Chapter 3 Foundations of Geometry 1: Points, Lines ..., Study notes of Geometry

Download Chapter 3 Foundations of Geometry 1: Points, Lines ... and more Study notes Geometry in PDF only on Docsity! Chapter 3 Foundations of Geometry 1: Points, Lines, Segments, Angles 3.1 An Introduction to Proof Syllogism: The abstract form is: 1. All A is B. 2. X is A 3. ∴ X is B Example: Let’s think about an example. Remark: Syllogism provides the basis for moving from the general to the particular, a process called deductive logic. On the other hand, Inductive logic, moves from the particular to the general. If-then statements, conditionals p → q read: ”p implies q, or If p, then q. The condition p is called the hypothesis, the Given part, and q the conclusion-the prove part. Negate a statement: if p is any condition, ∼ p means not p. Converse, Contrapositive: Given conditional its converse its contrapositive p → q q → p ∼ q →∼ p Remark: The contrapositive is logically equivalent to the original conditional, but the con- verse is not. Example 3.1 For example: Right angles are congruent. Find the if-then conditional of this statements, and find its converse and contrapositive. 9 Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 10 Example 3.2 If two sides of a triangle are congruent, then the angles opposite are congru- ent. Converse: In this case we have two statements are logically equivalent. Logically equivalent: if we have p → q and q → p i.e., p ↔ q we say that p and q are logically equivalent., or we can also say that q characterizes p. Direct Proofs direct proof, which is the embodiment of the propositional syllogism: 1. p implies q 2. q implies r 3. ∴ p implies r. Example 3.3 Given: ∠A and ∠B are right angles, Prove: ∠A is congruent to ∠B. Proof: Conclusions Justifications (1) The measures of ∠A and ∠B are each 900. Definition (of right angle) (2) m∠A = m∠B Algebra (3) ∴ ∠A ∼= ∠B Definition of congruence 2 Indirect Proof: Prove by contradiction Assume that the conclusion you are trying to prove is false, and then show that this leads to a contradiction of the hypothesis. In another words, to show that p implies q, one assumes p and notq, and then proceeds to show that not p follows, which is a contradiction. This means that we prove the contrapositive of the given proposition instead of the proposition itself. Example 3.4 Show that a prime number > 2 must be an odd number. Rule of Elimination Example 3.5 Trichotomy property of real numbers: For any real numbers a and b, either a < b, a = b, or a > b. For example: if we want to show x = 0. we can show it is impossible that x > 0 and x < 0. Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 13 Proof: Since C,D ∈ ←→AB, so ←→ AB passes C,D as well. By Axiom I, ←→ CD = ←→ AB. 2 Axiom 3.9 (I-2) Three noncollinear points determine a (unique) plane. Axiom 3.10 (I-3) If two points lie in a plane, then any line containing those two points lies in that plane. Axiom 3.11 (I-4) If two distinct planes meet, their intersection is a line. Axiom 3.12 (I-5) Space consists of at least four noncoplanar points, and contains three noncollinear points. Each plane is a set of points of which at least three are noncollinear, and each line is a set of at least two distinct points. Models for the axioms of incidence The model with the smallest number of points has four points, six lines, and four planes. Construction: Arrange four points A,B, C, D in ordinary xyz-space, where A is the origin, and B, C, D are at a unit distance from A on each of the three coordinate axes. This forms a tetrahedron ABCD. Theorem 3.13 If two distinct lines ` and m meet, their intersection is a single point. If a line meets a plane and is not contained by that plane, their intersection is a point. Proof: we prove the first part by contradiction. Assume ` ∩ m at two distinct points A 6= B. Then A, B are on both ` and m. Hence ` = m by Axiom I. This contradicts to the distinctness of ` and m. The second part can also be proved by contradiction. Assume a line ` does not lie in a plane P, i.e., ` * P, and A ∈ ` ∩ P. Assume point B 6= A also is a intersection point, i.e., B ∈ `∩P. Thus we have {A,B} ⊆ P and {A,B} ⊆ `. By Axiom I, ←→ AB = `, and by Axiom 3, ←→ AB ∈ P. This in turn gives ` ⊆ P, which is a contradiction. This completes the proof. 