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## Incidence Axioms

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**Last time we gave the following summary of where we have**been: Congruence Axioms Incidence Axioms Betweeneess Axioms Hilbert Plane**Last time we gave the following summary of where we have**been: Congruence Axioms Incidence Axioms Betweeneess Axioms Hilbert Plane Continuity Principles and Axioms Circle-circle Continuity Principle, Line-Circle Continuity Principle, Segment-Circle Continuity Principle, Archimedes’ Axiom Aristotle’s Unboundedness Axiom, Dedekind’s Axiom Neutral Geometry Because it makes no assumptions about parallel lines (remains “neutral”)**Then we proved:**Alternate Interior Angle (AIA) Theorem – Theorem 4.1 In any Hilbert plane, if two lines cut by a transversal have a pair of congruent alternate interior angles with respect to that transversal, then the two lines are parallel. Corollary 1: Two lines perpendicular to the same line are parallel. Corollary 2: If l is any line and P any point not on l, there exists at least one line m through P parallel to l. And we stated: Hilbert’s Euclidean Axiom of Parallelism: For every line l and every point P not lying on l, there is at mostone line m through P such that m is parallel to l.**Last time, we proved:**Alternate Interior Angle (AIA) Theorem – Theorem 4.1 In any Hilbert plane, if two lines cut by a transversal have a pair of congruent alternate interior angles with respect to that transversal, then the two lines are parallel. Corollary 1: Two lines perpendicular to the same line are parallel. Corollary 2: If l is any line and P any point not on l, there exists at least one line m through P parallel to l. And we stated: Hilbert’s Euclidean Axiom of Parallelism: For every line l and every point P not lying on l, there is at mostone line m through P such that m is parallel to l. Together, these are equivalent to the modern version of Euclid’s fifth postulate (the Parallel Postulate).**We then defined a Euclidean Plane.**Congruence Axioms Incidence Axioms Betweeneess Axioms Continuity Principles and Axioms Circle-circle Continuity Principle, Line-Circle Continuity Principle, Segment-Circle Continuity Principle, Archimedes’ Axiom Aristotle’s Unboundedness Axiom, Dedekind’s Axiom Neutral Geometry**We then defined a Euclidean Plane.**Congruence Axioms Incidence Axioms Betweeneess Axioms Circle-circle Continuity Principle Hilbert’s Euclidean Axiom of Parallelism: “at most” (implies “at least”) Euclidean Plane Neutral Geometry**Last time, we also proved:**Exterior angle theorem (EA) 4.2 In any Hilbert plane, an exterior angle of a triangle is greater than either remote interior angle. Corollary: If a triangle has a right or obtuse angle, the other two angles are acute. Proposition 4.1 (SAA Congruence Criterion) Given AC DF, A D, and B E. Then ABC DEF. Proposition 4.2 (Hypotenuse-Leg Criterion) Two right triangles are congruent if the hypotenuse and a leg of one are congruent, respectively, to the hypotenuse and leg of the other.**And we stated (but did not prove):**Proposition 4.3 (Midpoints) Every segment has a unique midpoint. Proposition 4.4 (Bisectors) (a) Every angle has a unique bisector. (b) Every segment has a unique perpendicular bisector. Proposition 4.5 In a triangle ABC, the greater angle lies opposite the greater side, and the greater side lies opposite the greater angle; i.e. AB > AC if and only if C > A Proposition 4.6 Given ABC and ABC, if we have AB AB and BC BC, then B < B if and only if AC < AC.**H-9. 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. Proposition 4.3 (Midpoints) Every segment has a unique midpoint. A l Choose a point C not on . Prop 2.3 B There exists a ray on the opposite side of from C such that CAE ABX . C-4 is parallel to . (AIA) Theorem There exists a point D on ray such that AC BD. C-1 D is on the opposite side of from C. H-9 C CD intersects at a point E (i.e. C * E * D). Def of opposite sides If you think that the two triangles are congruent, you are making a hidden assumption. What is that assumption? B A E D X**Proposition 4.3 (Midpoints)**Every segment has a unique midpoint. Claim: E is between A and B (in other words, A * E * B). E ≠ A since if they were the same point, AC and EC would be the same line. Then D would be on , but is parallel to so they have no points in common. Therefore, E ≠ A. Similarly, E ≠ B. Suppose E * A * B. Then intersects side EB of EBD and so it must also intersect either side BD or ED. C Pasch’s Theorem • can’t intersect BD because they are parallel. • can’t intersect ED since if it did, • and would have two points in • common and they would be the same line. B A E D Similarly we can prove A * B * E is not possible. X**Proposition 4.3 (Midpoints)**Every segment has a unique midpoint. Choose a point C not on . There exists a ray on the opposite side of from C such that CAE ABX . is parallel to . There exists a point D on ray such that AC BD. D is on the opposite side of from C. C CD intersects at a point E (E is between C and D.) Therefore E is between A and B. B A E Then CEA BED, making ACE BDE. Therefore, AE BE, making E the midpoint of AB. D Are we finished?? X**Proposition 4.3 (Midpoints)**Every segment has a unique midpoint. Suppose there is a second midpoint M, and say M is between A and E. Then ACM BDM, making CMA DMB. Let’s also remember CEA BED • But: • CMA > CEA Exterior Angle Theorem • DMB < BED Exterior Angle Theorem C Contradiction 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 < Q and Q R, then P < R. (c) If P > Q and Q R, then P > R. (d) If P < Q and Q < R, then P < R. B M A E D X**Proposition 4.4 (Bisectors)**(a) Every angle has a unique bisector. (b) Every segment has a unique perpendicular bisector.**Proposition 4.4 (Bisectors)**(a) Every angle has a unique bisector. (b) Every segment has a unique perpendicular bisector. Proof: Given angle A, there exist points B and C on either side of angle A such that AB AC. Segment BC has a unique midpoint M. 3. BAM CAM 4. BAM CAM so that the given angle has at least one bisector. 5. Suppose there was another angle bisector ≠ between and . A 6. P ≠ M. 7. Then BAP CAP, making BP PC. 8. But then P is the midpoint of BC. C 9. Contradiction – the midpoint of BC is unique. B P M**Measurement Theorem 4.3**A. There is a unique way of assigning a degree measure to each angle such that the following properties hold: (0) (A) is a real number such that 0 < (A) < 180. (1) (A) = 90 if and only if A is a right angle. (2) (A) = (B) if and only if A B. (3) If is interior to DAB, then(DAB) = (AC) + (CAB). (4) For every real number x between 0 and 180, there exists an A such that(A) =x. (5) If B is supplementary to , then (A) + (B) = 180. (6) (A) > (B) if and only if A > B. A B D C Note: The proof all parts of A except (4) requires Archimedes’ Axiom. The proof of (4) requires Dedekind’s Axiom. Archimedes’ Axiom: If CD is any segment, A any point, and r any ray with vertex A, then for ever point B ≠ A on r there is a number n such that when CD is laid off n times on r starting at A, a point E is reached such that n · CD AE and either B = E or B is between A and E. Hilbert called this the Axiom of Measurement**Definition: If (A) + (B) = 90, then A and B are**called complements of each other and are said to be complementary angles. Another Corollary to the Exterior Angle (EA) Theorem: The sum of the degree measures of any two angles of a triangle is less than 180. Proof: Construct the ray opposite C and are supplementary ) + ) = 180 < () ) + ) < 180 B A D Similarly for the other two pairs of angles (5) If B is supplementary to , then (A) + (B) = 180.**Measurement Theorem 4.3**B. Given segment OI, called a unit segment. Then there is a unique way of assigning a length to each segment AB such that the following properties hold: (7) is a positive real number and = 1. (8) = if and only if AB CD. (9) A * B * C if and only if = + (10) < If and only if AB < CD. (11) For every positive real number x, there exists a segment AB such that = x. Note: The proof all parts of B except (11) requires Archimedes’ Axiom. The proof of (11) requires Dedekind’s Axiom.**The second part of the Measurement Theorem can be used to**prove: Triangle Inequality: If , , If are lengths of the sides of a triangle ABC, then < + . • On a piece of paper, draw a triangle which has the following properties. • Include the measures of all sides and angles in your diagram. • Isosceles • One side of length 8 inches and one side of length 18 inches • One angle of measure 26 26 18 18 Proposition 4.5 In a triangle ABC, the greater angle lies opposite the greater side, and the greater side lies opposite the greater angle 77 77 8**Corollary: For any Hilbert plane, the converse of the**triangle inequality is equivalent to the circle-circle continuity principle. Hence the converse of the triangle inequality holds in Euclidean planes. Congruence Axioms Incidence Axioms Betweeneess Axioms Converse of the triangle inequality: If < + , then, If are lengths of the sides of a triangle ABC. Circle-circle Continuity Principle Hilbert’s Euclidean Axiom of Parallelism: “at most” (implies “at least”) Euclidean Plane**We have gone just about as far as we can go in neutral**geometry in our attempt to deduce all of Euclidean Geometry We are now going to explore those results in Euclidean Geometry that are equivalent to Euclid’s parallel postulate.