Download Math 135 Section 5.1 notes

Math 135
Section 5.1
Parallel and Perpendicular Lines
If two lines are intersected by a third line, the third line is called a transversal. In figure
1.1 below, there are lines l and m with transversal k.
1 2
l
3
4
k
5 6
m
7 8
Figure 1.1
In the figure above, there are also some angles which are numbered 1 through 8. Angles
∠1, ∠2, ∠7 and ∠8 are called exterior angles. Angles ∠3, ∠4, ∠5 and ∠6 are exterior
angles. In Euclid Geometry there are some special angle pairs.
Exterior angles that lie on opposite sides of the transversal are called alternate exterior
angles.
In figure 1.1, the pair of ∠1 and ∠8 are alternate exterior angles and the pair of ∠2 and
∠7 are alternate exterior angles.
Interior angles that lie on opposite sides of the transversal are called alternate interior
angles.
In figure 1.1, the pair of ∠3 and ∠6 are alternate exterior angles and the pair of ∠4 and
∠5 are alternate exterior angles.
Corresponding angles lie of the same side of the transversal where one angle is an
interior angle and the other angle is an exterior angle.
In figure 1.1, the angle pairs of ∠1 and ∠5 : ∠2 and ∠6 : ∠3 and ∠8 : ∠4 and ∠7 are
corresponding angles.
Recall: Theorem 4.11 (Exterior Angle Theorem)
A
1
3
2
C
B
The measure of an exterior angle of a triangle is greater than the measure of either remote
interior angle. Note: ∠3 > ∠1 and ∠3 > ∠2
Theorem 5.1
If two lines are cut by a transversal to form a pair of congruent alternate interior angles,
then the lines are parallel.
l
1
m
2
k
Proof:
Proof by contradiction: Assume that ∠1 ≅ ∠2 and lines l and m are not parallel.
If lines l and m are not parallel, then by theorem 4.11 (Exterior Angle Theorem) we can
conclude that ∠1 > ∠2 . This is a contradiction to the fact that ∠1 ≅ ∠2 . By indirect
reasoning, the theorem 5.1 is true.
l
1
k
m
2
Corollary 5.2
If two lines are both perpendicular to a transversal, then the lines are parallel.
l
1 3
k
2 4
m
If l ⊥ k and m ⊥ k , then l || m
Proof:
Given l ⊥ k and m ⊥ k , show that l || m
Statement
Reason
l ⊥ k and m ⊥ k
Given
m∠1 = 90°
m∠2 = 90°
m∠1 + m∠3 = 180°
Definition of perpendicular lines
90° + m∠3 = 180°
Supplementary angles have an angle sum
of 180 degrees.
Substitution property
Subtraction property of equality
90° − 90° + m∠3 = 180° − 90°
m∠3 = 90°
m∠1 = m∠3
Transitive property of equality
l || m
Theorem 5.1
Corollary 5.3
If two lines are cut by a transversal form a pair of congruent corresponding angles with
the transversal, then the lines are parallel.
Proof:
Given ∠1 ≅ ∠2 , show that l || m
Statement
∠1 ≅ ∠2
Reason
Given
∠1 ≅ ∠3
Vertical angles are congruent
∠2 ≅ ∠3
Transitive Property
l || m
Theorem 5.1
Corollary 5.4
If two lines cut by a transversal form a pair of supplementary angles on the same side of
the transversal, then the lines are parallel.
l
3
1
k
2
m
Proof: Given m∠1 + m∠2 = 180° , show that l || m
Let m∠1 + m∠2 = 180° . Since ∠1 and ∠3 are supplementary angles, m∠1 + m∠3 = 180°
By subtracting the two equations above we get the following result.
m∠1 + m∠2 = 180°
m∠1 + m∠2 = 180°
− (m∠1 + m∠3 = 180°)⇒
− m∠1 − m∠3 = −180°
m∠2 − m∠3 = 0°
If m∠2 − m∠3 = 0° , then m∠2 = m∠3
If two angles have the same angle measure they are congruent. Thus, ∠2 ≅ ∠3
Therefore, l || m by theorem 5.1.
