This handout gives a summary of topics and definitions in Chapter
|
|
- Collin Andrews
- 5 years ago
- Views:
Transcription
1 MA Notes on Chapter 6 This handout gives a summary of topics and definitions in Chapter 6. 1 Definitions. In the following, we assume that we have a Protractor Geometry (which means we have a Metric Geometry with Pasch Postulate and a protractor for all angles. We are about to add one more axiom and define the Neutral Geometry. 1. Comparisons of segments and angles. Given segments AB and CD, wesaythatab > CD if AB > CD. If AB = CD then we say AB = CD. Given angles ABC, DEF we say that ABC > DEF if m( ABC) > m( DEF) andifm( ABC) = m( DEF) thenwesay ABC = DEF. 2. Comparison of triangles. Given a triangle, ABC we shall find it convenient to abbreviate ABC to B and similarly for each of the three vertices. Given triangles ABC, DEF we say that they are congruent if there is a bijection F from S = {A, B, C} to T = {D, E, F} such that for each X S, wehave X = f(x) andalsoforx Y S, wehave XY = f(x)f(y ). Note that thus the notion of congruence involves checking six corresponding entities to be congruent (the three angles and the three sides). We shall write ABC = f(a)f(b)f(c). Warning. We shall not use the congruence symbol, unless we have the order of vertices arranged so that the corresponding entities are congruent. We will express the congruence by general words triangles ABC and DEF are congruent if we have not specified the correspondence between the vertices. Example. Thus, while it is true that triangles ABC and ACB are always congruent (when we match all vertices to themselves,) we can say ABC = ACB only if AB = AC and m( B) =m( C), i.e. the triangle is isoceles with AB = AC. 2 1 Preliminary Version.November 30, The theorem which says that AB = AC implies ABC = ACB is known as pons asimorum. It is amusing to look up what this means and why. This Theorem has a very easy proof due to Pappus, which is valid in a Neutral geometry. 1
2 3. The SAS axiom. In general, congruence of two triangles involves checking six congruences. In the Euclidean Geometry, it can be shown that a well chosen set of three of the six quantities is enough to deduce the congruence. We define: The SAS Axiom. Suppose that ABC, DEF are triangles. If AB = DE, B = E and BC = EF, then ABC = DEF. 4. Neutral (or Absolute) Geometry. A Protractor Geometry satisfying SAS axiom is said to be a Neutral Geometry. Both Euclidean and Poincare Geometries are Neutral. We shall deduce several congruence theorems and familiar facts assuming the Neutral Geometry hypothesis. It is tempting to think that a neutral Geometry shall have all the usual facts of Euclidean Geometry as true theorems, but that is not the case unless and until the so-called Parallel Postulate is added in. Thus we don t have the existence of a unique line M through a point P outside a line L so that L is parallel to M. We don t have the familiar theorem that the sum of the measures of the three angles of a triangle is a constant 180. Many theorems about angles between parallel lines and their transversals are also unavailable. Theorems in a Neutral Geometry. 1. Pons Asinorum. In a triangle ABC, ifab = AC then m( B) = m( C). Idea of proof. Argue ABC = ACB using SAS where the corresponding quantities are: AB = AC, m( A) =m( A),AC = AB. It follows that m( B) = m( C). 2. ASA Theorem. Given triangles ABC, DEF, if A = D, AB = DE and B = E then ABC = DEF. Idea of Proof. If AC = DF then we are done by SAS. Otherwise, assume without loss of generality, that AC < DF. By segment construction, there is a point G in DF with AC = DG. By SAS, ABC = DEG. 2
3 Then G is seen to be in the interior of DEF and hence DEG < DEF. But this is a contradiction since DEG = ABC = DEF. Here the first congruence follows from the proven congruence of triangles, while the second one is from hypothesis. 3. ASA implies SAS. It is possible to deduce the SAS statement as a theorem, if we assume the ASA statement as a hypothesis. Idea of the proof. Thus, assume we have triangles ABC, DEF with AB = DE, B = E,BC = EF. We wish to prove ABC = DEF using ASA as a known thoerem. If C = F, then we are done by ASA. We may assume without loss of generality that F < C and construct aray CD such that m( BCD) =m( F )andd is in the same side as A of the line BC. From hypothesis, the ray CD meets the crossbar AB, sayatsomepoint G, sothatwehavea G B. By ASA, we get GBC = DEF. But then, DE = GB < AB by betweenness and this contradicts the hypothesis AB = DE. 4. Converse of Pons Asinorum. In a triangle ABC, ifm( B) = m( C), then AB = AC. Idea of Proof. As before, argue that ABC = ACB by comparing the following ASA elements: B = C, BC = CB, C = B. The result follows from the congruence of triangles. 5. SSS Theorem. Given triangles ABC, DEF, if AB = DE, BC = EF,AC = DF then ABC = DEF. Idea of Proof. Choose a point Q on the opposite side of A from the line BC such that QB = DE and QBC = E. Then by SAS, deduce that DEF = QBC. Then it is enough to prove ABC = QBC. The segment AQ must then meet the line BC. 3
