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Chapter 4 Homework Read Pages 176 – 188 (top) Review Exercises 1 – 20 Exercises 3, 13, 20

Chapter 4 Homework Read Pages 176 – 188 (top) Review Exercises 1 – 20 Exercises 3, 13, 20. Review Exercises – Page 191. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. incorrect. 11. 12. 13. 14. 15. 16. 17. 18. 19. 2 0. correct. correct. incorrect. correct. incorrect. incorrect.

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Chapter 4 Homework Read Pages 176 – 188 (top) Review Exercises 1 – 20 Exercises 3, 13, 20

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  1. Chapter 4 Homework • Read Pages 176 – 188 (top) • Review Exercises 1 – 20 • Exercises 3, 13, 20

  2. Review Exercises – Page 191 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. incorrect 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. correct correct incorrect correct incorrect incorrect incorrect incorrect correct correct incorrect incorrect correct correct incorrect incorrect incorrect correct incorrect

  3. Proof of SAA criterion: F C E B D A G

  4. Proof of SAA criterion: • (1) Assume AB is not congruent to DE. • Then AB < DE or DE < AB. • If DE < AB, then there is a point G between A and B such that AG  DE. • Then CAG  FDE. • HenceAGC  E. • It follows that AGC  B. • This contradicts a certain theorem. • Therefore, DE is not less than AB. • By a similar argument using point H between D and E, AB is not less than DE. • Hence AB  DE. • Therefore, ABC DEF. F C E B D A G

  5. 13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn.

  6. 13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 1. Hilbert’s Euclidean parallel postulate if kl, m  k, and n  l, then either m= n or mn. n l Q  k  P m Proposition 4.7 Hilbert’s Euclidean parallel postulate  if a line intersects one of two parallel lines, then it also intersects the other.

  7. 13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 1. Hilbert’s Euclidean parallel postulate if kl, m  k, and n  l, then either m= n or mn. n l P  Q  k  Q  P m Proposition 4.8 Hilbert’s Euclidean parallel postulate  converse to the alternate interior angles theorem.

  8. 13. Prove Proposition 4.10 Proposition 4.10 Hilbert’s Euclidean parallel postulate  if kl, m  k, and n  l, then either m = n or mn. 2. if kl, m  k, and n  l, then either m= n or mn Hilbert’s Euclidean parallel postulate. Let PQlkbe the standard construction. Suppose a second line t kthrough P is parallel to l. Construct a perpendicular line n from Q to t. Since m l and n t, we have either m = n or mn by hypothesis. But since m and n both contain point Q, m = n. Then n must be perpendicular to t at point P. Since only one line can be perpendicular to m at point P, k and t are the same line - Contradiction P  k m t l  Q n

  9. Homework Read Chapter 5 (it’s fascinating) Review exercises 1 – 18 (they will test how well you read the chapter) Exercises 5, 7 PQlm– the standard construction P m Q l

  10. Review Exercises – Page 227 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. correct 11. 12. 13. 14. 15. 16. 17. 18. correct correct incorrect incorrect correct correct correct correct correct incorrect correct incorrect correct correct incorrect incorrect correct

  11. Exercise 5 - Find the hidden assumption in FarkasBolyai’s “proof” of the parallel postulate. C Given a triangle, the circumcenter (center of the circle passing through the vertices of the triangle) is the intersection of the perpendicular bisectors of the sides.  This also assures that a circle can be constructed through any three non-collinear points. A      B 

  12. Exercise 5 FarkasBolyai’s “proof” of the parallel postulate. P m Let PQlmbe the standard construction, and let n be any other line through P. We must show that n intersects l. n Q l

  13. C P m Choose a point A between P and Q. Let B be the unique point such that A*Q*B and AQ QB. Let R be the foot of the perpendicular from A to n. Let C be the unique point such that A*R*C and AR RC. R A n Q l B

  14. C P m Then A, B, and C are not collinear (If they were, R and P would be the same point). R A n Q l B

  15. C P Therefore, there is a unique circle  passing through A, B, and C. m R Since l is the perpendicular bisector of AB, and n is the perpendicular bisector of AC, l and n must meet at the center of circle . A n Q l B

