFsatisfy AB = DE, AC = DF and ∠BAC < ∠EDF . Then BC < EF .Proof. Let △ABC and △DEF be such that AB = DE and AC = DF ,with ∠BAC < ∠DEF (see figure 27.3).Define a point G ∈ H C,←→ ABsuch that △ABG ∼ = △DEF (theorem 23.3).Then BG = EF .C is in the interior of ∠BAG by the Betweenness Theorem for Rays (theorem16.10).Hence −→ AC must intersect BG at some point J (Crossbar Theorem, theorem17.1).Let −−→ AH ′ be the bisector of angle ∠CAG (existence theorem for angle bisection,theorem 19.2). Define the point H = JG ∩ −−→ AH ′ (crossbar theorem,theorem 17.1: H ′ is in the interior of ∠JAG it must intersect JG).By definition of the angle bisector α = β; by construction of AG, AG = AC.Since AH = AH, we have △CAH ∼ = △GAH by SAS.Hence HG = HC.Since B∗H ∗G (theorem 16.8 because −→ AC ∗ −−→ AH ∗ −→ AG), by the ruler postulateBG = BH + HG = BH + HCIf H ∈ ←→ BC then BC = BH − HC. HenceBC = BH − HC < HB + HC = BG = EFIf H ∉ ←→ BC then by the triangle inequalityBC < BH + HC = BG = EFRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

128 SECTION 27. SCALENE AND TRIANGLE INEQUALITYFigure 27.4: Hinge theorem (theorem 27.3) when B ∗ H ∗ G.Definition 27.4 The distance from a point P to a line l, denoted byd(P, l) is the distance from P to the foot of the perpendicular dropped fromP to l. (See figure 27.5.)Theorem 27.5 (Shortest Distance from a Point to a Line) Let l bea line, let P ∉ l a point, and let F be the foot of the perpendicular from Pto l. If R ∈ l such that R ≠ F , then P R > P F . (See figure 27.5.)Proof. (Exercise)Figure 27.5: The shortest distance from a point P to a line l is the lengthof the perpendicular segment dropped from P to l (theorem 27.5). Hence(F P < F R)(∀R ∈ l, R ≠ F ).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 28Characterization ofBisectorsTheorem 28.1 (Pointwise Characterization of Angle Bisector) LetA, B, C be non-collinear points and let P be in the interior of ∠BAC. ThenP lies on the angle bisector of ∠BAC if and only if d(P, ←→ AB) = d(P, ←→ AC).Proof. ( ⇒) Assume P is a point on the bisector of ∠BAC (see figure 28.1).Drop perpendiculars from P to ←→ AB and ←→ AC to their respective feet D ∈ ←→ ABand E ∈ ←→ AC.∠P AE = ∠P AD by definition of angle bisection, ∠P EA = ∠P DA becauseboth are right angles, and P A = P A. Hence by AAS, △ADP ∼ = △AEP .Hence DP = P E because they are corresponding sides of congruent trian-Figure 28.1: Theorem 28.1.129

130 SECTION 28. CHARACTERIZATION OF BISECTORSgles. Therefore, d(P, ←→ AB) = P D = P E = d(P, ←→ AC).( ⇐)Let P be a point in the interior of ∠BAC, and assume that d(P, ←→ AB) =d(P, ←→ AC).Drop perpendiculars and call the feet D and E as before. Hence P D = P E.We also have ∠P EA = ∠P DA = 90, and both triangles △AP D and△AP E share a common side AP . Hence by the hypoteneuse-leg theoremthey are congruent.Hence ∠P AE = ∠P AD, which means P is on the angle bisector of ∠BAC.Figure 28.2: Theorem 28.2.Theorem 28.2 (Pointwise Characterization of Perpendicular Bisector)Let A and B be distinct points. A point P lies on the perpendicularbisector of AB if and only if P A = P B.Proof. (⇒) Let P be a point on the perpendicular bisector of AB (figure28.2).Let M be the intersection of the bisector and AB.Then AM = BM by definition of bisector and ∠AMP = ∠BMP = 90.Since P M ∼ = P M, by SAS △AMP ∼ = △BMP .Hence P A = P B.(⇐) Let P be a point that satisfies P A = P B. Then △P AB is isoscelesand by the isosceles triangle theorem ∠P AB = ∠P BA.Drop a perpendicular from P to AB and call the foot M. Then∠P AM = ∠P AB = ∠P BA = ∠P BM« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 28. CHARACTERIZATION OF BISECTORS 131Since △P AM and △P BM share a side, are right triangles, and have congruenthypotenuses, they are congruent by the Hypotenuse Leg Theorem.Hence AM = BM, meaning that ←−→ P M is a bisector of AB. Since by construction←−→ P M ⊥ AB it is a perpendicular bisector.Figure 28.3: Theorem 28.3.xyf(x)f(y)Theorem 28.3 (Continuity of Distance) Let A, B, and C be noncollinearand let d = AB. For each x ∈ [0, d], there exists a unique pointD x ∈ AB such that AD x = x. Define the function f : [0, d] ↦→ [0, ∞) byf(x) = CD x . Then the function f is continuous.Proof. Let A, B, C, f be as defined in the statement of theorem 28.3 (seefigure 28.3), and let ɛ > 0 be given. Choose δ = ɛ.Suppose y ∈ [0, d] such thatBy the triangle inequalityD x D y = |x − y| < δ = ɛCD x < CD y + D x D y < CD y + ɛandHenceCD y < CD x + D x D y < CD x + ɛCD x − ɛ < CD y < CD x + ɛf(x) − ɛ < f(y) < f(x) + ɛ−ɛ < f(y) − f(x) < ɛ|f(y) − f(x)| < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

132 SECTION 28. CHARACTERIZATION OF BISECTORS« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 29TransversalsDefinition 29.1 Suppose that lines m = ←→ AC and n = ←→ DF are distinct.Suppose that a third line l = ←→ AB intersects m at a point B and line n at apoint E ≠ B. Then l is a transversal to lines m and n (see figure 29.1),and we say that m and n are cut by transversal l.Definition 29.2 Consider the transversal as illustrated in figure 29.1. Theangles α, β, γ, and δ are called interior angles. The pairs {α, δ} and{β, γ} are called alternate interior angle pairs.Figure 29.1: Lines ←→ AC and ←→ DF are cut by transversal ←→ BE. Angles α andδ form an alternating interior angle pair, as do angles γ and β.Definition 29.3 Suppose that lines ←→ AC and ←→ DF are cut by a transversal←→BE as illustrated in figure 29.2. Then angles α and β as shown are calledcorresponding angles. If µ(α) = µ(β) then α and β are called congruentcorresponding angles.133

134 SECTION 29. TRANSVERSALSFigure 29.2: Angles α and β are corresponding angles.Theorem 29.4 (Alternate Interior Angles Theorem) Suppose that land m are cut by transversal t in such a way that a pair of alternate interiorangles are congruent. Then l ‖ m.Proof. Suppose β = γ in figure 29.1. Suppose there is a point P that lies onboth l and m. Because t is a transversal, it intersects l and m at differentpoints; hence P ∉ t. Therefore by the plane separation postulate (axiom15.2) it must be on one side or the other of t, as illustrated in figure 29.3. IfFigure 29.3: Proof of alternate interior angle theorem (theorem 29.4). Left:case when P ∈ H C,t ; Right: case when P ∈ H A,t . Both cases violate theexterior angle theorem (theorem 24.4)P ∈ H C,t then γ is an exterior angle of triangle △P BE while β is a remoteinterior angle (corresponding to γ) of the same triangle.By the exterior angle theorem (theorem 24.4), γ > β; Since β = γ this is acontradiction, hence P ∉ H C,t .If P ∈ H A,t , then β is an exterior angle of triangle △P BE, while γ is aremote interior angle (corresponding to β) of the same triangle. This alsoviolates the exterior angle theorem.Hence there is no such point P , and that means the lines l and m neverintersect. So they are parallel.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 29. TRANSVERSALS 135Corollary 29.5 If m ⊥ l and n ⊥ l, then either m = n or m ‖ n.Proof. (Exercise)Corollary 29.6 If one pair of corresponding angles is congruent, so is theother pair.Proof. (Exercise)Corollary 29.7 (Corresponding Angles Theorem) If l and m are linescut by a transversal to t in such a way that two corresponding angles arecongruent, then l ‖ m.Proof. (Exercise)Figure 29.4: The corresponding angles theorem: lines a and b are parallelbecause they are cut by a transversal that has congruent correspondingangles.Corollary 29.8 If l and m are lines cut by a transversal t in such a way thattwo non-alternating interior angles on the same side of t are supplements,then l ‖ m.Proof. (Exercise)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

136 SECTION 29. TRANSVERSALSCorollary 29.9 (Existence of Parallel Line) If l is a line and P ∉ l isa point, then there is a line m ‖ l with P ∈ m.Proof. Drop a perpendicular from P to l (existence of perpendiculars, theorem21.5).Call the foot of the perpendicular Q and define t = ←→ P Q.By the protractor postulate (axiom 16.2), we can construct a line m ⊥ tthrough P .By the interior angle theorem, m ‖ l.Recall that the elliptic parallel postulate says that no parallel lines exist.Since we have just proven that parallel lines exist in neutral geometry, weknow that this postulate is false.Corollary 29.10 The elliptic parallel postulate is false in neutral geometry.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 30Triangles in NeutralGeometryDefinition 30.1 The angle sum of △ABC is defined asσ(△ABC) =µ(∠CAB) + µ(∠ABC) + µ(∠BCA)Figure 30.1: Theorem 30.2 says that β + γ < 180.Theorem 30.2 The sum of the measure of any two angles in a triangle isless than 180: For any triangle △ABC,µ(∠CAB) + µ(∠ABC) < 180Proof. Let D ∈ ←→ AB such that A ∗ B ∗ D (see figure 30.1).By the linear pair theorem (thm. 18.4), α + β = 180.137

138 SECTION 30. TRIANGLES IN NEUTRAL GEOMETRYBy the exterior angle theorem (theorem 24.4), γ < α, henceγ + β < α + β = 180Theorem 30.3 (Angle Sum Addition Theorem For Triangles) LetE be a point in the interior of segment BC of △ABC. Thenσ(△ABE) + σ(△ECA) = σ(△ABC) + 180Figure 30.2: Angle Sum addition theorem (theorem 30.3 ).Proof. See figure 30.2. By the linear pair theorem,ζ + η = 180and by the angle addition postulate,β = δ + ɛHence,σ(△ABE) + σ(△AEC)= (α + δ + ζ) + (γ + ɛ + η)= α + (δ + ɛ) + γ + (ζ + η)= α + β + γ + (ζ + η)= α + β + γ + 180= σ(△ABC) + 180« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 30. TRIANGLES IN NEUTRAL GEOMETRY 139Theorem 30.4 Let △ABC be a triangle. Then there exists a point D ∉←→AB such thatσ(△ABD) = σ(△ABC)and the angle measure of one of the interior angles of △ABD is less thanor equal to 1 2 µ(∠CAB).Proof. See figure 30.3.Let E be the midpoint of BC (existence of midpoints, theorem 14.29).By the ruler postulate we can define a point D ∈ −→ AE such that A ∗ E ∗ Dand AE = ED.By the vertical angle theorem (theorem 19.7), δ = γ, hence by SAS,△AEC ∼ = △DEB. Henceσ(△AEC) = σ(△DEB) (30.1)because the angle sums of congruent triangles are equal. By theorem 30.3,Figure 30.3: Proof of Theorem 30.4.σ(△ABE) + σ(△AEC) = σ(△ABC) + 180σ(△ABE) + σ(△BED) = σ(△ABD) + 180Subtracting the second equation from the first,σ(△AEC) − σ(△BED) =σ(△ABC) − σ(△ABD) (30.2)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

140 SECTION 30. TRIANGLES IN NEUTRAL GEOMETRYSubstituting equation 30.1 into equation 30.2 givesσ(△ABC) = σ(△ABD)which is the first result promised by theorem 30.4.To get the second result, observe that by the angle addition postulateζ = ɛ + αHence eitherɛ < 1 2 ζ or α < 1 2 ζBut because △AEC ∼ = △DEB (from above), β = α. Hence eitherɛ < 1 2 ζ or β < 1 2 ζwhich is the second conclusion of the theorem.Corollary 30.5 Let T be a triangle, and let one of its interior angles beα. Then there exists a triangle T ′ with an interior angle α ′ such thatσ(T ) = σ(T ′ ) and α ′ ≤ 1 2 α.Proof. Let T = △ABC and let α = ∠CAB. Then by theorem 30.4 thereexists a point D ∉ ←→ AB such that one of the interior angles of T ′ = △ABDhas measure less than or equal to 1 2 α. Call this angle α′ .Theorem 30.6 (Saccheri-Legendre Theorem) For any triangle△ABC,σ(△ABC) ≤ 180Proof. Let T = △ABC be given and let α = µ(∠CAB).Suppose σ(ABC) > 180 (RAA hypothesis).Then there is some number ɛ > 0 such thatσ(ABC) = 180 + ɛBy the Archimedian ordering property there is some integer n such that2 n ɛ > α. (Choose any n > log 2 ( α ɛ ).)By corollary 30.5 there is a triangle T 1 with an interior angle α 1 ≤ 1 2 α suchthat σ(T ) = σ(T 1 )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 30. TRIANGLES IN NEUTRAL GEOMETRY 141Similarly, there is a triangle T 2 with an interior angleα 2 ≤ 1 2 α 1 ≤ 1 4 αsuch that σ(T ) = σ(T 1 ) = σ(T 2 ).Repeating this n times there is a triangle T n with an interior angleα n =≤ 12 n α < 12 n (2n ɛ) = ɛsuch thatσ(T ) = σ(T n ) = 180 + ɛLet the other two interior angles of T n be β n and γ n . Thenα n + β n + γ n = σ(T n )= 180 + ɛ> 180 + α nβ n + γ n > 180But the last line contradicts theorem 30.2, which says that β n + γ n < 180.Thus we must reject the RAA hypothesis and conclude that σ(△ABC) ≤180.Corollary 30.7 The sum of the measures of two interior angles of a triangleis than or equal to the measure of their remote exterior angle.Proof. (Exercise)Figure 30.4: Corollary 30.7: α + β ≤ γ.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

142 SECTION 30. TRIANGLES IN NEUTRAL GEOMETRYCorollary 30.8 (Converse to Euclid’s Fifth Postulate) Let l and mbe two lines cut by a transversal t. If l and m meet on one side of t thenthe sum of the measures of the two interior angles on that side is strictlyless than 180.Proof. (exercise)Figure 30.5: The converse to Euclid’s 5th postulate, corollary 30.8: α+β

SECTION 30. TRIANGLES IN NEUTRAL GEOMETRY 143Figure 30.6: Additivity of defect for triangles.By definition 30.9,δ(△ECA) = 180 − (β + ɛ + ζ)δ(△ABE) = 180 − (η + γ + δ)henceδ(△ECA) + δ(△ABE)= 360 − (β + ɛ + ζ) − (η + γ + δ)= 360 − (ζ + (ɛ + δ) + η + (β + γ))By the angle addition postulate, α = β + γ.Since δ and ɛ form a linear pair, ɛ + δ = 180.Thusδ(△ECA) + δ(△ABE)= 360 − (ζ + 180 + η + α)= 180 − (ζ + η + α)= δ(△ABC)(2) Alternate proof. By theorem 30.3,σ(△ACE) + σ(△ABE) =σ(△ABC) + 180(180 − δ(△ACE))+(180 − δ(△ABE)) =(180 − δ(△ABC)) + 180δ(△ACE)δ(△ABE) =δ(△ABC)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

144 SECTION 30. TRIANGLES IN NEUTRAL GEOMETRYGo Geometry!« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 31Quadrilaterals in NeutralGeometryWe will see in a later section that there are no rectangles in neutralgeometry. In fact, the existence of a rectangle is equivalent to assumingthe Euclidean parallel postulate (theorem 33.3).Starting as early as the 11th century parallelograms with two right angles,and quadrilaterals with three right angles, were studied in many attemptsto prove that the remaining interior angles were right angles. Much of thework, which goes back to Umar al-Khayyami (1048-1131) and Nasir Eddin(1201-1274) is often attributed to the European geometers GiovanniSaccheri (1667-1733) and Johann Lambert (1728-1777) who rediscoveredthe earlier results. None of these attempts were able to successfully demonstratethat a rectangle could be constructed using only the axioms of neutralgeometry.Here we look at some of the results that can be obtained for quadrilateralsin Neutral Geometry.Definition 31.1 Let A, B, C, D be points, no 3 of which are collinear, suchthat any two of the segments AB, BC, CD, DA either have no point incommon or only have an endpoint in common. Then the points A, B, C, Ddetermine a quadrilateral, denoted by □ABCD.The points A, B, C, D are called the vertices of the quadrilateral.The segments AB, BC, CD, DA are called the sides of the quadrilateral.The diagonals of □ABCD are the segments AC and BD.145

146 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYTwo quadrilaterals are congruent if all four corresponding sides and allfour corresponding angles are congruent.Figure 31.1: □ABCD is a convex quadrilateral with diagonals AC andBD; □EF QH is a non-convex quadrilateral with diagonals EG and F H;□IJKL is not a quadrilateral, although □IKJL (not shown) is a quadrilateral.You should convince yourself that the following definition of a convexquadrilateral is consistent with definition 15.1.Definition 31.2 □ABCD is convex if each vertex is contained in theinterior of the angle formed by the three other vertices (in their cyclicorder around the quadrilateral).Figure 31.2: The angle sum of a quadrilateral is equal to the sum of theangle sums of the triangles defined by either diagonal (theorem 31.4).Definition 31.3 Let □ABCD be convex. Then its angle sum is given bythe sum of the measures of its interior angles:σ(□BACD) = µ(∠ABC) + µ(∠BCD)+ µ(∠CDA) + µ(∠DAC)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRY 147Theorem 31.4 (Additivity of Angle Sum) Let □ABCD be a convexquadrilateral with diagonal BD. Thenσ(□ABCD) = σ(△ABD) + σ(△BCD)Proof. See figure 31.2. Apply the angle addition postulate (axiom 16.2) toeach of the angles that are split by a diagonal to getσ(□ABCD) = α + β + ɛ + θ= α + γ + δ + ɛ + ζ + η= (α + γ + η) + (δ + ɛ + ζ)+ σ(△ABD) + σ(△BDC)Definition 31.5 The defect of a quadrilateral isδ(□ABCD) = 360 − σ(□ABCD)Theorem 31.6 (Additivity of Defect for Convex Quadrilaterals) If□ABCD is a convex quadrilateral, thenδ(□ABCD) = δ(△ABC) + δ(△ACD)Proof. Apply theorem 31.4.Corollary 31.7 If □ABCD is convex, thenσ(□ABCD) ≤ 360Proof. Apply theorem 31.4 and the Saccheri-Legendre Theorem (theorem30.6).Definition 31.8 □ABCD is called a parallelogram if AB ‖ CD and BC ‖AD.Theorem 31.9 Every parallelogram is convex.Proof. (Exercise)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

148 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYFigure 31.3: □ABCD is a parallelogram.Theorem 31.10 Let △ABC be a triangle and D and E points such thatA ∗ D ∗ B and A ∗ E ∗ C. Then □BCED is a convex quadrilateral.Proof. (Exercise)Figure 31.4: A ∗ D ∗ B ∧ A ∗ E ∗ C ⇒ □BCED is a quadrilateral.Theorem 31.11 □ABCD is convex if and only if the diagonals have aninterior point in common (i.e., they intersect, but not at an endpoint).Proof. (⇒) Assume □ABCD is convex. Then by definition of convexity,C is in the interior of ∠DAB.By the crossbar theorem (theorem 17.1) BD ∩ −→ AC ≠ ∅; call the point ofintersection E, where B ∗ E ∗ D (theorem 16.8).By a similar argument there is a point F = AC ∩ −−→ BD, where A ∗ F ∗ C.Since ←→ AC and ←→ BD are distinct (they correspond to opposite sides of aquadrilateral), they can meet in at most one point, we must have E = F .Hence the diagonals intersect at E.Since A ∗ F ∗ C and B ∗ E ∗ D, the intersection is not at an endpoint.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRY 149(⇐) Let □ABCD be a quadrilateral with E = AC ∩ BD with A ∗ E ∗ Cand B ∗ E ∗ D.Since A ∗ E ∗ C, A and E are on the same side of the line ←→ CD (theorem16.4.Similarly, since B ∗ E ∗ D, B and E are on the same side of ←→ CD.Hence A and B are on the same side of ←→ CD (Plane separation postulate,axiom 15.2), i.e, A ∈ H B,←→ CD.By a similar argument A and D are on the same side of ←→ BC, i.e, A ∈ H ←→ D, BC.Hence A ∈ H ←→ B, CD∩ H ←→ D, BC, and thus A is in the interior of ∠BCD.By a similar argument, each of the other vertices is in the interior of itsopposite angle. Hence by definition of convexity, the quadrilateral is convex.Theorem 31.12 If □ABCD and □ACBD are both quadrilaterals then□ABCD is not convex.Proof. LetP = “□ABCD is quadrilateral”Q = “□ACBD is quadrilateral”R = “□ABCD is not a convex quadrilateral”To prove the theorem we must show thatP ∧ Q ⇒ Ror equivalently, its contrapositive,¬R ⇒ ((¬P ) ∨ (¬Q))Assume □ABCD is a convex quadrilateral (i.e, assume that R is false).Then AC ∩ BD ≠ ∅, i.e., the diagonals of □ABCD share an internal point.Hence □ACBD is not a quadrilateral, i.e, ¬P is true. Hence ¬R ⇒ ¬P .By the rules of logic, ¬P ⇒ (¬P ) ∨ (¬Q).Hence ¬R ⇒ ((¬P ) ∨ (¬Q)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

150 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYTheorem 31.13 If □ABCD is a non-convex quadrilateral then □ACBDis a quadrilateral.Proof. Since □ABCD is a quadrilateral no three of the points A, B, C, Dare collinear.Since □ABCD is a quadrilateral BC ∩ AD = ∅.Since □ABCD is not convex then AC and BD are disjoint (the diagonalsdo not intersect).Thus segments AC, CB, BD, and DA share at most their endpoints.Hence □ACBD is a quadrilateral.Definition 31.14 □ABCD is a Saccheri Quadrilateral if ∠ABC =∠BAD = 90 and AD = BC. Segment AB is called the base and segmentDC is called the summit.Figure 31.5: A Saccheri Quadrilateral. It is not possible to prove that thisfigure is a rectangle using only the axioms of neutral geometry - one mustaccept Euclid’s fifth postulate to do so. Thus we sometimes draw the topand side edges as curves to remind ourselves of this.Theorem 31.15 The diagonals of a Saccheri Quadrilateral are congruent.Proof. Consider triangles △ABD and △ABC. Since BC = AD, AB =AB, and ∠A = 90 = ∠B, they are congruent. Hence BD ∼ = AC.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRY 151Figure 31.6: The shaded triangles are congruent, hence the correspondingdiagonals of the Saccheri Quadrilateral are congruent (theorem 31.15).DCDCABABTheorem 31.16 The summit angles of a Saccheri Quadrilateral are congruent.Proof. Repeat the argument in the previous proof, but with the upper-halftriangles. The triangles are congruent by SSS - they share the same top;the diagonals are congruent; and the sides are congruent. Hence the cornerangles are congruent.Theorem 31.17 Let □ABCD be a Saccheri quadrilateral. Then the segmentjoining the midpoints of the base and summit is perpendicular to thebase and summit.Proof. Let M be the bisector of AB and N the bisector of CD (see figure31.7).Figure 31.7: Proof of theorem 31.17. The line segment connecting the midpointsof the base and summit of a Saccheri Quadrilateral is perpendicularto both the base and the summit.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