4 Let the intersection point be named G and list the five cases for its position on the line. In each case, by repeated application of Pons Asinorum, deduce that A = Q and then apply SAS axiom using the known equalities: to finish of the congruence. AB = DE = QB, AC = DF = QC 6. Relation between SSS and SAS. Unlike the clean equivalence of ASA and SAS, it is not known if SSS alone is equivalent to SAS. While SAS implies SSS, the converse can only be proved, for now, with a few extra assumptions. No counterexamples are known where SSS holds but SAS does not. Thus, this is an outstanding problem. 7. Perpendiculars. Given a line L and a point P outside it, there is a line M perpendicular to L containing P. Recall that this was easy from the protractor axiom when P was on L. Later we shall show the line M to be unique for a given P and L. Idea of Proof. Choose two points A, B on L and construct a point Q on the opposite side of L with respect to P such that m( QAB) = m( PAB)andPA = QA. The segment PQ intersects L at some point M. First suppose A M. Then we use SAS to prove that PAM = QAM. from this, we deduce that AMP = AMQ and since they form a linear pair, each has measure 90 as needed. In case A = M we get the two congruent angles PAB and QAB to form a linear pair and hence each is 90 as needed. This perpendicular will turn out to be unique after the Exterior Angle Theorem below. Note that the perpendicular line does meet the line L. 8. Exterior Angle Theorem. Given a triangle ABC choose a point P such that B C P holds. Then the angle ACP forms a linear pair with C with respect to the line BC and we call it the linear mate of C along the line BC. There is a similar linear mate with respect to the line AC. Moreover, these two mates are congruent, since they form a vertical pair. The Exterior Angle Thoeorem states that the measure of a linear mate of C is bigger than the measures of either of the other two angles A, B. These two angles are called the remote interior angles of the mate. 4
5 Idea of Proof. 3 To prove the result for the mate of C along BC set M to be the midpoint of AC and choose a point E such that M is the midpoint of BE. Then we can argue that AMB = CME. This gives m( A) = m( ECM) and the angle m( ECM) isseentobesmallerthanour exterior angle ACP. 9. Uses of the Exterior Angle Theorem. Let L be a line and P a point outside it. Given points A B C on L, we see that for the triangle PAB the exterior angle PBC at B has bigger measure than that of the remote interior angle m( PAB). Fix the point B and for any point X B on L define a function ψ(x) =m( PXB). If we let X vary over the interior of the ray CB then we see that this function ψ(x) steadily decreases as the distance XB increases. Moreover, all the values of ψ(x) asx varies over the interior of CB are less than m( PBC). In contrast, if we take X to be in the interior of the ray BC, theneach value ψ(x) islessthanm( PBA). Again, ψ(x) decreases as XB increases. It follows that there is a unique B such that PB is perpendicular to L. We list, without proof, a few more applications. SAA Theorem. This is a generalization of the ASA theorem, where we allow the two angles to be any two of the three angles, not necessarily the ones at either end of the side. In Euclidean Geometry where the sum of all three angle measures is 180, this trivially equivalent to ASA. In general Neutral geometry, we need a construction. Comparison of Sides and Angles. This is a natural supplement to the result about isoceles triangles. Consider a triangle ABC. Then AB > AC iff C > B. 3 If unique parallel lines were to exist, then the measure of the exterior mate of C equals the sum of the measures of the remote interior angles, namely m( A)+m( B). However, this is not necessarily true for a Neutral Geometry, hence we have this weaker statement. 5
6 Triangle Inequality. Triangle inequality holds in a Neutral Geometry. Idea of Proof. Given a triangle ABC, wechoosed with D B C such that DC = AB+BC and deduce that m( D) <m( DAC). It follows that the opposite sides AC and DC satisfy AC < DC = AB + BC. This proves the triangle inequality. Distance from a point to various points of a chosen line. Let L be a line and P a point outside L. ChooseapointB on L such that the line PB is perpendicular to L. Let f be a ruler on L such that f(b) =0. LetPB = d. Define the function θ from L to R such that θ(x) =PX for each point X on L. The line is then a union of two rays R + = {X f(x) 0} and R = {X f(x) 0}. The only common point of the two rays is B. It can be shown that θ is an injective function on each of these rays with range [d, ). 4 Moreover, for any two points X 1,X 2 on L, wehave f(x 1 ) > f(x 2 ) iff θ(x 1 ) >θ(x 2 ). 4 Is it surjective? To be continued. 6
Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg ( )
Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg (2005-02-16) Logic Rules (Greenberg): Logic Rule 1 Allowable justifications.
More informationDefinitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg
Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg Undefined Terms: Point, Line, Incident, Between, Congruent. Incidence Axioms:
More informationDefinitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg ( )
Definitions, Axioms, Postulates, Propositions, and Theorems from Euclidean and Non-Euclidean Geometries by Marvin Jay Greenberg (2009-03-26) Logic Rule 0 No unstated assumptions may be used in a proof.
More informationNeutral Geometry. October 25, c 2009 Charles Delman
Neutral Geometry October 25, 2009 c 2009 Charles Delman Taking Stock: where we have been; where we are going Set Theory & Logic Terms of Geometry: points, lines, incidence, betweenness, congruence. Incidence
More informationright angle an angle whose measure is exactly 90ᴼ
right angle an angle whose measure is exactly 90ᴼ m B = 90ᴼ B two angles that share a common ray A D C B Vertical Angles A D C B E two angles that are opposite of each other and share a common vertex two
More informationFoundations of Neutral Geometry
C H A P T E R 12 Foundations of Neutral Geometry The play is independent of the pages on which it is printed, and pure geometries are independent of lecture rooms, or of any other detail of the physical
More informationChapter 3. Betweenness (ordering) A system satisfying the incidence and betweenness axioms is an ordered incidence plane (p. 118).
Chapter 3 Betweenness (ordering) Point B is between point A and point C is a fundamental, undefined concept. It is abbreviated A B C. A system satisfying the incidence and betweenness axioms is an ordered
More informationTriangle Geometry. Often we can use one letter (capitalised) to name an angle.
1) Naming angles Triangle Geometry Often we can use one letter (capitalised) to name an angle. A C B When more than two lines meet at a vertex, then we must use three letters to name an angle. Q X P T
More information10. Show that the conclusion of the. 11. Prove the above Theorem. [Th 6.4.7, p 148] 4. Prove the above Theorem. [Th 6.5.3, p152]
foot of the altitude of ABM from M and let A M 1 B. Prove that then MA > MB if and only if M 1 A > M 1 B. 8. If M is the midpoint of BC then AM is called a median of ABC. Consider ABC such that AB < AC.
More informationSOLUTIONS FOR. MTH 338 Final, June A 3 5 note card allowed. Closed book. No calculator. You may also use your list of the first 15 SMGS axioms.
SOLUTIONS FOR MTH 338 Final, June 2009 A 3 5 note card allowed. Closed book. No calculator. You may also use your list of the first 15 SMGS axioms. 1. In neutral geometry and without a compass, prove that
More informationCHAPTER 7 TRIANGLES. 7.1 Introduction. 7.2 Congruence of Triangles
CHAPTER 7 TRIANGLES 7.1 Introduction You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a
More information11. Prove that the Missing Strip Plane is an. 12. Prove the above proposition.
10 Pasch Geometries Definition (Pasch s Postulate (PP)) A metric geometry satisfies Pasch s Postulate (PP) if for any line l, any triangle ABC, and any point D l such that A D B, then either l AC or l
More informationtriangles in neutral geometry three theorems of measurement
lesson 10 triangles in neutral geometry three theorems of measurement 112 lesson 10 in this lesson we are going to take our newly created measurement systems, our rulers and our protractors, and see what
More informationThere are seven questions, of varying point-value. Each question is worth the indicated number of points.
Final Exam MAT 200 Solution Guide There are seven questions, of varying point-value. Each question is worth the indicated number of points. 1. (15 points) If X is uncountable and A X is countable, prove
More informationMAT 3271: Selected solutions to problem set 7
MT 3271: Selected solutions to problem set 7 Chapter 3, Exercises: 16. Consider the Real ffine Plane (that is what the text means by the usual Euclidean model ), which is a model of incidence geometry.
More informationPostulates, Definitions, and Theorems (Chapter 4)
Postulates, Definitions, and Theorems (Chapter 4) Segment Addition Postulate (SAP) All segments AB and BC have unique real number measures AB and BC such that: ABCBC = AC if and only if B is between A
More informationUNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).
