L are two lines cut by a transversal t in such a way

• Notes
• 26
• 100% (3) 3 out of 3 people found this document helpful

This preview shows page 16 - 20 out of 26 pages.

l’ are two lines cut by a transversal t in such a way that a pair of corresponding angles are congruent, then the two lines are parallel. Corollary 4.4.5 If l and l’ are two lines cut by a transversal t in such a way that two nonalternating interior angles on the same side of t are supplements, then the two lines are parallel. Corollary 4.4.6 If l is a line and P is an external point, then there is a line m such that P lies on m and m is parallel to l . Proof is mainly a famous construction “the double perpendicular construction of a parallel line”: Drop one perpendicular from P to l and call the point at the foot of the perpendicular Q. Construct QP uur s . Next construct a line perpendicular to QP uur s through P . (this is the second perpendicular in the construction, hence “double”). This is NOT the proof, though. That’s in the book. Please read it and learn it. Corollary 4.4.7 The Elliptic Parallel Postulate is false in any model of neutral geometry. Corollary 4.4.8 If l, m , and n are any three lines such that m l and n l then either m = n or m is parallel to n . 16
4.5 The Saccheri-Legendre Theorem Let A, B, and C be three noncollinear points. The angle sum, σ, for ABC is the sum of the measures of the three interior angles of this triangle. Theorem 4.5.2 Saccheri-Legendre Theorem If ABC is any triangle then σ( ABC ) 180°. Note: NOT equal. This covers the Hyperbolic case as well as the Euclidean case. Lemma 4.5.3 If ABC is any triangle, then ( ) ( ) 180 CAB ABC μ μ + < ° Let ABC be any triangle Let D be a point on AB uuur such that A*B*D. ( ) ( ) 180 CBD ABC μ μ + = ° because they are a linear pair. Further CBD is an exterior angle to the triangle so ( ) ( ) CAB CBD μ μ < . Substituted in the equation above, we find ( ) ( ) 180 CAB ABC μ μ + < ° . □ Lemma 4.5.4 If ABC is any triangle and E is a point on the interior of side BC , then ( ( ) ( ) 180 ABE ECA ABC σ σ σ + = + ° . Assume the hypothesis. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ABE ECA EAB ABE BEA CAE ECA AEC σ σ μ μ μ μ μ μ + = + + + + + By the Angle Addition Postulate (since E is in the interior of ACB , we know that ( ) ( ) ( ) EAB CAE BAC μ μ μ + = 17 A B C D A B C E
Now ( ) ( ) 180 BEA AEC μ μ + = ° because they are a linear pair. Combining these facts we get ( ) ( ) ( ) ( ) ( ) 180 ( ) 180 ABE ECA BAC ABE ECA ABC σ σ μ μ μ σ + = + + + ° = + ° Lemma 4.5.5 If A, B, and C are three noncollinear points, then there exists a point D that does not lie on AB uur s such that ( ) ( ) ABD ABC σ σ = and the angle measure of one of the interior angles in ABD is less than or equal to 0.5 ( ) CAB μ . Let A, B, and C be three noncollinear points. Construct ABC . Let E be the midpoint of BC . Let D be a point on AE uuur such that A*E*D and AE = ED. Why is AEC DEB 2245 ∆ ? Now since they are congruent, the angle sums will be equal. Now, why do we know that: ( ) ( ) ( ) 180 ABC ABE AEC σ σ σ = + - ° ? By similar reasoning ( ) ( ) ( ) 180 ABD ABE DEB σ σ σ = + - ° So how do we know that 18 A B C E E A B C D
( ) ( ) ABD BAC σ σ = ? This completes half of the proof.