152 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYSince M is a bisector, AM = MB; by definition of the Saccheri quadrilateralAD = BC and ∠A ∼ = ∠B. Hence △DAM ∼ = △CBM by SAS.Hence DM = CM; by definition of bisector, DN = CN; hence △DMN ∼ =△CMN by SSS (they share a common side).By congruence, ∠DNM ∼ = ∠CNM; since they form a linear pair, theymust each be right angles.To show that the angle with the base is right, observe that since △DAM ∼ =△CBM, then∠AMD ∼ = ∠BMCand since △DMN ∼ = △CMN thenHence∠DMN ∼ = ∠CMNm(∠AMN) = m(∠AMD) + m(∠DMN)= m(∠BMC) + m(∠CMN)= m(∠BMN)by angle addition. Since ∠AMN and ∠BMN are a linear pair then180 = m(∠AMN) + m(∠BMN)= 2m(∠AMN)hence the angles are right angles.Theorem 31.18 If □ABCD is a Saccheri quadrilateral, then it is a parallelogram.Proof. See figure 31.7. MN ⊥ AB and MN ⊥ DC. By the alternateinterior angles theorem (theorem 29.4), CD ‖ AB. Similarly, AD ‖ BC.Theorem 31.19 Saccheri quadrilaterals are convex.Proof. This follows because every Saccheri quadrilateral is a parallelogram(theorem 31.18) and every parallelogram is convex (theorem 31.9).Theorem 31.20 The summit angles of a Saccheri quadrilaterals are eitherright angles or acute.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRY 153Proof. Since □ABCD is convex (theorem 31.19), σ(□ABCD) ≤ 360 (Corollary31.7). By definition of a Saccheri Quadrilateral, ∠A = ∠B = 90. Hence∠C + ∠D ≤ 180By theorem 31.16 the summit angles are congruent, hence ∠C = ∠D.Hence ∠C ≤ 90 and ∠D ≤ 90.Definition 31.21 A Lambert quadrilateral is a quadrilateral in whichthree of the interior angles are right angles.Figure 31.8: A Lambert Quadrilateral. Like the Saccheri Quadrilateral, tis not possible to prove that this figure is a rectangle using only the axiomsof neutral geometry - one must accept Euclid’s fifth postulate to do so.Corollary 31.22 Let □ABCD be a Lambert quadrilateral. Then it is aparallelogram.Proof. By construction of the Lambert quadrilateral, DC ⊥ BC and AB ⊥BC hence either DC ‖ AB or DC = AB. Since the two sides are distinct,that means they are parallel.Similarly, by construction, AD ⊥ AB and AB ⊥ BC. Since AD and BCare distinct, they must therefore be parallel.Corollary 31.23 Let □ABCD be a Lambert quadrilateral.convex.Then it isProof. It is a parallelogram and all parallelograms are convex.Corollary 31.24 Let □ABCD be a Lambert quadrilateral with right anglesat vertices A, B, and C. Then ∠D is either a right angle or it isacute.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

154 SECTION 31. QUADRILATERALS IN NEUTRAL GEOMETRYProof. Since □ABCD is a convex quadrilateral then σ(□ABCD) ≤ 360.Since the sum of the first three angles is 270, the remaining one must be≤ 90.Corollary 31.25 Let □ABCD be a Lambert quadrilateral with right anglesat vertices A, B, and C. Then BC ≤ AD.Figure 31.9: In a Lambert quadrilateral, BC ≤ ADProof. Suppose BC > AD (RAA).Then there exists a point E with B ∗ E ∗ C such that BE = AD (rulerpostulate).□ABED is a Saccheri quadrilateral (def. of Saccheri quadrilateral).Hence ∠BED ≤ 90 (theorem 31.20).Angle ∠BED is an exterior angle for △ECD.Angle ∠C = 90, and it is a remote angle of ∠BED, By the exterior angletheorem (theorem 24.4), ∠C < ∠BED (strict inequality). This leads to theresult 90 < 90; therefore we must reject the RAA hypothesis and concludethat BC ≤ AD.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 32The Euclidean ParallelPostulateIn this section we consider some statements that are equivalent to Euclid’s5th postulate.When we say that two statements are equivalent in this sense we meanthat if we add either statement to the axioms of Neutral Geometry, we canprove the other statement. It does not mean that the two statements areprecisely logically equivalent.Axiom 32.1 .Euclidean Parallel Postulate (EPP). For every line l andfor every point P that does not lie on l there is exactly one line m suchthat p ∈ m and m ‖ l.Axiom 32.2 Euclid’s Fifth Postulate (E5P). Let l and m be two linescut by a transversal in such a way that the sum of the measures of the twointerior angles on one side of t is less than 180. Then l and m intersect onthat side of t.Theorem 32.3 Euclid’s 5th postulate ⇐⇒ the Euclidean Parallel Postulate.Proof. (⇒) (EPP ⇒ E5P). Assume that the EPP is true.Let l, m, t, α, β be as indicated in figure 32.1, i.e., construct the lines l,m, and n as shown, with then α + β < 180. We need to show that lines land m intersect on the same side of t as α and β.There is a line n through B such that γ + δ = 180 (by the protractor155

156 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATEpostulate). By the linear pair theorem, then ɛ + δ = 180 and α + γ = 180.HenceFigure 32.1: Euclid’s Fifth postulates states that if α + β < 180 then lintersects m at a point C that is on the same of t as α and β.α + ɛ = 180 − δ + 180 − γ= 360 − (δ + γ)= 360 − 180= 180 (32.1)Thus both pairs of non-alternating interior angles formed by t sum to 180.By assumptionSubstituting equation 32.1 givesα + β < 180180 − ɛ + β < 180β < ɛIn particular, since β ≠ ɛ, then m ≠ n.Since δ = 180 − γ = α, n ‖ l (alternate interior angles theorem).Since m ≠ n this means m is not parallel to l (this is because we are assumingthe Euclidean parallel postulate, that there is only one line throughB that is parallel to l).Since m is not parallel to l, they intersect at some point C. Either thatpoint is on the same side of t as α and β or on the opposite side.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 157Suppose they intersect on the opposite side. Then △P QC has exteriorangles α and β that form linear pairs with interior angles 180 − α and180 − β. PicK either linear pair, say the one formed with α. Then bythe exterior angle theorem α > 180 − β, which contradicts the fact thatα + β < 180. Hence the lines must intersect on the same side as α and β.This is Euclid’s 5th postulate.(⇐) (E5P ⇒ EPP). Assume that Euclid’s 5th postulate is true.Let l be a line and P be a point such that P ∉ l.Drop a perpendicular line from P to l, and call the foot of the line Q.Construct m through P such that m ⊥ ←→ P Q. By the alternate interior anglestheorem (theorem 29.4) l ‖ m.Assume n ≠ m is a second line through P such that n ‖ l.Then ←→ P Q is a transversal to n and l. Since n ≠ m, the interior anglesγ ≠ 90 and δ ≠ 90. Since they form a linear pair, γ + δ = 180.Hence one of γ, δ is less than 90 and the other is greater than 90.By Euclid’s fifth postulate, lines n and l meet on whichever side of ←→ P Q thesmaller of angles γ and δ lies (e.g., on the same side as δ in figure 32.2).Thus n ̸‖ l. Hence there is only one line through P that is parallel to l.Hence the Euclidean Parallel Postulate follows from Euclid’s Fifth Postulate.Figure 32.2: Euclid’s Fifth Postulate implies the Euclidean Parallel Postulate.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

158 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATEAxiom 32.4 (Converse of Alternate Interior Angles Theorem.) Iftwo parallel lines are cut by a transversal, then both pairs of alternateinterior angles are congruent.Theorem 32.5 The Converse of the Alternate Interior Angles Theorem isEquivalent to the Euclidean Parallel Postulate.Proof. (⇒) [ Assume that the converse of the Alternate Interior AnglesTheorem is true (Equivalent Axiom 32.4) and show that the EuclideanParallel Postulate (axiom 32.1) follows. ]Let l be a line and let P be a point such that P ∉ l.By theorem 21.5 we can drop a perpendicular line t from P to l and thenconstruct a second line m through P such that t ⊥ m.By the alternate interior angles theorem m ‖ l (because a pair of interiorangles are congruent; it just happens that by construction, all of the interiorangles are all right angles, figure 32.3).Suppose there is another line n ≠ m such that P ∈ n and n ‖ l.Then t is a transversal to l and n. Then by axiom 32.4 the alternate interiorangles δ = γ. Since γ = 90 then δ = 90, and consequently n ⊥ t.By the uniqueness part of the protractor postulate, there is only one linethrough P that is perpendicular to t. Hence n = m.Thus there is only one line parallel to l through P . Thus axiom 32.1 follows.Figure 32.3: The converse of the alternate interior angles theorem (converseto theorem 29.4) is equivalent to the Euclidean Parallel Postulate.(⇐) [ Assume the Euclidean Parallel Postulate (axiom 32.1) and show that« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 159that the converse to the Alternate Interior Angles Theorem (axiom 32.4)follows. ]Let l and m be parallel lines that are cut by a transversal t (hypothesis).We need to show (see figure 32.4) that α = β and γ = δ. Let n be the linethrough Q such that α = ɛ.By the alternate interior angles theorem, l ‖ n.By the Euclidean Parallel Postulate, there is only one line parallel to lthrough Q, and that line is m. Hence m = n.Figure 32.4: The Euclidean Parallel Postulate implies the converse of theAlternate Interior Angles Theorem.Hence β = ɛ = α.By the linear pair theorem (theorem 18.4), γ = 180 − β and δ = 180 − α,hence δ = γ. Thus both pairs of alternate angles are congruent, which isthe conclusion of the converse of the alternate interior angles theorem.Thus the Euclidean Parallel Postulate implies the converse of the AlternateInterior Angles Theorem.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

160 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATETheorem 32.6 The following are equivalent to the Euclidean Parallel Postulate.1. Hilbert’s Parallel Postulate. For every line l and for every externalpoint P there exists at most one line m such that P ∈ m andm ‖ l.2. Proclus’ Axiom. If l ‖ m and t ≠ l is a line such that t insersectsl then t also intersects m.3. If l ‖ m and t is a transversal such that t ⊥ l then t ⊥ m.4. if l, m, n and k are lines such that k ‖ l, m ⊥ k, and n ⊥ l, theneither m = n or m ‖ n.5. Transitivity of Parallels. If l ‖ m and m ‖ n then eitiher l = n orl ‖ n.Proof. (Exercise.)Transitivity of parallelism has the following two corollaries that we will uselater.Corollary 32.7 The diagonals of a parallelogram are congruent.Corollary 32.8 The diagonals of a parallelogram bisect one another.Proof. (Exercise.)Theorem 32.9 There exists a rectangle ⇐⇒ the EPP is true.Proof. (Given in a later section.)Figure 32.5: Illustration of lemma 32.11. There is a point T ∈ −→ QR suchthat α < ɛ.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 161Theorem 32.10 (Angle Sum Postulate.) σ(△ABC) = 180 ⇐⇒ EPPis true.The proof requires the following lemma.Lemma 32.11 Let P Q be a segment and R a point such that ∠P QR = 90.Then for every ɛ > 0 there exists a point T ∈ ←→ QR such that µ(∠P T Q) < ɛ.Proof. (of lemma) Let P, Q, R be given as described and let ɛ > 0 be given.Choose any S ∈ H ←→ R, P Qsuch that ←→ P S ⊥ ←→ P Q.Then every point T ∈ −→ QR is in the interior of ∠QP S.Define T 1 ∈ −→ QR such that P Q = QT1 (see figure 32.5).Chose T 2 such that Q ∗ T 1 ∗ T 2 and T 1 T 2 = P T 1 .Chose T 3 such that Q ∗ T 2 ∗ T 3 and T 2 T 3 = P T 2 .Inductively chose T n such that Q ∗ T n−1 ∗ T n and P T n−1 = T n−1 T n .Constructed in this way, by the isoceles triangle theorem,By the angle addition postulate,∠QT n P ∼ = ∠T n−1 P T n (32.2)∠QP T n = ∠QP T 1 + ∠T 1 P T 2 +∠T 2 P T 3 + · · · + ∠T n−1 P T n< ∠QP S = 90Suppose that ∠T i−1 P T i ≥ ɛ for all ɛ (RAA).By the Archimedian ordering property there is some number n such thatnɛ > 90.Hence∠QP T 1 +∠T 1 P T 2 +∠T 2 P T 3 + · · · + ∠T n−1 P T n≥ nɛ > 90(The first inequality follows from the RAA hypothesis; the second forArchimedes.) This is a contradiction; hence we conclude that for somei,∠T i−1 P T i < ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

162 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATEBut by equation 32.2∠QT i P < ɛThus we have constructed a point such that the angle α < ɛ, proving thelemma.Figure 32.6: Proof of lemma 32.11.Proof. (The Angle Sum Postulate (axiom 32.10) is equivalent to the EuclideanParallel Postulate (axiom 32.1).)(⇒) [Euclidean Parallel Postulate ⇒ Angle Sum Postulate]Assume the Euclidean Parallel Postulate.Let △ABC be a triangle, and construct point D such that α = β (see figure32.7).Figure 32.7: The Euclidean Parallel Postulate implies the Angle Sum Postulate(axiom 32.10).Then ←→ CD ‖ ←→ AB by the alternate interior angles theorem.Choose E ∈ ←→ CD such that E ∗C ∗D. Since the Euclidean Parallel Postulateimplies the converse to the alternate interior angles theorem, then γ = δ.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 163By the angle addition postulate and the linear sum theorem, α+γ+ɛ = 180.Hence (using α = β and γ = δ we haveσ(△ABC) = δ + β + ɛ = 180Hence the Euclidean Parallel Postulate ⇒ the Angle Sum Postulate.(⇐) [Angle Sum Postulate ⇒ Euclidean Parallel Postulate]Assume the angle sum postulate.Let l be a line and P ∉ l a point. Drop a perpendicular from P to l andcall the foot of the perpendicular Q (see figure 32.8).Let m be the line through P that is perpendicular to ←→ P Q. By the alternateinterior angle theorem m ‖ l.Suppose that there is a second line n through P such that n ≠ m and n ‖ l(RAA).Choose S on n such that S ∈ H Q,m .Choose R on m such that R ∈ H ←→ S, P Q.Choose T on l such that T ∈ H ←→ S, P Qand ɛ = ∠QT P < ∠SP R = γ. Sucha point exists by the lemma because n ‖ l.Point T is in the interior of ∠QP S (Otherwise, if S were in the interior of∠QP T , then n would intersect QT by the Crossbar theorem, which is notpossible since they are parallel.)Henceby the protractor postulate.α < δSince S is in the interior of ∠β, we also haveγ + δ = β = 90Since l ‖ m, the alternate interior angles are equivalent, i.e.,∠T P S = ɛ =⇒ α + ɛ < δ + γ = 90also by the protractor postulate, and consequentlyσ(△P QT ) = α + ɛ + 90< δ + γ + 90= 180Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

164 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATEThus we have shown that if there are multiple distinct lines (>1) throughP that are parallel to l, then there is a triangle whose measure is less than180.The contrapositive of this result, which is equivalent, is that if every trianglehas measure greater than or equal to 180, then there is at most one linethrough P that is parallel to l.But by the Saccheri-Legendre Theorem, σ(T ) > 180 is impossible. Hencewe conclude if if every triangle has measure 180, then the Euclidean ParallelPostulate follows.Figure 32.8: The angle sum postulate implies the Euclidean Parallel Postulate.Definition 32.12 (Similar Triangles) Triangles △ABC and △DEF aresaid to be similar if ∠ABC ∼ = ∠DEF , ∠BCA ∼ = ∠EF D, ∠CAB ∼ =∠F DE, and we write △ABC ∼ △DEF .Axiom 32.13 (Wallis’ Postulate) If △ABC is a triangle and DE is asegment there exists a point F such that △ABC ∼ △DEF .Theorem 32.14 Wallis’ Postulate ⇐⇒ the EPP is true.Proof. (⇒) (The Euclidean Parallel Postulate ⇒ Wallis’ Postulate). Assumethe Euclidean Parallel Postulate.See figure 32.9.Construct ray −−→ DG such that α = γ (protractor postulate).Construct ray −−→ EH with H ∈ H ←→ G, DE, such that δ = β (protractor postulate).Since γ + δ = α + β < 180, by the Euclidean Parallel Postulate, the tworays intersect at a point F in H G,←→ DE.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 32. THE EUCLIDEAN PARALLEL POSTULATE 165Figure 32.9: The Euclidean Parallel Postulate imples Wallis’ Postulate.Figure 32.10:Wallis’ Postulate implies the Euclidean Parallel Postulate.By the angle sum postulateHenceσ(ABC) = σ(DEF ) = 180ν = 180 − (γ + δ)= 180 − (α + β)= µand △ABC ∼ △DEF . Hence the Euclidean Parallel Postulate ⇒ Wallis’Postulate.(⇐) (Wallis’ Postulate ⇒ the Euclidean Parallel Postulate). Assume Wallis’Postulate is true.Let l be a line and P ∉ l a point. Drop a perpendicular from P to its footQ on l. Construct m through P perpendicular to ←→ P Q (see figure 32.10).Let n be the line through P that is parallel to l.Suppose n ≠ m (RAA).Chose S ∈ m such that S ∈ H Q,m . Let R be the foot of the perpendicularto ←→ P Q dropped from S.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

166 SECTION 32. THE EUCLIDEAN PARALLEL POSTULATEBy Wallis’ Postulate there exists a point T such that △P RS ∼ △P QT .Since ∠P QT = ∠P RS = 90 then T ∈ l.Since ∠QP T = ∠RP S then T ∈ n.Hence n ∩ l ≠ ∅ and therefore n ̸‖ l. Thus we must reject the RAAhypothesis and conclude that n = m, i.e., the parallel line is unique. ThusWallis’ Postulate ⇒ the Euclidean Parallel Postulate.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 33RectanglesDefinition 33.1 A rectangle is a quadrilateral each of whose angles areright angles.Definition 33.2 A square is a rectangle each of whose sides has equallength.Theorem 33.3 The following statements are equivalent to the Euclideanparallel postulate:1. There exists a triangle whose defect is 0.2. There exists a right triangle with defect 0.3. (Clairut’s Axiom.) There exists a rectangle.4. There exists arbitrarily large rectangles.5. The defect of every right triangle is 0.6. The defect of every triangle is 0.Lemma 33.4 Let △ABC be a triangle. Then at least two of its interiorangles are acute.Proof. Suppose that two of the interior angles are not acute. Call theseangles α and β, and call the third interior angle δ.Then α ≥ 90 and β ≥ 90. Henceσ(△ABC) = α + β + δ≥ 180 + δBut by the Saccheri-Legendre Theorem (theorem 30.6) σ(△ABC) ≤ 180.Hence δ = 0, which is impossible. Therefore at most one of the angles canbe non-accute, making the remaining two acute.167

168 SECTION 33. RECTANGLESLemma 33.5 Let A and B be the vertices of acute angles in △ABC. Thenthe foot of the perpendicular from C to ←→ AB is between A and B.Proof. Since the angles A and B are acute, neither one can be the foot ofthe perpendicular (otherwise the corresponding angle would be 90). Henceone of the following holds: A ∗ D ∗ B, D ∗ A ∗ B, or A ∗ B ∗ D (see figure33.1).Figure 33.1: Two possibilities for the foot of the perpendicular dropped forC to ←→ AB: A ∗ D ∗ B and D ′ ∗ A ∗ B. Lemma 33.5 says that only the firstcase is allowed.CD' A D BIf D ∗ A ∗ B, then the acute angle ∠CAD is an exterior angle for △ACD,and ∠CDA is a remote interior angle. But this contradicts the exteriorangle theorem (theorem 24.4) which says that the exterior angle must belarger than the corresponding remote interio angles.A similar argument shows that A ∗ B ∗ D is impossible.Hence only A ∗ D ∗ B is possible.Proof. (theorem 33.3.) We will show that (6) ⇒ (1) ⇒ (2) ⇒ (3) ⇒ (4) ⇒(5) ⇒ (6) and that (6) ⇐⇒ EPP. Since any one of the statementsimplies each of the others, they are all equivalent; and since one of them isequivalent to the Euclidean Parallel Postulate, they all are.[(6) ⇐⇒ EPP] We have already shown that the angle sum postulate(axiom 32.10) is equivalent to the Euclidean parallel postulate; by the definitionof defect, δ(△ABC) = 0 ⇐⇒ σ(△ABC) = 180, hence (6) is alsoequivalent.[(6) ⇒ (1)] By (6) all triangles have zero defect. Pick any triangle. Thenit has zero defect.[(1) ⇒ (2)] Let △ABC be a triangle with zero defect (hypothesis, statement(1)).By lemma 33.5 two interior angles are acute; label these vertices as in figure« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 33. RECTANGLES 16933.1 and drop the perpendicular from C to D as shown. Then triangles△ADC and △BDC are right.By additivity of defect (30.11),0 = δ(△ABC) = δ(△ADC) + δ(△BDC)Since defect is non-negative, each of the two sub-triangles also have zerodefect. Hence there is a right triangle with zero defect (statement (2)).[(2) ⇒ (3)] Let △ABC be a right triangle with zero defect (hypothesis,statement (2)).Designate B as the right angle. Then∠BAC + ∠ACB = 90Since it is possible to construct a congruent copy of any triangle on eitherside of a congruent base (see theorem 23.3), there exists a point D on theopposite side of ←→ AC from B such that△ABC ∼ = △ADCFigure 33.2: A rectangle constructed from two right triangles in proof of(2) ⇒ (3).(See figure 33.2.) By congruency,α = ɛγ = ζα + γ = 90 (Since δ(△ABC) = 0)ɛ + ζ = 90 (substitution)α + ζ = 90ɛ + γ = 90The last two lines complete the proof that □ABCD is a rectangle. Hencea rectangle exists (statement (3)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

170 SECTION 33. RECTANGLESFigure 33.3: Construct an arbitrariy large rectangle. Start with a givenrectangle, such as the one in the lower left corner, and then constructlarger rectangles by building congruent triangles on common edges. Therepeat the process, doubling the size of the rectangle at each step, until therequired size is reached.[(3) ⇒ (4)] (sketch of proof) By an arbitrarily large rectangle we mean thatfor any number M there is a rectangle all of whose sides are larger than M.From (3) there exists a rectangle. We can split this into two right triangles,and then proceed to duplicate the rectangle on one of the original rectangle’sedge by constructing congruent right triangles. This creates a rectangle oftwice the width of the first triangle (see figure 33.3).Then keep repeating the process until we have an edge longer than M(Archimedian property).Then repeat the process on the other direction, so that both dimensions arelarger than M. Thus by construction, an arbitrarily large triangle exists(statement (4)).[(4) ⇒ (5)] Let △ABC be a right triangle with right angle at C. By (4)there exists a rectangle □CDEF such that DC > AC and F C > BC.Since □CDEF is a parallelogram, it is convex (theorem 31.9).Since it is a rectangle, all four angles are 90 (definition of rectangle) andhence δ(□ABCD) = 0.By the additivity of defect (theorem 31.6),δ(△DEF ) = δ(△DCF ) = 0Similarly0 = δ(△CDF ) = δ(△ADF ) + δ(△ACF )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 33. RECTANGLES 171Figure 33.4: Every right triangle has zero defect.hence δ(△ACF ) = 0. Applying additivity of defect one more time,0 = δ(△ACF ) = δ(△ABF ) + δ(△ABC)hence δ(△ABC) = 0. Since △ABC was an arbitrary right triangle, we canconclude that every right triangle has zero defect (statement (5)).[(5) ⇒ (6)] By the lemma we can divide any triangle into right triangles.By (5) every right triangle has zero defect. By additivity of defect, everytriangle has zero defect (statement (6)).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

172 SECTION 33. RECTANGLES« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 34The Parallel ProjectionTheoremLemma 34.1 Let l, m, and n be distinct parallel lines; let t and t ′ betransversals that cut these three lines at points A, B, C, and A ′ , B ′ , C ′ withA ∗ B ∗ C. ThenAB = BC ⇒ A ′ B ′ = B ′ C ′Figure 34.1: Proof of theorem 34.1. This is a special case of the ParallelProjection Theorem (theorem 34.2) when AB = BC.Proof. By the Euclidean Parallel Postulate, we can construct lines t ′′ andt ′′′ parallel to t through A ′ and B ′ , as shown in figure 34.1.173

174 SECTION 34. THE PARALLEL PROJECTION THEOREMBy Proclus’ Axiom (Theorem 32.6), since m and n are parallel to l, linest ′′ and t ′′′ intersect m and n at points B ′′ and C ′′ . Define the angles α β,γ, δ and ɛ, as shown in figure 34.1.If A = A ′ then B ′′ = B and t = t ′′ hence AB = A ′ B ′′ .If B ′ = B then C = C ′′ , t ′′′ = t, and hence BC = B ′ C ′′ .If both A ≠ A ′ and B ′ ≠ B then □AA ′ B ′′ B and □BB ′ C ′′ C are parallelograms.Since the opposite sides of parallelograms are congruent (corollary32.7), AB = A ′ B ′′ and BC = B ′ C ′′ .By hypothesis, AB = BC, hence A ′ B ′′ = B ′ C ′′ .By transitivity of parallelism, t ′′ = t ′′′ or t ′′ ‖ t ′′′ . Hence γ and β arecorresponding angles. By the corresponding angles theorem (theorem 29.7),γ = δ.Similarly, by repeated application of the corresponding angles theorem,α = ɛ = β.Hence by ASA, △A ′ B ′′ B ′ ∼ = △B ′ C ′′ C ′ .Hence by the properties of congruent triangles A ′ B ′ = B ′ C ′ .Theorem 34.2 (The Parallel Projection Theorem) Let l, m, and nbe distinct parallel lines; let t and t ′ be transversals that cut these threelines at points A, B, C and A ′ , B ′ , C ′ , respectively, such that A ∗ B ∗ C.ThenABAC = A′ B ′A ′ C ′Proof. Either AB/AC is rational, or it is irrational.First suppose that AB/AC is rational. Then there exist positive integers pand q such thatABAC = p qAccording to the ruler postulate we can choose equally spaced points A 0 , A 1 , ..., A qsuch thatA i A i+1 = ACqThen A p = B (see figure 34.3). Define the lines l i ‖ l through each A i anddefine the intersection of each line with t ′ as A ′ i .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 34. THE PARALLEL PROJECTION THEOREM 175Figure 34.2: Parallel Projection Theorem (theorem 34.2).By the lemma,A ′ i A′ i+1A ′ C ′ = A iA i+1AC= 1 qThus A ′ B ′ = p × A ′ iA ′ i+1 = p × A′ C ′hence A′ B ′q A ′ C ′ = p q = AB , proving theACtheorem in the case where AB/AC is rational.To prove the theorem when x = AB/AC is irrational, define y = A ′ B ′ /A ′ C ′ .Let r be any rational number such that 0 < r < x and define D ∈ t suchthat AD/AC = r (see figure 34.4). Construct m ′ ‖ l through D and defineD ′ as the intersection of m ′ and t ′ . By the first caseSince l ‖ m ‖ m ′ , A ′ ∗ D ′ ∗ B ′ , henceA ′ D ′A ′ C ′ = rr < A′ B ′A ′ C ′ = yHence for every r < x, we have r < y. By a similar argument, for everyr < y, we have r < x. Hence x = y by the comparison theorem for realnumbers (theorem 6.5), x = y.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

