Proofs of Some EPP Equivalences - Proofs of Some EPP...

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Proofs o f Some EPP Equivalences (in Neutral Geometry) Recall the correct wording of the EPP (Euclidean Parallel Postulate): For every line l and every point P not on line l , there is one and only one line m that contains P and is parallel to l . Note that this wording reminds us that, in proving that the EPP is true under certain assumed conditions, we first define an arbitrary line l and an arbitrary point P not on l . Then, we have two tasks to complete: 1) We must prove that there exists at least one line m that contains P and is parallel to l . [This is accomplished by applying the (NIB) "Common Perpendicular" Theorem and then the "Common Perpendicular makes Parallel Lines" Theorem (Corollary 3.4.2) .] 2) We must prove that there does not exist a second line n that contains P and is parallel to l . [This is accomplished by using proof-by-contradiction: First, assume that there exists a second line n , n m , that contains P and is parallel to l ; then, use logical argument to prove that a contradictory situation exists under that assumption.] This process is illustrated in Part II of the proof of Theorem 3.4.6 directly below. Theorem 3.4.6: The Euclidean Parallel Postulate is equivalent to the Converse of the Alternate Interior Angle Theorem. Proof: [ Part I: Prove: If EPP is true, then the Converse of the A I A Theorem is true. ] Assume that the Euclidean Parallel Postulate is true. Let l and m be two parallel lines and let t be a transversal intersecting l and m at points A and B, respectively. [ We NTS that the alternate interior angles formed by t at points A and B are congruent . ] Let 1 and 2 be two alternate interior angles at A and B, respectively. Let point C on l and point D on m be positioned such that 1 BAC   and 2 ABD   . [We show that 1 2   using proof-by-contradiction.] Suppose, by way of contradiction, that 1 2 and are not congruent. Therefore, m( 1 ) m( 2 ) . By Postulate 12 (The Angle Construction Postulate), there is a unique ray BE  in the half-plane side of transversal t that contains point D such that m( ABE ) = m( 1 ) . Let line n = BE  and note that line m = BD  . Since m( ABD ) m( 1 ) , m( ABD ) m( ABE ) . Therefore, rays BD  and BE  are two distinct rays, and so, m and n are two distinct lines. Now, line t is a transversal of lines n and l which forms alternate interior angles 1 and ABE at A and B, resp. Also, 1 ABE   by definition. Therefore, by the Alternate Interior Angle Theorem (Theorem 3.4.1), lines l and n are parallel . Thus, lines m and n are two distinct lines, both containing B and parallel to line l , which contradicts the EPP (which was assumed to be true). Therefore, 1 2   . Therefore, if the EPP is true, then the Converse of the Alternate Interior Angle Theorem is true. [ End of Part I ] l m B A 2 1 t m l D C m( ABE) = m( 1 ) l m n B A 2 1 t m l E D C

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