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Chapter 3 Homework Review Exercises 2 – 7 2, 3, 6, 7 are correct.

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## Chapter 3 Homework Review Exercises 2 – 7 2, 3, 6, 7 are correct.

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**Chapter 3 Homework**Review Exercises 2 – 7 2, 3, 6, 7 are correct. If line m is parallel to line l, then all points of m lie on the same side of l.**B-3 If A, B, and C are three distinct points lying**on the same line, then one and only one of the points is between the other two. Chapter 3 Homework Given A * B * C and A * C * D (a) Prove that A, B, C, and D are four distinct points. B-1 If A * B * C, then A, B, and C are three distinct points lying on the same line, and C * B * A. D 1. A, B, and Care distinct collinear points. A 2. A, C, and D are distinct collinear points B-1 B B-1 C D 3. Suppose points D and B were the same point. Then D is between A and C. But A * C * D means C is between A and D. This contradicts axiom B-3. Thus A, B, C, and D are four distinct points. (b) Prove that A, B, C, and D are collinear. 4. Since the line containing A, B, and C, and the line containing A, C, and D have two points in common, they must be the same line, making all four points collinear. I-1**Chapter 3 Homework**1 (c) Prove the corollary to Axiom B-4. Corollary (iii) If A and B are on opposite sides of l and if B and C are on the same side as l, then A and C are on opposite sides of l. 1. Assume A and C are on the same side of l. 2. Since B and C are also on the same side of l, then A and B are on the same side of l. Axiom B-4, part (i) 3. Contradiction. Therefore, A and C are on opposite sides of l.**Chapter 3 Homework**2. (b) Finish the proof of Proposition 3.3 by showing that A * B * D. A E 1. We have already proven that A, B, C, and D are four distinct collinear points. B Prop 2.3 C 2. There exists a point E not on the line through A, B, C, and D. 3. Assume B is not between A and D. Then either B * A * D or A * D * B. D Axiom B-3 4. Consider line . Line meets in point B. Regardless of whether B * A * D or A * D * B, points A and D are on the same side of . Def of same side 5. Because B * C * D (by the first part of the proposition), points C and D are on the same side of . Def of same side 6. Therefore, A and C are on the same side of . Steps 4, 5 and Axiom B-4, part i 7. But A * B * C, so A and C are on opposite sides of . Def of opposite sides Contradiction Proposition 3.3 Given A * B * C and A * C * D. Then B * C * D and A * B * D.**Chapter 3 Homework**2. (d) Prove the corollary to Proposition 3.3 E A B 1. There exists a point E not on the line through A, B, C, and D. C D 2. Consider line . Line meets in point B. Because A * B * C, points A and C are on opposite sides of . Because B * C * D, points C and D are on the same side of . Therefore, A and D are on opposite sides of by the corollary to Axiom B-4. Therefore, A * B * D. Corollary Given A * B * C and B * C * D. Then A * B * D and A * C * D.**Chapter 3 Homework**2. (d) Prove the corollary to Proposition 3.3 E A B 1. There exists a point E not on the line through A, B, C, and D. C D 2. Consider line . Line meets in point B. Because A * B * C, points A and C are on opposite sides of . Because B * C * D, points C and D are on the same side of . Therefore, A and D are on opposite sides of by the corollary to Axiom B-4. Therefore, A * B * D. 3. Assume C is not between A and D. Then either C * A * D or A * D * C. 4. Consider line . Line meets in point C. Regardless of whether C * A * D or A * D * C, points A and D are on the same side of . 5. Because B * C * D, points B and D are on opposite sides of . 6. Then A and B are on opposite sides of by the corollary to Axiom B-4. But we are given A * B * C, so points A and B are on the same side of Contradiction Corollary Given A * B * C and B * C * D. Then A * B * D and A * C * D.**Chapter 3 Homework**4. (a) Given A * B * C. If P is a fourth point collinear with A, B, and C, use Proposition 3.3 and an axiom to prove that A * B * P A * C * P. Assume A * C * P. Then A * B * C and A * C * P A * B * P by Prop 3.3. But we are given A * B * P. Contradiction. Proposition 3.3 Given A * B * C and A * C * D. Then B * C * D and A * B * D.