176 SECTION 34. THE PARALLEL PROJECTION THEOREMFigure 34.3: The Parallel Projection Theorem (theorem 34.2) whenAB/AC = p/q for positive integers p, q.Figure 34.4: The Parallel Projection Theorem (theorem 34.2) when AB/ACis irrational.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 35Similar TrianglesLet us recall the definition of similar triangles, previously stated as definition32.12:Definition 35.1 (Similar Triangles) Triangles △ABC and △DEF aresaid to be similar if ∠ABC ∼ = ∠DEF , ∠BCA ∼ = ∠EF D, ∠CAB ∼ =∠F DE, and we write △ABC ∼ △DEF .Theorem 35.2 (Fundamental Theorem on Similar Triangles)△ABC ∼ △DEF ⇒ ABAC = DEDFProof. Let △ABC ∼ △DEF .If AB = DE then △ABC ∼ = △DEF by ASA, and hence AC = DF by theproperties of congruent triangles, so that AB/AC = DE/DF .Suppose that AB ≠ DE. Then either AB > DE or AB < DE.Suppose that AB > DE. Choose B ′ ∈ AB such that AB ′ = DE.Construct a line m through B ′ that is parallel to ←→ BC. By Pasch’s theorem(theorem 15.12), m intersects AC. Call the point of intersection C ′ (seefigure 35.1).By the corresponding angles theorem η = β = ɛ; hence by ASA (α = δ,AB ′ = DE, and η = ɛ), △AB ′ C ′ ∼ = △DEF . Hence AC ′ = DF .Construct line n through A such that n ‖ m ‖ ←→ BC.177

178 SECTION 35. SIMILAR TRIANGLESBy the parallel projection theorem (theorem 34.2),AB ′AB = AC′ACSubstituting AC ′ = DF and AB ′ = DE,DEAB = DFACCross multiplying gives the result of the theorem.If AB < DE, then relabel the two triangles (exchange all the labels onthe first triangle with the corresponding item on the second triangle); thenrepeat the above argument to complete the proof.Figure 35.1: The fundamental theorem on similar triangles (theorem 35.2)tells us that AB/AC = DE/DF .Corollary 35.3 If △ABC ∼ △DEF then there is a number r > 0 suchthatDE = r × ABDF = r × ACEF = r × BCThe number r is called the common ratio of the sides of the triangle.Proof. Suppose that △ABC ∼ △DEF . Then by the fundamental theoremon similar triangles,DefineABAC = DEDFABandBC = DEEFr = DEAB« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 35. SIMILAR TRIANGLES 179Then DE = r × AB andDFAC = r = EFBCHence DF = r × AC and EF = r × BCCorollary 35.4 (SAS Criterion for Similarity) If △ABC and △DEFare two triangles such that ∠CAB = ∠F DE and AB/AC = DE/DF then△ABC ∼ △DEF .Proof. Refer again to figure 35.1; suppose that α = δ and ABAC = DEDF .If AB = DE, the △ABC ∼ = △DEF and the theorem follows.Suppose that AB ≠ DE. We can assume without loss of generality thatAB > DE (otherwise we just relabel the two triangles).Pick B ′ such that AB ′ = DE and construct line m parallel to BC as before.As in the earlier proof, m intersects BC by Pasch’s theorem at some pointC ′ .Since B and C are on one side of m and A is on the other side, A ∗ C ′ ∗ C.By the corresponding angles theorem η = β and γ = ∠AC ′ B ′ . HenceBy the parallel projection theorem△ABC ∼ △AB ′ C ′ (35.1)AB ′AB = AC′AC ⇒ AB′AC ′ = ABAC = DEDFwhere we have used the hypotheses for the theorem in the last equality.But by construction AB ′ = DE, so thatDEAC ′ = DEDFBy SAS (DF = AC ′ , α = δ, AB ′ = DE),Substituting equation 35.1 gives⇒ DF = AC′△AB ′ C ′ ∼ = △DEF△ABC ∼ △DEFThe following theorem is also the converse of the Fundamental Theorem ofSimilar Triangles.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

180 SECTION 35. SIMILAR TRIANGLESCorollary 35.5 (SSS Criterion for Simlarity) If △ABC and △DEFare two triangles such that AB/DE = AC/DF = BC/EF then △ABC ∼△DEF .Proof. Suppose AB/DE = AC/DF = BC/EF in triangles △ABC and△DEF .If AB = DE then AC = DF and BC = EF , so by SSS the triangles arecongruent.Suppose AB ≠ DE. Assume AB > DE (if not, relabel). Construct△AB ′ C ′ as in the proof of the previous theorem; as we saw in that proof△AB ′ C ′ ∼ △ABC.By the fundamental theorem of similar triangles,ABAB ′ = ACAC ′ = BCB ′ C ′ (35.2)Since AB ′ = DE by construction, and by hypothesis AB/DE = AC/DF ,ACDF = ABDE = ABAB ′ = ACAC ′and therefore AC ′ = DF . Also by hypothesis AB/DE = BC/EF in 35.2givesBCEF = ABDE = ACAC ′ = BCB ′ C ′hence B ′ C ′ = EF . Thus △AB ′ C ′ ∼ = △DEF by SSS. Since △AB ′ C ′ ∼△ABC, we have △ABC ∼ △DEF .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 36Triangle CentersDefinition 36.1 Three distinct lines are concurrent at point P if P lieson all three lines, and P is called the point of concurrency of the threelines.Theorem 36.2 The three perpendicular bisectors of the sides of any trianglemeet at a single point we call the circumcenter that is equidistantfrom the vertices of the triangle.Proof. Let △ABC be a triangle. Denote the midpoint of AB by P , themidpoint of BC by Q, and the midpoint of CA by R.Let l be the bisector of AB and m the bisector of BC. Then either l ‖ m,l = m, or l intersects m.Suppose l ‖ m. Then since AB ⊥ l, this means AB ⊥ m. Since we alsohave BC ⊥ m this means that either BC ‖ AB or ←→ BC = ←→ AB. Since BCand AB are two distinct sides of the same triangle, neither case is possible.Hence l ̸‖ m.Similarly, if l = m, either BC ‖ AB or ←→ BC = ←→ AB, which we have just saidis impossible. Hence l ≠ m.Therefore l must intersect m. Define the point O = l ∩ m. Then l = ←→ P Oand m = ←→ QO by definition of m and l.By SAS, △AOP ∼ = △BOP , hence AO = BO.Also by SAS, △BOQ ∼ = △COQ, hence BO = CO.Since R is the midpoint of AC, then by SSS, △AOR ∼ = △COR.181

182 SECTION 36. TRIANGLE CENTERSFigure 36.1: The intersection of the perpendicular bisectors of the sides ofa triangle is called the orthocenter of the triangle.Thus ∠ARO = ∠CRO. But these angles are supplements, hence RO ⊥AC.Thus all three side bisectors pass through the point O. We have alreadyshown that O is equidistant from A, B, and C.Definition 36.3 Altitude. Let l be a line that is constructed perpendicularto any side of a triangle that is concurrent with the vertex V oppositethat side. Then l is said to be the line containing the altitude of thetriangle. The altitude is the distance from V to l.Theorem 36.4 The lines containing the altitudes of any triangle intersectat a point called the orthocenter.Proof. The construction is illustrated in figure 36.2. Construct line l ‖ ABthrough C; line m ‖ BC through A; and line n ‖ AC through B.Define the intersections l ∩ n = A ′ , l ∩ n = B ′ , and m ∩ n = C ′ as shownin figure 36.2.By construction, □ABA ′ C is a parallelogram, so that BA ′ = AC.By construction □C ′ BCA is a parallelogram, so that BC ′ = AC.Hence BA ′ = BC ′ , so that B is the midpoint of A ′ C ′ .Since the altitude through B is perpendicular to AC, and AC ‖ ←−→ A ′ C ′ , weconclude that the altitude through B is the perpendicular bisector of A ′ C ′ .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 36. TRIANGLE CENTERS 183By a similar argument, the altitude through C is the perpendicular bisectorof A ′ B ′ , and the altitude through A is the perpendicular bisector of B ′ C ′ .By theorem 36.2, these three bisectors of the sides of triangle △A ′ B ′ C ′must meet at a common point O.Thus the altitudes of the original triangle △ABC meet at a common pointO.Figure 36.2: The orthocenter is the intersection of the three altitudes of atriangle.Figure 36.3: The three medians of a triangle all meet at the centroid of thetriangle.Definition 36.5 A median of a triangle is a line segment whose endpointsare one vertex and the midpoint of the side opposite that vertex.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

184 SECTION 36. TRIANGLE CENTERSTheorem 36.6 (Median Concurrence Theorem) The three mediansof any triangle are concurrent, and meet at a point called the centroidthat is 2/3 the length from each vertex to the midpoint of the opppositeside.Proof. (Outline). Define the usual Cartesian coordinate system on theEuclidean plane and let the coordinates of the vertices be A = (x a , y a ), B =(x b , y b ), C = (x c , y c ). The midpoints are (see figure 36.3)D = (1/2)(A + B) = (1/2)(x a + x b , y a + y b )E = (1/2)(B + C) = (1/2)(x b + x c , y b + y c )F = (1/2)(A + C) = (1/2)(x a + x c , y a + y c )Then the point O on F B that is 1/3 of the way from F to B isO = F + (1/3)(B − F ) = (1/3)(A + B + C)By a similar argument the point on segment CD that is 1/3 of the wayfrom D to C, and the point on segment AE that is 1/3 of the way from Eto A is also at (A + B + C)/3.Theorem 36.7 (Euler Line Theorem) The orthocenter O, circumcenterC, and centroid M of any triangle are collinear. If the triangle is equilateralthe three points coincide; otherwise, O ∗ M ∗ C.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 36. TRIANGLE CENTERS 185Figure 36.4: Illustration of the Euler Line Theorem. P , Q and R aremidpoints of sides AB, BC, and CA. The perpendicular bisectors areshown as thin solid lines; the altitudes as thin dashed lines; and the mediansas thick dotted segments. K denotes the circumcenter; M the centroid; andO the orthocenter. Euler’s Line Theorem states that O, M and K line onthe same line, and that O ∗ M ∗ K, unless △ABC is equilateral, in whichcase all three points coincide.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

186 SECTION 36. TRIANGLE CENTERSCenters« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 37AreaDefinition 37.1 The interior of △ABC denoted by Int(△ABC), is theintersection of the interiors of the three angles, ∠A, ∠B, ∠C.Definition 37.2 The Associated Triangular Region of △ABC, denotedby (ABC), is(ABC) = △ABC ∪ Int(△ABC)Definition 37.3 A Polygonal Region is a subset R of the plane that canbe written as the union of a finite number of triangular regions in such away that if two of the triangular regions intersect than the intersection iscontained in an edge of each.Definition 37.4 Two regions are non-overlapping if they only intersectalong their edges.Figure 37.1: Illustration of a triangulation.187

188 SECTION 37. AREADefinition 37.5 A Triangulation of a polygonal region is a completedecomposition of the region into non-overlapping triangular regions suchthat every point in the the region is contained in at least one triangularregion.Axiom 37.6 (Neutral Area Postulate) Associate with each polygonalregion R there is a nonnegative number α(R) called the area of R such that1. (Congruence) If two triangles are congruent then their associated areasare equal.2. (Additivity) If R is the union of two non-overlapping polygonal regionsR 1 and R 2 then α(R) = α(R 1 ) + α(R 2 ).Definition 37.7 The Rectangular Region ABCD is the union of thetriangular regions of the four triangles formed by the intersecting diagonalsof the rectangle:ABCD =AED ∪ DEC∪ CEB ∪ BEAwhere E is the intersection of the diagonals.We also definelength(R) = ABwidth(R) = BCFigure 37.2: A rectangular region is the union of the four triangular regionsdefined by the edges two intersecting diagonals.Axiom 37.8 (Euclidean Area Postulate) Let R be a rectangle. Thenα(R) = length(R) × width(R)Theorem 37.9 Let T = ABC where △ABC is a right triangle with rightangle at C. Thenα(T ) = 1 AC × BC2« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 37. AREA 189Proof. (Exercise.)Definition 37.10 Let T be a triangular region corresponding to △ABC.Drop a perpendicular from C to ←→ AB and call the foot of the perpendicularD. Then we definebase(T ) = ABheight(T ) = CDFigure 37.3: Definition of base and height of any triangle. Pick any edgeof the triangle (such as AB). Then the length of AB is called the base ofthe triangle, and the distance from AB to its opposite vertex C is calledthe height of the triangle.Theorem 37.11 Let T be a triangle. ThenÅ 1α(T ) = base(T ) × height(T )2ãProof. (Exercise.)Theorem 37.12 Suppose that △ABC ∼ △DEF . Thenα(△DEF ) = r 2 α(△ABC)where r = DE/AB.Proof. (Exercise.)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

190 SECTION 37. AREAToday’s Lesson: Pythagorean Theorem« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 38The PythagoreanTheoremThe Pythagorean theorem is perhaps the mostfamous theorem in all of Mathematics; in canbe proven in hundreds of different ways. 1 Itsorigins go back to antiquity, and it is notknown by whom it was first discovered, or evenwho wrote the first proof, although Pythagoras(c. 580 BCE - c.500 BCE) is given credit. 2There is evidence that is already known by theBabylonians and ancient Egyptians by 2500BC. In China it is called the Gougu Theorem,and in India it is known as Bhaskara’s Theorem.A special case was first given (withoutproof) by Baudhayana Baudhayana Sulba Sutra(India, c800 BCE)The rope which is stretched acrossthe diagonal of a square producesan area double the size of the original square.Roman coin with a figureof Pythagoras, madeduring the reign of Decius(249-251) [Wikimedia Commons/PublicDomain Image].1 Elisha Loomis (The Pythagorean Proposition, 1907. Reissued by the NCTM, 1968)contains 367 different proofs of the theorem, one for every day of the year, plus an coupleof extras. A website maintained by Alexander Bogomolny (http://www.cut-the-know.org/pythagoras") contains over 80 proofs.2 For a detailed history see the book by Eli Maor, The Pythagorean Theorem: A 4000Year History, Princeton University Press, 2007.191

192 SECTION 38. THE PYTHAGOREAN THEOREMwhile a later sutra by Katyayana (c 200 BCE) is more general 3A rope stretched along the length of the diagonal produces anarea which the vertical and horizontal sides make together.It has inundated common culture as well. Gilbert and Sullivan gave theMajor-General “many cheerful facts about the square of the hypotenuse”while the scarecrow in the Wizard of Oz mistakenly quotes it as:The sum of the square roots of any two sides of an isoscelestriangle is equal to the square root of the remaining side. Oh,joy! Oh, rapture! I’ve got a brain! 4Even Homer Simpson has had a word to say about it, and at least onepresident of the United States (James Garfield) is credited with finding anew proof. 5Theorem 38.1 (Pythagorean Theorem) Let △ABC be a right trianglewith right angle at vertex C. Then if a, b, and c are the sides oppositevertices A,B, and C, respectively,a 2 + b 2 = c 2Figure 38.1: Notation for the Pythagorean Theorem.Proof. Drop a perpendicular from C to ←→ AB and call its foot D. Label thesegments as shown in figure 38.2.By lemma 33.5, A ∗ D ∗ B.By the angle sum theorem 180 = m(∠A) + m(∠B) + m(∠C), hence since3 Mac Tutor History of Math, http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Indian_sulbasutras.html.4 Internet Movie Data Base (IMDB), http://www.imdb.com/title/tt0032138/quotes?qt04099235 See Venema Problem 9.12.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 38. THE PYTHAGOREAN THEOREM 193m(∠C) = 90,m(∠A) + m(∠B) = 90Also by the angle sum theorem,m(∠A) + m(∠ACD) = 90m(∠B) + m(∠BCD) = 90Hencem(∠A) = m(∠BCD)m(∠B) = (∠ACD)Thus△ABC ∼ △ACD ∼ △CBDBy the fundamental theorem of similar triangles,HenceAdding,pb = b c and q a = a cb 2 = pc and a 2 = qca 2 + b 2 = pc + qc = c(p + q) = c 2Figure 38.2: Notation for proof of the Pythagorean Theorem.Corollary 38.2 Then h = √ pq (see figure 38.2).Proof. (Exercise.)Corollary 38.3 a = √ qc and b = √ pc.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

194 SECTION 38. THE PYTHAGOREAN THEOREMProof. (Exercise.)Theorem 38.4 (Converse of Pythagorean Theorem) If a 2 + b 2 = c 2then ∠C is a right angle.Proof. We are given △ABC with a 2 + b 2 = c 2 .Construct a right angle at point F on rays −→ −−→F G and F H.Define point E ∈ −−→ F H such that F E = a, and define point D ∈ −→ F G suchthat F D = b. Then △DEF is a right triangle. By the PythagoreanTheoremf 2 = d 2 + e 2 = a 2 + b 2 = c 2This means that f = c and hence by SSS, △ABC ∼ = △DEF .∠C = ∠F = 90.HenceFigure 38.3: Converse of the Pythagorean Theorem.Figure 38.4: Definition of sin θ = BC/AB and cos θ = AC/AB.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 38. THE PYTHAGOREAN THEOREM 195Definition 38.5 (Trigonometry) Let △ABC be a right triangle withright angle at vertex C, and let θ = ∠CAB. Then if θ is acute, we definesin θ = BCABand cos θ =ACABIf θ is obtuse, then let θ ′ = 180 − θ and definesin θ = sin θ ′ and cos θ = − cos θ ′Also, definesin 0 = 0 and cos 0 = 1sin 90 = 1 and cos 90 = 0Theorem 38.6 (Pythagorean Identity)Proof. (Exercise.)sin 2 θ + cos 2 θ = 1Theorem 38.7 (Law of Sines) Let △ABC be any triangle with sidesa, b, c opposite vertices A, B, C. ThenProof. (Exercise.)asin ∠A =bsin ∠B =csin ∠CTheorem 38.8 (Law of Cosines) Let △ABC be any triangle with sidesa, b, c opposite vertices A, B, C. ThenProof. (Exercise.)c 2 = a 2 + b 2 − 2ab cos ∠CRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

196 SECTION 38. THE PYTHAGOREAN THEOREMEuclid, of course, did not state the Pythagorean theorem in terms of thesum of the squares of the edges; to do so would have required algebra,which was not invented for another thousand years after Euclid. Instead,the theorem was expressed in terms of area.Theorem 38.9 (Euclid’s Version of the Pythagorean Theorem) Thearea of the square on the hypoteneuse of a right triangle is equal to the sumof areas of the squares on the legs.Proof. (Second Proof of the Pythago- rean Theorem, due to Euclid6 )Let △ABC be a right triangle with right angle at C. We construct thesquares on each edge and define the points as indicated in figure 38.5.Point J is the foot of the perpendicular to AB dropped from C, and pointK is the intersection of the same perpendicular with DE.From the Euclidean Area Postulate,α(□ABED) = α(□AJKD) + α(□JBEK) (38.1)Since ∠ACB is a right angle and □ACHI is a square, H ∈ ←→ BC and AI ‖←→BC.Thus, since triangles △IAC and △IAB each have base IA and height AC,they both have the same area:Furthermore, since IA = AC, AB = AD, andthen by SASα(△IAC) = α(△IAB) (38.2)∠IAB = 90 + ∠CAB = ∠CAD△IAB ∼ = △CADand since congruent triangles have equal areas (by the neutral area postulate),α(△IAB) = α(△CAD) (38.3)Since △CAD and △JAD each have base AD and height AJ, they haveequal area,α(△CAD) = α(△JAD) (38.4)Combining equations 38.2, 38.3 and 38.4 givesα(△IAC) = α(△JAD) (38.5)6 Version given in Book 1 of the Elements; another proof due to Euclid is in Book VI.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 38. THE PYTHAGOREAN THEOREM 197Figure 38.5: Euclid’s Proof of the Pythagorean Theorem.Butα(□ACHI) = 2α(△IAC)= 2α(△JAD)= α(□AJKD) (38.6)A similar argument is used to show thatα(□CBF G) = α(□JBEK) (38.7)Substituting equations 38.6 and 38.7 into 38.1 givesα(□ABED) = α(□ACHI) + α(□CBF G)Illustration of the proof the Guogu thoerem (third proof) from Zhoubi suanjing (c 100BCE). The illustrations were not in the original text but were added sometime beforethe 16th century, when this version was printed. [MAA Digitial Sciences Library]Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

198 SECTION 38. THE PYTHAGOREAN THEOREMProof. (Third Proof of the Pythagorean Theorem 7 ) Let △ABC bea right triangle with vertices A, B, C opposite sides a, b, c, and right angleat C. Label the vertices so that a ≤ b. See figure 38.6.On the left hand side of figure 38.6, the large square of area c 2 is equalto four times the area of triangle △ABC plus the area of the small squarewith sides of length b − a. HenceÅ ã abc 2 = (4) + (b − a) 2 (38.8)2= 2ab + b 2 − 2ab + a 2 (38.9)= a 2 + b 2Proof. (Fourth Proof of the Pythagorean Theorem.) This is a varianton the previous proof. Rearrange the triangles as shown on the right handside of figure 38.6. The four triangles plus the small square are rearranged7 Attributed to the Indian mathematician Bhaskara Acharya (1114-1185), one of thegreatest mathematicians of all time. Bhaskara is also attributed with the first proofsof the general solutions of the quadratic, cubic, and quartic equations; as well as thesolution of several Diophantine equations of the second order that were later posed byFermat, not knowing that they had been solved centuries earlier; and the developmentof the differential calculus and theory of infinitesimals, some 500 years prior to Newton.Essentially the same proof is also given in the Zhoubi Suanjing, published in Chinaapproximately 1000 years prior to that, the the logic by which (38.8) is reached isslightly different.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 38. THE PYTHAGOREAN THEOREM 199Figure 38.6: Bhaskara’s Proof of the Pythagorean Theorem.Figure 38.7: Proof by rearrangement of the Pythagorean Theorem.to cover a small a × a square (above the dashed line) and a larger b × bsquare (beneath the dashed line).Proof. Fifth Proof of the Pythagorean Theorem. See figure 38.7. Oneach side of the figure are two large squares, of size (a + b) × (a + b). On theleft, we see that this is equal to four times the area of the triangle plus thearea of the square on the hypoteneuse. On the right we see that this is equalto four times the area of the triangle plus the sum of the areas of the squareon each of the sides. Hence the area of the square on the hypoteneuse isequal to the sum of the areas of the squares on the sides.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

200 SECTION 38. THE PYTHAGOREAN THEOREMHow Many Proofs Can You Find?« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 39CirclesDefinition 39.1 The circle C(O, r) with center O and radius r > 0 isthe set of all points P that are a distance r from O,C(O, r) = {P |OP = r}Definition 39.2 A chord of Γ = C(O, r) is a segment P Q joining twopoints P and Q on Γ. A line l that contains a chord of Γ is called a secantline of Γ.Definition 39.3 Points P and Q on Γ = C(O, r) are called antipodal ifP ∗ O ∗ Q, in which case the chord P Q is called a diameter.We will also use the term diameter to mean the length of the diameter; itshould be clear from the context whether we mean P Q or the length of P Q.Definition 39.4 Let Γ = C(O, r). Then point A is said to be outside ofΓ if OA > r, and inside of Γ if OA < r.Definition 39.5 A line l is tangent to a circle Γ = C(O, r) if Γ∩l containsprecisely one point. A segment AB ∈ l is tangent to Γ if l is tangent to Γand the point of tangency lies in the interior of AB.The following theorem tells us that given any line and any circle, the lineeither is a secant line, a tangent line, or it does not intersect the circle.Theorem 39.6 Let Γ = C(O, r) be a circle and l be a line.number of points in Γ ∩ l is either 0, 1, or 2.Then theProof. Suppose there are three distinct points A, B, C that all lie in Γ ∩ l(figure 39.2).201