  16. 77 statements equivalent to Euclid’s fifth postulate

  17. 77 statements equivalent to Euclid’s fifth postulate

  18. Exercise 7 J. D. Gergonne’s “proof” of the parallel postulate. Let PQlmbe the standard construction with point A ≠ P on m. Let be the last ray between and that intersects l, B being the point of intersection. There exists a point C on l such that Q*B*C (by axioms B-1 and B-2). It follows that is not the last ray between and that intersects l, and hence all rays between and meet l . Thus m is the only parallel to l through P. A P m  Q B C l

  19. Up until now, we have been working in neutral geometry – in the absence of any postulate about parallels. “at most one parallel” Now let’s replace Euclid V with the following negation of Hilbert’s Euclidean parallel postulate: There exist a line l and a point P not on l such that at least two distinct lines parallel to l pass through P. We will also assume Aristotle’s axiom. Aristotle’s Angle Unboundedness Axiom: Given any side of an acute angle and any segment AB, there exists a point Y on the given side of the angle such that if X is the foot of the perpendicular from Y to the other side of the angle, then XY > AB. What we know so far: Basic Theorem 6.1 A non-Euclidean plane satisfying Aristotle’s axiom satisfies the acute angle hypothesis. From the acute angle hypothesis alone, the following properties follow: The angle sum of every triangle is < 180, the summit angles of all Saccheri quadrilaterals are acute, the fourth angle of every lambert quadrilateral is acute, and rectangles do not exist. The summit of a Saccheri quadrilateral is greater than the base. The segment joining the midpoints of the summit and the base is perpendicular to both, is the shortest distance between the base line and the summit line, and is the only common perpendicular between those lines. A side adjacent to the acute angle of a Lambert quadrilateral is greater than the opposite side.

  20. Up until now, we have been working in neutral geometry – in the absence of any postulate about parallels. Now let’s replace Euclid V with the following negation of Hilbert’s Euclidean parallel postulate: There exist a line l and a point P not on l such that at least two distinct lines parallel to l pass through P. We will also assume Aristotle’s axiom. Aristotle’s Angle Unboundedness Axiom: Given any side of an acute angle and any segment AB, there exists a point Y on the given side of the angle such that if X is the foot of the perpendicular from Y to the other side of the angle, then XY > AB. What we know so far: Basic Theorem 6.1 A non-Euclidean plane satisfying Aristotle’s axiom satisfies the acute angle hypothesis. From the acute angle hypothesis alone, the following properties follow: The angle sum of every triangle is < 180, the summit angles of all Saccheri quadrilaterals are acute, the fourth angle of every lambert quadrilateral is acute, and rectangles do not exist. The summit of a Saccheri quadrilateral is greater than the base. The segment joining the midpoints of the summit and the base is perpendicular to both, is the shortest distance between the base line and the summit line,and is the only common perpendicular between those lines. A side adjacent to the acute angle of a Lambert quadrilateral is greater than the opposite side.

  21. Note: We proved in class that Euclid V  Rectangles exist.

  22. Remember we proved Euclid V  Hilbert’s Euclidean parallel axiom (Theorem 4.4) Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Given Point P not on line l R Construct the perpendicular from P to l at point A, the perpendicular to at P, and the perpendicular to lat point B on l ( ). P  is parallel to , so the perpendicular at P must intersect at some point C (if not, there would be two lines through P parallel to , violating Hilbert’s // axiom).   B  A l

  23. Remember we proved Euclid V  Hilbert’s Euclidean parallel axiom (Theorem 4.4) Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Given Point P not on line l R Construct the perpendicular from P to l at point A, the perpendicular to at P, and the perpendicular to lat point B on l ( ). P C     Q Q is parallel to , so the perpendicular at P must intersect at some point C (if not, there would be two lines through P parallel to , violating Hilbert’s // axiom).   B A l Let’s examine how AP and BC compare. is parallel to line l, so C is on ray . If BC  AP, then ABCP is a Saccheri quadrilateral, making C  . Then ABCP would be a rectangle and we are finished. Let Q be the unique point on such that AP  BQ. Assume that BC and AP are not congruent. Then either B * Q * C or B * C * Q.