**Chapter 3 Homework**8. Using the incidence and betweenness axioms, prove that every line has at least five points. • Given line l. • Line l contains two distinct points , call them B and D. (Axiom I-2) • There exist points A, C, and E such that A * B * D, B * C * D and B * D * E. (Axiom B-2) • A, B, and D are collinear; B, C, and D are collinear; B, D, and E are collinear. (Axiom B-1) • Since every pair of lines in step 4 shares points B and D, their lines are all line l. (Axiom I-1). • Therefore, l has at least 5 points.**Chapter 3 Homework**9. Prove: Given l, a point A on l, and a point B not on l. Then every point of the ray (except A) is on the same side of l as B. 1. A is the only point common to line l and line . (Prop 2.1) B P 2. Choose a point P on ray , P ≠ A or B. P A 3. By definition of ray, either A * P * B or A * B * P, so that A is not between B and P. l 4. Therefore, segment BP does not intersect line l, and so B and P are on the same side of l. Let’s call this Proposition H-9.**Prove the following proposition:**If point D is in the interior of CAB and a ray emanating from D intersects , then the ray does not intersect 1. Points B and C are on opposite sides of . Corollary to the Crossbar Theorem Corollary to the Crossbar Theorem: If point D is in the interior of CAB, then B and C are on opposite sides of . A B D C**Prove the following proposition:**If point D is in the interior of CAB and a ray emanating from D intersects , then the ray does not intersect 1. Points B and C are on opposite sides of . Corollary to the Crossbar Theorem 2. Let the given ray from D intersect ray at point E. By hypothesis 3. Assume ray intersects at point R. RAA Hypothesis A 4. Either D * R * E or D *E * R. Definition of ray E B 5. In either case, R and E are on the same side of . def same side R D 6. Points C and E are on opposite sides of . Cor to Crossbar Theorem C 7. From steps 1 and 6, points B and E are on the same side of . Axiom B-4 (ii) 8. From steps 5 and 6, points R and C are on opposite sides of . Corollary to Axiom B-4 9. But both R and C are on ray and all points on are on the same side of . Contradiction Proposition H-9**A** B Proposition 3.8: If D is in the interior of CAB, then (a) so is every other point on ray except A; (b) no point on the opposite ray to is in the interior of CAB; and (c) if C * A * E, then B is in the interior of DAE. D C • (c) Since and are the same line and D is in • the interior of CAB, we know that B and D • are on the same side of (). • All that remains is to prove that B and E • are on the same side of . • SinceC * A * E, we know C and E are on opposite sides of . • Since ray is between rays and , B and C are on opposite sides of • Then B and E are on the same side of • Therefore, B is in the interior of DAE. Def of angle interior Def of opposite sides Corollary to Crossbar Thm E Axiom B-4 (ii) Def of angle interior What is the flaw in this “proof” of part (c)?**Chapter 3 Homework**15. Find an interpretation in which the incidence axioms and the first two betweenness axioms hold, but B-3 fails in the following way: There exist three collinear points, no one of which is between the other two. Hint: In the usual Euclidean model, introduce a new betweenness relation A * B * C to mean that B is the midpoint of AC). First, verify that the Incidence Axioms and Axioms B-1 and B-2 hold in this interpretation of “betweenness.” Let l be a line with points A and B. Then by Axiom B-2, there exists a point C on l such that A * C * B. Again using Axiom B-2, there exists a point D on l such that C * D * B. A C D Using this interpretation of “betweenness,” C is the midpoint of AB, and D is the midpoint of CB. Consider points A, C, and D. None of the three is “between” the other two in this interpretation. But Axiom B-3 says “one and only one of the points is between the other two.” So Axiom B-3 does not hold. B l**Axioms of Congruence**Congruence Axiom 1 (C-1) If A and B are distinct points and if A is any point, then for each ray r emanating from A there is a unique point B on r such that B ≠ A and AB AB. r B B A A **Axioms of Congruence**Congruence Axiom 1 (C-1) If A and B are distinct points and if A is any point, then for each ray r emanating from A there is a unique point B on r such that B ≠ A and AB AB. Congruence Axiom 2 (C-2) If AB CD and AB EF, then CD EF. Moreover, every segment is congruent to itself.