202 SECTION 39. CIRCLESFigure 39.1: The segment BC is a chord of of Γ, and the lines m and l aresecants of Γ. Line n is a tangent, with point of tangency D. Line l containschord AB, which is a diameter of Γ. Points A and P are antipodel.Since the points all lie on l they are collinear and hence can be ordered.Rename them so thatA ∗ B ∗ CBecause all three points are on Γ, we haver = OA = OB = OCHence △ABO, △BCO and △ACO are all isosceles triangles with sides oflength r.By the isosceles triangle theorem (theorem 21.4), the base angles of anisosceles triangle are equal to one another, hencelet α = ∠CAO = ∠ACO (triangle △ACO)let β = ∠BAO = ∠ABO (triangle △ABO)let γ = ∠BCO = ∠CBO (triangle △BCO)By the angle sum theorem,α + α < 180 =⇒ α < 90β + β < 180 =⇒ β < 90γ + γ < 180 =⇒ γ < 90« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 39. CIRCLES 203Figure 39.2: A circle cannot intersect a line at three distinct points, becauseotherwise all four of the indicated angles would have to be equal and smallerthan 90.Henceα + γ < 180But all three points A, B, C ∈ l, henceα + γ = 180This is a contradiction. Hence there cannot be three distinct points thatlie on both a line and a circle.Hence there cannot be more than 3 points that lie on both the line and thecircle; otherwise, if there were, pick any three of them. By the argumentgiven above, they cannot exist.Consequently the largest number of points that can lie in Γ ∩ l is two.Theorem 39.7 (Tangent Line Theorem) A line l is tangent to a circleΓ if and only if it is perpendicular to the diameter at the point of tangency.Proof. (⇒) Let Γ = C(O, r). Suppose that l is tangent to Γ at P . We needto show that l ⊥ OP .Drop a perpendicular line from O to l and call the foot Q. Suppose thatP ≠ Q (RAA Hypothesis).Then we can chose a point R ∈ l such that P ∗ Q ∗ R and P Q = QR..Triangles △OP Q and △ORQ share a common side (OQ); and since OQ ⊥ l(by hypothesis), ∠OQR = ∠OQP . Since R was constructed such thatP Q = QR, then two triangles are congruent by SAS.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

204 SECTION 39. CIRCLESFigure 39.3: The tangent line theorem.By congruency, corresponding sides are congruent, hence OP = OR. Thismeans that R lies on Γ by definition of a circle.Since P ≠ R, this contradicts the fact that P is the only point that lies onboth l and Γ (since l is tangent to Γ at P ). Hence the RAA hypothesis iswrong and P = Q.Since OQ ⊥ l, then OP ⊥ l.(Proof of ⇐) Assume that l intersects Γ at P and that OP ⊥ l.Choose any point Q ∈ l that is distinct from P .Since OP ⊥ l at P , the shortest segment from O to l is OP . HenceOQ > OPBy definition of outside, Q is outside of Γ. Since Q was chosen arbitrarily,every point on l that is distinct from P is outside of Γ.Hence Γ ∩ l contains precisely one point P , which means l is tangent to Γat P .Corollary 39.8 Let Γ be a circle and l a line tangent to Γ at P . Thenevery point on l that is distinct from P is outside of l.Proof. This was proven by the next to the last line of the proof of theprevious theorem.Theorem 39.9 (Secant Line Theorem) Let l be a secant line that intersectscircle Γ = C(O, r) at distinct points P and Q. Then O lies on theperpendicular bisector of chord P Q.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 39. CIRCLES 205Proof. Since P, Q ∈ Γ, OP = OQ = r. By the point-wise characterizationof perpendicular bisectors (theorem 28.2), O is on the perpendicularbisector of P Q.Figure 39.4: Proof of the Secant Line Theorem. Since points P and Q areequidistant from O, O must lie on their perpendicular bisector.Theorem 39.10 Let l be a secant line that intersects circle Γ = C(O, r)at distinct points P and Q. Then every point on the interior of segmentP Q is inside Γ, and every point on ←→ P Q − P Q lies outside of Γ.Proof. (Exercise)Theorem 39.11 (Weak Circular Continuity a ) Let Γ = C(O, r) be acircle and let l be a line with points A, B ∈ l such that A is inside Γ andB is outside Γ. Then l is a secant line of Γ.a We call this result weak circular continuity to distinguish it from the the similar result(theorem 39.17) that deals with the intersection of two circles, which we will call strongcircular continuity.Proof. Since A ∈ l is inside Γ, then l cannot be a tangent line (corollary39.8). Hence the number of intersections is either 0 or 2, but not 1 (theorem39.6). If we can show that any intersection exists, then since a singleintersection is not possible, we know that there must be two intersections.Let d = AB.By the ruler postulate, ∀x ∈ [0, d], ∃ a unique point D(x) ∈ AB such thatAD(x) = x.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

206 SECTION 39. CIRCLESFigure 39.5: Weak circular continuity (theorem 39.11): If A is inside of Γand B is outside of Γ, then ←→ AB must intersect Γ.Define f : [0, d] ↦→ [O, ∞) by f(x) = OD(x).By theorem 28.3 (Continuity of distance), f(x) is a continuous function.Since A is inside Γ, OA < r. HenceSince B is outside Γ, OB > r. Hencef(0) = OA < rf(d) = OB > rBy the intermediate value theorem there is some number u ∈ (0, d) suchthatf(u) = rSince the mapping between AB and [0, d] is one-to-one, there is some pointQ in AB such that OQ = r. Since f(u) = r, this means OQ = r, which inturn means Q is on Γ.Since 0 < r < d, then A ∗ Q ∗ B. Hence Q is in AB, and therefore ABintersects Γ.Corollary 39.12 If Γ is a circle and l is a line that contains a point A thatis inside Γ, then l is a secant line of Γ.Theorem 39.13 (Tangent Circles Theorem) If circles Γ = C(O 1 , r 1 )and Λ = C(O 2 , r 2 ) are tangent at P then points O 1 , O 2 , P are distinct andcollinear, and the two circles share a common tangent line at P .Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 39. CIRCLES 207Theorem 39.14 Let △OCQ be a right triangle with right angle at C. IfP is a point such that O ∗ P ∗ Q and OP = OC, then CP < CQ.Proof. (Exercise; see figure 39.6.)Theorem 39.15 Let C be the point of tangency of line l to Γ = C(O, r).Pick any point B ∈ l and let β = µ(∠BOC). Define P (α) : (0, β) ↦→ Γsuch that µ(∠P (α)OC) = α. Thenlim CP (α) = 0α→0 +Figure 39.6: See theorem 39.15.Proof. See figure 39.6.For any x, by the crossbar theorem ←−−−→ OP (x) intersects BC at some pointQ(x).By theorem 39.14, 0 < CP (x) < CQ(x)By the continuity axiom (theorem 20.2), the function mapping the distanceCQ(x) to α is continuous. Hencelim CQ(α) = 0α→0 +The result of the theorem follows by the Sandwich theorem for limits fromCalculus.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

208 SECTION 39. CIRCLESTheorem 39.16 Let Γ = C(A, r) be a circle. Choose any point B ∈ Γ andany point C not in Γ. Define the functionbyf(α) : (0, 180) ↦→ [0, ∞)f(α) = CD(x)where D(x) ∈ H C,←→ ABis chosen so that µ(∠DAB) = α (see figure 39.7).Then f(x) is continuous.Figure 39.7: The function f(α) : (0, 180) ↦→ [0, ∞) is continuous.Proof. Let α ∈ (0, 180).By the triangle inequality,HenceCD(x) < CD(a) + D(a)D(x)CD(a) < CD(x) + D(a)D(x)By the definition of f(x),Taking limits,CD(a) − D(a)D(x) < CD(x)< CD(a) + D(a)D(x)f(a) − D(a)D(x) < f(x) < f(a) + D(a)D(x)lim (f(a) − D(a)D(x)) < lim f(x)x→α x→α< lim (f(a) + D(a)D(x))x→α« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 39. CIRCLES 209Since D(a)D(x) → 0 as x → a, we getlim f(x) = f(α)x→αTheorem 39.17 (Strong Circular Continuity) Let Γ = C(O, r) andΓ ′ = C(O ′ , r ′ ) be circles, let A be a point on Γ that is outside Γ ′ , and let Bbe a point on Γ that is inside Γ ′ . Then Γ ∩ Γ ′ contains precisely two points.Proof. Assume that A and B are not antipodal. If they are, by the continuityof f described in the previous theorem we can replace A with anotherpoint A ′ close to A ′ so that A is still outside Γ ′ .Chose a point C ∈ Γ such that A, B are on the same side of ←→ OC. Use thispoint to define a function f(x) like the one in theorem 39.16:f(x) : [0, µ(∠BOC)) ↦→ [0, ∞)byf(α) = O ′ Q(α)where Q(α) is the point on Γ such that α = µ(∠COQ(α)) (see figure 39.8).There exist numbers a, b ∈ (0, 180) such that Q(a) = A and Q(b) = B. ByFigure 39.8: Strong circular continuity.the previous theorem, f is continuous, with f(a) > r ′ and f(b) < r ′ .Hence by the intermediate value theorem there is a number x such thatf(x) = r ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

210 SECTION 39. CIRCLESLet D be the point on Γ corresponding to x. Since f(x) = r ′ we haveD ∈ Γ ′ , hence Γ and Γ ′ intersect at D.This proves that there is at least one point in Γ ∩ Γ ′ .Figure 39.9: If Γ and Γ ′ share a common tangent line then they are eitheron the same side of the line (top) or the opposite sides (bottom).Suppose that D is the only point in Γ ∩ Γ ′ (RAA hypothesis).Then the two circles share a common tangent line l (figure 39.9).Either they are on the same side of l or on different sides of l.If the two circles are on opposite sides of the l then every point on Γ (exceptfor D) is outside of Γ ′ , contradicting the fact that B is inside Γ ′ .Hence the two circles cannot be on opposite sides of l. Then every pointof one circle (except for D) is inside the other circle. This contradicts thefact that A is outside Γ ′ and B is inside Γ ′ .Since the two circles cannot either be on the same side or the oppositesides of the tangent line, the tangent line cannot exist. This means we« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 39. CIRCLES 211must reject the RAA hypothesis, and conclude that there is at least oneother point besides D in Γ ∩ Γ ′ .Figure 39.10: Proof that there must be precisely two points in the intersection.Since there are at least two points in the intersection, define R ≠ D as asecond point in the intersection (see figure 39.10).Suppose that there is a third point S ≠ D.Since OR = OD, O is on the perpendicular bisector of DR.Since O ′ R = O ′ D, O ′ is on the perpendicular bisector of DR.Hence ←−→ OO ′ is the perpendicualr bisector of DR.By a similar argument OS = OD and O ′ S = O ′ D hence ←−→ O ′ O is the perpendicularbisector of DS.Since there is only perpendicular line to ←−→ O ′ O through D, this means that←→DS = ←→ DR.Let F be the foot of the perpendicular from D to ←−→ OO ′ . ThenD ∗ F ∗ RD ∗ F ∗ Shence R and S are on the same side of ←−→ OO ′ .FurthermoreDF = DR = DSby definition of perpendicular bisector.Hence R = S by the ruler postulate. This means that there is no thirdpoint in Γ ∩ Γ ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

212 SECTION 39. CIRCLES« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 40Circles and TrianglesDefinition 40.1 A circle that contains all three vertices of a triangle issaid to circumscribe the triangle. The circle is called the circumcircle,and its center the circumcenter of the triangle.Theorem 40.2 (Circumscribed Triangle Theorem) A triangle can becircumscribed if and only if the perpendicular bisectors of the sides of thetriangle are concurrent. If a triangle can be circumscribed, then the circumcenterand the circumcircle are unique.Proof. Let △ABC be a triangle with perpendicular bisectors l, m, and nfor segments AB, AC, and BC(⇒) We have already shown in theorem 36.2 that l, m, and n meet a singlepoint O that is equidistant from the vertices. Hence circle C(O, r) wherer = AO circumscribes the triangle.(⇐) Suppose △ABC is circumscribed by some circle Γ = C(O, r).Since Γ contains the vertices A, B, and C, we know that they are equidistantfrom O, i.e,r = AO = BO = COSince AO = BO, we know that O lies on the perpendicular bisector of AB(pointwise characterization of perpendicular bisectors, theorem 28.1).Since CO = BO, we know that O lies on the perpendicular bisector of BC(pointwise characterization of perpendicular bisectors, theorem 28.1).Since AO = CO, we know that O lies on the perpendicular bisector of AC(pointwise characterization of perpendicular bisectors, theorem 28.1).213

214 SECTION 40. CIRCLES AND TRIANGLESHence the three perpendicular bisectors are concurrent at point O.(Uniqueness) Suppose that Γ circumscribes △ABC. By the argument in(⇐), the center of the circle occurs at the point of concurrence of the threeedge bisectors. since this point of concurrence is unique, any other circlethat circumscribes the triangle must also have a center at this same point,and be of the same radius. Hence the circumcircle is unique.Theorem 40.3 The Euclidean Parallel Postulate is equivalent ot the assertionthat every triangle can be circumscribed.This theorem says two things: (1) If the Euclidean Parallel Postulate holdsthen every triangle can be circumscribed; and (2) if the Euclidean ParallelPostulate fails, then there exists a triangle that cannot be circumscribed.We will restate and prove these statements as separate theorems 40.4 and40.6.Figure 40.1: Illustration of the proof of the Circumscribed Triangle Theorem.Theorem 40.4 If the Euclidean Parallel Postulate holds then every trianglecan be circumsribed. aa This is Euclid Book 4 Proposition 5.Proof. Assume the Euclidean Parallel Postulate and let △ABC be a triangle.Let m = ←→ AB and let k be the perpendicular bisector of AB.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 40. CIRCLES AND TRIANGLES 215let n = ←→ AC and let l be the perpendicular bisector of ACIf k ‖ l the either m = n or m ‖ n. Since the lines through any two sidesof a triangle are neither equal nor parallel we conclude that k ̸‖ l.Hence k and l intersect. Call this point O.Let r = OA. By the point-wise characterization of perpendicular bisectors,since O ∈ k, it is equidistant from A and B, and since O ∈ l, it is equidistantfrom A and C.Hence OA = OB = OC. Therefore the circle C(O, r) passes through allthree vertices.Corollary 40.5 If the Euclidean Parallel Postulate holds then all threeperpendicular bisectors of the sides of any triangle are concurrent and meetat the circumcenter of the triangle.Theorem 40.6 If the Euclidean Parallel Postulate fails then there exists atriangle that cannot be circumscribed.Figure 40.2: The triangle △CDE cannot be circumbscribed if the EuclideanParallel Postulate fails, because the lines l and m are parallel; if the EuclideanParallel Postulate held, they would intersect at the center of thecircumscribing circle.Proof. See figure 40.2.Assume that the Euclidean Parallel Postulate fails.This is equivalaent to assuming that the hyperbolic parallel postulate holds.Let □ABCD be any Saccheri quadrilateral with base AB and summit CD,and define M and N as the midpoints of AB and CD.By a result from Hyperbolic Geometry (that we have not yet proven),MN < AD.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

216 SECTION 40. CIRCLES AND TRIANGLESChoose point E ∈ −−→ MN such that ME = AD = BC.Then by construction □AMED and □BMEC are Saccheri quadrilaterals.Let l be the line through the midpoints of AM and DE.Let m be the line through the midpoints of BM and CE.Then by theorem 31.17 l ⊥ DE and m ⊥ CE.By the alternate interior angles theorem (theorem 29.4), l ‖ m.Hence l and m do not intersect. By theorem 40.2 (the circumscribed triangletheorem), triangle △CDE does not have a circumscribed circle.Definition 40.7 A circle is said to be inscribed in a triangle if each of theedges of the triangle is tangent to the circle. The center of the inscribedcircle is called the incenter of the triangle.Theorem 40.8 (Inscribed Circle Theorem a ) Every triangle has aunique inscribed circle. The bisectors fo the interior angles of any triangleare concurrent and the point of concurrency is the incenter of the triangle.a Euclid Book 4 Proposition 4.Proof. (Proof of existence.) By the crossbar theorem, the angle bisector of∠BAC intersects side BC at some point D (figure fig 40.3).By the crossbar theorem, the angle bisector of ∠ABD must intersect segmentAD at some point E.Since A ∗ E ∗ D, E is in the interior of angle ∠ACD.Let F , G, and H be the feet of the perpendiculars dropped from E to eachof the three sides of the triangle (bottom of figure 40.3).Since the interior angle of any triangle is less than 180, we have∠EAB = 1 180∠CAB

SECTION 40. CIRCLES AND TRIANGLES 217By a similar argument, C ∗ G ∗ B.By AASHence△AEH ∼ = △AEF△BEG ∼ = △BEFEH = EF = EGLet r = EH; then points F , G, and H are equidistant from E and lie on acircle ΓC(E, r).Figure 40.3: Proof of the Inscribed Circle TheoremFurthermore, the sides of the triangle are tangent to the circle because thethree radii are perpendicular to the corresponding sides. Hence Γ is aninscribed circle.Definition 40.9 Let P 1 , P 2 , ...P n , n ≥ 3, be points such that no three ofpoints are collinear and that the segments P i P i+1 (including P n P 1 ) shareat most an endpoint. Then a polygon is the union of the segmentsP 1 P 2 ∪ P 2 P 3 ∪ · · · ∪ P n−1 P n ∪ P n P 1The points P i are called the vertices of the polygon and the segments arecalled the edges.Definition 40.10 A polygon is called a regular polygon if all of its sidesare congruent and all of its angles are congruent.Definition 40.11 A polygon is said to be inscribed in a circle if all of itsvertices lie on the circle.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

218 SECTION 40. CIRCLES AND TRIANGLESTheorem 40.12 Let Γ be circle and P a point on Γ. Then for each n ≥ 3there is a regular polygon inscribed in Γ with a vertex at P .Proof. Let P 1 be defined on Γ such that m(∠P OP 1 ) = 360/n.Continue to construct points P 2 , P 3 , ...P n−1 such that m(∠P i OP i+1 ) =360/n.Then all of the triangles P i OP i+1 are congruent to one another.Hence all of the sides P i P i+1 are congruent to one another.Since each interior angle of P P 1 P 2 ...P n−1 P satisfiesm(∠P i−1 P i P i+1 ) = m(∠P i−1 P i O) + m(∠P i OP i+1 )= 180 − 360nthen each of the interior angles are congruent. Hence the polygon P P 1 P 2 ...P n−1 Pis regular.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 41Euclidean CirclesIn this section we assume that the Euclidean Parallel Postulate holds forall theorems.Theorem 41.1 Let △ABC be a triangle and let M be the midpoint ofAB. If AM = MC then ∠ACB is a right angle.Proof. See figure 41.1. Since AM = MC, by the isoceles triangle theoremα = β.Since BM = MC, them by the isosceles triangle theorem δ = γ.Henceσ(△ABC) = α + (β + γ) + δ= 2β + 2γSince the Euclidean Parallel Postulate holds, σ(△ABC) = 180, henceβ + γ = 90Corollary 41.2 If the vertices of a triangle △ABC lie on a circle and ABis a diameter of the circle then ∠ACB is a right angle. aa Euclid book 3 proposition 31: An angle inscribed in a semicircle is a right angle.Proof. This follows immediately from the previous theorem because themidpoint of AB is the center of the circle, hence AO = BO = CO.219

220 SECTION 41. EUCLIDEAN CIRCLESFigure 41.1: Proof of theorem 41.1.Theorem 41.3 Let △ABC be a triangle and let M be the midpoint ofAB. If ∠ACB is a right angle, then AM = MC.Proof. Construct a point D on −−→ MC such that MD = MA. By theorem41.1, ∠ADB = 90.Suppose that C ≠ D. Then either M ∗ D ∗ C or M ∗ C ∗ D. Suppose thatM ∗ D ∗ C.Then (see figure 41.2), γ is an exterior angle of △ADC, hence γ > ɛ(exterior angles theorem).Similarly, β is an exterior angle of △BDC, hence β > δ (also by the exteriorangles theorem).Hence,which is a contradiction (90 ¿90).A similar result holds if M ∗ C ∗ D.90 = ∠BDA = β + γ > δ + ɛ = 90Hence C = D and therefore MC = MD = MA.Corollary 41.4 If ∠ACB is a right angle, then AB is a diameter of thecircle that circumscribes △ABC.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 41. EUCLIDEAN CIRCLES 221Figure 41.2: Proof of theorem 41.3.Theorem 41.5 (30-60-90 Theorem) If the interior angles of triangle△ABC measure 30, 60, and 90, then the length of the side opposite the 30angle is one half the length of the hypotenuse.Proof. (Exercise.)Theorem 41.6 (Converse to 30-60-90 Theorem) If △ABC is a righttriangle such that the length of one leg is one-half the length of the hypotenusethen the interior angles of the triangle measure 30, 60, and 90.Proof. (Exercise.)Definition 41.7 Let Γ be a circle. An angle ∠P QR is said to be inscribedin Γ if P, Q, R ∈ Γ. The arc intercepted by ∠P QR is the set of all pointsin Γ in the interior of ∠P QR.Definition 41.8 Let Γ = C(O, r) be a circle. An angle ∠P OR (where Ois the center of Γ), is called a central angle.Definition 41.9 Let ∠P QR be inscribed in Γ = C(O, r) such that Q, R lieon opposite sides of ←→ OP or O, P lie on opposite sides of ←→ QR. Then ∠P ORis called the corresponding central angle.Theorem 41.10 (Central Angle Theorem) The measure of an inscribedangle for a circle is one half the measure of the corresponding central angle.Proof. Let ∠P QR be inscribed in Γ = C(O, r). We will consider threecases: (1) O lies on ∠P QR; (2) O lies in the interior of ∠P QR; (3) O liesin the exterior of ∠P QR.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

222 SECTION 41. EUCLIDEAN CIRCLESFigure 41.3: An inscribed angle (α) and its corresponding central angle (β).(Case 1) O lies on ∠P QR. We can assume it lies on Segment QR; otherwiserelable the points. Then Q ∗ O ∗ R (figure 41.4).Let α = µ(∠OQP ) and β = µ(∠ORP ).By theorem 41.1 ∠RP Q = 90. By the Euclidean Parallel Postulate, α +β + 90 = 180; henceα + β = 90By the isosceles triangle theorem on △P OR, β = δ. Henceγ = 180 − β − δ= 180 − 2β= 180 − 2(90 − α)= 2α(Case 2) Suppose that O lies in the interior of ∠P QR.Let S ∈ Γ be chosen so that S ∗ O ∗ Q. Then (see figure 41.5).Applying case 1 to ∠P QS givesApplying case 1 to ∠RQS givesα = β + δγ = ɛ + ζɛ = 2βζ = 2δ« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 41. EUCLIDEAN CIRCLES 223Figure 41.4: Proof of Central Angle theorem when O lies on an edge of theinscribed angle.Henceγ = ɛ + ζ = 2β + 2δ = 2α(Case 3) O is neither on ∠P QR nor in its interior.Let S ∈ Γ such that Q ∗ O ∗ S.Either R is in the interior of ∠P QS or P is in the interior of angle ∠RQS.We can assume the first case (R ∈ interior(∠P QS)); if the second case istrue, exchange the labels of R and P .Then (see figure 41.6)α = β − δBy Case 1Henceɛ = 2δγ + ɛ = 2βγ + 2δ = 2βγ = 2(β − δ) = 2αCorollary 41.11 (Inscribed Angle Theorem) If two inscribed anglesintercept the same arc then they are congruent.Proof. This follows form the central angle theorem because if two anglesinscribe the same arc then they share the same corresponding central angle.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

224 SECTION 41. EUCLIDEAN CIRCLESFigure 41.5: Proof of the Central Angle Theorem when the center of thecircle is in the interior of the inscribed angle.Definition 41.12 Let B be a circle and let O be a point that does not lieon B. The power of O with respect to B is defined as follows: Choose anyline l through O that intersects B. If l is a secant line that intersects B attwo points P, Q, then the power is the product (OP )(OQ). If l is tangentto B at P then the power of O is defined as (OP ) 2 .Theorem 41.13 The power of a point is well defined, i.e, the same valueis obtained regardless of which line is used, so long as at least one point onthe line intersects the circle. aa Euclid book 3, Proposition 36.Proof. (Case 1) Suppose that O is center of Γ(A, r). Then any line throughO intersects Γ at two points P and Q such that OP = OQ = r. Hence(OP )(OQ) = r 2 regardless of which line is chosen through O.(Case 2) O is inside Γ and is not the center of Γ = C(A, r).Let l be any line through O. Then it intersects Γ in two poins Q and R.Let S and T be the points where line ←→ AO intersects Γ.Then α = β since both inscribed angles intercept the same arc (correspondingto central angle γ).Similarly, η = θ because both inscribed angles intercept the same arc (cor-« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 41. EUCLIDEAN CIRCLES 225Figure 41.6: Proof of the central angle theorem when the center of the circleis exterior to the inscribed angle.responding to central angle ζ).Also, ɛ = δ because they are vertical angles.Hence (AAA) △OQT ∼ △OSR.By the similar triangle theoremOROS = OTOQ(OR)(OQ) = (OS)(OT )Since the right-hand side (OS)(OT ) does not depend upon which line l ischosen, so long as it passes through O, then the left hand side, which givesthe power of l, is the same for all lines l that pass through O.(Case 3) O is any point ouside Γ and l is a secant line that intersects Γ atpoints Q and R.Construct line ←→ OA. Since one pont is inside Γ and one point is outside Γ,it intersects Γ at two points, which we denote by S and T .Then α = β (see figure 41.8) because they both intercept the same arc.Similarly δ = γ = 90 because the both interscept the same arc (and adiameter).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

