Last time we defined a bi-right quadrilateral

1. Last time we defined a bi-right quadrilateral summit D C   summit angles base   B A And we defined a Saccheriquadrilateral as a bi-right quadrilateral ABDC in which CA  DB

2. We proved two important results about Saccheri quadrilaterals the line joining the midpoints of the summit and the base is perpendicular to both the summit and the base. The summit angles are congruent C  ), and D C   MN is called the midline segment of the quadrilateral N     B A M Together, this is Proposition 4.12

3. Proposition 4.12 (a) The summit angles of a Saccheri quadrilateral are congruent to each other. (b) The line joining the midpoints of the summit and the base is perpendicular to both the summit and the base. C D  We also proved B A  Proposition 4.13 In any bi-right quadrilateral ABDC, C > D  BD > AC. In other words, the greater side is opposite the greater summit angle. D C B A

4. We also discussed this list from the textbook. 77 statements equivalent to Euclid’s fifth postulate This says Euclid V  the existence of rectangles and the existence of rectangles  Euclid V

5. We also discussed this list from the textbook. 77 statements equivalent to Euclid’s fifth postulate This says Euclid V  the existence of rectangles and the existence of rectangles  Euclid V

6. Remember we proved Euclid V  Hilbert’s 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

7. Remember we proved Euclid V  Hilbert’s 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.

8. Remember we proved Euclid V  Hilbert’s 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.

9. Remember we proved Euclid V  Hilbert’s 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.

10. Remember we proved Euclid V  Hilbert’s 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.

11. 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.

12. Proposition 4.12 (a) (SaccheriI) The summit angles of a Saccheri quadrilateral are congruent to each other. (b) (SaccheriII) The line joining the midpoints of the summit and the base is perpendicular to both the summit and the base. C D  B A  Proposition 4.13 In any bi-right quadrilateral ABDC, C > D  BD > AC. In other words, the greater side is opposite the greater summit angle. D C B A

13. Corollary 1 Given any acute angle with vertex V. Let Y be any point on one side of the angle, let Y be any point farther out on that side , i.e. V * Y * Y. Let X, X be the feet of the perpendiculars from Y, Y, respectively, to the other side of the angle. Then YX > YX. Y Proof: 1. VYX and VYX are acute. Y 2. VYX and XYY are supplementary. 3. Since (VYX) < 90, (XYY) > 90, making XYY obtuse. V X X 4. Therefore, XYY > VYX. 5. In bi-right quad XXYY, YX > YX.

14. Corollary 1 Given any acute angle with vertex V. Let Y be any point on one side of the angle, let Y be any point farther out on that side , i.e. V * Y * Y. Let X, X be the feet of the perpendiculars from Y, Y, respectively, to the other side of the angle. Then YX > YX. Y Y Corollary 2 Euclid V implies Aristotle’s axiom Y V V X X X 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.  B A  In other words, the lengths of the perpendicular segments from one side of an acute angle to the other have no limit.

15. Our main goal tonight is to prove the Uniformity Theorem Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse).

16. Consider the following construction that is valid in neutral geometry D C     B A

17. Can we justify each step in the following construction using only those results that are valid in neutral geometry? Wouldn’t it be great if we could prove that this quadrilateral was a Saccheri quadrilateral? Then we would have a rectangle. And we would make history!! D C     B A

18. Definition A quadrilateral with at least three right angles is called a Lambert quadrilateral. The fourth angle  is referred to as “the fourth angle.” CD is adjacent to the fourth angle D C    CA is opposite the fourth angle BD is adjacent to the fourth angle BA is opposite the fourth angle   B A

19. Corollary 3 A side adjacent to the fourth angle  of a Lambert quadrilateral is, respectively, greater than, congruent to, or less than its opposite side if and only if  is acute, right, obtuse, respectively. D C      B A Proposition 4.13 In any bi-right quadrilateral ABDC, C > D  BD > AC. In other words, the greater side is opposite the greater summit angle.