**Hilbert’s Euclidean Axiom of Parallelism:**For every line l and every point P not lying on l, there is at mostone line m through P such that m is parallel to l. Corollary to the Alternate Interior Angle Theorem: If l is any line and P any point not on l, there exists at leastone line m through P parallel to l. Together, these are equivalent to the modern version of Euclid’s fifth postulate (the Parallel Postulate): Given a line and a point not on the line, there exists exactly one line through the given point parallel to the given line.**Actual statement of Euclid’s fifth Postulate**If two lines are intersected by a transversal in such a way that the sum of the degree measures of the two interior angles on one side of the transversal is less than 180, then the two lines meet on that side of the transversal. Theorem 4.4 Hilbert’s Euclidean parallel postulate Euclid’s fifth postulate. t 2 m 1 l**Part I:**Hilbert’s Euclidean parallel postulate Euclid’s fifth postulate. 1. Assume Hilberts’ postulate and assume 1) + ) < 180. We must prove l and m intersect on the same side of line t as the two angles. 2. 1) + ) = 180 Definition of supplementary angles and Theorem 4.3, Part 5 3. 2) < 180) = ) Steps 1 and 2 t 4. There is a unique ray on the opposite side of t from such that and are congruent. Axiom C-4 B C 2 5. is parallel to line l. Theorem 4.1 (AIA) m 6. Since 2 < , then 2 < Prop 3.21 (b) 3 1 B 7. is not on line m l 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 < Q and Q R, then P < R. (c) If P > Q and Q R, then P > R. (d) If P < Q and Q < R, then P < R. and so is not parallel to l. Hilbert’s Postulate Are we finished with Part I?**Part I:**Hilbert’s Euclidean parallel postulate Euclid’s fifth postulate. Suppose m and l meet at a point A on the opposite side of t from C. Consider triangle BAB. Then 2 is an exterior angle of triangle BAB. But 2 is smaller than 3, which contradicts the exterior angle theorem. t B C A 2 m 3 1 B l**Part II**Euclid’s fifth postulate Hilbert’s Euclidean parallel postulate 1. Given line l and point P not on l. , and let m be the perpendicular to line t through P. 2. Let t be the perpendicular from P to line l. t Prop 3.16 P m 3. Certainly, m is parallel to l. Theorem 4.1 (AIA) 1 n 4. Suppose line n ≠ m is another line through P. We must prove that n is not parallel to l. l R 5. Let 1 be the acute angle line n makes with line t. Q 6. 1) + PQR) < 90 + 90 = 180 Theorem 4.3 A(1) and definition of acute angle 7. But then according to Euclid’s fifth postulate (which is our hypothesis), line n meets line m.**Proposition 4.7**Hilbert’s Euclidean parallel postulate if a line intersects one of two parallel lines, then it also intersects the other. Proposition 4.8 Hilbert’s Euclidean parallel postulate converse to the alternate interior angles theorem. Proposition 4.9 Hilbert’s Euclidean parallel postulate if t is a transversal to l and m, lm, and t l, then t m. Proposition 4.10 Hilbert’s Euclidean parallel postulate if kl, m k, and n l, then either m = n or mn.**If lines are parallel, then AIA are congruent**Proposition 4.8 Hilbert’s Euclidean parallel postulate converse to the alternate interior angles theorem. Part (i) Hilbert’s Euclidean parallel postulate converse to the AIAtheorem. Proof: 1. Begin with parallel lines l and m and line t intersecting l at P. t R P m 2. Line t intersects m at some point Q. Prop 4.7 X 3. Choose points R and S on m and l, respectively, and on opposite sides of t. l Q S 4. Suppose RPQ is not congruent toSQP. C-4 5. There exists a unique ray on the opposite side of line t from R such that RPQ PQX. 6. Then PQX cannot be congruent toSQP. Steps 4 and 5 7. Hence, does not lie on line l. 8. But is parallel to line m. AIA theorem 9. Thus, there are two lines through point Q parallel to m, contradicting Hilbert’s Euclidean parallel postulate.**If lines are parallel, then AIA are congruent**Proposition 4.8 Hilbert’s Euclidean parallel postulate converse to the alternate interior angles theorem. Part (ii) converse to the AIA theorem Hilbert’s Euclidean parallel postulate. Proof: 1. Begin with parallel lines l and m and transversal t so that by hypothesis, RPQ and PQS are congruent alternate interior angles. t R P m X RAA Hypothesis 2. Suppose a second line ≠ l is parallel to m through Q. l Q S 3. Then by the converse of AIA (which is our hypothesis), RPQ PQX. 4. But then ray and ray must be the same ray by axiom C-4 (uniqueness). 5. Hence, point X is on l, which implies line = l. Contradiction.**Q**P PCA) QCB) ACB) + + Proposition 4.11 In any Hilbert plane, Hilbert’s Euclidean parallel postulate implies that for every triangle ABC, A) + B) + ) = 180. C B A B) A) = 180°**Homework**Read Chapter 4 through page 176 (and the rest if you want ) Continue work on the Problem Set. (There will be time in class next time to work on it.) Complete the Midterm Exam (due March 12)