  24. Remember we proved Euclid V  Hilbert’s Euclidean parallel axiom (Theorem 4.4) Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Given Point P not on line l R Construct the perpendicular from P to l at point A, the perpendicular to at P, and the perpendicular to lat point B on l ( ). P C     Q Q is parallel to , so the perpendicular at P must intersect at some point C (if not, there would be two lines through P parallel to , violating Hilbert’s // axiom).   B A l Let’s examine how AP and BC compare. is parallel to line l, so C is on ray . If BC  AP, then ABCP is a Saccheri quadrilateral, making C  . Then ABCP would be a rectangle and we are finished. Let Q be the unique point on such that AP  BQ. Assume that BC and AP are not congruent. Then either B * Q * C or B * C * Q. (i) Assume B * Q * C.

  25. Remember we proved Euclid V  Hilbert’s Euclidean parallel axiom (Theorem 4.4) Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Given Point P not on line l R Construct the perpendicular from P to l at point A, the perpendicular to at P, and the perpendicular to lat point B on l ( ). P C    Q is parallel to , so the perpendicular at P must intersect at some point C (if not, there would be two lines through P parallel to , violating Hilbert’s // axiom). X    B A l Let’s examine how AP and BC compare. is parallel to line l, so C is on ray . If BC  AP, then ABCP is a Saccheri quadrilateral, making C  . Then ABCP would be a rectangle and we are finished. Let Q be the unique point on such that AP  BQ. Then either B * Q * C or B * C * Q. Assume that BC and AP are not congruent. Contradiction must intersect lat some point X by Hilbert’s parallel axiom. (i) Assume B * Q * C. SInce is between and APQ is acute. Thus (APQ) + (PAB) < 180 so by Euclid V, meets l at a point X on the same side of as Q. Since ABQP is a Saccheri quadrilateral, PQB  APQ,making PQBacute. But PQB must be greater than right angle QBX by the EA theorem applied to BQX.

  26. Remember we proved Euclid V  Hilbert’s Euclidean parallel axiom (Theorem 4.4) Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Given Point P not on line l Q Construct the perpendicular from P to l at point A, the perpendicular to at P, and the perpendicular to lat point B on l ( ). P C R    Since is parallel to , the perpendicular at P must intersect at some point C (if not, there would be two lines through P parallel to , violating Hilbert’s // axiom).    B T A l is parallel to line l, so C is on ray. Assume B * C* Q. Then APQ is obtuse. Therefore, its supplement is acute, so that (APR) + (PAT) < 180. Then by Euclid V, intersects lat some point X on the opposite side of from C. Since ABQP is a Saccheri quadrilateral, PQB  APQ,making PQB obtuse. But then BQX has an obtuse angle and a right angle, which contradicts a corollary to the EA theorem.

  27. Prove that Euclid V implies the existence of rectangles i.e. Prove: Euclid V the existence of rectangles Q (Q) P C   Therefore, the only possibility is that point Q and point C are the same point.   B A l Thus, because the summit angles of Saccheri quadrilateral ABCP must be congruent, ABCP is a rectangle.

  28. Universal Non-Euclidean Theorem In a Hilbert plane in which rectangles do not exist, for every line l and every point P not on l, there are at least two parallels to l through P. Note: We proved in class that Euclid V  Rectangles exist. The contrapositive is: Rectangles do not exist  not Euclid V

  29. Universal Non-Euclidean Theorem In a Hilbert plane in which rectangles do not exist, for every line l and every point P not on l, there are at least two parallels to l through P. Proof: Let PQlm be the standard configuration. Let R be another point on l and erect perpendicular t to l through R. Let S be the foot of the perpendicular from P to t. Then is parallel to l because they are perpendicular to the same line t. ≠ m since otherwise S would lie on m and PQRS would be a rectangle AND P m rectangles don’t exist!!! Therefore, there are two parallels to l through P. S Q l R t

  30. Universal Non-Euclidean Theorem In a Hilbert plane in which rectangles do not exist, for every line l and every point P not on l, there are at least two parallels to l through P. Corollary In a Hilbert plane in which rectangles do not exist, for every line l and every point P not on l, there are infinitely many parallels to l through P. P m S S R R Q l t t