2 3.4 Distance, Ruler Postulate, Segments, Rays, and Angles We assume that the distance (or metric) between any two points can be measured. Therefore, following axioms are assumed: Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 14 Axiom 3.14 (Metric Axioms) D-1: Each pair of points A and B is associated with a unique real number, called the distance from A to B, denoted by AB. D-2 For all points A and B, AB ≥ 0, with equality only when A = B. D-3: For all points A and B, AB = BA. Remark: 1.) the distance between two points is not affected by the order in which they occur, i.e., AB = BA 2. Triangle inequality(will be proved in Chapter 3): Given three points A,B, C, then AB + BC ≥ AC the equality holds when B “between”(defined later) A and C. Definition and properties of betweenness Definition 3.15 (Betweenness) For any three points A,B and C, we say that B is be- tween A and C, and we write A − B − C, iff A, B, and C are distinct, collinear points, and AB + BC = AC. Theorem 3.16 If A−B − C then C −B − A, and neither A− C −B nor B − A− C. Proof: The first statement is trivial (using definition). Let’s prove A− C − B is not true. Assume A − C − B, then AC + CB = AB. Since A − B − C, AB + BC = AC. So we have AC + CB + BC = AB + BC = AC, i.e., 2BC = 0. We then have B = C, which is a contradiction. Similarly one can prove the last part of the theorem. 2 Definition 3.17 (Ordering of four points on a line) If A,B,C,D are four distinct collinear points, then we write A − B − C − D iff the composite of all four betweenness relations A−B − C, B − C −D, A−B −D and A− C −D are true. Theorem 3.18 If A−B − C, B − C −D, and A−B −D hold, then A−B − C −D. Proof: By definition, only need to show A− C −D. 2 Example 3.19 Suppose that in a certain metric geometry satisfying Axioms D1-D3, points A,B,C and D are collinear and AB = 2, AC = 3, AD = 4, BC = 5 BD = 6, and CD = 7. What betweenness relations follow, by definition, among these points? Can these four points be place in some order such that a betweenness relation for all four points holds for them, as in the above definition? Solutions: Note 1)the geometry given here is not the ordinary Euclidean geometry. 2) Use the definition of betweenness to check the four possibilities: for collections of {A, B, C}, {A,B, D}, {A,C,D}, {B, C, D} 3) Concludes there is no “quadruple” betweenness. Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 15 Segments, Rays, and Angles The concept of betweenness can be used to define segments and rays, and angles. Definition 3.20 Segment AB: AB = {X : A−X −B, X = A, orX = B} Ray AB: −→ AB = {X : A−X −B, A−B −X, X = A, orX = B} Angle ABC: ∠ABC = −→ BA ∪ −−→BC (A,B, and C noncollinear) One also can give a similar formula to the line AB (although a line is an undefined object in the Euclidean Geometry):←→ AB = {X : X − A−B,A−X −B, A−B −X, X = A, orX = B} The point A and B are called the end points of segment AB, and point A is the end point (or origin) of ray −→ AB. Point B in ∠ABC is called the vertex of the angle, and rays −→ BA and −−→ BC its sides. Extended segments, rays Definition 3.21 The extension of segment AB is either the ray −→ AB (in the direction B) , the ray −→ BA (in the direction A), or the line ←→ AB ( in both directions). The extension of ray −→ AB is just the line ←→ AB. Definition 3.22 A segment without its end points is called an open segment, and a ray without its point (origin) is called an open ray. Any point of a segment (or ray) that is not an end point is referred to as an interior point of that segment (or ray). Ruler Postulate Axiom 3.23 (D4: Ruler Postulate) The points of each line ` may be assigned to the en- tire set of real numbers x, −∞ < x < ∞, called coordinates, in such a way that (1) each point on ` is assigned to a unique coordinate; (2) no two points are assigned to the same coordinate; (3) any two points on ` may be assigned the coordinates zero and a positive real number, respectively; (4) if points A and B on ` have coordinates a and b, then AB = |a− b| ordering of the geometric points on a line and the ordering of their coordinates Theorem 3.24 For any line ` and any coordinate system under the Ruler Postulate, if A[a], B[b], and C[c] are three points on line `, with their coordinates, then A − B − C iff either a < b < c or c < b < a. Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 18 Remark: this definition is only valid in Euclidean geometry. Axiom 3.32 (A2: Angle addition postulate) If D lies in the interior of ∠ABC, then m∠ABD + m∠DBC = m∠ABC. Conversely, if m∠ABD + m∠DBC = m∠ABC, then ray −−→ BD passes through an interior point of ∠ABC. Definition 3.33 (Betweenness For Rays) For any three rays −→ BA, −−→ BD and −−→ BC (having the same end point), we say that ray −−→ BD lies between rays −→ BA and −−→ BC, and we write−→ BA- −−→ BD- −−→ BC, iff the rays are distinct and m∠ABD + m∠DBC = m∠ABC Example 3.34 Suppose that the betweenness relation C − E − D − B holds on line ←→ BC. Show that −→ AC- −→ AE- −−→ AD- −→ AB holds (where the definition of betweenness for four rays is directly analogous to that of four points), and that m∠CAB = m∠1 + m∠2 + m∠3. , Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 19 Solution: Need to show the betweenness for −→ AC- −→ AE- −−→ AD, −→ AC- −→ AE- −→ AB, −→ AC- −−→ AD- −→ AB, and −−−→ ACE- −−→ AD- −→ AB. Since E is interior to ∠CAD, −→ AC- −→ AE- −−→ AD follows from Axiom A-2. Similarly for the other three betweenness. It follows by definition−→ AC- −→ AE- −−→ AD- −→ AB. Now m∠CAB = m∠CAD + m∠3 = (m∠1 + m∠2) + m∠3. Remark: Similar results hold concurrent rays (having the same end point) and collinear points. We say there is a duality between them. Axiom 3.35 (A3: Protractor postulate) The set of rays −−→ AX lying on one side of a given line −→ AB, including ray −→ AB, may be assigned to the entire set of real numbers x, 0 ≤ x < 180, called coordinates, in such a manner that (1) each ray is assigned to a unique coordinate (2) no two rays are assigned to the same coordinates (3) the coordinate of −→ AB is 0 (4) if rays −→ AC and −−→ AD have coordinates c and d, then m∠CAD = |c− d|. we state a dual result of Theorem 3.24 next without proof. Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 20 if e < f < g, then −→ AE- −→ AF - −→ AG. Angle construction Theorem Definition 3.36 (Angle Bisector) We define an angle bisector for ∠ABC to be any ray −−→ BD lying between the sides −→ BA and −−→ BC such that m∠ABD = m∠DBC. Theorem 3.37 (Angle Construction Theorem) For any two angles ∠ABC and ∠DEF such that m∠ABC < m∠EDF , there is a unique ray −−→ EG such that m∠ABC = m∠GEF and −→ Ed- −−→ EG- −→ EF Corrolary 3.38 The bisector of any angle exists and is unique. Proof of the theorem (Outline): (1) Let a = m∠ABC and b = ∠EDF , then 0 < a < b < 180 (By Axiom A1) (2) Set up a coordinate system for the rays from E on the D-side of line ←→ EF , with 0 assigned to the ray −→ EF . (Protractor Postulate) (3) If the coordinate of ray −−→ ED is x, then b = m∠DEF = |0− x| = x. (4) Let ray −−→ EG be the unique ray having coordinate a. Then −→ Ed- −−→ EG- −→ EF . (5) m∠GEF = |a− 0| = a = m∠ABC Perpendicularity Definition 3.39 If A−B − C, then −−→ BC and −→ BAare called opposite or opposing rays. Theorem 3.40 Given a ray −→ BA, then its opposing ray −−→ BC exists and is unique. Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 23 Questions: Think about how many cases can we have for two angles with a common side. Which of the cases give two adjacent angles? Theorem 3.49 If line ←→ BD meets segment AC at an interior point B on that segment, then←→ BD⊥AC iff the adjacent angles at B have equal measures. Proof: Need to prove both forward and converse statements. 2 Theorem 3.50 Given a point A on line `, there exists a unique line m perpendicular to ` at A. Proof: Need to show existence and uniqueness. existence: construct a line ←→ AB such that the adjacent two angles at A have the equal measure, then by Theorem 3.50, ←→ AB⊥ell. uniqueness: Assume ←→ AC⊥ell, want to show ←→ AC = ←→ AB. m∠DAB = m∠DAC = 90 Now use Protractor postulate, set up a coordinate system, such that the coordinate of−→ AB = 0, then by the protractor postulate, the coordinate of −→ AB is 90, and so is −→ AC, so we must have −→ AC = −→ AB. One then similarly can show the other two halves of −→ AC and −→ AB are equal. That concludes ←→ AC = ←→ AB. 2 Vertical angles Yi Wang Chapter 3. Foundations of Geometry 1: Points, Lines, Segments, Angles 24 Definition 3.51 Vertical angles: are two angles having the sides of one opposite the sides of the other. Theorem 3.52 (Vertical Pair Theorem) Vertical angles have equal measures. Proof: Use the Linear Pair Axiom twice and Theorem 3.42. 2 Chapter 4 Foundations of Geometry 2 4.1 Triangles, Congruence Relations, SAS Hypothesis 25