EOCT Practice Items 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1). The dilation is Which statement is true? A. B. C. D. AB B' C' A' B' BC AB BC A' B'
More informationTriangle Congruence and Similarity Review. Show all work for full credit. 5. In the drawing, what is the measure of angle y?
Triangle Congruence and Similarity Review Score Name: Date: Show all work for full credit. 1. In a plane, lines that never meet are called. 5. In the drawing, what is the measure of angle y? A. parallel
More informationExercises for Unit V (Introduction to non Euclidean geometry)
Exercises for Unit V (Introduction to non Euclidean geometry) V.1 : Facts from spherical geometry Ryan : pp. 84 123 [ Note : Hints for the first two exercises are given in math133f07update08.pdf. ] 1.
More informationEUCLIDEAN AND HYPERBOLIC CONDITIONS
EUCLIDEAN AND HYPERBOLIC CONDITIONS MATH 410. SPRING 2007. INSTRUCTOR: PROFESSOR AITKEN The first goal of this handout is to show that, in Neutral Geometry, Euclid s Fifth Postulate is equivalent to the
More informationExercise 2.1. Identify the error or errors in the proof that all triangles are isosceles.
Exercises for Chapter Two He is unworthy of the name of man who is ignorant of the fact that the diagonal of a square is incommensurable with its side. Plato (429 347 B.C.) Exercise 2.1. Identify the error
More informationSection 5-1: Special Segments in Triangles
Section 5-1: Special Segments in Triangles Objectives: Identify medians, altitudes, angle bisectors, and perpendicular bisectors. perpendicular bisector C median altitude Vocabulary: A B Perpendicular
More informationUnit 1: Introduction to Proof
Unit 1: Introduction to Proof Prove geometric theorems both formally and informally using a variety of methods. G.CO.9 Prove and apply theorems about lines and angles. Theorems include but are not restricted
More informationLogic, Proof, Axiom Systems
Logic, Proof, Axiom Systems MA 341 Topics in Geometry Lecture 03 29-Aug-2011 MA 341 001 2 Rules of Reasoning A tautology is a sentence which is true no matter what the truth value of its constituent parts.
More informationGeometry Problem Solving Drill 08: Congruent Triangles
Geometry Problem Solving Drill 08: Congruent Triangles Question No. 1 of 10 Question 1. The following triangles are congruent. What is the value of x? Question #01 (A) 13.33 (B) 10 (C) 31 (D) 18 You set
More informationNAME: Mathematics 133, Fall 2013, Examination 3
NAME: Mathematics 133, Fall 2013, Examination 3 INSTRUCTIONS: Work all questions, and unless indicated otherwise give reasons for your answers. If the problem does not explicitly state that the underlying
More informationExistence Postulate: The collection of all points forms a nonempty set. There is more than one point in that set.
Undefined Terms: Point Line Distance Half-plane Angle measure Area (later) Axioms: Existence Postulate: The collection of all points forms a nonempty set. There is more than one point in that set. Incidence
More informationGeometry Midterm Exam Review 3. Square BERT is transformed to create the image B E R T, as shown.
1. Reflect FOXY across line y = x. 3. Square BERT is transformed to create the image B E R T, as shown. 2. Parallelogram SHAQ is shown. Point E is the midpoint of segment SH. Point F is the midpoint of
More informationMathematics 2260H Geometry I: Euclidean geometry Trent University, Fall 2016 Solutions to the Quizzes
Mathematics 2260H Geometry I: Euclidean geometry Trent University, Fall 2016 Solutions to the Quizzes Quiz #1. Wednesday, 13 September. [10 minutes] 1. Suppose you are given a line (segment) AB. Using
More informationSuggested problems - solutions
Suggested problems - solutions Parallel lines Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 4.1, pp 219-223.
More informationUNIT 1: SIMILARITY, CONGRUENCE, AND PROOFS. 1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1).
1) Figure A'B'C'D'F' is a dilation of figure ABCDF by a scale factor of 1. 2 centered at ( 4, 1). The dilation is Which statement is true? A. B. C. D. AB B' C' A' B' BC AB BC A' B' B' C' AB BC A' B' D'
More informationLesson. Warm Up deductive 2. D. 3. I will go to the store; Law of Detachment. Lesson Practice 31
Warm Up 1. deductive 2. D b. a and b intersect 1 and 2 are supplementary 2 and 3 are supplementary 3. I will go to the store; Law of Detachment Lesson Practice a. 1. 1 and 2 are. 2. 1 and 3 are. 3. m 1
More informationExercises for Unit I I (Vector algebra and Euclidean geometry)
Exercises for Unit I I (Vector algebra and Euclidean geometry) I I.1 : Approaches to Euclidean geometry Ryan : pp. 5 15 1. What is the minimum number of planes containing three concurrent noncoplanar lines
More informationHigher Geometry Problems
Higher Geometry Problems (1 Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement
More information1) If AB is congruent to AC, then B is congruent to C.