  31. We have already encountered the defect of a triangle. Definition: In ABC, the defect, denoted (ABC), is the difference between the triangle’s angle sum and 180. In other words, (A) + (B) + (C) + (ABC) = 180 And we examined Saccheri’s Angle Theorem Part (a) For any Hilbert plane, if there exists a triangle whose angle sum is < 180, then every triangle has an angle sum < 180, and this is equivalent to the fourth angles of Lambert quadrilaterals and the summit angles of Saccheri quadrilaterals being acute. C Last time, we proved a version of the following theorem: Proposition 6.1 (Additivity of the Defect) If D is any point between A and B, then (ABC) = (ACD) + (BCD) In a Hilbert plane in which rectangles do not exist, (ABC) > 0. B A D A convex quadrilateral is one in which each vertex lies in the interior of the opposite angle. Question: If two sides of a quadrilateral are parallel, is it convex?

  32. Definition: Given an angle CAB, define point D to be in the interior of CAB if D is on the same side of as B and if D is also on the same side of as C. D C B A

  33. We have already encountered the defect of a triangle. Definition: In ABC, the defect, denoted (ABC), is the difference between the triangle’s angle sum and 180. In other words, (A) + (B) + (C) + (ABC) = 180 And we examined Saccheri’s Angle Theorem Part (a) For any Hilbert plane, if there exists a triangle whose angle sum is < 180, then every triangle has an angle sum < 180, and this is equivalent to the fourth angles of Lambert quadrilaterals and the summit angles of Saccheri quadrilaterals being acute. C Last time, we proved a version of the following theorem: Proposition 6.1 (Additivity of the Defect) If D is any point between A and B, then (ABC) = (ACD) + (BCD) In a Hilbert plane in which rectangles do not exist, (ABC) > 0. B A D A convex quadrilateral is one in which each vertex lies in the interior of the opposite angle. In a Hilbert plane satisfying the acute angle hypothesis, the angle sum of a convex quadrilateral is < 360. The defect of a convex quadrilateral is defined to be 360 minus its angle sum.

  34. Definition (from Chapter 5) Two triangles are similar if their vertices can be put in one-to-one correspondence in such a way that corresponding angles are congruent (AAA). Chapter 5 presented a “proof” by John Wallis of Hilbert’s Euclidean parallel postulate. In the proof, he introduced and used a new postulate. His postulate basically said that given any triangle, a similar triangle can be constructed having any line segment as a side. The fact that Wallis needed to introduce a new postulate is an indication that the existence of similar, non-congruent triangles is equivalent to Hilbert’s Euclidean parallel postulate. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles.

  35. Definition (from Chapter 5) Two triangles are similar if their vertices can be put in one-to-one correspondence in such a way that corresponding angles are congruent (AAA). Chapter 5 presented a “proof” by John Wallis of Hilbert’s Euclidean parallel postulate. In the proof, he introduced and used a new postulate. His postulate basically said that given any triangle, a similar triangle can be constructed having any line segment as a side. The fact that Wallis needed to introduce a new postulate is an indication that the existence of similar, non-congruent triangles is equivalent to Hilbert’s Euclidean parallel postulate. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles.

  36. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles. Proof: Begin with similar, non-congruent triangles ABC and ABC. Since the triangles are not congruent, one of the triangles must have two sides larger than the corresponding sides of the other. Let’s say it is ABC, so that AB > AB, and AC > AC. A Then there exists points B and C such that AB  AB and AC  AC making  ABC  ABC by SAS. Therefore, A Hence is parallel to . B B B C  C C

  37. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles. Proof: Begin with similar, non-congruent triangles ABC and ABC. Since the triangles are not congruent, one of the triangles must have two sides larger than the corresponding sides of the other. Let’s say it is ABC, so that AB > AB, and AC > AC. A Then there exists points B and C such that AB  AB and AC  AC making  ABC  ABC by SAS. Therefore, A Hence is parallel to . Therefore, quadrilateral BCCB is convex. B B B C sum = 180  C C