20. Corollary 3 A side adjacent to the fourth angle  of a Lambert quadrilateral is, respectively, greater than, congruent to, or less than its opposite side if and only if  is acute, right, obtuse, respectively. D C      B A

21. Corollary 4 The summit of a Saccheri quadrilateral is, respectively, greater than, congruent to, or less than the base if and only if its summit angles are acute, right, or obtuse, respectively. D C        B A What we have just seen is thatcutting a Saccheri quadrilateral in “half” (connecting the midpoints of the base and summit) always yields 2 Lambert quadrilaterals. We call this “halving” the Saccheri quadrilateral.

22. True or False: “Doubling” a Lambert quadrilateral yields a Saccherri quadrilateral. C C D D         B B A

23. True or False: “Doubling” a Lambert quadrilateral yields a Saccherri quadrilateral. C C D D         B B A

24. Here is a “proof” given by a student that a rectangle must exist in any Hilbert plane (thus “proving” Hilbert’s Euclidean parallel postulate). Yell “STOP” when we come to the flaw in the proof. Then CNM and AMN are right angles by proposition 4.12. Now join the midpoint M of CN with the midpoint N of AM. Begin with a Saccheri quadrilateral. Join the midpoint of the summit and base. D C   Then CNM and AMN are right angles by proposition 4.12, N N makingMMNN a rectangle.     B A   M M

25. Earlier this class, we proved Euclid V  the existence of rectangles Hilbert’s // axiom  Hilbert’s Euclidean parallel postulate implies that every Lambert quadrilateral and every Saccheri quadrilateral is a rectangle. If we could prove the existence of rectangles  Hilbert’s // axiom Then the existence of a Saccheri quadrilateral that is also a Lambert quadrilateral would prove Hilbert’s Euclidean parallel postulate.

26. Our main goal tonight is to prove the Uniformity Theorem: Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Proposition 4.12 (a) (SaccheriI) The summit angles of a Saccheri quadrilateral are congruent to each other. (b) (SaccheriII) The line joining the midpoints of the summit and the base is perpendicular to both the summit and the base. Proposition 4.13 In any bi-right quadrilateral ABDC, C > D  BD > AC. In other words, the greater side is opposite the greater summit angle.

27. Lemma 1: Given any Saccheri quadrilateral ABDC and a point P between C and D. Let Q be the foot of the perpendicular from P to base AB. Then a) PQ < BD iff the summit angles of ABDC are acute. b) PQ  BD iffthe summit angles of ABDC are right angles. c) PQ > BD iffthe summit angles of ABDC are obtuse D C   P    B A  Q

28. Lemma 1: Given any Saccheri quadrilateral ABDC and a point P between C and D. Let Q be the foot of the perpendicular from P to base AB. Then a) PQ < BD iff the summit angles of ABDC are acute. D C Suppose PQ < BD.   In bi-right quadrilateral QBDP, D < QPD by prop 4.13. Similarly, in bi-right quadrilateral AQPC, PQ < AC implies C < PC. But since PC and QPD are supplementary, either they are both right angles or one of them is acute. Either way, since C D, C and D are acute. P    B A  Q

29. Lemma 1: Given any Saccheri quadrilateral ABDC and a point P between C and D. Let Q be the foot of the perpendicular from P to base AB. Then a) PQ < BD iff the summit angles of ABDC are acute. Suppose C D are both acute. D C   Since PC and QPD are supplementary, they are either both right angles, or one of them is obtuse. If they are right angles, then quadrilateral PQBD is a Lambert quadrilateral. Thus, the side adjacent to D (BD) is greater than the side opposite D (PQ). If one of the angles (say QPD) is obtuse, then in bi-right quadrilateral QBDP, acute D < QPD making PQ < BD by prop 4.13. P    B A  Q

30. Lemma 2: Given any Saccheri quadrilateral ABDC and a point P such that C * D * P. Let Q be the foot of the perpendicular from P to . Then a) PQ > BD iff the summit angles of ABDC are acute. b) PQ  BD iffthe summit angles of ABDC are right angles. c) PQ < BD iffthe summit angles of ABDC are obtuse P C   D    Q A  B