226 SECTION 41. EUCLIDEAN CIRCLESFigure 41.7: Proof that the power is well defined when O is any point insideof Γ besides the center of the circle.Figure 41.8: Proof that the power is well defined when O is any pointoutside of Γ and l is a secant line of Γ.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 41. EUCLIDEAN CIRCLES 227Figure 41.9: Proof that the power is well defined when O is any pointoutside of Γ and l is a tangent line of Γ.Since triangles △ROS and △T OS share a common angle their remainingangles are also equal:δ = 180 − (β + γ)= 180 − (α + γ)= ɛThus △ORS ∼ △OT Q. By the similar triangles theorem, as in case 2,OROS = OT ⇒ (OR)(OQ) = (OS)(OT )OQSince (OS)(OT ) depends only the point O and not on the line l, we concludethat (OR)(OQ) is the same for all lines that pass through O and intersectΓ at two points.(Case 4) Let O be outside Γ and l a line through O that is tangent to Γ atP .Then OP ⊥ AP , hence △OP A is a right triangle. Let S = Γ ∩ AO.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

228 SECTION 41. EUCLIDEAN CIRCLESBy the Pythagorean theorem(OP ) 2 = (OA) 2 − (AP ) 2= (OA − AP )(OA + AP )= (OA − AS)(OA + AS)because AS = r = AP= OS(OA + AT )because AT = r = AS= (OS)(OT )Thus by case 3 the power is well defined.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 42Area and Circumferenceof Circles in EuclideanGeometryDefinition 42.1 The perimeter L of polygon P 1 P 2 · · · P n isP = P 1 P 2 + P 2 P 3 + · · · + P n−1 P n + P n P 1We can inscribe a regular hexagon within within circle as follows:1. Choose any point on P 1 ∈ Γ = C(O, r).2. Define a ray r = −−→ OP 1 .3. On one side of r we can define a points P 2 ∈ Γ such thath µ(∠P 1 OP 2 ) =60.4. On the opposite side of −−→ OP 2 from P 1 determine a a point P 3 on Γsuch that ∠P 1 OP 2 = 60.5. Continue to determine points P 4 , P 5 , P 6 in the same manner, so thatP 3 OP 4 = P 4 0P 5 = P 5 OP 6 = 60.6. Then P 1 P 2 P 3 P 4 P 5 P 6 is a regular hexagon.Each △P i OP i+1 (including P 6 OP 1 ) constructed in this manner is a congruentequilateral triangle.229

230 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESFigure 42.1: Construction of inscribed regular hexagon H 1 = P 1 P 2 · · · P 6 .We will define L 1 to be the perimeter of H 1 :L 1 = P 6 P 1 +5∑P i P i+1We will then construct a sequence of inscribed regular polygons H 2 , H 3 , . . .as follows:i=1a. Each vertex of H i is a vertex of H i+1 ;b. For k = 1, .., i − 1, choose point Q k ∈ Γ to be the intersection of theangle bisector of ∠P k OP k+1 and Γ;c. Choose point Q i to be the intersection of the angle bisector of ∠P n OP 1 .and Γ;d. Define H i+1 = P 1 Q 1 P 2 Q 2 · · · P n Q nEach H i+1 has twice as many sides as the H i ; hence H k has 3 × 2 k sides.H 1 has 6 sides; H 2 has 12 sides; H 3 has 24 sides; etc.Define L n as the perimeter of the 3 × 2 n -gon defined in this manner.Theorem 42.2 The sequence of perimeters L 1 , L 2 , L 3 , ... is strictly increasing.Proof. (Exercise.)Theorem 42.3 The perimeter of a circumscribed square exceeds theperimeter of each of the 3 × 2 n -gons defined above.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLES 231Figure 42.2: Construction of inscribed regular 12-gon from regular hexagonby bisecting each central angle.Theorem 42.4 The sequence of perimeters L 1 , L 2 , ... converges to somenumber C.Proof. The sequence is bounded above and increasing. From a theorem ofcalculus, the sequence converges.Definition 42.5 Let Γ = C(O, r). The circumference of Γ is defined tobeC = limn→∞ L nTheorem 42.6 The value of the circumference only depends on the radiusand not the center point.Proof. Let Γ = C(O, r) and Γ ′ = C(O ′ , r).We can construct the perimeter of each sequence of inscribed 3 × 2 n -gonsH n and H n.′ Applying SAS to each sector, the corresponding triangles△P k OP k+1∼ = △P′kO ′ Pk+1′Since each triangle is congruent, the corresponding edges have equal length.Hence the L n = L ′ n for each n. Hence each sequence must converge to thesame limit.Theorem 42.7 Let r, r ′ > 0 be any positive numbers; and let C and C ′be the circumferences of concentric circles Γ = C(O, r) and Γ ′ = C(O, r ′ ).Then C/r = C ′ /r ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

232 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESProof. Within each circle construct the sequence of 3×2 n -gons as describedabove. Then each triangle △P k OP k+1 ∼ △P ′ k OP ′ k+1By the similar triangles theoremfor each k and for each H n .Since OP k = r and OP ′ k = r′ ,Hence for each n,P k P k+1OP k= P ′ k P ′ k+1OP ′ kP k P k+1rL nr= P ′ k P ′ k+1r ′=L′ nr ′Since L n → C and L ′ n → C ′ as n → ∞, the result follows.Definition 42.8 π = C/2r.Definition 42.9 The interior of circle Γ = C(O, r) is the set of all pointsinside Γ. The circular region of Γ is the union of Γ with Int(Γ).Recall that we denote the area of polygon H n by α(H n ). Then we have thefollowing.Theorem 42.10 Let H 1 , H 2 , ... be the sequence of inscribed polygons asdefined before. Then the areas of the polygonal regions forms an increasingsequence:α(H n ) < α(H n+1 )Proof. Each polygon H n can be dissected into a union of non-overlappingtrianglesn⋃H n = (△P (n)kOP (n)k+1 )k=1where we define P (n)n+1 = P (n)1 , andH n+1 =n⋃k=1Ä(n) △PkOQ (n)kwhere the Q k are as defined previously.∪ △Q (n)k(n)OPk+1ä« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLES 233Sinceα(△P (n)kOQ (n)k) + α(△Q(n) k> α(△P (n)kOP (n)k+1 )(n)OPk+1 )(see figure 42.2) the sequence of areas is increasing.Since the sequence of areas is increasing an bounded (say, for example, bythe area of a circumscribed square), the sequence converges to a limit A.Definition 42.11 (Area of a Circle) Let Γ = C(O, r) be a circle. Thenα(Γ) = limn→∞ α(H n)Theorem 42.12 (Archimedes’ Theorem) Let Γ = C(O, r) be a circlewith circumference C. ThenA = 1 2 rCProof. Construct the sequence of inscribed hexagons as before, and definethe sequence of areas as A n .Since each H n is composed of n isosceles triangles of side length r andcentral angle 360/n, the area of each triangle is a n = b n h n /2 where b n isthe base of the triangle and h n is its height. HenceA n = n 2 b nh n = b nL n2where L n is the perimeter of H n . HenceA = limn→∞ A n= 1 2 lim b nL nn→∞= 1 ( ) ( )lim2b n lim L nn→∞ n→∞= 1 2 rCCorollary 42.13 Let Γ = C(O, r). Thenα(Γ) = πr 2Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

234 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLESProof. This follows by substituting the C = 2πr from the definition of π inthe Archimedes’ formula,A = 1 2 rC = 1 r(2πr) = πr22« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLES 235Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

236 SECTION 42. AREA AND CIRCUMFERENCE OF CIRCLES« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 43Indiana Bill 246Indiana Bill #246 of 1897 would have effectively set π = 3.2. 1 .A Bill for an act introducing a new mathematical truth and offeredas a contribution to education to be used only by the State of Indianafree of cost by paying any royalties whatever on the same, providedit is accepted and adopted by the official action of the Legislature of1897.Section 1 Be it enacted by the General Assembly of the State of Indiana:It has been found that a circular area is to the square on a lineequal to the quadrant of the circumference, as the area of an equilateralrectangle is to the square on one side. The diameter employedas the linear unit according to the present rule in computing the circle’sarea is entirely wrong, as it represents the circle’s area one andone-fifth times the area of a square whose perimeter is equal to thecircumference of the circle. This is because one fifth of the diameterfails to be represented four times in the circle’s circumference. Forexample: if we multiply the perimeter of a square by one-fourth ofany line one-fifth greater than one side, we can in like manner makethe square’s area to appear one-fifth greater than the fact, as is doneby taking the diameter for the linear unit instead of the quadrant ofthe circle’s circumference.Section 2 It is impossible to compute the area of a circle on thediameter as the linear unit without trespassing upon the area outsideof the circle to the extent of including one-fifth more area than iscontained within the circle’s circumference, because the square on1 see http://www.agecon.purdue.edu/crd/localgov/second%20level%20pages/Indiana_Pi_Story.htm237

238 SECTION 43. INDIANA BILL 246the diameter produces the side of a square which equals nine whenthe arc of ninety degrees equals eight. By taking the quadrant of thecircle’s circumference for the linear unit, we fulfill the requirementsof both quadrature and rectification of the circle’s circumference.Furthermore, it has revealed the ratio of the chord and arc of ninetydegrees, which is as seven to eight, and also the ratio of the diagonaland one side of a square which is as ten to seven, disclosing the fourthimportant fact, that the ratio of the diameter and circumference isas five-fourths to four; and because of these facts and the further factthat the rule in present use fails to work both ways mathematically, itshould be discarded as wholly wanting and misleading in its practicalapplications.Section 3 In further proof of the value of the author’s proposed contributionto education and offered as a gift to the State of Indiana, isthe fact of his solutions of the trisection of the angle, duplication ofthe cube and quadrature of the circle having been already acceptedas contributions to science by the American Mathematical Monthly,the leading exponent of mathematical thought in this country. Andbe it remembered that these noted problems had been long sincegiven up by scientific bodies as insolvable mysteries and above man’sability to comprehend.Legislative HistoryHouse: Passed, 67-0, 5 Feb. 1897Senate: Postponed indefinitely, 12 Feb. 1897.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 44Estimating πInscribed PolygonsWhile there are many ways of calculating π using calculus (e.g., as the sumof an infinite series), here we outline a procedure based purely on geometriccalculations. It is iterative, in the sense that we need to calculate a sequenceof estimates based on previous estimates, slowly converging to the desiredvalue. We will define π n as the ratio of the perimeter to the diameter ofa succession of inscribed polygons with 2 n sides, starting with n = 2. Theidea is that as more sides are added the inscribed polygon looks more andmore like a circle (rigorously, the limit of the perimeters is the circumferenceof the circle). In general, the perimeter of the n th polygon isP n = 2 n H nwhere H n is the length of the outer edge. Since the diameter of the unitcircle is always 2, we have an estimate for π ofπ n = 1 2 P n = 2 n−1 H nFor n = 2, we inscribe a square in a circle of radius 1, and divide the squareinto four right triangles with hypotenuse H 2 .ThenH 2 = √ 2π 2 = 2 2−1 H 2 = 2 √ 2 ≈ 2.82843For n = 3 we divide each central angle in half. Define the distance a asshown in figure 44.2.239

240 SECTION 44. ESTIMATING πFigure 44.1: A square inscribed in a unit circle.Triangle △ABC can be decomposed into two right triangles.Pythagorean theorem we haveFrom the(1 − a) 2 +Solving for a,ÅH22a = 1 − 1 −Also from the Pythagorean theorem,H3 2 = a 2 +Substituting the expression for a,[H3 2 = 1 − 1 −= 1 − 2 1 −+ 1 −ÅH2= 2 − 2 1 −ÅH2ã 2= 1Å ã 2 H22Å ã 2 H222Å ã 2 H222ã 2+Å ã 2 H22ã 2] 2+Å ã 2 H22Å ã 2 H22« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 44. ESTIMATING π 241Figure 44.2: One segment of an octagon inscribed in a unit circle is obtainedby spliting each of the triangles in figure 44.1 in half.Therefore (since H 2 = √ 2),By a similar argumentH 2 3 = 2 − 2Ã1 −Ç√ å 22= 2 − √ 22»H 3 = 2 − √ 2 ≈ 0.765367»π 3 = 2 2 H 3 = 4 2 − √ 2 ≈ 3.06147H 2 4 = 2 − 21 −Å ã 2 H32and by induction, we obtainH 2 n = 2 − 21 −Å ã 2 Hn−12so our n th estimate of π is π n = 2 n−1 H n .Our successive estimates of H n also giveRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

242 SECTION 44. ESTIMATING πso thatand so forth.H 2 = √ 2»H 3 = 2 − √ 2√ »H 4 = 2 − 2 + √ 2… √ »H 5 = 2 − 2 + 2 + √ 2… √ »H 6 = 2 − 2 + 2 + 2 + √ 2One could, for example, calculate the first i values of π i with the followingcode in Mathematica:f[x_]:= Sqrt[2*(1-Sqrt[1-(x/2)^2])];pi[i_] := NestList[f, Sqrt[2], i]*Table[2^n, {n, 1, i + 1}];To print the first 25 estimates to 20 significant figures, we enterN[pi[25],20]The resulting guesses would be{2.8284271247461900976, 3.0614674589207181738,3.1214451522580522856, 3.1365484905459392638,3.1403311569547529123, 3.1412772509327728681,3.1415138011443010763, 3.1415729403670913841,3.1415877252771597006, 3.1415914215111999740,3.1415923455701177423, 3.1415925765848726657,3.1415926343385629891, 3.1415926487769856695,3.1415926523865913458, 3.1415926532889927653,3.1415926535145931202, 3.1415926535709932089,3.1415926535850932311, 3.1415926535886182366,3.1415926535894994880, 3.1415926535897198008,3.1415926535897748791, 3.1415926535897886486,3.1415926535897920910, 3.1415926535897929516}The correct value of π to 20 figures is 3.1415926535897932385.Method of ArchimedesArchimedes estimated the value of π by assuming that the perimeter I n ofany inscribed regular n-sided polygon was less than the circumference ofa circle, and the perimeter C n of any regular-sided circumscribed polygonwas greater than the circumference of the same circle:I n < C < C n« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 44. ESTIMATING π 243As the number of sides of the corresponding polygons gets larger and larger,these two numbers approach one another.If we let Γ = C(O, r) be any circle with radius 1/2, then C = 2πr = π.HenceI n < π < C nArchimedes obtained the following recurrence relations: 1C 2n =2C nI nC n + I n(circumscribed polygon)I 2n = √ C 2n I n (inscribed polygon)where C 2n and I 2n are the circumferences of the 2n-sided regular circumbscribedand inscribed polygons.Archimedes obtained the bounds22371 < π < 227by setting n = 6·2 k for k = 0, 1, .., 4 (i.e., triangle, hexagon, 12-gon, 24-gon,48-gon, 96-gon). His limits were rational numbers rather than real numbersbecause he approximated all of his square roots with real numbers by somemethod that he did not describe. 2Taylor SeriesUsing the Taylor Series expansion for the arctangent,tan −1 x = x − 1 3 x3 + 1 5 x5 − 1 7 x7 + ...gives (using tan π/4 = 1),π = 4Å1 − 1 3 + 1 5 − 1 ã7 + ...(44.1)Unfortunately this converges very slowly - here are the results after up toa million iterations:1 If you are interested, you can find a derivation of these formulas using trigonometry athttp://mathworld.wolfram.com/ArchimedesRecurrenceFormula.html.2 A likely method that was known by the ancients to estimate square roots was therecursion relation x n+1 = (x n + a/x n)/2 whic converges to √ a as n → ∞. A perfectlygood starting guess is x 0 = a.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

244 SECTION 44. ESTIMATING πn π n10 3.2323158100 3.15149341,000 3.142591710,000 3.1416926100,000 3.14160271,000,000 3.1415937Clearly this method is not very practical: even after a million iterations itis only correct to the 5th decimal place!It turns out, however, that the Taylor series converges much more quicklyfor some values of x, particularly for 1/5 and 1/239. This is useful becauseof the following lemma, which is known as Machin’s formula. 3Lemma 44.1Proof. Use the trigonometric formulaThen using a = b,Å ã Å ãπ1 14 = 4 tan−1 − tan −15239tan −1 a + tan −1 b = tan −12 tan −1 a = tan −1 2a1 − a 2and using 2a/(1 − a 2 ) for both a and b,Setting a = 1/5,a + b1 − ab4 tan −1 a = tan −1 4a(1 − a2 )a 4 − 6a 2 + 14 tan −1 1 5 = 4(1/5)(1 − tan−1 (1/5)2 )(1/5) 4 − 6(1/5) 2 + 1−1 120= tan119−1 1 + 1/239= tan1 − 1/239= tan −1 1 + tan −1 1239The result of the lemma then follows because tan −1 1 = π/4.3 for John Machin(1680-1752) who used it to calculate the first 100 digits of pi in 1706.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 44. ESTIMATING π 245ThuswhereThis converges very quicklyÅ Å ã Å ãã1 1π = lim 4 4fn→∞ 5 , n − f239 , nf(x, n) =n∑k=0x 2k+1 (−1) k2k + 1nπ n0 3.18326359832635983261 3.14059702932606031432 3.14162102932503442503 3.14159177218217729504 3.14159268240439951725 3.14159265261530860816 3.14159265362355476207 3.14159265358860222878 3.14159265358983584759 3.141592653589791696910 3.1415926535897932947Another version of this method is to sum the terms in equation 44.1 usingan Euler transform. A version of this known as Van Wijngararden Transformation,was invented in the 1960’s, tells us that we do not have to sumthe whole series to get a better approximation. 4 .Theorem 44.2 (Euler Transform) Let a k be a converging sequence ofpositive numbers, such that the partial sumss k,0 =k∑(−1) n a nn=0coverge to some number S. Then the sequence s k,j+1 → S, wheres k,j+1 = s k,j + s k+1,j2Setting π k,0 = f(x, k) for all k, then computing the partial sums as π k,j+1 =(π k,j + π k+1,j )/2, we can compute approximations of π. Van Wijngaardendeveloped an efficient algorithm for computing the sum without having torecompute earlier values in the sequence of partial sums. Table 44.1 showsthe number of digits accuracy as indicated by log 10 |(π i,j − π)/π|.4 A. van Wijngaarden, in: Cursus: Wetenschappelijk Rekenen B, Process Analyse, StichtingMathematisch Centrum, (Amsterdam, 1965) pp. 51-60.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

246 SECTION 44. ESTIMATING πTable 44.1: Error in π k,j+1 using Van Wijngaarden’s method. The numbersgive the base-ten logarithm of the relative error.0 1 2 3 4 5 6 7 8 9 100 0.6 0.8 1. 1.1 1.2 1.3 1.3 1.4 1.5 1.5 1.51 1.2 1.6 1.9 2.1 2.3 2.4 2.6 2.7 2.8 2.8 2.92 1.7 2.2 2.6 2.9 3.1 3.3 3.5 3.7 3.8 3.9 4.03 2.2 2.8 3.2 3.6 3.9 4.1 4.3 4.5 4.7 4.9 5.04 2.6 3.3 3.8 4.2 4.5 4.8 5.1 5.3 5.5 5.7 5.95 3.0 3.8 4.3 4.8 5.1 5.5 5.8 6.0 6.3 6.5 6.76 3.4 4.2 4.8 5.3 5.7 6.1 6.4 6.7 7.0 7.2 7.57 3.8 4.6 5.3 5.8 6.3 6.7 7.0 7.3 7.6 7.9 8.28 4.2 5.0 5.7 6.3 6.8 7.2 7.6 8.0 8.3 8.6 8.89 4.5 5.4 6.2 6.8 7.3 7.7 8.2 8.5 8.9 9.2 9.510 4.9 5.8 6.6 7.2 7.8 8.3 8.7 9.1 9.5 9.8 10.1BBP FormulasThe BBP formula, discovered in 1995, allows the computation of the n thdigit of π without knowing any of the first n − 1 digits. 5 The formula is:π =∞∑n=0Å 48n + 1 − 28n + 4 −18n + 5 − 1 ã Å ã 1 n8n + 6 16The algorithm is implemented as follows. LetThenS n =∞∑k=0116 k (8k + j)π = 4S 1 − 2S 4 − S 5 − S 65 Bailey, D. H.; Borwein, P. B.; and Plouffe, S. ”On the Rapid Computation of VariousPolylogarithmic Constants.” Math. Comput. 66, 903-913, 1997, which is availableat http://crd.lbl.gov/~dhbailey/dhbpapers/digits.pdf. For details on the numericalimplementation on computing π, see http://crd.lbl.gov/~dhbailey/dhbpapers/bbp-alg.pdf« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 44. ESTIMATING π 247Define the notation{x} = x mod 1, the fractional part of x. Then{ n}∑{16 n 16 n−k ∞S p } =8k + p +∑ 16 n−k8k + pk=0k=n+1{ n{∑ 16 n−k , 8k + p }=8k + pk=0}∞∑ 16 n−k+8k + pDefines p (n) =k=n+1{ n∑ 16 n−k , 8k + p }k=0Then the n th hex-digit of π is given by8k + pd n = ⌊16 {4{s 1 (n)}2{s 4 (n)} − {s 5 (n)}−{s 6 (n)}}⌋where ⌊x⌋ denotes the greatest integer less than or equal to x. The followinggives a Mathematica implementation:s[p_, n_]:= Sum[Mod[16^(n-k), 8k+p]/(8k+p),{k,0,n}];r[p_, n_]:= Mod[s[p,n],1];hexit[n_]:= Floor[16.0*Mod[4r[1,n]-2r[4,n]-r[5,n]-r[6,n], 1]];Thenhexit/@Range[0,11]gives the first 12 hex figures to the right hand side of the radix point:{2, 4, 3, 15, 6, 10, 8, 8, 8, 5, 10, 3}which corresponds toπ − 3 ≈ .243F 6A88 85A3 16The correct value of in hex to 16 figures isπ ≈ 3.243F 6A88 85A3 08D3 16To convert the number back to decimal, divide each integer by its correspondingpower of 16.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

248 SECTION 44. ESTIMATING πnums = hexit/@Range[0,11];dens = (16.^#)&/@Range[1,12];Plus@@(nums/dens)which returns the valueπ − 3 ≈ 0.14159265358979312Buffon’s NeedleThis method calculates π experimentally via a Monte-Carlo simulation; itwas invented in 1777 by GL Leclerc de Buffon (1707-1788). The idea isto drop an object of length L on a surface that contains parallel lines adistance S units apart, where S > L (Buffon used needles). If the object isdropped n times and x times it falls crossing a line, thenπ ≈ 2nLxSThis approximation is technically only correct in the limint as n → ∞, andis derived from a calculation of the probability of hitting the line. 6 Theproblem with this method is that it takes a very long time to get veryfar; calculating every 3 digits accurately requires over a million throws,and the number of throws required increases exponentially with the numberof digits. One mischievous implementation describes this technique as“calculating pi by throwing frozen hot dogs.” 7Modern ComputationsMost modern computer languages have some approximate value implementedfor π. In most cases the value is a preset constant that is limited tomachine accuracy, typically to around 16 digits. To obtain more digits onemust implement some algorithm to actually calculate π. A few computerlanguages have such an algorithm built it.For example you can compute π to n digits in Mathematica with N[Pi,n]; for example, N[Pi,10^6] will compute the first one-million digits inapproximately 3.7 seconds (on a 2.4 GHz core 2 quad processor). Printingit will take a bit longer; assuming you can fit 80 digits per line and 50lines per page it will fill up 150 pages. Mathematica uses the Chudnuvsky6 See the Mathworld article at http://mathworld.wolfram.com/BuffonsNeedleProblem.html for a derivation.7 http://www.wikihow.com/Calculate-Pi-by-Throwing-Frozen-Hot-Dogs« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 44. ESTIMATING π 249algorithm,whereS =π = k √6 k3S∞∑(−1) n (6n)!(k 2 + nk 1 )n! 3 (3n)!(8k 4 k 5 ) nn=0and the constants are k 1 = 545140134, k 2 = 13591409, k 3 = 640320, k 4 =100100025, k 5 = 327843840 and k 6 = 53360. The Chudnuvskys used thisformula to calculate over a billion digits in the 1980’s. The original formulais actually due to Srinvasa Ramaujan.The first 100 billion digits of ı are posted at http://ja0hxv.calico.jp/pai/epivalue.html.An iterative based on results of Gauss and Legendre was developed byRichard Brent and Eugene Salamin in 1975 and forms the basis of thepopular superpi program that runs under Windows. The Salamin-BrentAlgorithm sets initial values a 0 = 1, b 0 = 1/ √ 2, t 0 = 1/4, and p 0 = 1, thenrepeats the following calculation until the desired accuracy is reached:a n+1 = a n + b n2b n+1 = √ a n b nt n+1 = t n − p n (a n − a n+1 ) 2p n+1 = 2p nπ n+1 = (a n + b n ) 24t nThis algorithm converges very quickly, with the number of correct digitsapproximately doubling with every iteration. Several versions of thisprogram have been posted on Yasumasa Kanada’s website (http://www.super-computing.org). has used supercomputers to calculate over a trilliondigits.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

