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Chapter 3 Parallel and Perpendicular Lines. Sec. 3-1 Properties of Parallel Lines Objective: a) Identify Angles formed by Two Lines & a Transversal. b)

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Presentation on theme: "Chapter 3 Parallel and Perpendicular Lines. Sec. 3-1 Properties of Parallel Lines Objective: a) Identify Angles formed by Two Lines & a Transversal. b)"— Presentation transcript:

1 Chapter 3 Parallel and Perpendicular Lines

2 Sec. 3-1 Properties of Parallel Lines Objective: a) Identify Angles formed by Two Lines & a Transversal. b) To Prove & Use Properties of Parallel Lines.

3 Parallel Lines – Two lines in the same plane which never intersect. Symbol: // Transversal – A line that intersects two // lines. 8 Special Angles are formed. 1 2 34 56 87 Interior Portion of the // Lines t m n

4 Corresponding Angles Most Important Angle Relationship Most Important Angle Relationship Always Congruent Always Congruent Cut the Transversal & lay the top part onto the bottom part. Overlapping Angles are Corresponding. Cut the Transversal & lay the top part onto the bottom part. Overlapping Angles are Corresponding. 12 34 5 6 78 Corresponding Angles 1 & 5 2 & 6 3 & 7 4 & 8

5 P(3 – 1) Corresponding Angle Postulate If a Transversal Intersects two // lines, then the corresponding angles are Congruent. 12 34 65 87

6 12 34 65 87 4 Pairs of Vertical Angles Are Congruent 1 4 2 3 5 8 6 7 3 & 6 4 & 5 Alternate Interior Angles 3 & 5 4 & 6 Same-Sided Interior Angles Special Interior Angles Are congruent Are Supplementary(= 180)

7 Th (3-1) Alternate Interior Angle Theorem If a Transversal intersects two // lines, then the alternate interior angles are congruent. 12 3 4 5 6 78 Statements 1.l // m 2. 3 7 3. 7 6 4. 3 6 Given: l // m Prove: 3 6 Reasons 1.Given 2.Corrsp. Angles are Congruent 3.Def. of Vertical Angles 4.Subs l m

8 Th (3-2) Same-Sided Interior Angle Theorem If a Transversal intersects two // lines, then the same- sided interior angles are supplementary. Given: l // m Prove: 4 & 6 are Supplementary 1 2 3 4 56 78 Statements 1.l // m 2.m 4 + m 2 = 180 3.m 2 = m 6 4.m 4 + m 6 = 180 5. 4 & 6 are Supplementary Reasons 1.Given 2. Add. Postulate 3.Corrsp. s are 4.Subs 5.Def of Supplementary l m

9 Examples 1 & 2 Solve for the missing s Solve for the missing s Solve for x, then for each. 5x - 20 3x l m 14x - 5 13x l m 5x – 20 +3x = 180 8x = 200 x = 25 14x – 5 = 13x -5 = -x 5 = x

10 Use what you have learned! 1. Find m 2 if l//m. 42 12 m 1 = 42 (Corrsp. ) m 1 + m 2 m 1 + m 2 = 180 m 2 = 180 42 + m 2 = 180 m 2 = 138 2. Solve for angles a, b, c if l//m l m m l 6540 a b c m a = 65 (Alt. Inter. ) m c = 40 (Alt. Inter. ) m a + m b + m c = 180 65 + m b + 40 = 180 m = 75

11 Solve for x and find the measure of each angle if l//m. (x + 40) x l m x + x +40 = 180 2x + 40 = 180 -40 -40 2x = 140 x = 70


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