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Unformatted text preview: Chapter 4 Neutral Geometry We have now accepted six axioms. We will explore what we can prove WITHOUT making a choice of Parallel Axiom. In Chapter 5 well choose the Euclidean parallel postulate and then in Chapter 6 well choose the Hyperbolic parallel postulate. Note that we are using Euclids approach. He, too, proved as much as possible without his Parallel postulate. 4.1 The Exterior Angle Theorem and the Existence of Perpendiculars What, exactly, is an exterior angle to a triangle? A D B E C Note that the pair of exterior angles are congruent because they are vertical angles. Note the remote interior angles for the exterior angles shown in the illustration. Theorem 4.1.2(Exterior Angle Theorem) The measure of an exterior angle for a triangle is strictly greater than the measure of either remote interior angle. This is NOT the Euclidean Exterior Angle theorem in which the sum of the measures of the remote interior angles EQUALS the measure of the exterior angle! For ease in proving, lets restate this in implication form: If ABC is a triangle and D is a point such that ray CD uuur is opposite to ray CB uuur , then ( DCA ) > ( BAC ) and ( DCA ) > ( ABC ). 1 Proof [Assume the hypothesis]: Let ABC be a triangle and let D be a point such that ray CD uuur is opposite to ray CB uuur . [Now well begin a construction] Let E be the midpoint of segment AC . And choose F to be the point on ray FE so that BE = EF . Note that BEA 2245 FEC because they are vertical angles. So that BEA 2245 FEC by SAS. E F B D A C What we want to do is demonstrate that point F is in the interior of ACD so we can then invoke the Betweenness Theorem for Rays (3.4.5): Let A, B, C, and D are four distinct points such that C and D are on the same side of AB uur s . Then ( BAD ) < ( BAC ) if and only if ray AD uuur is between rays AB uuur and AC uuur . So we use the Plane Separation Axiom and a couple of theorems from that section. Points F and B are on opposite sides of line AC uur s , which puts F and D on the same side of AC uur s . Now points A and E are on the same side line CD uur s , along with F and since segment EF is wholly contained in the halfplane determined by line CD uur s , F is on the same side of CD uur s as A is. This puts F in the interior of ACD . NOW by Betweenness for Rays (3.4.5), ( DCA ) > ( FCA ) and from above FCA ABC 2245 . So their measures are the same, thus ( DCA ) > ( ABC ) Now lets look at the Exterior Angles on the Sphere! 2 Where does this Theorem work and NOT work? Theorem 4.1.3 Existence and Uniqueness of Perpendiculars For every line l and for every point P , there exists a unique line m such that P lies on m and m l ....
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This note was uploaded on 02/22/2012 for the course MATH 3379 taught by Professor Staff during the Spring '08 term at University of Houston.
 Spring '08
 Staff
 Geometry

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