Angles Formed by Parallel Lines and Transversals 21.1 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Geometry Holt Geometry

Identify each angle pair. 1. 1 and 3 2. 3 and 6 3. 4 and 5 21.1 Warm Up Identify each angle pair. 1. 1 and 3 2. 3 and 6 3. 4 and 5 4. 6 and 7 corr. s alt. int. s alt. ext. s same-side int s

21.1 Objective Prove and use theorems about the angles formed by parallel lines and a transversal.

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Example 1: Using the Corresponding Angles Postulate 21.1 Example 1: Using the Corresponding Angles Postulate Find each angle measure. A. mECF x = 70 Corr. s Post. mECF = 70° B. mDCE 5x = 4x + 22 Corr. s Post. x = 22 Subtract 4x from both sides. mDCE = 5x = 5(22) Substitute 22 for x. = 110°

21.1 Example 2 Find mQRS. x = 118 Corr. s Post. mQRS + x = 180° *Def. of Linear Pair* mQRS = 180° – x Subtract x from both sides. = 180° – 118° Substitute 118° for x. = 62°

21.1 If a transversal is perpendicular to two parallel lines, all eight angles are congruent. Helpful Hint

21.1 Remember that postulates are statements that are accepted without proof. Since the Corresponding Angles Postulate is given as a postulate, it can be used to prove the next three theorems.

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21.1 Example 3 Find each angle measure. A. mEDG mEDG = 75° Alt. Ext. s are Congruent. B. mBDG x – 30° = 75° Alt. Ext. s are congruent. x = 105 Add 30 to both sides. mBDG = 105°

21.1 Example 4 Find x and y in the diagram. By the Alternate Interior Angles Theorem, (5x + 4y)° = 55°. By the Corresponding Angles Postulate, (5x + 5y)° = 60°. 5x + 5y = 60 –(5x + 4y = 55) y = 5 Subtract the first equation from the second equation. Substitute 5 for y in 5x + 5y = 60. Simplify and solve for x. 5x + 5(5) = 60 x = 7, y = 5

21.1 Lesson Quiz State the theorem or postulate that is related to the measures of the angles in each pair. Then find the unknown angle measures. 1. m1 = 120°, m2 = (60x)° 2. m2 = (75x – 30)°, m3 = (30x + 60)° Alt. Ext. s Thm.; m2 = 120° Corr. s Post.; m2 = 120°, m3 = 120° 3. m3 = (50x + 20)°, m4= (100x – 80)° 4. m3 = (45x + 30)°, m5 = (25x + 10)° Alt. Int. s Thm.; m3 = 120°, m4 =120° Same-Side Int. s Thm.; m3 = 120°, m5 =60°

Proving Lines Parallel 21.2 Proving Lines Parallel Warm Up Lesson Presentation Lesson Quiz Holt McDougal Geometry Holt Geometry

21.2 Warm Up State the converse of each statement. 1. If a = b, then a + c = b + c. 2. If mA + mB = 90°, then A and B are complementary. 3. If AB + BC = AC, then A, B, and C are collinear. If a + c = b + c, then a = b. If A and  B are complementary, then mA + mB =90°. If A, B, and C are collinear, then AB + BC = AC.

21.2 Objective Use the angles formed by a transversal to prove two lines are parallel.

21.2 Recall that the converse of a theorem is found by exchanging the hypothesis and conclusion. The converse of a theorem is not automatically true. If it is true, it must be stated as a postulate or proved as a separate theorem.

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21.2 The Converse of the Corresponding Angles Postulate is used to construct parallel lines. The Parallel Postulate guarantees that for any line ℓ, you can always construct a parallel line through a point that is not on ℓ.

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21.2 Lesson Quiz: Part I Name the postulate or theorem that proves p || r. 1. 4  5 Conv. of Alt. Int. s Thm. 2. 2  7 Conv. of Alt. Ext. s Thm. 3. 3  7 Conv. of Corr. s Post. 4. 3 and 5 are supplementary. Conv. of Same-Side Int. s Thm.

Perpendicular Lines Warm Up Lesson Presentation Lesson Quiz 21.3 Holt McDougal Geometry Holt Geometry

21.3 Objective Prove and apply theorems about perpendicular lines.

21.3 Vocabulary perpendicular bisector distance from a point to a line

21.3 The perpendicular bisector of a segment is a line perpendicular to a segment at the segment’s midpoint. The shortest segment from a point to a line is perpendicular to the line. This fact is used to define the distance from a point to a line as the length of the perpendicular segment from the point to the line.

Example 5: Distance From a Point to a Line 21.3 Example 5: Distance From a Point to a Line A. Name the shortest segment from point A to BC. AP B. Write and solve an inequality for x. AC > AP AP is the shortest segment. x – 8 > 12 Substitute x – 8 for AC and 12 for AP. + 8 Add 8 to both sides of the inequality. x > 20

21.3 HYPOTHESIS CONCLUSION

21.3 Example 6 Solve to find x and y in the diagram. x = 9, y = 4.5