Chapter 3Chapter 3

Parallel and Parallel and Perpendicular Perpendicular

LinesLines

Sec. 3-1Sec. 3-1Properties of Properties of Parallel LinesParallel Lines

Objective: a) Identify Angles formed by Two Lines & a Transversal.

b) To Prove & Use Properties of Parallel Lines.

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 23 4

5 6

87

Interior Portion of the // Lines

t

m

n

Corresponding AnglesCorresponding AnglesMost Important Angle RelationshipMost Important Angle Relationship

Always CongruentAlways Congruent

Cut the Transversal & lay the top part Cut the Transversal & lay the top part onto the bottom part. Overlapping Angles onto the bottom part. Overlapping Angles are Corresponding.are Corresponding.

1 2

3 4

5 67 8

Corresponding Angles

1 & 5

2 & 6

3 & 7

4 & 8

P(3 – 1) Corresponding P(3 – 1) Corresponding Angle Angle PostulatePostulate

If a Transversal Intersects two // lines, If a Transversal Intersects two // lines, then the corresponding angles are then the corresponding angles are Congruent.Congruent.

1 2

3 4

65

87

1 2

3 4

6587

4 Pairs of Vertical Angles

Are Congruent

1 4 2 3

5 8 6 7

3 & 6

4 & 5Alternate Interior Angles

3 & 5

4 & 6

Same-Sided Interior Angles

Special Interior Angles

Are congruent

Are Supplementary

(= 180)

Th (3-1) Alternate Interior Angle Theorem

If a Transversal intersects two // lines, then the alternate interior angles are congruent.

1 2

3 4

5 6

7 8

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

Th (3-2) Same-Sided Interior Angle Theorem

If a Transversal intersects two // lines, then the same-sided interior angles are supplementary.Given: l // mProve: 4 & 6 are Supplementary

1 23 4

5 67 8

Statements

1. l // m

2. m4 + m2 = 180

3. m2 = m6

4. m4 + m6 = 180

5. 4 & 6 are Supplementary

Reasons

1. Given

2. Add. Postulate

3. Corrsp. s are

4. Subs

5. Def of Supplementary

l

m

Examples 1 & 2Examples 1 & 2

Solve for the Solve for the missing missing ss

Solve for x, then for Solve for x, then for each 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

Use what you have Use what you have learned!learned!1. Find m1. Find m2 if l//m.2 if l//m.

42

1 2

mm1 = 42 1 = 42 (Corrsp. (Corrsp. ))

mm1 + m1 + m22 = 180

42 + mm2 = 1802 = 180

mm2 = 1382 = 138

2. Solve for angles a, b, c if l//m

l

m

m

l

65 40

a bc

ma = 65 (Alt. Inter. )

mc = 40 (Alt. Inter. )

ma + mb + mc = 180

65 + mb + 40 = 180

m = 75

Solve for x and find the measure of each angle if l//m.

(x + 40)

x

l

mx + x +40 = 180

2x + 40 = 180

-40 -40

2x = 140

x = 70