233 1) If is congruent to, then is congruent to. Proof of 1). 1) ssume ". (We must prove that ".) 2) ", because the identity is a rigid motion that moves to. 3) Therefore, Δ " Δ by the xiom. (The correspondence
More informationMathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions
Mathematics 2260H Geometry I: Euclidean geometry Trent University, Winter 2012 Quiz Solutions Quiz #1. Tuesday, 17 January, 2012. [10 minutes] 1. Given a line segment AB, use (some of) Postulates I V,
More informationQuestion 1 (3 points) Find the midpoint of the line segment connecting the pair of points (3, -10) and (3, 6).
Geometry Semester Final Exam Practice Select the best answer Question (3 points) Find the midpoint of the line segment connecting the pair of points (3, -0) and (3, 6). A) (3, -) C) (3, -) B) (3, 4.5)
More informationMidpoint M of points (x1, y1) and (x2, y2) = 1 2
Geometry Semester 1 Exam Study Guide Name Date Block Preparing for the Semester Exam Use notes, homework, checkpoints, quizzes, and tests to prepare. If you lost any of the notes, reprint them from my
More informationChapter 7. Geometric Inequalities
4. Let m S, then 3 2 m R. Since the angles are supplementary: 3 2580 4568 542 Therefore, m S 42 and m R 38. Part IV 5. Statements Reasons. ABC is not scalene.. Assumption. 2. ABC has at least 2. Definition
More information3. In the Poincaré Plane let l = 0 L and. 4. Prove the above Theorem. [Th 7.1.4, p173] 5. Show that in the Poincaré Plane there is
22 The Existence of Parallel Lines The concept of parallel lines has led to both the most fruitful and the most frustrating developments in plane geometry. Euclid (c. 330-275 B.C.E.) defined two segments
More information6 CHAPTER. Triangles. A plane figure bounded by three line segments is called a triangle.
6 CHAPTER We are Starting from a Point but want to Make it a Circle of Infinite Radius A plane figure bounded by three line segments is called a triangle We denote a triangle by the symbol In fig ABC has
More informationGeometry 3 SIMILARITY & CONGRUENCY Congruency: When two figures have same shape and size, then they are said to be congruent figure. The phenomena between these two figures is said to be congruency. CONDITIONS
More informationUCLA Curtis Center: March 5, 2016
Transformations in High-School Geometry: A Workable Interpretation of Common Core UCLA Curtis Center: March 5, 2016 John Sarli, Professor Emeritus of Mathematics, CSUSB MDTP Workgroup Chair Abstract. Previous
More informationSOLUTIONS TO ADDITIONAL EXERCISES FOR II.1 AND II.2
SOLUTIONS TO ADDITIONAL EXERCISES FOR II.1 AND II.2 Here are the solutions to the additional exercises in betsepexercises.pdf. B1. Let y and z be distinct points of L; we claim that x, y and z are not
More informationGeometry GENERAL GEOMETRY
Geometry GENERAL GEOMETRY Essential Vocabulary: point, line, plane, segment, segment bisector, midpoint, congruence I can use the distance formula to determine the area and perimeters of triangles and
More informationGeometry Honors Review for Midterm Exam
Geometry Honors Review for Midterm Exam Format of Midterm Exam: Scantron Sheet: Always/Sometimes/Never and Multiple Choice 40 Questions @ 1 point each = 40 pts. Free Response: Show all work and write answers
More informationHigher Geometry Problems
Higher Geometry Problems (1) Look up Eucidean Geometry on Wikipedia, and write down the English translation given of each of the first four postulates of Euclid. Rewrite each postulate as a clear statement
More informationThe Crossbar Theorem and Coordinates for Rays
The Crossbar Theorem and Coordinates for Rays Theorem (The Crossbar Theorem: If a point D lies in the interior of pc, then ray Dmeets segment C at some point G interior to C. Note: This is the first of
More informationI I I : Basic Euclidean concepts and theorems
I I I : Basic Euclidean concepts and theorems The purpose of this unit is to develop the main results of Euclidean geometry using the approach presented in the previous units. The choice of topics reflects
More informationGeometry First Semester Exam Review
Geometry First Semester Exam Review Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Name three points that are collinear. a. points T, Q, and R c. points
More informationTriangles. 3.In the following fig. AB = AC and BD = DC, then ADC = (A) 60 (B) 120 (C) 90 (D) none 4.In the Fig. given below, find Z.