  38. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles. Proof: Begin with similar, non-congruent triangles ABC and ABC. Since the triangles are not congruent, one of the triangles must have two sides larger than the corresponding sides of the other. Let’s say it is ABC, so that AB > AB, and AC > AC. A Then there exists points B and C such that AB  AB and AC  AC making  ABC  ABC by SAS. Therefore, A Hence is parallel to . Therefore, quadrilateral BCCB is convex. B B B C sum = 180  C C

  39. Proposition 6.2 In a plane satisfying the acute angle hypothesis, if two triangles are similar, then they are congruent. In other words AAA is a valid criterion for congruence of triangles. Proof: Begin with similar, non-congruent triangles ABC and ABC. Since the triangles are not congruent, one of the triangles must have two sides larger than the corresponding sides of the other. Let’s say it is ABC, so that AB > AB, and AC > AC. Put another way, the existence of similar triangles that are not congruent is equivalent to Hilbert’ Euclidean parallel axiom. A Then there exists points B and C such that AB  AB and AC  AC making  ABC  ABC by SAS. Therefore, A Hence is parallel to . Similarly, sum = 180 Therefore, quadrilateral BCCB is convex. B B B C  C Hence, the angle sum of quadrilateral BCCB= 360 C Contradiction The angle sum of a convex quad is < 360 in a Hilbert plane satisfying the acute angle hypothesis . Therefore, any triangles that are similar must be congruent in a Hilbert plane satisfying the acute angle hypothesis.

  40. 77 statements equivalent to Euclid’s fifth postulate

  41. Proposition 6.3 In a plane in which rectangles don’t exist, if ll , then any set of points on l equidistant from l has at most two points in it. l C D X Y P Q l B A

  42. Proposition 6.3 In a plane in which rectangles don’t exist, if ll , then any set of points on l equidistant from l has at most two points in it. l C D l B A

  43. The following proposition summarizes what we have just observed and some past results. Proposition 6.4 In a Hilbert plane satisfying the acute angle hypothesis, if ll  and if there exists a pair of points A and B on l equidistant from l , then l and l  have a common perpendicular segment MM dropped from the midpoint M of AB. MM is the shortest segment joining a point of l to a point of l, and the segments AA and BB increase as A and B recede from M. Proposition 6.5 In a Hilbert plane in which rectangles do not exist, , if l and l  have a common perpendicular segment MM, then they are parallel and that common perpendicular segment is unique. Moreover, if A and B are any points on l such that M is the midpoint of AB, then A and B are equidistant from l .

  44. The following proposition summarizes what we have just observed and some past results. Proposition 6.4 In a Hilbert plane satisfying the acute angle hypothesis, if ll  and if there exists a pair of points A and B on l equidistant from l , then l and l  have a common perpendicular segment MM dropped from the midpoint M of AB. MM is the shortest segment joining a point of l to a point of l, and the segments AA and BB increase as A and B recede from M. Proposition 6.5 In a Hilbert plane in which rectangles do not exist, , ifl and l  have a common perpendicular segment MM, then they are parallel and that common perpendicular segment is unique. Moreover, if A and B are any points on l such that M is the midpoint of AB, then A and B are equidistant from l . Are there any parallel lines that do not have a common perpendicular?

  45. From Page 269 in the textbook

  46. The Pythagorean Theorem  Hilbert’s Euclidean parallel postulate Since we know the Pythagorean Theorem is valid in Euclidean geometry, all we really need to show is that the Pythagorean Theorem  Hilbert’s Euclidean parallel postulate. Proof: Let and be two perpendicular rays emanating from point A and choose B as the unique point on such that AC  AB and call there length a. Then by the Pythagorean Theorem, the length of BC is a. Let M be the unique midpoint of segment BC, so that = = ½a. a ACM  ABM (SSS). Therefore, AMC  AMB, and since they are also supplementary, they are right angles. C  ½a a  M Using the Pythagorean Theorem on right ACM, the length of AM is also ½a. ½a   A a P B Therefore, all three triangles are isosceles andMACACM  CBA  BAM. C  C, CMA  CAB, CAM  CBA , yet they are not congruent since their sides have unequal lengths. But this means that CAB is similar to CMA. And we proved earlier that the existence of similar, non-congruent triangles is equivalent to Hilbert’s Euclidean parallel postulate.

  47. From Page 269 in the textbook

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