**Axioms of Congruence**Congruence Axiom 1 (C-1) If A and B are distinct points and if A is any point, then for each ray r emanating from A there is a unique point B on r such that B ≠ A and AB AB. Congruence Axiom 2 (C-2) If AB CD and AB EF, then CD EF. Moreover, every segment is congruent to itself. Congruence Axiom 3 (C-3) If A * B * C, A * B * C, AB AB, and BC BC, then AC AC. C A C B A B**Axioms of Congruence**Congruence Axiom 1 (C-1) If A and B are distinct points and if A is any point, then for each ray r emanating from A there is a unique point B on r such that B ≠ A and AB AB. Congruence Axiom 2 (C-2) If AB CD and AB EF, then CD EF. Moreover, every segment is congruent to itself. Congruence Axiom 3 (C-3) If A * B * C, A * B * C, AB AB, and BC BC, then AC AC. C A C B A B**Axioms of Congruence**Congruence Axiom 4 (C-4) Given any BAC (where, by the definitionof “angle,” is not opposite ) and given any ray emanating from a point A, then there is a unique ray on a given side of line such that BAC BAC. C C B B A A**Axioms of Congruence**Congruence Axiom 4 (C-4) Given any BAC (where, by the definitionof “angle,” is not opposite ) and given any ray emanating from a point A, then there is a unique ray on a given side of line such that BAC BAC. Congruence Axiom 5 (C-5) If A B, and B C, then . Moreover, every angle is congruent to itself.**Definition of Congruent Triangles**Triangles ABC and DEF are congruent if there exists a one-to-one correspondence between their vertices such that corresponding sides are congruent and corresponding angles are congruent. (SAS) Congruence Axiom 6 (C-6) If two sides and an included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then the two triangles are congruent. Corollary to SAS Given ABC and segment DE AB, there is a unique point F on a given side of line such that ABC DEF.**C**F B A E D (SAS) Congruence Axiom 6 (C-6) If two sides and an included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then the two triangles are congruent. Corollary to SAS Given ABC and segment DE AB, there is a unique point F on a given side of line such that ABC DEF.**C**F B A E D (SAS) Congruence Axiom 6 (C-6) If two sides and an included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then the two triangles are congruent. Corollary to SAS Given ABC and segment DE AB, there is a unique point F on a given side of line such that ABC DEF.**P **C By C-6 By C-1 By C-4 F B A E D (SAS) Congruence Axiom 6 (C-6) If two sides and an included angle of one triangle are congruent, respectively, to two sides and the included angle of another triangle, then the two triangles are congruent. Corollary to SAS Given ABC and segment DE AB, there is a unique point F on a given side of line such that ABC DEF.**Proposition 3.10: If in a ABC we have AB AC, then B** C. A A B C C B ABC CBA by C-6 (SAS) Therefore,B C**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. B C Proof: Assume BC is not congruent to EF. A RAA Hypothesis D P E 2. There is a point P on such that BC EP. F Axiom C-1 3. If P = F, we would be finished, so assume P F. 4. AC DP. Axiom C-3 5. Therefore, DF DP. Axiom C-2 6. Thus, F = P. Axiom C-1 (uniqueness) Contradiction**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. Proposition 3.12:GivenAC DF, then for any point B between A and C, there is a unique point E between D and F such that AB DE.**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. Proposition 3.12:GivenAC DF, then for any point B between A and C, there is a unique point E between D and F such that AB DE. Definition:AB < CD (or CD > AB) means that there exists a point E between C and D such that AB CE. D E B A C**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. Proposition 3.12:GivenAC DF, then for any point B between A and C, there is a unique point E between D and F such that AB DE. Definition:AB < CD (or CD > AB) means that there exists a point E between C and D such that AB CE. Proposition 3.13 (Segment Ordering): (a) Exactly one of the following three conditions holds (trichotomy): AB < CD, AB = CD, or AB > CD. (b) If AB < CD and CD EF, then AB < EF. (c) If AB > CD and CD EF, then AB > EF. (d) If AB < CD and CD < EF, then AB < EF (transitivity).**Definition: If two angles have a common side and the other**two sides form opposite rays, the angles are supplements of each other. Definition: An angle is a right angle if it has a supplement to which it is congruent. Definition: Two lines l and m are perpendicular if they intersect at a point A and if there is a ray on l and a ray on m such that BAC is a right angle. Note: The first two definitions are on page 19 in the textbook and the third is on page 42.