Triangles 1.Two sides of a triangle are 7 cm and 10 cm. Which of the following length can be the length of the third side? (A) 19 cm. (B) 17 cm. (C) 23 cm. of these. 2.Can 80, 75 and 20 form a triangle?
More informationGeometry Arcs and Chords. Geometry Mr. Austin
10.2 Arcs and Chords Mr. Austin Objectives/Assignment Use properties of arcs of circles, as applied. Use properties of chords of circles. Assignment: pp. 607-608 #3-47 Reminder Quiz after 10.3 and 10.5
More informationHonors 213 / Math 300. Second Hour Exam. Name
Honors 213 / Math 300 Second Hour Exam Name Monday, March 6, 2006 95 points (will be adjusted to 100 pts in the gradebook) Page 1 I. Some definitions (5 points each). Give formal definitions of the following:
More information2) Are all linear pairs supplementary angles? Are all supplementary angles linear pairs? Explain.
1) Explain what it means to bisect a segment. Why is it impossible to bisect a line? 2) Are all linear pairs supplementary angles? Are all supplementary angles linear pairs? Explain. 3) Explain why a four-legged
More informationSOLUTION. Taken together, the preceding equations imply that ABC DEF by the SSS criterion for triangle congruence.
1. [20 points] Suppose that we have ABC and DEF in the Euclidean plane and points G and H on (BC) and (EF) respectively such that ABG DEH and AGC DHF. Prove that ABC DEF. The first congruence assumption
More informationTHE FIVE GROUPS OF AXIOMS.
2 THE FIVE GROUPS OF AXIOMS. 1. THE ELEMENTS OF GEOMETRY AND THE FIVE GROUPS OF AXIOMS. Let us consider three distinct systems of things. The things composing the first system, we will call points and
More informationSegment Measurement, Midpoints, & Congruence
Lesson 2 Lesson 2, page 1 Glencoe Geometry Chapter 1.4 & 1.5 Segment Measurement, Midpoints, & Congruence Last time, we looked at points, lines, and planes. Today we are going to further investigate lines,
More informationSOLUTIONS TO EXERCISES FOR. MATHEMATICS 133 Part 4. Basic Euclidean concepts and theorems
SOLUTIONS TO EXERCISES FOR MATHEMATICS 133 Part 4 Winter 2009 NOTE ON ILLUSTRATIONS. Drawings for several of the solutions in this file are available in the following file: http://math.ucr.edu/ res/math133/math133solutions04.figures.f13.pdf
More informationSUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-2) MATHEMATICS CLASS IX
SUMMATIVE ASSESSMENT-1 SAMPLE PAPER (SET-) MATHEMATICS CLASS IX Time: 3 to 3 1 hours Maximum Marks: 80 GENERAL INSTRUCTIONS: 1. All questions are compulsory.. The question paper is divided into four sections
More information2 Homework. Dr. Franz Rothe February 21, 2015 All3181\3181_spr15h2.tex
Math 3181 Dr. Franz Rothe February 21, 2015 All3181\3181_spr15h2.tex Name: Homework has to be turned in this handout. For extra space, use the back pages, or blank pages between. The homework can be done
More informationSOLUTIONS SECTION A [1] = 27(27 15)(27 25)(27 14) = 27(12)(2)(13) = cm. = s(s a)(s b)(s c)
1. (A) 1 1 1 11 1 + 6 6 5 30 5 5 5 5 6 = 6 6 SOLUTIONS SECTION A. (B) Let the angles be x and 3x respectively x+3x = 180 o (sum of angles on same side of transversal is 180 o ) x=36 0 So, larger angle=3x
More information5.5 Inequalities in Triangles
5.5 Inequalities in Triangles Learning Objectives Determine relationships among the angles and sides of a triangle. Understand the Triangle Inequality Theorem. Understand the Hinge Theorem and its converse.
More informationChapter 6 Summary 6.1. Using the Hypotenuse-Leg (HL) Congruence Theorem. Example
Chapter Summary Key Terms corresponding parts of congruent triangles are congruent (CPCTC) (.2) vertex angle of an isosceles triangle (.3) inverse (.4) contrapositive (.4) direct proof (.4) indirect proof
More informationMA 460 Supplement: Analytic geometry
M 460 Supplement: nalytic geometry Donu rapura In the 1600 s Descartes introduced cartesian coordinates which changed the way we now do geometry. This also paved for subsequent developments such as calculus.
