Download 200 300 400 500 100 200 300 400 500 100 200 300 400 500 100

Properties,
Conditionals,
Postulates, & Biconditionals,
Etc.
Theorems
Deductive
Reasoning
Algebraic
Reasoning
Proofs
100
100
100
100
100
200
200
200
200
200
300
300
300
300
300
400
400
400
400
400
500
500
500
500
500
Properties, Postulates, & Theorems 100
Why does
3  x  6  3x  18
Properties, Postulates, & Theorems 100
Distributive Property
Properties, Postulates, & Theorems 200
If BY = XD and XD = ZC
then why does BY = ZC?
Properties, Postulates, & Theorems 200
Transitive Property
Properties, Postulates, & Theorems 300
Directions: Determine if the following is always,
sometimes, or never true.
Vertical angles are supplementary.
Properties, Postulates, & Theorems 300
Sometimes.
(Only when perpendicular lines.)
Properties, Postulates, & Theorems 400
Directions: Determine if the following is always,
sometimes, or never true.
Every obtuse angle has a supplement.
Properties, Postulates, & Theorems 400
Always.
Properties, Postulates, & Theorems 500
Directions: Determine if the following is always,
sometimes, or never true.
Adjacent angles form linear pairs.
Properties, Postulates, & Theorems 500
Sometimes.
(If the two angles are supplementary.)
Conditionals, Biconditionals, etc. 100
Identify the hypothesis and conclusion of the
following statement:
If two lines intersect at right angles, then
the two lines are perpendicular.
Conditionals, Biconditionals, etc. 100
Hypothesis: Two lines intersect at right
angles.
Conclusion: The two lines are perpendicular.
Conditionals, Biconditionals, etc. 200
Write the conditional and converse of the following:
Exercising keeps you in good shape!
Conditionals, Biconditionals, etc. 200
Conditional: If you exercise, then you are
in good shape.
Converse: If you are in good shape, then
you exercise.
Conditionals, Biconditionals, etc. 300
What two conditions make up the following
biconditional:
Two lines are perpendicular if and only if
they intersect to form four right angles.
Conditionals, Biconditionals, etc. 300
• If two lines are perpendicular then, they
intersect to form four right angles.
• If two lines form four right angles, then
they are perpendicular.
Conditionals, Biconditionals, etc. 400
Write the converse. If the converse is true, then
write the biconditional. If the converse is false, then
write a counter example.
If x = 3, then x  9
2
Conditionals, Biconditionals, etc. 400
Converse: If x 2  9 , then x = 3
False.
Counter example: x  3
Conditionals, Biconditionals, etc. 500
Is the following a statement a good definition?
Prove it. If not, find a counterexample.
A ray that divides an angle into two congruent
angles is an angle bisector.
Conditionals, Biconditionals, etc. 500
Conditional: If a ray divides an angle into two
congruent angles, then it is an angle bisector. (True)
Converse: If a ray is an angle bisector, then it divides
an angle into two congruent angles. (True)
Biconditional: A ray divides an angle into two
congruent angles if and only if it is an angle bisector.
Deductive Reasoning 100
Use the Law of Detachment to draw a conclusion.
If a team wins on Sunday, then practice is
cancelled Monday.
The Bears won on Sunday!
Deductive Reasoning 100
The Bears practice is cancelled on Monday!
Deductive Reasoning 200
Use the Law of Detachment to draw a conclusion.
If the measure of two angles is 90 degrees,
then they are complementary.
mA  mB  90
Deductive Reasoning 200
Angle A and Angle B are complementary
Deductive Reasoning 300
Use the Law of Syllogism to draw a conclusion.
If you study, then you will earn an A.
If you earn an A, then you are HAPPY!
Deductive Reasoning 300
If you study, then you are HAPPY!
Deductive Reasoning 400
Use the Law of Detachments to draw a conclusion.
If a dog eats Doggie Delights, then it is a happy dog.
Fido is a happy dog.
Deductive Reasoning 400
No Conclusion!
Deductive Reasoning 500
Use the Law of Syllogism to draw a conclusion.
If two lines intersect, then they intersect at a point.
If two lines are not parallel, then they intersect.
Deductive Reasoning 500
If two lines are not parallel, then they intersect
at a point.