**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. Proposition 3.12:GivenAC DF, then for any point B between A and C, there is a unique point E between D and F such that AB DE. Definition:AB < CD (or CD > AB) means that there exists a point E between C and D such that AB = CE. Proposition 3.13 (Segment Ordering): (a) Exactly one of the following three conditions holds (trichotomy): AB < CD, AB = CD, or AB > CD. (b) If AB < CD and CD EF, then AB < EF. (c) If AB > CD and CD EF, then AB > EF. (d) If AB < CD and CD < EF, then AB < EF (transitivity). Proposition 3.14: Supplements of congruent angles are congruent.**Proposition 3.10: If in a ABC we have AB AC, then B** C. Proposition 3.11 (Segment Subtraction): If A * B * C, D * E * F, AB DE, and AC DF, then BC EF. Proposition 3.12:GivenAC DF, then for any point B between A and C, there is a unique point E between D and F such that AB DE. Definition:AB < CD (or CD > AB) means that there exists a point E between C and D such that AB = CE. Proposition 3.13 (Segment Ordering): (a) Exactly one of the following three conditions holds (trichotomy): AB < CD, AB = CD, or AB > CD. (b) If AB < CD and CD EF, then AB < EF. (c) If AB > CD and CD EF, then AB > EF. (d) If AB < CD and CD < EF, then AB < EF (transitivity). Proposition 3.14: Supplements of congruent angles are congruent. Proposition 3.15: (a) Vertical angles are congruent to each other. (b) An angle congruent to a right angle is a right angle.**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l.**P**Proposition 3.16: For every line l and every point P there exists a line through P perpendicular to l. l • Proof: • Case (i) P is not on line l. • Let A and B be any two points on l. • On the opposite side of l from P there exists a ray • such that XAB PAB. • There is a point P on such that AP AP. • PP intersects l in a point Q so that P * Q * P. • If Q = A, then and are opposite rays, making XAB and PAB supplementary by definition. Since they are also congruent, they are right angles by definition. Therefore l by definition of perpendicular lines. • If Q ≠ A then PAQ PAQ . • Hence, PQA PQA. • Since P * Q * P, then and are opposite rays so that PQA and PQA • are supplementary. • 9. Therefore, PQA and PQA are right angles, making l. A Q B I-2 P X C-4 C-1 Def of opposite sides C-6 (SAS) Def of s Def of supplementary angles Def of right angle and def of perpendicular**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l. Case (ii) P is on line l. Since there is a point C not lying on l, case (i) allows us to drop a perpendicular from point C to l, thereby obtaining a right angle. By axiom C-4, there exists a ray emanating from P such that the angle formed by ray and ray is congruent to the right angle we just constructed. This ray and its opposite lie on a perpendicular to l through P. P l B A X C**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l. Proposition 3.17 (ASA Criterion for Congruence): Given ABC and DEF with , , and AC DF. Then ABC DEF. P Proof: B E 1. Suppose ABCis not congruent toDEF. RAA Hyp 3. There exists a point P E on such the AB DP. Then DE and AB are not congruent, since if they were, the triangles would be congruent by SAS. 4. Claim: Ray ray . If they were equal, then line FE and line FP would intersect line DP in the same point (by Prop 2.1) . But P E (step 2). A C-1 D C F C-6 (SAS) 5. ABC DPF Congruence Axiom 4 (C-4) Given any BAC and given any ray emanating from a point A, then there is a unique ray on a given side of line such that BAC BAC. Def of s 6. BCA PFD. 7. Then PFD EFD. C-5 Contradiction 8. Therefore ray =ray . C-4 (uniqueness)**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l. A A Proposition 3.17 (ASA Criterion for Congruence): Given ABC and DEF with , , and AC DF. Then ABC DEF. Proposition 3.18 (Converse of Proposition 3.10): If in a ABC we have B C, then AB AC and ABC is isosceles. B C C B**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l. Proposition 3.17 (ASA Criterion for Congruence): Given ABC and DEF with , , and AC DF. Then ABC DEF. Proposition 3.18 (Converse of Proposition 3.10): If in a ABC we have B C, then AB AC and ABC is isosceles. Proposition 3.19 (Angle Addition): Given between , between , , and . Then .**Proposition 3.16: For every line l and every point P there**exists a line through P perpendicular to l. Proposition 3.17 (ASA Criterion for Congruence): Given ABC and DEF with , , and AC DF. Then ABC DEF. Proposition 3.18 (Converse of Proposition 3.10): If in a ABC we have B C, then AB AC and ABC is isosceles. Proposition 3.19 (Angle Addition): Given between , between , , and . Then . Proposition 3.20 (Angle Subtraction): Given between , between , , and , then .