holt mcdougal geometry angles formed by parallel lines and transversals angles formed by parallel...

Holt McDougal Geometry

Angles Formed by Parallel Lines and Transversals


Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz


21.1



Warm UpIdentify each angle pair.

1. 1 and 3

2. 3 and 6

3. 4 and 5

4. 6 and 7 same-side int s

corr. s

alt. int. s

alt. ext. s

21.1



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

Objective

21.1



21.1



Find each angle measure.

Example 1: Using the Corresponding Angles Postulate

A. mECF

x = 70

B. mDCE

mECF = 70°

Corr. s Post.

5x = 4x + 22 Corr. s Post.

x = 22 Subtract 4x from both sides.

mDCE = 5x

= 5(22) Substitute 22 for x.

= 110°

21.1



Example 2

Find mQRS.

mQRS = 180° – x

x = 118

mQRS + x = 180°

Corr. s Post.

= 180° – 118°

= 62°

Subtract x from both sides.

Substitute 118° for x.

*Def. of Linear Pair*

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.

21.1



21.1



Find each angle measure.

Example 3

A. mEDG

B. mBDG

mEDG = 75° Alt. Ext. s are

Congruent.

mBDG = 105°

x – 30° = 75° Alt. Ext. s are congruent.

x = 105 Add 30 to both sides.

21.1



Find x and y in the diagram.

Example 4

By the Alternate Interior AnglesTheorem, (5x + 4y)° = 55°.

By the Corresponding Angles Postulate, (5x + 5y)° = 60°.

5x + 5y = 60–(5x + 4y = 55) y = 5

5x + 5(5) = 60

Subtract the first equation from the second equation.

x = 7, y = 5

Substitute 5 for y in 5x + 5y = 60. Simplify and solve for x.

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. m1 = 120°, m2 = (60x)°

2. m2 = (75x – 30)°, m3 = (30x + 60)°

Corr. s Post.; m2 = 120°, m3 = 120°

Alt. Ext. s Thm.; m2 = 120°

3. m3 = (50x + 20)°, m4= (100x – 80)°

4. m3 = (45x + 30)°, m5 = (25x + 10)°Alt. Int. s Thm.; m3 = 120°, m4 =120°

Same-Side Int. s Thm.; m3 = 120°, m5 =60°

21.1


Angles Formed by Parallel Lines and TransversalsProving Lines Parallel

Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz


21.2



Warm UpState the converse of each statement.

1. If a = b, then a + c = b + c.

2. If mA + mB = 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 mA + mB =90°.

If A, B, and C are collinear, then AB + BC = AC.

21.2



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

Objective

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.

21.2



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.

21.2



21.2



Lesson Quiz: Part I

Name the postulate or theoremthat 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.

21.2


Angles Formed by Parallel Lines and TransversalsPerpendicular Lines

Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz


21.3



Prove and apply theorems about perpendicular lines.

Objective

21.3



perpendicular bisectordistance from a point to a line

Vocabulary

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.

21.3



Example 5: Distance From a Point to a Line

AP

B. Write and solve an inequality for x.

AC > AP

x – 8 > 12

x > 20

Substitute x – 8 for AC and 12 for AP.

Add 8 to both sides of the inequality.

A. Name the shortest segment from point A to BC.

AP is the shortest segment.

+ 8 + 8

21.3



HYPOTHESISCONCLUSION

21.3



Example 6

Solve to find x and y in the diagram.

x = 9, y = 4.5

21.3

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