More informationSegment Measurement, Midpoints, & Congruence
Lesson 2 Lesson 2, page 1 Glencoe Geometry Chapter 1.4 & 1.5 Segment Measurement, Midpoints, & Congruence Last time, we looked at points, lines, and planes. Today we are going to further investigate lines,
More informationSection 2-1. Chapter 2. Make Conjectures. Example 1. Reasoning and Proof. Inductive Reasoning and Conjecture
Chapter 2 Reasoning and Proof Section 2-1 Inductive Reasoning and Conjecture Make Conjectures Inductive reasoning - reasoning that uses a number of specific examples to arrive at a conclusion Conjecture
More information7. m JHI = ( ) and m GHI = ( ) and m JHG = 65. Find m JHI and m GHI.
1. Name three points in the diagram that are not collinear. 2. If RS = 44 and QS = 68, find QR. 3. R, S, and T are collinear. S is between R and T. RS = 2w + 1, ST = w 1, and RT = 18. Use the Segment Addition
More informationMath 1230, Notes 2. Aug. 28, Math 1230, Notes 2 Aug. 28, / 17
Math 1230, Notes 2 Aug. 28, 2014 Math 1230, Notes 2 Aug. 28, 2014 1 / 17 This fills in some material between pages 10 and 11 of notes 1. We first discuss the relation between geometry and the quadratic
More informationCircle Theorems Standard Questions (G10)
Circle Theorems Standard Questions (G10) Page 1 Q1.(a) A, B and C are points on the circumference of a circle with centre O. Not drawn accurately Work out the size of angle x. (1) Page 2 (b) P, Q and R
More information4 Arithmetic of Segments Hilbert s Road from Geometry
4 Arithmetic of Segments Hilbert s Road from Geometry to Algebra In this section, we explain Hilbert s procedure to construct an arithmetic of segments, also called Streckenrechnung. Hilbert constructs
More informationChapter 3 Cumulative Review Answers
Chapter 3 Cumulative Review Answers 1a. The triangle inequality is violated. 1b. The sum of the angles is not 180º. 1c. Two angles are equal, but the sides opposite those angles are not equal. 1d. The
More informationIB MYP Unit 6 Review
Name: Date: 1. Two triangles are congruent if 1. A. corresponding angles are congruent B. corresponding sides and corresponding angles are congruent C. the angles in each triangle have a sum of 180 D.
More informationThe Crossbar Theorem and Coordinates for Rays
The Crossbar Theorem and Coordinates for Rays Theorem (The Crossbar Theorem: If a point D lies in the interior of pc, then ray Dmeets segment C at some point G interior to C. Note: This is the first of
More information9 th CBSE Mega Test - II
9 th CBSE Mega Test - II Time: 3 hours Max. Marks: 90 General Instructions All questions are compulsory. The question paper consists of 34 questions divided into four sections A, B, C and D. Section A
More informationMTH 250 Graded Assignment 4
MTH 250 Graded Assignment 4 Measurement Material from Kay, sections 2.4, 3.2, 2.5, 2.6 Q1: Suppose that in a certain metric geometry* satisfying axioms D1 through D3 [Kay, p78], points A, B, C and D are
More information5200: Similarity of figures. Define: Lemma: proof:
5200: Similarity of figures. We understand pretty well figures with the same shape and size. Next we study figures with the same shape but different sizes, called similar figures. The most important ones
More informationEuclidean Geometry Proofs
Euclidean Geometry Proofs History Thales (600 BC) First to turn geometry into a logical discipline. Described as the first Greek philosopher and the father of geometry as a deductive study. Relied on rational
More informationGeometry S1 (#2211) Foundations in Geometry S1 (#7771)
Instructional Materials for WCSD Math Common Finals The Instructional Materials are for student and teacher use and are aligned to the Course Guides for the following courses: Geometry S1 (#2211) Foundations
More information2016 State Mathematics Contest Geometry Test
2016 State Mathematics Contest Geometry Test In each of the following, choose the BEST answer and record your choice on the answer sheet provided. To ensure correct scoring, be sure to make all erasures
More informationGeometry Advanced Fall Semester Exam Review Packet -- CHAPTER 1
Name: Class: Date: Geometry Advanced Fall Semester Exam Review Packet -- CHAPTER 1 Multiple Choice. Identify the choice that best completes the statement or answers the question. 1. Which statement(s)
More informationLesson 14: An Axiom System for Geometry
219 Lesson 14: n xiom System for Geometry We are now ready to present an axiomatic development of geometry. This means that we will list a set of axioms for geometry. These axioms will be simple fundamental
