Use angles formed by a transversal to show that lines are parallel

1. Proving Lines Parallel
The student is able to (I can):
• Use angles formed by a transversal to show that lines are
parallel.

2. Recall that the converse of a theorem is found by exchanging
(flipping) it around.
The converses of the parallel line theorems can be used to
prove lines parallel.
• Corresponding Angles
• Alternate Interior Angles
• Alternate Exterior Angles
• Consecutive Interior Angles
– If these angles are supplementary, then the lines are
parallel.
If these angles are
congruent, then the lines
are parallel.

3. Example Find values of x and y that make the red
lines parallel and the blue lines parallel.
If the blue lines are parallel, then the con-
secutive interior angles must be supple-
mentary.
x  40 + x + 40 = 180
2x = 180
x = 90
(x40) (x+40)
y

4. Example Find values of x and y that make the red
lines parallel and the blue lines parallel.
If the red lines are parallel, then the con-
secutive interior angles must be supple-
mentary.
90  40 + y = 180
50 + y = 180
y = 130
(x40) (x+40)
y

5. • One of the enduring concepts of advanced mathematics is
the idea of justification – why are we able to do such-and-
such.
• Over the years this has progressed from a practical
“because it just works” to taking specific cases and making
general statements from them.
– For example, going from physically measuring the area
of a triangle to figuring out the formula for the area of
any triangle.
• The way we do this in Geometry is to use properties and
postulates in a logical argument called a proof.
• In addition to the geometric properties, we will also make
use of algebraic properties that you have used, although
you weren’t necessarily told what they were.

6. Algebraic Properties of Equality
These are properties that let us solve equations:
• Addition Property of Equality (“add. prop. =“)
– If a = b, then a + c = b + c
• Subtraction Property of Equality (“subtr. prop. =“)
– If a = b, then a  c = b  c
• Multiplication Property of Equality (“mult. prop. =“)
– If a = b, then ac = bc
• Division Property of Equality (“div. prop. =“)
– If a = b and c  0, then
• Symmetric Property of Equality (“sym. prop. =“)
– If a = b, then b = a

a b
c c

7. Algebraic Properties of Equality (cont.)
• Substitution Property of Equality (“subst. prop. =“)
– If a = b, then b may be substituted for a.
• Transitive Property of Equality (“trans. prop. =“)
– If a = b and b = c, then a = c
Take a look at https://www.ancient.eu/article/606/greek-
mathematics/ for an interesting history of early mathematics.

8. Example: Solve the equation 21 = 4x – 7. Write a
justification for each step.
21 = 4x – 7 Given equation
21 + 7 = 4x – 7 + 7 Add. prop. =
28 = 4x Simplify
7 = x Simplify
x = 7 Sym. prop. =
28
4 4
4x
 Div. prop. =

9. Line segments with equal lengths are congruent, and angles
with equal measures are also congruent. Therefore, the
symmetric and transitive properties of equality have
corresponding properties of congruence.
• Symmetric Property of Congruence (“sym. prop. ”)
– If fig. A  fig. B, then fig. B  fig. A
• Transitive Property of Congruence (“trans. prop. ”)
– If fig. A  fig. B and fig. B  fig. C, then fig. A  fig. C
In addition, we also have the following property that may
seem rather silly, but it ends up being useful in proofs:
• Reflexive Property of Congruence (“refl. prop. ”)
– fig. A  fig. A (everything is congruent to itself)

10. Given: bisects DBA; 3  1
Prove:
Plan of proof: Because bisects DBA, 2  3. Because
3  1, I can set 2 1 using substitution.
1 and 2 are alternate interior angles, and
since they are congruent, the lines are parallel.
BE
CD BE
C
32
1
ED A
B
BE

11. Given: bisects DBA; 3  1
Prove:
BE
CD BE
C
32
1
ED A
B
Statements Reasons
1. bisects DBA 1. GivenBE

12. Given: bisects DBA; 3  1
Prove:
BE
CD BE
C
32
1
ED A
B
Statements Reasons
1. bisects DBA 1. Given
2. 2  3 2.  bisector creates 2  s
BE

13. Given: bisects DBA; 3  1
Prove:
BE
CD BE
C
32
1
ED A
B
Statements Reasons
1. bisects DBA 1. Given
2. 2  3 2.  bisector creates 2  s
3. 3  1 3. Given
BE

14. Given: bisects DBA; 3  1
Prove:
BE
CD BE
C
32
1
ED A
B
Statements Reasons
1. bisects DBA 1. Given
2. 2  3 2.  bisector creates 2  s
3. 3  1 3. Given
4. 2  1 4. Substitution prop. 
BE

15. Given: bisects DBA; 3  1
Prove:
BE
CD BE
C
32
1
ED A
B
Statements Reasons
1. bisects DBA 1. Given
2. 2  3 2.  bisector creates 2  s
3. 3  1 3. Given
4. 2  1 4. Substitution prop. 
5. 5. If alt. int. s , then lines 
BE
CD BE