31. Lemma 2: Given any Saccheri quadrilateral ABDC and a point P such that C * D * P. Let Q be the foot of the perpendicular from P to . Then a) PQ > BD iff the summit angles of ABDC are acute. Suppose PQ > BD. P C   Then there exists a point E such that P * E * Q and QE  BD  AC. AQEC and BQED are Saccheri quads , so the marked angles are congruent by prop 4.12. D  Therefore, BDE > ACE. Also, exteriorPDE > DCE. Then by addition, BDP > ACD. But ACD  BDC, so BDP > BDC. Since they are supplementary, BDC must be acute.   Q A E   B

32. Lemma 2: Given any Saccheri quadrilateral ABDC and a point P such that C * D * P. Let Q be the foot of the perpendicular from P to . Then a) PQ > BD iff the summit angles of ABDC are acute. Suppose the summit angles of ABDC are acute. P C   If PQ  BD, then we have three Saccheri quadrilaterals. In one, P  C, in another P BDP, and in a third, C  BDC. Therefore, BDC  BDP. But then they are right angles, and BDC is given acute. Contradiction D    Q A  B

33. Lemma 2: Given any Saccheri quadrilateral ABDC and a point P such that C * D * P. Let Q be the foot of the perpendicular from P to . Then a) PQ > BD iff the summit angles of ABDC are acute. Suppose the summit angles of ABDC are acute. P C   If PQ < BD, then there exists a point E such that Q * P * E and QE  BD. AQEC and BQED are Saccheri quads , so the marked angles are congruent by prop 4.12. D  Then ACE > EDB and since EDP > ECD by EA theorem, ACD > PDB by subtraction.  E   Q A Since ACD is given acute, so is PDB, making CDB obtuse. But CDB is a summit angle of Saccheri quadrilateral ABDC, and so it is acute. Contradiction  B

34. Loosely stated, these two lemmas say that in a Saccheri quadrilateral with acute summit angles, the further you move away from the midline, the longer the perpendicular from the summit to the base. D C B A

35. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). First we will use the lemmas to prove a special case of the uniformity theorem: The case where the midline segments of two Saccheri quadrilaterals are congruent Q Y Z Begin with two Saccheri quadrilaterals ABCD and WXYZ with congruent midline segments NM and PQ. We are given that D in Saccheri quadrilateral ABDC is acute. We must show Y is acute. X W P D C M B N A

36. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). First we will use the lemmas to prove a special case of the uniformity theorem: The case where the midline segments of two Saccheri quadrilaterals are congruent Q Y Z On , construct the unique points X and W such that NX = PX and NW = PW. Construct the perpendiculars to at W and X, intersecting CD at points Z and Y. X W P Y Z D C M Claim: quad WXYZ is congruent to quad WXYZ     B N A X W

37. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). First we will use the lemmas to prove a special case of the uniformity theorem: The case where the midline segments of two Saccheri quadrilaterals are congruent Q Y Z WNM  WQP ZWM  WPQ X W P Z Y D C M Claim: quad WXYZ is congruent to quad WXYZ     B N A X W

38. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). First we will use the lemmas to prove a special case of the uniformity theorem: The case where the midline segments of two Saccheri quadrilaterals are congruent Q Y Z Using lemma 1 on Saccheri quad ABDC, YX < BD But, looking at Saccheri quad WXYZ, that means XYZmust be acute by lemma 2. X W P But Y XYZ, therefore, Y is acute. Z Y D C M Note: We have also proved a uniformity result for Lambert quadrilaterals MNBD and MNXY (which “overlap” and share a side connecting two right angles in each).     B N A W X

39. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Q Y Z Now suppose PQ > MN Construct point L, such that N * M * L and NL  PQ. Q Let X be the unique point on ray such that NX  PX Y Z and Let G be the point on the perpendicular to on the same side as D such that LG  QY. LNX  QPX LGX  QYX L G X W P So GY and by addition, NXG PXY, so thatNXG is also right angle.   X W P D C M  B N A X

40. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Q Y Z Now suppose PQ > MN Since is parallel to , G and X are on opposite sides of . Let Y be the point at which GX meets . Y is on ray since G and X are on the same side of So MYX and D are the same type. L G   X W P D C M Y Note: We have also proved a uniformity result for Lambert quadrilaterals MNBD and MNXY (which “overlap” and share a side connecting two right angles in each).   B N A X

41. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). If MN > PQ, just reverse their roles in the proof. Q Y Z Now suppose PQ > MN Since is parallel to , G and X are on opposite sides of . Let Y be the point at which GX meets . Y is on ray since G and X are on the same side of So MYX and D are the same type. L G AND MYX and G are the same type.   X W P Therefore, G and D are the same type. D C M Y Note: We have also proved a uniformity result for Lambert quadrilaterals MNBD and MNXY (which “overlap” and share a side connecting two right angles in each).  So Y and D are the same type.  B N A X

42. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Corollary 1 For any Hilbert plane , if one Lambert quadrilateral has an acute (respectively, right, obtuse) fourth angle, then so do all Lambert quadrilaterals. Furthermore, the type of the fourth angle is the same as the type of the summit angles of Saccheri quadrilaterals Lemma: If one Lambert quadrilateral has an acute fourth angle, then all Saccheri quadrilaterals have acute summit angles. Begin with Lambert quadrilateral BACD with acute fourth angle D. P  Then BD > CA by prop 4.13. Choose the unique point P on AC with A * C* P such that AP  BD. Thus, quad ABDP is a Saccheri quadrilateral. D C Right ACD > P by the exterior angle theorem, making P acute. B A Then by the Uniformity Theorem, all Saccheri quads have acute summit angles.

43. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Corollary 1 For any Hilbert plane , if one Lambert quadrilateral has an acute (respectively, right, obtuse) fourth angle, then so do all Lambert quadrilaterals. Furthermore, the type of the fourth angle is the same as the type of the summit angles of Saccheri quadrilaterals Nowsuppose one Lambert quadrilateral has an acute fourth angle. Take any other Lambert quadrilateral BACD and assume its fourth angle is not acute. The fourth angle cannot be a right angle, since it would then be a rectangle making it a Saccheri quadrilateral with non-acute summit angles. But that contradicts the lemma. D C Suppose the fourth angle D is obtuse. Then AC > BD by prop 4.13. P  Choose the unique point P on AC with A * P * C such that AP  BD. B A Then ABDP is a Saccheriquadrilateral. So by the lemma, APD is acute. But by the exterior angle theorem, APD > right C. Contradiction

44. Uniformity Theorem For any Hilbert plane, if one Saccheri quadrilateral has acute summit angles, then so do all Saccheri quadrilaterals (the same for right or obtuse). Corollary 1 For any Hilbert plane , if one Lambert quadrilateral has an acute (respectively, right, obtuse) fourth angle, then so do all Lambert quadrilaterals. Furthermore, the type of the fourth angle is the same as the type of the summit angles of Saccheri quadrilaterals Definition: A Hilbert plane is called semi-Euclidean if all Lambert quadrilaterals and all Saccheri quadrilaterals are rectangles. If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis(similarly for obtuse). Corollary 2: There exists a rectangle in a Hilbert plane iff the plane is semi-Euclidean. Opposite sides of a rectangle are congruent to each other. Corollary 3: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, a side of a Lambert quadrilateral adjacent to the acute angle (respectively, obtuse) is greater than (respectively, less than) its opposite side.

45. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). D C B A

46. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis. D C B A

47. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis. D C B A

48. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis. D C B A

49. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis. D C B A

50. Corollary 4: In a Hilbert plane satisfying the acute (respectively, obtuse) angle hypothesis, the summit of a Saccheri quadrilateral is greater than (respectively, less than) the base. The midline segment MN is the only common perpendicular to the summit line and the base line. If P is any point ≠ M on the summit line, and Q is the foot of the perpendicular from P to the base line, the PQ > MN (respectively, PQ < MN). As P moves away from M along a ray of the summit line emanating from M, PQ increases (respectively, decreases). If the fourth angle of every Lambert quadrilateral is acute, we say the plane satisfies the acute angle hypothesis. D C B A