Algebraic Reasoning 100
Justify each step:
7x  4 10
7x 14
x2
Given
Algebraic Reasoning 100
7x  4 10
7x 14
x2
Given
Addition Property of Equality
Division Property of Equality
Algebraic Reasoning 200
The measure of two verticals angles are
 7x  8  and  6x 11
Solve for x. Justify each step.
Algebraic Reasoning 200
 7x  8 
 6x 11
7x  8  6x 11
x  8  11
x  19
Algebraic Reasoning 300
Given: mAOC 150
 2x   B
Prove: x  21
A
6  x  3 
O
C
Algebraic Reasoning 300
 2x   B
A
6  x  3 
O
C
mAOB  mBOC  mAOC Angle Addition Post.
Substitution
 2x   6  x  3  150
2x  6x 18 150
8x 18 150
8x  168
x  21
Distributive Prop.
Simplify.
Addition Prop.
Division Prop.
Algebraic Reasoning 400
Given: 11x  6y  1 when x  8
89
Prove: y 
6
Algebraic Reasoning 400
11x  6y  1 when x  8
11 8   6y  1
88  6y  1
6y  89
89
y
6
Given
Substitution
Multiplication Prop. of =
Subtraction Prop. of =
Division Prop. of =
Algebraic Reasoning 500
Solve for x and justify each step:
Segment KL is bisected by point M,
KM  2x  3, KL 18, find x.
Algebraic Reasoning 500
2x  3
2x  3
K
M
KM  ML  KL
KM  ML
 2x  3   2x  3  18
4x  6  18
4x  12
x 3
L
Segment Addition Post.
bisect  2  segments
Substitution
Simplify
Subtraction Prop. of =
Division Prop. of =
Proofs 100
Given: AB  BC
A
1
B
m2  m3
Prove: m1 m3  90
C
2
3
D
Reasons
Statements
1. AB  BC
1.
2. ABC is a right angle
2.
3. mABC  90
4. mABC  m1 m2
5. m1 m2  90
3.
4.
5.
6. m2  m3
7. m1 m3  90
6.
7.
Proofs 100
Given: AB  BC
A
1
B
m2  m3
Prove: m1 m3  90
C
2
3
D
Reasons
Statements
1. AB  BC
1.
2. ABC is a right angle
2.
3. mABC  90
4. mABC  m1 m2
5. m1 m2  90
3.
4.
5.
6. m2  m3
7. m1 m3  90
6. Given
7. Substitution Pr op.
Given
  right 
right   90
Angle Addition Post.
Substitution Pr op.
Proofs 200
D
Given: EDF  HDG
E
Prove: EDG  HDF
H
F
Reasons
Statements
1.
1.
2.
2.
3.
3.
G
Proofs 200
D
Given: EDF  HDG
E
Prove: EDG  HDF
Statements
H
F
G
Reasons
1. EDF  HDG
1. Given
2. FDG  FDG
2. Re flexive Pr operty
3. EDG  HDF
3. Addition Pr operty
Proofs 300
G
E
O
F
U
N
Given: GO  FN
EO  FU
Prove: GE  UN
Reasons
Statements
1.
1.
2.
2.
3.
3.
Proofs 300
G
E
O
F
U
N
Given: GO  FN
EO  FU
Prove: GE  UN
Statements
Reasons
1. GO  FN
1. Given
2. EO  FU
2. Given
3. GE  UN
3. Subtraction Pr operty
Proofs 400
Given: B is the midpoint of AC
AB  BD
Prove: BD  BC
D
A
Reasons
Statements
1.
1.
2.
2.
3.
3.
4.
4.
B
C
Proofs 400
Given: B is the midpoint of AC
D
AB  BD
Prove: BD  BC
Statements
A
B
C
Reasons
1. B is the midpoint of AC
1. Given
2. AB  BC
2. midpoint  2  segments
3. AB  BD
3. Given
4. BD  BC
4. Transitive Pr operty
Proofs 500
Given: 1  2
Prove: 3  4
Statements
2
4
Reasons
1
3
Proofs 500
Given: 1  2
2
4
Prove: 3  4
Statements
1
3
Reasons
1. 1  2
1. Given
2. 1  3
2. vertical s  
3. 2  3
3. transitive prop.
4. 2  4
4. same as 2
5. 3  4
5. same as 3