250 SECTION 44. ESTIMATING π« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 45Euclidean ConstructionsThis section outlines the basic Euclidean constructions that use a straightedgeand a compass (or the straight-edge and circle tools in GeoGebra).Construction 45.1 (Copy a Line Segment) Given a line segment AB,and a ray −−→ CD, find a point E ∈ −−→ CD such that AB = CE.1. Place the ends of a the compass at points A and B to determine itslength AB. [ In GeoGebra, use the line segment tool to create asegment AB and observe its name (say it is called b).]2. With the compass fixed, place one end at C and draw a circular arcintersecting −−→ CD of radius AB. [ In GeoGebra use the “Circle withcenter and given radius,” then click on point C. When the pop-upmenu asks you for the value of the radius, type the letter b, which isthe name of segment AB.]3. Define E as the intersection of the new circle with −−→ CD. [ In GeoGebra,using the “Intersection of two objects” tool, click first anywhere onthe circle, and then anywhere on ray −−→ CD. Point E will appear.]4. Construct segment CE. [In Geogebra, use the “Segment between twopoints” and click first on C and then on E.]Construction 45.2 (Copy an Angle) Given an angle α = ∠BAC and aray −−→ DE, construct an angle ∠EDF that is congruent to α.1. Construct a circle d = C(A, AB).2. Use the compass to measure the distance AB.3. Construct a circle h = C(D, AB). This circle intersects −→ AE at Q.251

252 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.1: Detail of Raphael’s The School of Athens showing eitherEuclid or Archimedes doing a construction with a compass.4. Circle d intersects −→ AC at P .5. Use the compass to measure the distance BP6. construct a circle k = C(Q, AB).7. Circle k intersects circle h at some point F .8. Then ∠EBF ∼ = ∠BAC.Construction 45.3 (Angle Bisection) Given and angle ∠BAC, find aray −→ AD that bisects ∠BAC.1. Construct a circle c = C(A, r) of any radius centered at A.2. Let P = −→ AC ∩ c and Q = −→ AB ∩ c.3. Use the compass to measure distance P Q.4. Construct circles α = C(P, P Q) and β = C(Q, P Q).5. Let G = α ∩ β (the circles actually intersect at two points, use eitherpoint).Construction 45.4 (Perpendicular Line) Given a line l and a pointA ∈ l, construct a line through A that is perpendicular to l.1. For any radius r, construct circle c = C(A, r).2. Let D and E be the two points where c intersects l.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 45. EUCLIDEAN CONSTRUCTIONS 253Figure 45.2: Construction of a line segment on −−→ CD that is congruent tosegment AB.3. Let s > r be any number larger than r. Construct circles δ = C(D, s)and ɛ = C(E, s).4. δ and ɛ at two points; call them P and Q.5. Line ←→ P Q is perpendicular to l.Construction 45.5 (Drop a Perpendicular to a Line) Given a line land a point P ∉ l, construct a line through P that is perpendicular to l.1. Let A be any point on l.2. Construct a circle c = C(P, r) that passes through A.3. The circle c intersects l at a second point; call this point B.4. Construct the perpendicular bisector of ∠BP A. This line is perpendicularto l and passes through P , as required.Construction 45.6 (Perpendiculare Bisector) Given a line segmentAB, construct its perpendicular bisector.1. Draw circle C 1 = C(A, AB) and circle C 2 = C(B, AB).2. Circles C 1 and C 2 intersect at two points C and D.3. Construct line ←→ CD, which is the perpendicular bisector of AB.Construction 45.7 (Parallel to a Line) Given a line l, and a pointP ∉ l, construct a line through P that is parallel to l.1. Let A be any point on l.2. Construct the line m = ←→ AP . Let r = AP .3. Draw an arc of c 1 = C(A, r) that intersects l at some point B.4. Draw an arc of c 2 = C(P, r).Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

254 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.3: Construction 45.2. A copy of ∠EDF on ray −−→ DE is constructedsuch that ∠EDF ∼ = ∠BAC.5. Draw an arc of c 3 = C(B, r).6. Circles c 2 and c 3 intersect at two points: one of these points is A; callthe other point Q.7. Draw line n = ←→ P Q. Then n ‖ l.Construction 45.8 (Partition a Segment) Given a line segment AB anda positive integer n, divide the segment into n segments of equal length.1. Pick any point Q that is not on the line ←→ AB.2. Construct ray −→ AQ.3. Pick any point Q 1 ∈ −→ AQ.4. Define r = AQ 1 .5. Define points Q 2 , Q 3 , ..., Q n ∈ −→ AQ such that Q i Q i+1 = r.6. Construct line m n = ←−→ Q n B.7. Construct lines m 1 , m 2 , ..., m n−1 through points Q 1 , Q 2 , ..., Q n−1 thatare parallel to to m n .8. Define the points P 1 = m 1 ∩ −→ AB, P2 = m 2 ∩ −→ AB,..., Pn−1 = m n−1 ∩−→AB.9. Then the segments AP 1 , P 1 P 2 ,...,P n−2 P n−1 , P n−1 B are the desiredsegments of equal length.Construction 45.9 (Tangent to a Circle) Given a circle γ = C(O, r)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 45. EUCLIDEAN CONSTRUCTIONS 255Figure 45.4: Illustration of construction 45.3. Ray −→ AG bisects angle ∠BAC.Figure 45.5: Construction 45.4. Line ←→ QP passes through A and is perpendicularto l.and a point P that is outside γ, construct a line through P that is tangentto γ.1. Let γ = C(O, r) and let P be any point oustide γ.2. Draw the line segment OP .3. Find the midpoint M of OP .4. Draw the circle δ = C(M, OM).5. Let Q and R be the two points where the two circles intersect. Lines←→P R and ←→ P Q are both tangent to γ.Constructible NumbersGiven a segment of length 1, we can ask the question of which other segmentswe can construct. For example, we know that by duplicating theRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

256 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.6: Construction 45.5: dropping of a perpendicular to a line froma point not on the line. The bisector of ∠AP B is perpendicular to line l.segment n times we can construct a segment of any integer length n.Definition 45.10 A number x is said to be constructible if we can constructa segment of length x given a segment of length 1, with a straightedgeand compass.Are all numbers constructible? If not, which numbers are?1. Any integer length n can be constructed by duplicating the unit segmentn times.2. Given any segments of length p and q, where p > q, both integers, wecan find segments of length p + q, p − q, p ∗ q (construction ??), andany combination thereof.3. By finding midpoints repeatedly, we can find a segment that is 1/2 mtimes as long as any of these other segments.4. By adding these segments together, we can find segments of lengthk/2 m for k = 1, 2, ..., 2 m .5. We can find segments of any rationl length p/q, where p, q are anyintegers (construction 45.11).6. For any segment of length x, we can find a segment of length √ x(construction 45.12).7. We can construct elemnts of the extension field having the field ofrational numbers as a subset.8. We can construct elements of the extension field of a + b √ c, where a,« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 45. EUCLIDEAN CONSTRUCTIONS 257Figure 45.7: Illustration of Construction 45.6, the perpendicular bisectorof a line segment.b, and c are constructable.Construction 45.11 (Rational Construction) Give a segment u = P Qof length 1, and any two integers p and q, construct a segment b that haslength p/q.1. Copy segment u a total of p times on ray −→ P Q.2. Define point A as the p th copy of point Q, so that P A = p.3. Construct any other ray −→ r emanating from P such that r is notparallel to or equal to −→ P A.4. Copy segment u a total of q times on ray −→ r .5. Define point U as the first copy of point Q on −→ r .6. Define point B as the p th copy of point Q on ray −→ r , so that P B = p.7. Construct a line l through Q 1 that is parallel to ←→ AB.8. Define point V = −→ P A ∩ l.9. UV = q/pConstruction 45.12 (Square Root) Given a segment u = P Q of length1, a second segment a = AB of length a, construct a segment that haslength √ a.1. Copy segment u onto line ←→ AB as segment u ′ = BC such that A∗B∗C.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

258 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.8: Construction 45.7, construction of a parallel line through apoint that is not on the line.2. Find the midpoint M of AC.3. Construct circle γ = C(M, AM).4. Construct a line l through B that is perpendicular to AC.5. γ intersects l at two points. Name one of these points D. (It doesnot matter which one).6. Let c = BD. Then c = √ a.Theorem 45.13 All constructible numbers are algebraic numbers, i.e., everyconstructible number is a root of the algebraic equationwhere a 0 , a 1 , ..., a n are integers.a n x n + a n−1 x n−1 + ... + a 0 = 0A number is not algebraic is said to be transcendental. Since π is transcendental1 , we have the following result.Theorem 45.14 π is not constructible.Corollary 45.15 The construction of squaring the circle with a straightedgeand a compass is impossible.Theorem 45.16 The general construction of trisecting an angle cannot beperformed with a straight-edge an compass alone.1 This result was proven by T. Lindemann in 1882.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 45. EUCLIDEAN CONSTRUCTIONS 259Figure 45.9: Construction 45.8, the division of a segment into n equal lengthsub-segments, illustrated for n = 3.Proof. Otherwise, we would be able to construct a segment of length π.Proof. (outline) Suppose we can trisect the angle 60 into thirds.cos 60 = 1/2, using the trigonometric identitySincecos 3θ = 4 cos 3 θ − 3 cos θif we set x = cos θ then8x 3 − 6x − 1 = 0Then one must show that (a) there are no rational solutions 2 ; and (b)if there is any solution of the form a + b √ c as a solution, we obtain acontradiction. 3Other ResultsTheorem 45.17 (Poncelet-Steiner Theorem) All constructions thatcan be performed with the straight-edge and compass can be performedwith a straight-edge alone, given a single circle and its center.2 Assume p/q is a solution; plug it in and you get that q must be a factor of 8. Byexhaustive checking one eliminates all possibilities.3 Since a + b √ c is a solution, so is a − b √ c as well as a third solution r; the sum of thesethree solutions must be zero. We add them up and solve for r and obtain the resultthat r = 2a is a solution, which contradicts the assumption that a + b √ c represents theleast-inclusive extension field of solutions.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

260 SECTION 45. EUCLIDEAN CONSTRUCTIONSFigure 45.10: Construction of a tangent line to a circle, drawn through apoint that is outside the circle (construction 45.9).Figure 45.11: Construction of a rational number (construction ).SegmentUV has length q/p.Theorem 45.18 (Mohr-Mascheroni Theorem) All constructions thatcan be performed with the straight-edge and compass can be performedwith the compass alone.Theorem 45.19 (Paper Folding Theorem) All constructions that canbe performed with a straight-edge and compass can be performed by foldingand creasing paper.The following result tells us that the following regular polygons are constructible:those with 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24 sides, amongTheorem 45.20 (Polygon Construction, Gauss) A regular polygoncan be inscribed in a circle by means of a straightedge and a compass ifand only if the number of sides can be written asn = 2 x p 1 p 2 ...p k« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 45. EUCLIDEAN CONSTRUCTIONS 261Figure 45.12: Construction of a square root (construction 45.12).c 2 = a, where c = BD and a = AB.Herewhere x is a non-negative integer and each p i is a Fermat number. 44 A Fermat number is a number of the form F n = 2 2n + 1 for some non-negative integer.The first five Fermat numbers are 3, 5, 17, 257, and 65,537, and 4,294,967,297.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

262 SECTION 45. EUCLIDEAN CONSTRUCTIONSAn Apollonian Gasket, a fractal generated by triples of circles that aretangent to one another. Different gaskets are formed by choosing differentdiameters for the starting circles. The gasket forms a complete tessellationof the hyperbolic plane, and all Apollonian gaskets are congruent in hyperbolicgeometry. Named for Apollonius of Perga (262 - 190 BCE). [Imagegenerated with Mathematica.]« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 46Hyperbolic GeometryThe Hyperbolic Parallel Postulate (HPP) is mutually exclusive of the morefamiliar Euclidean parallel postulate. We restate them both here for reference.Axiom 46.1 (Hyperbolic Parallel Postulate) For every line l and everypoint P ∉ l there are at least two distinct lines m and n (m ≠ n) such thatP ∈ m and P ∈ n and m ‖ l and n ‖ l. (Note that m ̸‖ n because theyintersect at P !)Axiom 46.2 (Euclidean Parallel Postulate) For every line l and forevery point P that does not lie on l there is exactly one line m such thatp ∈ m and m ‖ l.One consequence of the failure of rectangles to exist in neutral geometry isthat the absence of Euclidean geometry implies the necessity of hyperbolicgeometry. For reference we recall the hyperbolic parallel postulate here. Inany model for neutral geometry, either the Euclidean parallel postulate orthe Hyperbolic parallel postulate will hold. This is a consequence of thefollowing theorem.Theorem 46.3 (Universal Hyperbolic Theorem (UHT)) If the EuclideanParallel Postulate fails, then the Hyperbolic Parallel Postulate istrue, i.e,¬(EP P ) ⇒ (HP P )The Universal Hyperbolic Theorem is sometimes stated differently: In anymodel of neutral geometry in which rectangles do not exist, the Hyperbolic263

264 SECTION 46. HYPERBOLIC GEOMETRYFigure 46.1: Illustration of the hyperbolic parallel postulate. Lines m, n,p, and q all pass through the point P and all are parallel to the line l butnone of them are parallel to each other.parallel postulate is true.Proof. Let l be a line and let P be a point not on l. Assume that theEuclidean parallel postulate is false. Then no rectangle can exist (theorem33.3).We construct a line m that is parallel to l as follows: drop a perpendiculart from P to its foot Q on l. Then construct a perpendicular to t throughP . By the alternatie interior angles theorem, m ‖ l.Pick R ∈ l such that R ≠ Q and construct a perpendicular line u throughR.Drop a perpendicular line from P to u and call the foot S. By the alternateinterior angle theorem applied to transversal u, ←→ P S ‖ l.□P QRS is a Lambert Quadrilateral. It cannot be a rectangle because wehave already stated that no rectangle can exist.Hence ∠QP S is not a right angle. Therefore ←→ P S ≠ m.Hence there are two distinct lines, ←→ P S and m, both through P , both parallelto l. Hence the Hyperbolic parallel postulate is true.In this section we assume that the hyperbolic parallel postulate holds. Asconsequences of the Universal Hyperbolic Theorem, we can then immediatelyaccept as true any state that is equivalent to the negation of theEuclidean parallel postulate.Theorem 46.4 For every triangle △ABC, σ(△ABC) < 180.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 46. HYPERBOLIC GEOMETRY 265Figure 46.2: Proof of the universal hyperbolic theorem (theorem 46.3).Proof. HPP + UHT + theorem 32.10.Theorem 46.5 For every convex quadrilateral □ABCD, σ(□ABCD)

266 SECTION 46. HYPERBOLIC GEOMETRYFigure 46.3: A Saccheri quadrilateral (top) and Lambert quadrilateral (bottom)in Hyperbolic geometry.Proof. In neutral geometry we know that the fourth angle is either acuteor right. It it were right, the quadrilateral would be a rectangle, which isnot possible in Hyperbolic geometry. Hence the angle is acute.Theorem 46.10 In a Lambert quadrilateral, the length of a side betweentwo right angles is less than the length of the opposite sides.Proof. Let A, B, and C be the vertices with the right angles, as in figure46.3 (bottom).By corollary 31.25 we already know that BC ≤ AD.Suppose that BC = AD (RAA).This would make the □ABCD a Saccheri quadrilateral with summit CD.Then by theorem 31.16 the summit angles of a Saccheri Quadrilateral arecongruent, meaning ∠D = ∠C = 90. Thus □ABCD is a rectangle, whichviolates theorem 46.6.Hence BC < AD.Definition 46.11 The altitude of a Saccheri quadrilateral is the length ofthe segment joining midpoint of its base and the midpoint of its summit.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 46. HYPERBOLIC GEOMETRY 267Figure 46.4: In hyperbolic geometry similar triangles are always congruent.Theorem 46.12 The altitude of a Saccheri quadrilateral is shorter thanthe length of its sides.Proof. (Exercise.)Theorem 46.13 The length of the summit of a Saccheri quadrilateral isgreater than the length of its base.Proof. (Exercise.)Theorem 46.14 (AAA in Hyperbolic Geometry) Similar triangles arecongruent:△ABC ∼ △DEF =⇒ △ABC ∼ = △DEFProof. Suppose that △ABC ∼ △DEF but AB ≠ DE, BC ≠ EF andAC ≠ DF . (If any one of the equalities holds the by ASA the triangles arecongruent, so we are in effect assuming that the triangles are not congruent).At least two edges of one triangle are longer than two edges of the secondtriangle. Assume that AB > DE and AC > DF (if not, relabel the verticesaccordingly.)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

268 SECTION 46. HYPERBOLIC GEOMETRYThen we can choose a point B ′ ∈ AB and C ′ ∈ AC such that AC ′ = DFand AB ′ = DE.By SAS △AB ′ C ′ ∼ = △DEF . Hence∠AB ′ C ′ = ∠E = ∠B∠AC ′ B ′ = ∠F = ∠CSince ∠BB ′ C ′ + ∠AB ′ C ′ = 180 then∠BB ′ C ′ + ∠B = 180Similarly, since ∠B ′ C ′ A + ∠B ′ C ′ C = 180 then∠B ′ C ′ C + ∠C = 180Hence σ(□BB ′ C ′ C) = 360, which contradicts theorem 46.5 which says thatthe angle sum must be strictly less than 360.Hence our RAA hypothesis, which is that hte triangles are not congruent,is false.Theorem 46.15 Let □ABCD and □A ′ B ′ C ′ D ′ be Saccheri quadrilateralswith equal defect and congruent summits. Then □ABCD ∼ = □A ′ B ′ C ′ D ′ .Proof. Let □ABCD and □A ′ B ′ C ′ D ′ be Saccheri quadrilaterals such thatδ(□ABCD) = δ(□A ′ B ′ C ′ D ′ )CD ∼ = C ′ D ′Since σ(□) = 360 − δ(□), for any quadrilateral,σ(□ABCD) = σ(□ ′ B ′ C ′ D ′ )Since the summit angles in any Saccheri Quadrilateral are congruent∠C = ∠D, ∠C ′ = ∠D ′Since both quadrilaterals have the same measure and their base angles areall right angles, then the sum of the summit angles are equal.∠C + ∠D = ∠C ′ + ∠D ′=⇒ 2∠C = 2∠C ′=⇒ ∠C = ∠D = ∠C ′ = ∠D ′« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 46. HYPERBOLIC GEOMETRY 269Figure 46.5: If the summits of two Saccheri quadrilaterals with equal defectare congruent, then the two quadrilaterals are congruent.Chose points F ∈ −→ DA and G ∈−→ CB such that DF = D ′ A ′ = CG.By SAS, △F DC ∼ = △A ′ D ′ C ′ . Henceand therefore∠F CD = ∠A ′ C ′ D ′=⇒ ∠GCF = ∠C − ∠F CDBy SAS, △GCF ∼ = △B ′ C ′ A ′ , Hence= ∠C ′ − ∠A ′ C ′ D ′= ∠B ′ C ′ A ′∠G = ∠B ′ = 90Similarly,∠F = ∠A = 90Hence □F GCD ∼ = □A ′ B ′ C ′ D ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

270 SECTION 46. HYPERBOLIC GEOMETRYSuppose that F ≠ A (RAA). Then AB ‖ F G by the alternate interiorangles theorem. This means that B ≠ G (otherwise AB would intersectF G which is not possible for parallel segments).Hence □ABGF is a rectangle, which is not possible in hyperbolic geometry.Hence the RAA hypothesis is false. Hence F = A and thus B = G. Thismeans that□ABCD ∼ = □A ′ B ′ C ′ D ′« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 47Perpendicular Lines inHyperbolic GeometryIn Euclidean geometry, if two lines l and m are parallel, and if for any pointP ∈ l we have d(P, m) = D then for every point Q ∈ l, d(Q, m) = D. Inother words, parallel lines are a “fixed distance” apart. Contrary to ourusual intuitive feeling for what parallelism means, this result is not true inhyperbolic geometry; in fact, there are at most two such points on l thatare the same distance from m!Theorem 47.1 If l is a line, P ∉ l is a point, and m is a line such thatP ∈ m, then there exists at most one point Q ∈ m, Q ≠ P , such thatd(Q, l) = d(P, l).Figure 47.1: In hyperbolic geometry, parallel lines are not a fixed distanceapart; there are at most two points on any given line that are the samedistance from any other (distinct) line.271

272SECTION 47.PERPENDICULAR LINES IN HYPERBOLICGEOMETRYFigure 47.2: Lines m and n admit a common perpendicular because thetwo poitns D and F satisfy d(D, m) = d(F, m).Proof. Suppose that there are, in fact, three distinct points P , Q, and Ron m that are a fixed distance D from l,D = d(P, l) = d(Q, l) = d(R, l) > 0Let P ′ , Q ′ , and R ′ be the feet of perpendiculars dropped from P , Q, andR to l from m.None of the points P , Q or R lie on l because they are each a distance Daway. Hence at least two of them must lie on the same side of l. Supposethese points are P and Q (else, relable the points accordingly).Then □P P ′ Q ′ Q is a Saccheri Quadrilateral. Hence m ‖ l (because allSaccheri Quadrilaterals are parallelograms by theorem 31.18).Since m ‖ l, then all points on m are on the same side of l. Hence P , Q,and R are all on the same side of l.Relabel the points, if necessary, so that P ∗ Q ∗ R. Then both □P P ′ Q ′ Qand □QQ ′ R ′ R are Saccheri quadrilaterals. Since the summit angles of aSaccheri quadrilateral are acute in (theorem 31.20), ∠P QQ ′ and ∠RQQ ′are acute.Since ∠P QQ ′ and ∠RQQ ′ are supplements, this is a contradiction.Hence the RAA hypothesis is false, and we conclude that there are at mosttwo points P and Q a on m that are a fixed distance from l.Theorem 47.2 (Existence of Common Perpendiculars) If l ‖ m areparallel lines and there are two points on m that are equidistant from lthen l and m admit a common perpendicular.Proof. (Exercise.)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 47. PERPENDICULAR LINES IN HYPERBOLICGEOMETRY 273Figure 47.3: The alternate interior angles δ = γ iff P Q is a common perpendicularto l and m and P M = QM.Theorem 47.3 (Uniqueness of Common Perpendiculars) If two linesadmit a common perpendicular that perpendicular is unique.Proof. (Exercise.)Theorem 47.4 Suppose that lines l and m are parallel and are cut by acommon transversal. Then the alternate interior angles are congruent ifand only iff the following two conditions hold: (1) lines l and m admit acommon perpendicular; (2) the transversal intersects the common perpendicularsegment at its midpoint.Proof. (⇒)Suppose δ = γ (see figure 47.3). If δ = 90 and γ = 90 thenthe result of the theorem follows immediately. Hence we assume that theseangles are not right angles.Let M be the midpoint of RS, and drop perpendiculars from M to m and l,calling their feet Q and P as shown in fig 47.3. By AAS △SQM ∼ = △RP M.Hence ∠SMQ = ∠P MR. Thus −−→ MP and −−→ MQ are opposite rays thereforeP M and P Q form a single segment P Q.By congruence of triangles, P M = QM, and P M is a common perpendicular.(⇐) Assume that the transversal t intersects a common perpendicular QPat the its midpoint M, as shown in figure 47.3. We need to show that∠δ = ∠γ.Since ∠QMS = ∠P MR by the vertical angle theorem, triangles △MQS ∼ =△MP R by ASA.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

274SECTION 47.PERPENDICULAR LINES IN HYPERBOLICGEOMETRYBy congruence ∠MRP = ∠MSQ. Since γ and δ are their respective supplements,we conclude that δ = γ.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 48Parallel Lines inHyperbolic GeometryDefinition 48.1 (Intersecting Set) Let l be a line, P ∉ l a point, Athe foot of the perpendicular dropped from P to l, and B ∈ l any otherpoint in l. Then for each θ ∈ (0, 90] there is some point D(θ) such thata µ(AP Dθ) = θ. The critical set K is the set of all θ such that −−−−→ P D(θ)intersects −→ AB:K = {θ| −−−−→ P D(θ) ∩ −→ AB ≠ ∅}Since K is a bounded set of real numbers (it is bounded because 90 is anupper bound) it has a least upper bound.Definition 48.2 (Critical Number) The critical number θ C of P for−→AB is the leqst upper bound of the critical set K:θ C = lub KThe critical number defines an angle such that any rays that make a largerangle with P (on the same side of ←→ AP as B) do not intersect ←→ AB and henceare parallel to ←→ AB. The angle with measure given by the critical number iscalled the angle of parallelism. Rays that make a smaller angle with −→ P Aare not parallel to ←→ AB and intersect it. There are two angles of parallelism,one on each side of P A; these two angles are congruent.Definition 48.3 (Angle of Parallelism) Let D be a point of the sameside of ←→ P A as B such that µ(∠P AD) = θ C . Then angle ∠P AD is called275