Postulate 5.1
Given a line l and point P not on l , there is only one line m containing P such that
l || m
Theorem 5.5
If a pair of parallel lines is cut by a transversal, then the alternate interior angles formed
are congruent.
k
l
P
1
2
m
Proof:
Given l || m , show that ∠1 ≅ ∠2
Let l || m , and assume m∠1 ≠ m∠2 . If m∠1 ≠ m∠2 , then there are two cases
m∠1 > m∠2 and m∠1 < m∠2
Case 1: m∠1 > m∠2
If m∠1 > m∠2 , then exist a line n through P that form an angle ∠3 where ∠3 ≅ ∠2
This implies that l || n . This is a contradiction to the parallel postulate because there are
two lines, m and n, parallel to l through a point P.
k
l
n
P
1
3
2
m
Case 2:
If m∠1 < m∠2 , then exist a line t through P that form an angle ∠4 where ∠4 ≅ ∠2
This implies that l || t . This is a contradiction to the parallel postulate because there are
two lines, m and t, parallel to l through a point P.
k
l
t
1
P
4
2
m
Thus, by indirect reasoning, we have ∠1 ≅ ∠2
Corollary 5.6
If two lines are parallel and a line is perpendicular to one of the two lines, then it is
perpendicular to the other line.
l
1 3
k
2 4
m
Proof:
Given l || m and l ⊥ k , show that k ⊥ m
Statement
l || m
k⊥m
∠2 ≅ ∠3 ⇒ m∠2 = m∠3
m∠3 = 90°
m∠2 = 90°
k⊥m
Reason
Given
If two lines are parallel, then the alternate
interior angles are congruent.
Theorem 5.5
Two lines are perpendicular iff the angles
formed are right angles.
Substitution property
Two lines are perpendicular iff the angles
formed are right angles.
Corollary 5.7
If a pair of parallel lines is cut by a transversal, then each pair of corresponding angles
formed is congruent.
k
5
l
3
6
1
4
2
m
Proof:
Given l || m , show that ∠1 ≅ ∠2
Statement
l || m
Reason
Given
∠2 ≅ ∠3
Theorem 5.5
∠1 ≅ ∠3
If two lines intersect, the vertical angles
formed are congruent.
Transitive property of congruence
∠1 ≅ ∠2
Corollary 5.8
If two parallel lines are cut by a transversal, then both pairs of interior angles on the same
side of the transversal are supplementary.
k
5
l
3
6
1
4
2
m
Proof:
Given l || m , show that m∠2 + ∠4 = 180°
Statement
l || m
Reason
Given
∠2 ≅ ∠3
⇒ m∠2 = m∠3
∠3 and ∠4 are supplementary
Theorem 5.5
m∠3 + ∠4 = 180°
The sum of two supplementary angles is
180 degrees.
Substitution property
m∠2 + ∠4 = 180°
Example 1
Use the following figure to exercises 1-5
2
1
4
7
9
3
6
5
8
10
11
12
13
14
1) Indentify one pair of alternate interior angles.
Solution: ∠4 and ∠8
2) Indentify two pairs of corresponding angles.
Solution:
∠4 and ∠9
∠1 and ∠11
∠6 and ∠14
∠3 and ∠8
3) Find the measure of ∠1 + ∠12
m∠1 = m∠11
m∠11 + m∠12 = 180°
m∠1 + m∠12 = 180°
4) Find the measure of ∠3 and ∠8 , if m∠4 = 60°
Solution:
m∠3 = m∠4 = 60°
m∠8 = m∠3 = 60°
5) Find the measure of all of the numbered angles given m∠4 = 40° and m∠1 = 65°
Solution:
65°
40°
7
9
2
3
6
5
8
10
11
12
13
14
m∠2 = 180° − (40° + 65°) = 180° − 105° = 75°
m∠3 = m∠1 = 40°
m∠6 = m∠4 = 65°
m∠5 = m∠2 = 75°
Corresponding Angles:
m∠9 = m ∠4 = 40°
m∠11 = m∠1 = 65°
m∠14 = m ∠6 = 65°
m∠8 = m∠3 = 40°
Supplementary angles give:
m∠7 = 140° : m∠100 = 140° : m∠12 = 115° : m∠13 = 115°