More informationHIGHER GEOMETRY. 1. Notation. Below is some notation I will use. KEN RICHARDSON
HIGHER GEOMETRY KEN RICHARDSON Contents. Notation. What is rigorous math? 3. Introduction to Euclidean plane geometry 3 4. Facts about lines, angles, triangles 6 5. Interlude: logic and proofs 9 6. Quadrilaterals
More information7.5 Proportionality Relationships
www.ck12.org Chapter 7. Similarity 7.5 Proportionality Relationships Learning Objectives Identify proportional segments when two sides of a triangle are cut by a segment parallel to the third side. Extend
More informationChapter 2. Reasoning and Proof
Chapter 2 Reasoning and Proof 2.1 Use Inductive Reasoning Objective: Describe patterns and use deductive reasoning. Essential Question: How do you use inductive reasoning in mathematics? Common Core: CC.9-12.G.CO.9
More informationANSWERS STUDY GUIDE FOR THE FINAL EXAM CHAPTER 1
ANSWERS STUDY GUIDE FOR THE FINAL EXAM CHAPTER 1 N W A S Use the diagram to answer the following questions #1-3. 1. Give two other names for. Sample answer: PN O D P d F a. Give two other names for plane.
More informationExterior Angle Inequality
xterior ngle Inequality efinition: Given Î, the angles p, p, and p are called interior angles of the triangle. ny angle that forms a linear pair with an interior angle is called an exterior angle. In the
More informationGeometry Practice Midterm
Class: Date: Geometry Practice Midterm 2018-19 1. If Z is the midpoint of RT, what are x, RZ, and RT? A. x = 19, RZ = 38, and RT = 76 C. x = 17, RZ = 76, and RT = 38 B. x = 17, RZ = 38, and RT = 76 D.
More information0110ge. Geometry Regents Exam Which expression best describes the transformation shown in the diagram below?
0110ge 1 In the diagram below of trapezoid RSUT, RS TU, X is the midpoint of RT, and V is the midpoint of SU. 3 Which expression best describes the transformation shown in the diagram below? If RS = 30
More informationName: Date: Period: ID: REVIEW CH 1 TEST REVIEW. 1. Sketch and label an example of each statement. b. A B. a. HG. d. M is the midpoint of PQ. c.
Name: Date: Period: ID: REVIEW CH 1 TEST REVIEW 1 Sketch and label an example of each statement a HG b A B c ST UV d M is the midpoint of PQ e Angles 1 and 2 are vertical angles f Angle C is a right angle
More informationTopic 2 [312 marks] The rectangle ABCD is inscribed in a circle. Sides [AD] and [AB] have lengths
Topic 2 [312 marks] 1 The rectangle ABCD is inscribed in a circle Sides [AD] and [AB] have lengths [12 marks] 3 cm and (\9\) cm respectively E is a point on side [AB] such that AE is 3 cm Side [DE] is
More informationHonors Geometry Mid-Term Exam Review
Class: Date: Honors Geometry Mid-Term Exam Review Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. Classify the triangle by its sides. The
More informationClass IX Chapter 8 Quadrilaterals Maths
Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between
More informationClass IX Chapter 8 Quadrilaterals Maths
1 Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Let the common ratio between the angles
More informationSimilarity of Triangle
Similarity of Triangle 95 17 Similarity of Triangle 17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree
More informationNozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1-Complete 1. in the parallelogram, each two opposite
More informationGeometry Honors: Midterm Exam Review January 2018
Name: Period: The midterm will cover Chapters 1-6. Geometry Honors: Midterm Exam Review January 2018 You WILL NOT receive a formula sheet, but you need to know the following formulas Make sure you memorize
More informationProofs. by Bill Hanlon
Proofs by Bill Hanlon Future Reference To prove congruence, it is important that you remember not only your congruence theorems, but know your parallel line theorems, and theorems concerning triangles.
More informationExercises for Unit I I I (Basic Euclidean concepts and theorems)
Exercises for Unit I I I (Basic Euclidean concepts and theorems) Default assumption: All points, etc. are assumed to lie in R 2 or R 3. I I I. : Perpendicular lines and planes Supplementary background
More informationCONGRUENCE OF TRIANGLES
Congruence of Triangles 11 CONGRUENCE OF TRIANGLES You might have observed that leaves of different trees have different shapes, but leaves of the same tree have almost the same shape. Although they may
More information