276 SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRYFigure 48.1: Illustration of the intersecting set. The rays −−−−→ P D(α) and −−−−→ P D(β)intersect −→ AB, and hence are members of the intersecting set, whereas theray P D(γ) and any ray that makes an angle between than γ and 90 with−→P A does not intersect −→ AB, and hence are not members of the intersectingset. There is some limiting angle, beneath which all rays are members ofthe intersecting set, and beyond which all rays are not members of theintersecting set. This limiting angle is the critical number of P and −→ AB.Pαβ γD(α)D(β)D(γ)ABthe angle of parallelism, −−→ P D is called the limiting parallel ray for −→ AB,and we write −−→ P D| −→ AB.Theorem 48.4 If 0 < θ < θ C then θ ∈ K, otherwise, if θ C ≤ θ ≤ 90, thenθ ∉ K.Proof. Let θ ∈ (0, 90] be given.(1) Suppose tht θ < θ C = lub K.Then there is some number β ∈ K such that θ < β < θ C (otherwise θwould be an upper bound of K which contradicts the assumption θ < θ C ).Since β ∈ K there is some point D β and some point T ∈ −→ AB such that−−−−→P D(β) ∩ −→ AB = TSince θ < β then D θ is in the interior of ∠AP T . Hence by the crossbartheorem −−→ P D θ intersects AT at some point U.Hence θ ∈ K, proving the first part of the theorem.(2) Suppose that θ ≥ θ C (hypothesis).Suppose (RAA) that θ ∈ K. Then there is some point U such that −−→ P D θintersects −→ AB at U.Pick any T ∈ −→ AB such that A ∗ U ∗ T .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRY 277Figure 48.2: Proof of theorem 48.4.Figure 48.3: There are two critical angles, one on each side of the perpendiculardropped from P to l.Define β = µ(∠AP T ).Since P −−→ D β ∩ −→ AB ≠ ∅, then β ∈ K. But β > θ because A ∗ U ∗ T . Henceβ ≥ θ > θ C which contradicts the RAA. Hence θ ∉ K.Theorem 48.5 The critical number θ C depends only on the distance d(P, l).Proof. Let l be a line, let P ∉ l be a point, let A be the foot of theperpendicular from P to l, and let B ≠ A be another point in l.Let l ′ be a second line, let P ′ ∉ l ′ , let A ′ be the foot of the perpendicularfrom P ′ to l ′ such that P ′ A ′ = P A, and let B ′ ≠ A ′ be another point in l ′ .Let the intersecting sets of the two points and their respective lines be Kand K ′ , and suppose that θ ∈ K.Then there is a point D(θ) such that −−−−→ P D(θ) intersects ←→ AB at some pointT .Choose T ′ ∈ −−→ A ′ B ′ so that A ′ T ′ = AT .By SAS, △P AT ∼ = △P ′ A ′ T ′ . Therefore there is a point D ′ (θ) such that−−−−−→P ′ D ′ (θ) intersects ←−→ A ′ T ′ (pick any point on ←−→ A ′ T ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

278 SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRYFigure 48.4: The critical number depends only on the distance of the pointfrom the line (theorem 48.5).Hence θ ∈ K ′ .By a similar argument, for any θ ∈ K ′ then θ ∈ K.Hence the K = K ′ , which means the critical number depends only ond(P, l).Since the critical number θ C depends only on the distance x = d(P, l) wecan write it as θ C = κ(x) for some function κ which we call the criticalfunction.Theorem 48.6 The critical functionis a decreasing function.κ(x) : (0, ∞) ↦→ (0, 90]Proof. Let a, b ∈ R such that 0 < a < b.Choose points P, A, B, D such that P A = a and θ = µ(∠AP D) is the angleof parallelism for P and −→ AB; and choose point A such that QA = b.We need to show that κ(b) ≤ κ(a) = θ.Choose E on the same side of ←→ AQ as D such that µ(∠AQE) = θ (see figure48.5).« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRY 279Figure 48.5: The angle of parallelism is a decreasing function (Theorem48.6).By the corresponding angles theorem ←→ QE ‖ ←→ P D.Hence ←→ QE does not intersect ←→←→P D.AB (because otherwise it would have to crossHence θ is not in the intersecting set of Q and −→ AB. Hence κ(b) ≤ θ. Henceκ(b) ≤ κ(a). which means κ is non-increasing.Theorem 48.7 Every angle of parallelism is acute, and every critical numberis less than 90.Proof. Let l be a line P ∉ l a point.Drop the perpendicular from P to its foot A ∈ l.Let m be a line through P such that m ⊥ ←→ P A. Hence m ‖ l.By the hyperbolic parallel postulate there is at least one additional line nthorugh P such that n ‖ l.Since n is distinct from m it is not perpendicular to ←→ P AHence it must make an acute angle with ←→ P A on one side of P A. Call themeasure of this angle θ.Since n ‖ l, θ ∉ K. Hence the critical angle must be less than θ, and henceit must be less than 90.Theorem 48.8 Suppose −−→ P D| −→ AB, Q ∈−−→ P D, and let C be the foot of theperpendicular from Q to ←→ AB. If B ′ ∈ −→ AB such that A∗C ∗B ′ and D ′ ∈ −→ P Qsuch that P ∗ Q ∗ D ′ , then −−→ QD ′ | −−→ CB ′ .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

280 SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRYFigure 48.6: The angle of parallelism is acute (theorem 48.7).Proof. (Exercise.)Theorem 48.9 −−→ P D| −→ ←→ ←→AB =⇒ P D ‖ ABProof. (Exercise.)Definition 48.10 If −−→ P D| −→←→ ←→AB then the lines P D and AB are called asymptoticallyparallel in the direction −→ AB.Theorem 48.11 (Symmetry of Limiting Parallels) If −−→ P D| −→ AB, Q is thefoot of the perpendicular from A to −−→ P D, and D ′ ∈ −−→ P D such that P ∗Q∗D ′ ,then −→ AB|−−→ QD ′ .Proof. (Exercise.)Corollary 48.12 (Symmetry of Asymptotic Parallelism) If l is asymptoticallyparallel to m then m is asymptotically parallel to l.Proof. (Exercise.)Theorem 48.13 If l and m admit a common perpendicular, then they arenot asymptotically parallel.Proof. Let l and m be two lines that admit a common perpendicular, whichwe will call t. Let the intersections of t with l and m be S and R.Since m is perpendicular to RS, no subray of M starting at R can beasymptotically parallel to l, because the angle of parallelism must be acute.Let P ∈ m such that P ≠ R. Drop a perpendicular from P to its footA ∈ l.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRY 281Figure 48.7: If two lines admit a common perpendicular then thay are notasymptotically parallel (theorem 48.13).Let D ∈ m such that R ∗ P ∗ D.Since □SAP R is a Lambert quadrilateral, then ∠AP R is acute; hence,∠AP D is obtuse.Hence −−→ P D is not a limiting parallel because the angle is not acute.Let B ∈ l on the same side of P A as D.Let C ∈ l such that C ∗ S ∗ BLet E ∈ m such that E ∗ R ∗ D.Then the ray −→−→P E cannot be a limiting parallel ray of AC, because if itwere, that would mean that −→−→RE was a limiting parallel ray of SC, and wehave already ruled that out.Hence there can be no limiting parallel ray to l in m originating at P . SinceP was chosen arbitrarily, that means that no subray of m can be a limitingparallel ray to l.Theorem 48.14 If l and m are asymptotically parallel lines, then l andm do not admit a common perpendicular.Proof. This is the contrapositive of theorem 48.13.Theorem 48.15 Suppose that l ‖ m; P, Q, R are points on m with P ∗Q∗R;and A, B, C are the feet of the perpendiculars from P, Q, R to l, respectively.Then1. ←→ P A ⊥ m =⇒ P Q < QB < RC2. −→ P A| −→ AB =⇒ P A > QB > RCProof. (Exercise.)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

282 SECTION 48. PARALLEL LINES IN HYPERBOLIC GEOMETRYTheorem 48.16 If l ‖ m are parallel lines that admit a common perpendicular,the for every d 0 > 0 there exists a point P ∈ m such that d(P, l) > d 0 ;and P may be chosen to line on either side of the common perpendicular.Proof. (omitted)Theorem 48.17 (Transitivity of Limiting Parallels) If l is asymptoticallyparallel to m in the direction −→ AB and l is asymptotically parallel ton in the direction −→ AB, then either m = n or m is asymptotically parallelto n.Proof. (omitted)Theorem 48.18 If l ‖ m then either l and m admit a common perpendicularor l and m are asymptotically parallel.We have already shown that the critical function is decreasing. In fact, itis strictly decreasing.Theorem 48.19 The critical function κ(x) is strictly decreasing.Proof. (See Venema.)Theorem 48.20 If l and m are asymptotically parallel lines then for everyɛ > 0 there exists a point T ∈ m such that d(T, l) < ɛ.Proof. (See Venema.)Theorem 48.21 limx→∞ κ(x) = 0Proof. (See Venema.)Theorem 48.22 limx→0 + κ(x) = 90Proof. (See Venema.)Theorem 48.23 The critical funnction κ(x) is onto.Proof. (Exercise.)Theorem 48.24 The critical functionis continuous.Proof. (Exercise.)κ(x) : (0, ∞) ↦→ (0, 90)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 49Triangles in HyperbolicGeometryTheorem 49.1 For every ɛ > 0 there exists an isosceles triangle △BP Cwith σ(△BP C) < ɛ and δ(△BP C) > 180 − ɛ.Proof. Let ɛ > 0 be given.Since lim x→∞ κ(x) = 0 there exists some y such that ∀a > y, κ(a) < ɛ/4.Choose one such a.Let △P AB be a triangle with right angle at A and both legs of length a.Since −→−→P B intersects AB then µ(∠AP B) < K(a).Since ∠ABP = ∠AP B,andσ△BP C = µ(∠ACP ) + µ(∠ABP )+ 2µ(∠AP B)= 4µ(∠AP B)< 4κ(a)< ɛδ(△BP C) = 180 − σ(△BP C)< 180 − ɛTheorem 49.2 For every ɛ > 0 there is a right triangle △ABC such thatσ(△ABC) > 180 − ɛ and δ(△ABC) < ɛ.283

284 SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRYFigure 49.1: Construction of an isoceles triangle with measure less that ɛ.Proof. Let ɛ > 0 be given.Construct a right triangle △ABC with one interior angle 90 − ɛ as follows:Let A and B be points, and define Q so that ∠ABQ = 90 − ɛ. The dropa perpendicular from A to BQ and call its foot C. Then △ABC is a righttriangle with one interior angle of 90 − ɛ.Thenandσ(△ABC) = 90 + 90 − ɛ + µ(∠ACB)> 180 − ɛδ(△ABC) = 180 − σ(△ABC)< 180 − (180 − ɛ) = ɛTheorem 49.3 For every ɛ > 0 there is a number d > 0 such that if leg oftriangle △ABC has length less than d then δ(△ABC) < ɛ.Proof. Let ɛ > 0 be given.By theorem 49.2 there is a right triangle △EF G with right angle at E suchthat δ(△EF G) < ɛ/2.Let d be the length of the shortest side of △EF G.Let △ABC be any triangle with sides shorter d. Let D be the foot ofthe perpendicdular from C to AB. Then △ADC and △BDC are righttriangles with right angles at D (if D does not satisfy A ∗ D ∗ B relabel thevertices and choose another edge as AB).(See figure 49.2.)By construction we chose AC < d, BC < d, and AB < d.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRY 285Figure 49.2: Left:Any triangle can be divided into two right triangles. IfD is not between A and B then relabel the vertices to use a different edge.Right: △A ′ EC ′ ∼ = △ADC if A ′ E = AC and AD = C ′ E. This is possiblebecause the shortest side of △EF G has length d and all the sides of △ABChave length less than d.Since AB = AD + BD then BD < d and AD < d.By the scalene inequality CD < AC < d.As shown in figure 49.2, we can embed a triangle that is congruent to eachof △BDC and △ADC in △EF G.By additivity of defecthenceδ(△BDC) < δ(△EF G) < ɛ/2δ(△ADC) < δ(△EF G) < ɛ/2δ(△ABC) = δ(△BDC) + δ(△ADC)< ɛ 2 + ɛ 2 = ɛTheorem 49.4 For every pair of points A and B and for every ɛ > 0 thereexists a d > 0 such that if C ∉ ←→ AB and AC < d then δ(△ABC) < ɛ.Proof. Let ɛ > 0 be given.Choose a point P such that P ∗ A ∗ B, and choose D such thatµ(∠BP D) = 90 − ɛRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

286 SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRYDrop a perpendicular from B to −−→ P D and call its foot Q. Sinceσ(△BP Q) = µ(∠P BQ) + 90 + 90 − ɛ=⇒ δ(△BP Q) < ɛ> 180 − ɛDefined = min[d(A, ←→ P Q), d(A, ←→ BQ)]Let C be any point in the interior of △BP Q not on ←→ AB such that AC < d.By the crossbar theorem there is a point E such that −→ AC intersects P Q.By repeated applications of additivity of defect,δ(△P QB) = δ(△BQE) + δ(△ABC)+ δ(△AP C) + δ(△P EC)> δ(△ABC)Since δ(△P QB) < ɛ then δ(△ABC) < ɛ.Figure 49.3: Proof of theorem 49.4.Theorem 49.5 (Continuity of Defect) Let △ABC be a triangle and letc = AB. For every number x ∈ [0, c] define P (x) to be the unique pointP (x) ∈ AB such that AP (x) = x. Defineand define f(0) = 0. Thenis continuous.f(x) = δ(△AP (x)C)f(x) : [0, c] ↦→ R« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRY 287Figure 49.4: Continuity of Defect.Proof. Let ɛ > 0 be given and suppose that x ∈ [0, c].By theorem 49.4 there is a d > 0 such then whenever P (x)P (y) thenδ(△CP (x)P (y)) < ɛ (see figure 49.4).Choose y ∈ [0, c] such that |x − y| < d.Then P (x)P (y) = |x − y| < d which implies that δ(CP (x)P (y)) < ɛ.By the additivity of defect,|f(x) − f(y)| = |δ(△AP (x)C) − δ(△AP (y)C)|= δ(△CP (x)P (y))< ɛTheorem 49.6 The defect is onto. Specifically, ∀y ∈ (0, 180) there existsa triangle △ABC such that δ(△ABC) = y.Proof. Let y ∈ (0, 180) be given.Let ɛ = 180 − y.Since 0 < y < 180, then 0 < ɛ < 180.By theorem 49.1, there exists a triangle △ABC such thatδ(△ABC) > 180 − ɛ = yBy theorem 49.2, there exists a triangle △A ′ B ′ C ′ such thatδ(△A ′ B ′ C ′ ) < yBy continuity of defect and the intermediate value theorem, there existssome triangle with defect y.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

288 SECTION 49. TRIANGLES IN HYPERBOLIC GEOMETRY« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 50Area in HyperbolicGeometryDefinition 50.1 We say that two polygonal regions R and R ′ are Equivalentby Dissection if they can each be triangulatedR = T 1 ∪ T 2 ∪ · · · ∪ T nwhereR ′ = T ′ 1 ∪ T ′ 2 ∪ · · · ∪ T ′ nT 1∼ = T′1 , . . . , T n∼ = T′nTheorem 50.2 (Fundamental Theorem of Dissection Theory) If Rand R ′ are polygonal regions such that α(R) = α(R ′ ) then R ≡ R ′ inneutral geometry.Proof. (Beyond the scope of this class.)Theorem 50.3 Equivalence by Dissection is reflexive, symmetric, and transitive,i.e., it is an equivalence relation:1. ∀R, R ≡ R2. R 1 ≡ R 2 =⇒ R 2 ≡ R 13. If R 1 ≡ R 2 and R 2 ≡ R 3 then R 1 ≡ R 3Proof. (Beyond the scope of this class.)Theorem 50.4 (Bolyai’s Theorem) If △ABC and △DEF satisfy δ(△ABC) =δ(△DEF ) then △ABC ≡ △DEF .289

290 SECTION 50. AREA IN HYPERBOLIC GEOMETRYDefinition 50.5 Let △ABC be a triangle. The Associated SaccheriQuadrilateral (see figure 50.1) □ABDE is constructed as follows: Let Mand N be the midpoints of AC and BC. Drop perpendiculars from A andB to ←−→ MN and call their feet D and E.Figure 50.1: The associated Saccheri Quadrilateral.Theorem 50.6 The Associated Saccheri Quadrilateral is a Saccheri Quadrilateral.Proof. We need to show that BE = AD.Drop a perpendicular from C to MN and call the foot F .By AAS △BNE ∼ = △CNF (∠BNE = ∠CNF , ∠BEN = ∠CF N andNC = BN).By congruency, BE = F C. By a similar argument, AD = F C. HenceAD = BE.Theorem 50.7 Let △ABC have an associated Saccheri Quadrilateral □ABDE.Then in neutral geometry they are equivalent by dissection, i.e., △ABC ≡□ABDE.Proof. (See Venema.)Lemma 50.8 Let △ABC have an associated Saccheri Quadrilateral □ABDE,then δ(□ABDE) = δ(△ABC).Proof. (Exercise.)Lemma 50.9 Suppose that δ(△ABC) = δ(△DEF ) and that AB = DE.Then △ABC ≡ △DEF .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 50. AREA IN HYPERBOLIC GEOMETRY 291Figure 50.2: Proof that the associated quadrilateral is a Saccheri Quadrilateral.Proof. Suppose that δ(△ABC) = δ(△DEF ) and that AB = DE (hypothesis).Let □ABB ′ A ′ and DEE ′ D ′ be their corresponding Saccheri quadrilaterals.By lemma 50.8δ(□ABB ′ A ′ ) = δ(△ABC)= δ(△DEF )= δ(□DEE ′ D ′ )By hypothesis the two quadrilaterals have congruent summits. Then bytheorem 46.15□(ABB ′ A ′ ) ∼ = □DEE ′ D ′Hence by transitivity of equivalence,△ABC ≡ □ABB ′ A ′≡ □DEE ′ D ′≡ △DEFProof. (Boylai’s theorem, theorem 50.4.)Suppose that δ(△ABC) = δ(△DEF ).Assume that DF ≥ AC (if not, relabel).Let M be the midpoint of AC and let N be the midpoint of BC.Choose G ∈ ←−→ MN such that AG = (1/2)DF .Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

292 SECTION 50. AREA IN HYPERBOLIC GEOMETRYFigure 50.3: Proof of Boylai’s theorem.Choose H ∈ ←→ AG such that A ∗ G ∗ H and GH = AG.Then △ABC and △ABH share the same Saccheri Quadrilateral □AA ′ B ′ B.Hence by transitivity of equivalence and lemma 50.8 thenSince△ABC ≡ △ABHδ(△ABH) = δ(△ABC) = δ(△DEF )by hypothesis, and DF = AH, then by lemma 50.9,. By transitivity of equivalence,△ABH ≡ △DEF△ABC ≡ △DEFTheorem 50.10 For any two triangles △ABC and △DEF ,α(△ABC)δ(△ABC) = α(△DEF )δ(△DEF )« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 50. AREA IN HYPERBOLIC GEOMETRY 293Figure 50.4: Proof that area and defect are proportional.Proof. If δ(△ABC) = δ(△DEF ), by Boylai’s theorem △ABC ≡ △DEF ,hence they have equal area, and α(△ABC) = α(△DEF ).Now suppose that δ(△ABC)/δ(△DEF ) is rational. Then there exist positiveintegers a and b such thatδ(△ABC)δ(△DEF ) = a bAssume that a < b (if not, relabel the triangles).By the additivity of defect and continuity of defect we can define pointsP 0 , P 1 , . . . , P b on DE such thatD = P 0 , P i−1 ∗ P i ∗ P i+1 , P b = Eandδ(△F P i P i+1 ) = 1 b δ(△DEF )(By additivity of defect the sum of the defects of the smaller triangles mustadd up to the defect of the larger triangle. Hence by choosing n sufficientlylarge we can ensure that each triangle has a defect that is sufficiently small.)By the first part of the proof, all the smaller triangles have the same area.δ(△ABC) = a b δ(△DEF )= aδ(△F P i P i+1 )= δ(△F P 0 P a )Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

294 SECTION 50. AREA IN HYPERBOLIC GEOMETRYwhere the last step follows by additivity of defect. Henceby Boylai’s theorem and thererforehence△ABC ≡ △F P 0 P aα(△ABC) = α(△F P 0 P a )= aα(△F P i P i+1 )= a b α(△DEF )α(△ABC)α(△DEF ) = a b = δ(△ABC)δ(△DEF )which proves the theorem when the ratio is rational.Now suppose that x,δ(△ABC)δ(△DEF ) = xα(△ABC)α(△DEF ) = ywhere x and y are real numbers. Let r be any rational number such that0 < r < x. Choose C ′ ∈ BC such thatThenand by the first part of the theorem,r = δ(△ABC′ )δ(△DEF )α(△ABC ′ ) < α(△ABC)r = α(△ABC′ )α(△DEF )< α(△ABC)α(△DEF ) = ySince r was chosen as an arbitrary rational such that r < x, then for everyrational r < x, we have r < y.By a similar argument, for every rational number r < y, we can show thatr < x.Hence by the comparison theorem for real numbers (theorem 6.5), x =y.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 50. AREA IN HYPERBOLIC GEOMETRY 295Theorem 50.11 Area is proportional to defect: there exists a constant ksuch thatα(△ABC) = kδ(△ABC)for every triangle △ABC (same k for every triangle).Corollary 50.12 There is an upper bound to the area of triangles.Proof. Since δ(△ABC) < 180 for every triangle, thenα(△ABC) < 180kCorollary 50.13 If α(△ABC) = α(△DEF ) then △ABC ≡ △DEF .Proof.δ(△ABC) = 1 k α(△ABC)By Boylai’s theorem △ABC ≡ △DEF .= 1 k α(△DEF )= δ(△DEF )Theorem 50.14 Let R 1 and R 2 be polygonal regions such that α(R 1 ) =α(R 2 ). Then R 1 ≡ R 2 .Theorem 50.15 There exist triangles with the same base and height butdifferent areas.Proof. Let l be a line and P a point with d(P, l) = 1.Let Q 0 be the foot of the perpendicular from P to l.Define Q 1 , Q 2 , . . . in l such that Q i Q i+1 = 1.Define T i = △P Q i−1 Q i .equal to 1.Then the base and height of each T i are eachSuppose that all the T i have the same area, α(T i ) = a for all i.Then all the T i have the same defect, δ(T i ) = d for all i. By additivity ofdefectδ(△P Q 0 Q n ) = nd < 180Since we can choose n arbitrarily large, this violates the Archimedian orderingproperty. Hence the defects cannot all be equal, and hece the areascannot all be equal.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

296 SECTION 50. AREA IN HYPERBOLIC GEOMETRYFigure 50.5: All of the triangles have base 1 and height 1.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 51The Poincare Disk ModelIn the Poincare Disk Model of hyperbolic geometry the plane is representedby the unit disk (the disk of radius 1 centered at the origin). Linesare diameters of the unit circle or the portions of arcs of circles that areorthogonal to the unit disk that fall within the unit disk.Two intersecting circles are said to be orthogonal if their tangents meetat 90 ◦ angles (figure 51.1).We illustrate this in figure 51.2. In this illustration, lines l, m and n areall parallel to k. Lines l and m both intersect at the point A, and lines mand n intersect at point B, given two examples of multiple parallel lines tothe same line passing through a single point. The figure also illustrates thefailure of the transitivity of parallelism; although l and m are parallel to k,they are not parallel to each other; furthermore, while l is parallel to n, mis not parallel to n.Angle measure is given by the Euclidean angle measure of the angleformed by the tangent lines at the point of intersection (see figure 51.3).To prove that there is a unique line (given by the circular arc) through anytwo points, we will use analytic geometry to find the general equation fora circle that is orthogonal to the unit circle at the origin. Let the circlehave radius a and center (x 0 , y 0 ). Then from analytic geometry we havethe equation of the circle is(x − x 0 ) 2 + (y − y 0 ) 2 = a 2 (51.1)From figure 51.1 we see by an application of the Pythagorean theorem thatx 2 0 + y 2 0 = a 2 + 1 (51.2)297

298 SECTION 51. THE POINCARE DISK MODELFigure 51.1: Orthogonal circles. The tangents (or radii) of each circle meetat a 90 ◦ angle at the points of intersection.O=(0,0)r=1aSubstituting equation 51.2 into equation 51.1 givesx 2 − 2xx 0 + y 2 − 2yy 0 + 1 = 0 (51.3)Suppose now that we want the circle to pass through the pointsP = (x 1 , y 1 ), Q = (x 2 , y 2 )This gives two equations in the unknowns x 0 and y 0 :x 2 1 − 2x 1 x 0 + y1 2 − 2y 1 y 0 + 1 = 0x 2 2 − 2x 2 x 0 + y2 2 − 2y 2 y 0 + 1 = 0Rearranging, Å ã Å ãx1 y 1 2x0x 2 y 2 2y 0Å ã 1 + r2= 11 + r22where r i = x 2 i + y2 i , for i = 1, 2. HenceÅ ã 2x0= 1 Å ã Å ãy2 −y 1 1 + r212y 0 ∆ −x 2 x 1 1 + r22where ∆ is the determinant(51.4)(51.5)∆ = x 1 y 2 − x 2 y 1« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 51. THE POINCARE DISK MODEL 299Figure 51.2: The Poincare Disk represents the plane as a unit circle centeredat the origin of the normal Euclidean plane. Lines are either diameters orarcs of circles that are orthogonal to the edge of the disk.The above matrix inverse holds so long as ∆ ≠ 0. Thereforex 0 = (1 + r2 1)y 2 − (1 + r 2 2)y 12∆y 0 = (1 + r2 2)x 1 − (1 + r 2 1)x 22∆(51.6)(51.7)Theorem 51.1 Let P = (x 1 , y 1 ) and Q = (x 2 , y 2 ) be the Euclidean coordinatesof two points in the Poincare Disk. Then the unique line connectingthese two points is an arc of the circle with center at (x 0 , y 0 ) (from equations51.6 and 51.7), and radius x 2 0 + y 2 0 − 1.We can define a distance measure as follows.Definition 51.2 Let A and B be any two points in the Poincare Disk, andlet P and Q be the endpoints of the line that connects them. Then theCross-Ratio is defined as[AB, P Q] =(AP )(BQ)(BP )(AQ)where the distances AP , BQ, BP , and AQ are the ordinary Euclideansegment lengths.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