**Definition: < DEF means there is a ray between and**such that GEF D A G B E C F Proposition 3.21 (Angle Ordering): (a) Exactly one of the following three conditions holds (trichotomy): P < Q, P Q, or Q < P. (b) If P < Qand QR, then P < R. (c) If P > Q and Q R, then P > R. (d) If P < Q and Q < R, then P < R.**Proposition 3.22 (SSS Criterion for Congruence): If AB **DE, BC EF, and AC DF, then ABC DEF. You will supply the reasons for the proof of this proposition on the midterm exam.**Proposition 3.22 (SSS Criterion for Congruence): If AB **DE, BC EF, and AC DF, then ABC DEF. Proposition 3.23 (Euclid’s Fourth Postulate): All right angles are congruent to each other.**Proposition 3.22 (SSS Criterion for Congruence): If AB **DE, BC EF, and AC DF, then ABC DEF. Proposition 3.23 (Euclid’s Fourth Postulate): All right angles are congruent to each other. Definition:An angle is acute if it is less than a right angle, obtuse if it is greater than a right angle.**Definition:A model of our incidence, betweenness, and**congruence axioms is called a Hilbert Plane.**A well known theorem from Euclidean Geometry:**An inscribed angle is half the measure of its intercepted arc. m APB = (m ACB) As a result, an angle inscribed in a semicircle is a right angle. P B A C**** **Euclid’s Postulate I**For every point P and for every point Q not equal to P there exists a unique line that passes through P and Q. Euclid’s Postulate II For every segment AB and for every segment CD there exists a unique point E on line AB such that B is between A and E and segment CD is congruent to segment BE. Euclid’s Postulate III For every point O and every point A not equal to O, there exists a circle with center O and radius OA. Euclid’s Postulate IV All right angles are congruent to one another. The Euclidean Parallel Postulate For every line l and for every point P that does not lie on l, there exists a unique line m through P that is parallel to l.**Euclid’s First Proposition: Given any segment, there is**an equilateral triangle having the given segment as one of its sides. 1. Let AB be the given segment. 2. There exists a circle with center A and radius AB. (Euclid phrased this as “With center A and radius AB, let the circle BCD be described.”) Euclid III Euclid’s Postulate III For every point O and every point A not equal to O, there exists a circle with center O and radius OA. D B A**Euclid’s First Proposition: Given any segment, there is**an equilateral triangle having the given segment as one of its sides. 1. Let AB be the given segment. 2. There exists a circle with center A and radius AB. (Euclid phrased this as “With center A and radius AB, let the circle BCD be described.”) Euclid III 3. With center B and radius AB, let the circle ACE be described.” Euclid III Euclid I 4. From point C in which the circles cut one another, draw the segments CA and CB. D C E Euclid’s Postulate I For every point P and for every point Q not equal to P there exists a unique line that passes through P and Q. B A**Euclid’s First Proposition: Given any segment, there is**an equilateral triangle having the given segment as one of its sides. Where is the hidden assumption? 1. Let AB be the given segment. 2. There exists a circle with center A and radius AB. (Euclid phrased this as “With center A and radius AB, let the circle BCD be described.”) Euclid III 3. With center B and radius AB, let the circle ACE be described.” Euclid III Euclid I 4. From point C in which the circles cut one another, draw the segments CA and CB. D C E 5. Since A is the center of circle BCD, AB is congruent to AC. Def. of circle B A 6. Since B is the center of circle ACE, AB is congruent to BC. Def. of circle 7. Since AC and BC are each congruent to AB, they are congruent to each other. C-2 (Euclid called it a “common notion.”) Definition of equilateral 8. Therefore, ABC is an equilateral triangle having AB as one of its sides.**Euclid’s First Proposition: Given any segment, there is**an equilateral triangle having the given segment as one of its sides. 1. Let AB be the given segment. 2. There exists a circle with center A and radius AB. (Euclid phrased this as “With center A and radius AB, let the circle BCD be described.”) Euclid III 3. With center B and radius AB, let the circle ACE be described.” Euclid III Euclid I 4. From point C in which the circles cut one another, draw the segments CA and CB. D C E B A

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