300 SECTION 51. THE POINCARE DISK MODELFigure 51.3: The measure of ∠P AQ in the Poincare Disk is given by theEuclidean measure of ∠MAL.Definition 51.3 The Poincare Distance d(A, B) between points A andB is defined asd(A, B) = |ln[AB, P Q]|Theorem 51.4 The Poincare Distance is a metricProof. Let A and B be any points in the Poincare Disk. Then:(1) Reflexivity. Since ln x = − ln(1/x), we know that | ln x| = | ln 1/x| andtherefored(A, B) =(AP )(BQ)∣ln (BP )(AQ) ∣=(BP )(AQ)∣ln (AP )(BQ) ∣= d(B, A)(2) Positivity. d(A, B) ≥ 0 because of the absolute value.(3) Identity. If A = B then d(A, B) = | ln 1| = 0. For the converse, suppose« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 51. THE POINCARE DISK MODEL 301that d(A, B) = 0. Then(AP )(BQ)∣ln (BP )(AQ) ∣ = 0(BP )(AQ)=⇒(AP )(BQ) = 1=⇒ (BP )(AQ) = (AP )(BQ)=⇒ BPBQ = APAQSuppose A ≠ B. Let C be the intersection of AQ and BP . If they do notintersect, interchange the labels on A and B so that they do intersect.By the vertical angles theorem ∠ACP ∼ = ∠BCQ.The power of point C is independent of the line chosen, henceso thatPower(C) = (CA)(CQ) = (CB)(CP )CACP = CBCQBy the SAS criterion for triangle similarity,△ACP ∼ △BCQHenceTherefore by SAS again,∠P AC ∼ = ∠QBC△P AQ ∼ △P BQBy the converse to the similar triangles theorem,BQAQ = BPAP = P QQP = 1Hence BQ = AQ and BP = AP . Thus A = B.If the points fall on a diameter, then suppose that P ∗ A ∗ B ∗ Q. Then(BP )(AQ) = (AP )(BQ)(P A + AB)(AB + BQ) = (AP )(BQ)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

302 SECTION 51. THE POINCARE DISK MODEL(P A)(AB) + (P A)(BQ)+(AB)(AB)+(AB)(BQ) = (AP )(BQ)(P A)(AB) + (AB)(AB) + (AB)(BQ) = 0(AB) [(P A) + (AB) + (BQ)] = 0(AB)(P Q) = 0Hence either AB = 0 or P Q = 0. Since P and Q are distinct, then A =B.Example 51.1 Suppose that Q ∗ A ∗ B ∗ C ∗ P on a line in the Poincaredisk. Then since AP > BP and BQ > AQ, the ratio(AP )(BQ)(AQ)(BP ) > 1Similarly, since BP > CP and CQ > BQ, the ratio(BP )(CQ)(CP )(BQ) > 1and hence we can drop the absolute values around the logarithms:(AP )(BQ)d(AB) + d(AC) = ln(AQ)(BP )(BP )(CQ)+ ln(CP )(BQ)(AP )(BQ)(BP )(CQ)= ln(AQ)(BP )(CP )(BQ)(AP )(CQ)= ln(AQ)(CP )= d(AC)Theorem 51.5 Let d = d(OA). Thenwhere d is the Poincare distance.OA = ed − 1e d + 1« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 51. THE POINCARE DISK MODEL 303Proof. Let P and Q be the ends of the diameter containing A such thatQ ∗ A ∗ O ∗ P . Thend = ln(AP )(OQ) AP= ln(AQ)(OP ) AQe d = APAQ = 1 + AO1 − AOe d (1 − AO) = 1 + AOe d − (AO)e d = 1 + AOe d − 1 = (AO)(e d + 1)Lemma 51.6 (Euclidean Geometry) Let AB be a chord of a circle Γ, andlet C be the intersection of the tangent lines to Γ at A and B. Then △ABCis isoceles.Theorem 51.7 (Boylai Lobachevsky Formula) The angle of parallelismθ(d) in the Poincare Disk model of Hyperbolic geometry satisfiestan θ(d)2= e −dProof. Since θ(d) depends only on the distance we are free to choose any lineand any point. Let l be a diameter and let P be a point on a perpendiculardiameter. We will computer θ(d) = θ(OP ), as in fig. 51.4.Figure 51.4: Proof of the Boylai-Lobachevsky Formula. The edge of thePoincare Disk is the dotted line.The limiting parallel ray through P will be an arc of circle that is tangentto the diameter at of the Poincare disk at C. From lemma 51.6 △P CR isRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

304 SECTION 51. THE POINCARE DISK MODELisoceles and β = β ′ . Hence ∠CRP = π − 2β (measured in radians). Thus∠P RO = 2β. Henceθ = π 2 − 2βtan θ 2 = tan ( π4 − β )= 1 − tan β1 + tan βSince tan β = OP = (e d − 1)/(e d + 1) by theorem 51.5,tan θ 2 = 1 − ed − 1e d + 11 + ed − 1e d + 1= e −d« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 52Arc LengthIn this section we begin our consideration of geometry on the surface of asphere. We first need to figure out what we mean by a “line” on the surfaceof a sphere. Since we would like out geometry to satisfy as m much ofneutral geometry as possible (we will find that it is not possible to satsifyall the postulates of neutral geometry in spherical geometry), we will beginwith the result that the distance between two points is the length of a linecontaining those two points. By the triangle inequality, we know that wecan not construct any shorter path between those two points by any pairof line segments. Applying this concept repeatedly, any path composed ofline segments will have an overall length at least as long as the length of asingle segment.Figure 52.1: The segment AB is the shortest path between A and B inneutral geometry.305

306 SECTION 52. ARC LENGTHAs we see in figure 52.1,AB ≤ AP 1 + P 1 P 2 + P 2 P 3 + · · · + P n−1 Bfor any points A = P 0 , P 1 , P 2 , . . . , P n−1 , P n = B. Now suppose we have any“smooth” curve γ connecting P 0 and P 1 , as shown in figure 52.2.Figure 52.2: The segment P 0 P n is the shortest path between P 0 and P n .Is it possible for the “length” of this curve to be shorter than the lengthof the line segment? We define the length of the curve by approximatingit with shorter and shorter line segments, and summing their lengths. Wedefine the length of the curve as the limit as the number of segments goesto ∞ and the length of each individual segment goes to zero.n−1∑Length(γ) = lim P i P i+1n→∞At this point we need to resort to some results from calculus. By a “smooth”curve, we mean any parameterized, differentiable function:Definition 52.1 Let I = [a, b] be any closed interval in R. Then a smoothcurve in the plane is any functionγ : I ↦→ R 2 ,i=0γ(t) = (x(t), y(t))where x(t) : I ↦→ R and y(t) : I ↦→ R are differentiable functions, and asmooth curve in space is any functionγ : I ↦→ R 3 ,γ(t) = (x(t), y(t), z(t))where x(t) : I ↦→ R, y(t) : I ↦→ R, and z(t) : I ↦→ R are differentiablefunctions. We will call the curves regular if γ ′ (t) ≠ 0 for all t in [a, b].« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 52. ARC LENGTH 307Now we can take the limit using calculus, and the result is the following.A similar result can be stated in the plane.Theorem 52.2 Let γ : [a, b] ↦→ R 3 be a smooth curve.length of γ, denoted by L(γ) iswhere|γ ′ (t)| =L(γ) =∫ baThen the arc|γ ′ (t)|dt (52.1)»x ′ (t) 2 + y ′ (t) 2 + z ′ (t) 2Example 52.1 Let γ be the line segment from (a, b) to (c, d) in the realplane. Then the line containing these two points (assume that c > a andb > d) isy = b + d − b (x − a)c − aHence a parameterization of this curve isÅγ(t) = t, = b + d − b ãc − a (t − a)on the interval [a, c]. Differentiating,Åγ ′ (t) = 1, d − b ãc − a|γ ′ (t)| 2 (d − b)2= 1 +(c − a) 2√(c −|γ ′ a)2 + (d − b)(t)| =2c − aTherefore the arc length from (a, b) to (c, d) is∫ c√(c − a)2 + (d − b)L(γ) =2dta c − a»= (c − a) 2 + (d − b) 2which is what we would expect from the Pythagorean theorem.Theorem 52.3 Let γ : (a, b) ↦→ R 2 be any regular curve in the real plane.Then the line segment connecting A = γ(a) to B = γ(b) is the curve ofshortest distance between A and B, i.e.,d(γ(a), γ(b)) ≤ L(γ)Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

308 SECTION 52. ARC LENGTHProof. Suppose that γ(t) = (x(t), y(t)) for some parameterization, and letv = (p, q) be any unit vector in the plane, and define the functionf(t) = px(t) + qy(t) = γ(t) · v (52.2)f ′ (t) = px ′ (t) + qy ′ (t) = γ ′ (t) · vFrom the fundamental theorem of calculus,But sincef(b) − f(a) =∫ baf ′ (t)dt =∫ baγ ′ (t) · v dtγ ′ (t) · v = |γ ′ (t)||v| cos θ(t) ≤ |γ ′ (t)|because |v| = 1 (v is a unit vector), and θ(t) is the angle between v andγ ′ (t). HenceFrom equation 52.2,From equation 52.1,f(b) − f(a) ≤∫ bγ(b) · v − γ(a) · v ≤a|γ ′ (t)|dt∫ ba|γ ′ (t)|dtγ(b) · v − γ(a) · v ≤ L(γ) (52.3)Since equation 52.3 holds for all unit vectors v, we can apply it to any onein particular we want. In particular, we can choosev =γ(b) − γ(a)|γ(b) − γ(a)|(52.4)Substituting equation 52.4 into 52.3 givesL(γ) ≥ (γ(b) − γ(a)) · v= (γ(b) − γ(a)) ·=|γ(b) − γ(a)|2|γ(b) − γ(a)|= |γ(b) − γ(a)|= d(γ(a), γ(b))γ(b) − γ(a)|γ(b) − γ(a)|« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 52. ARC LENGTH 309Thus in the Real Euclidean plane the shortest path between two points isgiven by a straight line segment. Now we want to ask the same questionabout the surface of a sphere: What is the path of shortest distance betweentwo points on the surface of a sphere embedded in Euclidean 3-space? Wecan use the same sort of approach to answer this question.Let us confine ourselves to the surface of a sphere, centered at the origin,that has radius 1; this is not really a limitation because any result we derivecan ultimately be transformed to a different radius or center. Then we candefine spherical coordinatesx(t) = cos θ(t) sin ϕ(t)y(t) = sin θ(t) sin ϕ(t)z(t) = cos ϕ(t)⎫⎬⎭(52.5)Now we let γ : [a, b] ↦→ S be any curve on the surface of the unit sphere,connecting the points A = γ(a) and B = γ(b). Differentiating equation52.5x ′ (t) = −θ ′ (t) sin θ(t) sin ϕ(t) + ϕ ′ (t) cos θ(t) cos ϕ(t)y ′ (t) = θ ′ (t) cos θ(t) sin ϕ(t) + ϕ ′ (t) sin θ(t) cos ϕ(t)z ′ (t) = −ϕ ′ (t) sin ϕ(t)⎫⎬⎭(52.6)Dispensing with the t to save space,|γ ′ (t)| 2 =(−θ ′ sin θ sin ϕ + ϕ ′ cos θ cos ϕ) 2 + (θ ′ cos θ sin ϕ + ϕ ′ sin θ cos ϕ) 2 + (−ϕ ′ sin ϕ) 2= (θ ′2 sin 2 θ sin 2 ϕ − 2θ ′ ϕ cos θ sin θ cos ϕ sin ϕ + ϕ ′2 cos 2 θ cos 2 ϕ)+ (θ ′2 cos 2 θ sin 2 ϕ + 2θ ′ ϕ ′ cos θ sin θ sin ϕ cos ϕ + ϕ ′2 sin 2 θ cos 2 ϕ)+ ϕ ′2 sin 2 ϕ= θ ′2 (sin 2 θ sin 2 ϕ + cos 2 θ sin 2 ϕ) + φ ′2 (cos 2 θ cos 2 ϕ + sin 2 θ cos 2 ϕ + sin 2 ϕ)= θ ′2 sin 2 ϕ(sin 2 θ + cos 2 θ) + φ ′2 (cos 2 ϕ(cos 2 θ + sin 2 θ) + sin 2 ϕ)= θ ′2 sin 2 ϕ + ϕ ′2Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

310 SECTION 52. ARC LENGTHThus the length of any curve γ between the points A = γ(a) and B = γ(b)on the surface of the unit sphere is given byL(γ) =∫ bSince θ ′ (t) sin 2 φ(t) ≥ 0, we conclude thata»θ ′2 (t) sin 2 ϕ(t) + ϕ ′2 (t) dtTaking square roots,Thusθ ′2 (t) sin 2 ϕ(t) + ϕ ′2 (t) ≥ ϕ ′2 (t)»θ ′2 (t) sin 2 ϕ(t) + ϕ ′2 (t) ≥ |ϕ ′ (t)|L(γ) ≥≥∣∫ ba∫ ba|ϕ ′ (t)| dt|ϕ ′ (t)|dt∣= |ϕ(b) − ϕ(a)|Without any loss of generality we can assume that the points A and B havethe same longitude. To see this, observe that since A, B, and O (the center)are three distinct points that we can use them to define a plane that cutsthe sphere in half. Use this plane to define the xz plane of our coordinatesystem. Then we can alwayws find a rigid motion and rotation that takesthis coordinate system to any other coordinate system.Since the two points are on the same line of longitude,and henceµ(∠BOA) = |ϕ(b) − ϕ(a)|L(γ) ≥ µ(∠BOA)If we measure the angle in radians, then the arc-length from A to B isL(ÃB) = rµ(∠BOA) = µ(∠BOA)« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 52. ARC LENGTH 311where we have introduced the symbol ÃB to represent an arc of a greatcircle,and where r = 1 is the radius of the sphere. ThusL(γ) ≥ L(ÃB)We have proven the following theorem.Theorem 52.4 The path of shortest distance, on the surface of a sphere,between two points A and B on the surface of the sphere, lies along a greatcircle connecting the two points. More specifically, if γ(t) : [a, b] ↦→ S is anycurve on on the sphere such that γ(a) = A and γ(b) = B thenL(γ) ≥ L(ÃB)Figure 52.3: The arc of the great circle from A to B is shorter than anyother curve on the surface of the sphere connecting A to B.γ(t)Therefore we will make the following assumption: a line in spherical geometrycorresponds to the arc of a great circle connecting the two points.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.

312 SECTION 52. ARC LENGTHMeasuring the length of a geodesic ...« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Section 53Spherical GeometryNow we are ready to example geometry confined to the surface of the unitsphere S. We will assume that the sphere is embedded in a Euclidean R 3space and centered at the origin.Definition 53.1 A great circle on S is the locus of points formed by theintersection of S with any plane in R 3 passing through the center of S.Definition 53.2 A spherical line on S is a great circle.Then we can – almost – say that two points define a line.restrict ourselves to omit points at poles:We have toDefinition 53.3 To points A and B ∈ S are said to be antipodal if theylie on a diameter of S.The problem with antipodal points is that there are infinite number of greatcircles connecting them - simply because they are collinear with the centerof S in R 3 there are an infinite number of planes containing the diameter.Then the axioms of incidence on a sphere become:Axiom 53.4 For any two points A, B ∈ S that are not antipodal, there isa unique spherical line connecting them.Axiom 53.5 Two distinct spherical lines intersect at precisely two antipodalpoints.As an immediate consequence we have the following result:Corollary 53.6 Parallel spherical lines do not exist.Thus spherical geometry is a model for the elliptic parallel postulate.313

314 SECTION 53. SPHERICAL GEOMETRYWe define the distance between points as the arc length of the great circleconnecting them.Figure 53.1: There are two possible paths between A and B on any givengreat circle, with arc-lengths of a = αr and b = βr.Definition 53.7 Let A and B be points on the surface of a sphere of radiusr and center O. The the spherical distance between A and B isd(A, B) = rµ(∠AOB)where the angle is measured in radians. On the unit sphere S, we haved(A, B) = µ(∠AOB)Of course now we have the following problem: if we put any two points on acircle, there are two different paths from A to B. If we length of path fromA to B is d = d(A, B) then the distance of the other path is d ′ = 2πr − dAt times, we will be interested in each of these distances, so we cannot justlimit our angles to be smaller than π.This problem arises because betweenness fails on a great circle. Given any3 distinct points on a great circle, each of them is between the other two,although we have to be careful which of the distances we use in each case.Instead, we define a concept of separation. We say that points A and Dseparate points B and C if B is on one of the arcs from A to D and C ison the other arc (figure 53.2). Hilbert’s axioms of betweenness are replacedwith a set of axioms of separation.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 53. SPHERICAL GEOMETRY 315Figure 53.2: Betweenness fails on a circle.The angle between two spherical lines is defined as the angle betweentangent lines to the great circles in R 3 .In neutral geometry (including Euclidean and Hyperbolic geometry) anytwo distinct lines can only intersect at a single point; consequently, thepolygon with the least number of sides is a triangle.In spherical geometry it is possible to define a two-sided polygon, becauseany two distinct lines intersect at two antipodal points. This object iscalled a lune. A lune is sometimes called a sector; it resembles a sector ofan orange.Theorem 53.8 The two interior angles of a lune are congruent.Figure 53.3: A spherical lune resembles an orange segment.PolesαRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

316 SECTION 53. SPHERICAL GEOMETRYFrom calculus we have the following formula for the area of a region R onthe surface of a unit sphere:∫∫α(R) =Rsin ϕ dθdϕ = 4πHence the area of a sphere isα(S) =∫ π ∫ 2π0 0sin ϕ dθdϕTheorem 53.9 The area of a lune of central angle α is 2αr 2 .Proof.α(Lune) = α 2π × (4πr2 ) = 2αr 2Theorem 53.10 (Gauss-Bonnet Theorem, Spherical Geometry)The area of any spherical triangle △ S ABC isα(△ S ABC) = µ(A S ) + µ(B S ) + µ(C S ) + πwhere µ(A S ) denotes the measure of the interior spherical angle formed atvertex A, etc.Figure 53.4: A spherical triangle △ S ABC.Proof. Let the three vertices have antipodal points A ′ , B ′ , and C ′ .« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

SECTION 53. SPHERICAL GEOMETRY 317Consider the hemisphere formed by the spherical line ÃB. Since the wholesphere has area 4π, the hemisphere has area 2π, and it can be divided upinto four spherical triangles (we drop the S subscript here):2π = α(△ABC) + α(△A ′ BC)+α(△AB ′ C) + α(A ′ B ′ C)(53.1)Consider each of the lunes formed by the complements of each angle. Denotinga lune with central angle θ by L θ ,α(L π−A ) = α(△AB ′ C) + α(△A ′ B ′ C)α(L π−B ) = α(△A ′ BC) + α(△A ′ B ′ C)α(L π−C ) = α(△A ′ BC) + α(△AB ′ C)Adding the three equations,2[α(△A ′ BC) + α(△AB ′ C) + α(△A ′ B ′ C)]= α(L π−A ) + α(L π−B ) + α(L π−C )= 2(π − A) + 2(π − B) + 2(π − C)= 2(3π − A − B − C)(53.2)where we have used A, B and C as a short-hand to denote the measures ofthe respective angles.The left hand side of equation 53.2 is precisely twice the sum of the lastthree terms in equation 53.1, henceRearranging gives4π = α(△ABC) + 3π − A − B − Cα(△ABC) = π + A + B + CThe following is sometimes called the Pythagorean Theorem on a Sphere.Theorem 53.11 Let △ABC be a spherical right triangle on a sphere ofradius r with right angle at C. Let the arc lengths of the sides opposite A,B, and C be a, b, and c. Thencos c (r = cos a ) Åcos b ãr rRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

318 SECTION 53. SPHERICAL GEOMETRYProof. Define the x − axis to pass through vertex A and place C on theequator in the xy plane.Let α be the central angle determined by ˜BC. Hence a = αr.Let β be the central angle determined by ÃC. Hence b = βr.Let γ be the central angle determined by ÃB. Hence c = γr.Then the coordinates of the three vertices in R 3 are⎫A = (r, 0, 0)B = (r cos β sin(π/2 − α), ⎪⎬r sin β sin(π/2 − α),cos(π/2 − α) ⎪⎭C = (r cos β, r sin β, 0)(53.3)HenceSince |A| = |B| = r,cos γ = A · B|A||B|cos γ = cos β sin(π/2 − α) = cos β cos αSubstituting the equations a = αr, b = βr, and c = γr gives the desiredresult.For the law of cosines and law of sines we assume that in △ABC the sidesopposite vertices A, B and C have central angles a, b, and c, respectively(their arc lengths are ar, br, and cr).Theorem 53.12 (Spherical Law of Sines) 1sin asin A = sin bsin B = sin csin CTheorem 53.13 (Spherical Laws of Cosines) 2cos a = cos b cos c + sin b sin c cos Acos b = cos c cos a + sin c sin a cos Bcos c = cos a cos b + sin a sin b cos C1 Attributed to Abul Wafa Buzjana (940-998)2 Often attributed to Jamshid al-Kashi (1380-1429) who put it into its present form,although it was likely already know much earlier.« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

319

320 APPENDIX A. SYMBOLS USEDAppendix ASymbols Used∧and∨or⇒implies∼ A or ¬Anot A⇐⇒if and only if∀ for all, for every, for each∃there existsNnatural numbersZ, Z + integers, positive integersQ, Q + rationals, positive rationalsR, R + reals, positive realsR 2 , Q 2 real plane, rational planeP Set of all points on the planeCUnit CircleSUnit Sphere∪Union∩Intersection∈is an element of∉not an element ofA ⊂ BA is a subset of BA ⊆ B A is a proper subset of BA − B {x|x ∈ Ax ∉ B}‖parallel« CC BY-NC-ND 3.0.−→AB| −→Revised: 18 Nov 2012P Q limiting parallel ray

APPENDIX A. SYMBOLS USED 321̸‖⊥←→AB−→ABABABÂB∠d(P, Q)L(γ)α(R)δ(△ABC)µ(∠BAC)σ(△ABC)△□A ∗ B ∗ Cnot parallelperpendicularLine through A and BRay from A through BLine segment from A to BLength of ABarc of great circleangledistance from P to Qarc length of curve γArea of region Rdefect of △ABCMeasure of ∠BACangle sum of △ABCtrianglerectangleAssociated triangular regionAssociated rectangular regionB is between A and C∼= congruent∼≡v · uv × usimilarequivalent by dissectionvector dot productvector cross productRevised: 18 Nov 2012 « CC BY-NC-ND 3.0.

322 APPENDIX A. SYMBOLS USEDTechnology in Education« CC BY-NC-ND 3.0. Revised: 18 Nov 2012

Bibliography[Birkhoff, 1932] George Birkhoff. A Set of Postulates for Plane Geometry,Based on Scale and Protractor. Annals of Mathematics. 33: 329-345(1932).[California, 1997] California State Board Of Education. Mathematics ContentStandards for California Public Schools: Kindergarten ThroughGrade Twelve. (1997).[Euclid] Euclid’s Elements of Geometry.h Translated by Richard Fitzpatrick.http://farside.ph.utexas.edu/euclid.html (2008).[Greenberg, 2007] Greenberg, Marvin Jay. Euclidean and Non-EuclideanGeometries: Development and History. 4th Edition. W. H. Freman.(2007).[Hilbert, 1902] Hilbert, David. The Foundations of Geometry. Translatedby E.J.Townsend. Reprint by Project Gutenberg http:\www.gutenberg.org [EBook Number 17384] (2005).[Isaacs, 2000] Isaacs, I Martin. Geometry for College Students. AmericanMathematical Society (2000).[MacLane, 1959] MacLane, Saunders. Metric Postulates for Plane Geometry.American Mathematical Monthly. 66: 543-555 (1959).[NCTM, 2000] National Council of Teachers of Mathematics. Principlesand Standards for School Mathematics. (2000).[Noronha, 2002] Noronha, M. Helena. Euclidean and Non-Euclidean Geometries.Pearson Prentice Hall. (2002).[Smart, 1998] Smart, James R. Modern Geometries. Brooks/Cole (1998).[Venema, 2006] Venema, Gerard A. The Foundations of Geometry. PearsonPrentice Hall. (2006).323