Download Math 2 Geometry Based on Elementary Geometry

Math 2 Geometry
Based on Elementary Geometry, 3rd ed, by Alexander & Koeberlein
1.5
Introduction to Geometric Proof
Properties of Equality
Addition property of equality
If a = b, then a + c = b + c
Subtraction property of equality
If a = b, then a – c = b – c
Multiplication property of equality
If a = b, then a·c = b·c
Division property of equality
If a = b and c  0, then a/c = b/c
Properties of Inequality
Addition property of inequality
If a > b, then a + c > b + c
Subtraction property of inequality
If a > b, then a – c > b – c
Multiplication property of inequality
If a > b, and c > 0, then a·c > b·c
Division property of inequality
If a > b and c > 0, then a/c > b/c
More Algebra Properties
Distributive property
a(b + c) = a·b + a·c
Substitution property
If a = b, then a replaces b in any
equation
Transitive property
If a = b and b = c, then a = c
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1. 2(x – 3) + 4 = 10
2.
3.
4.
5. x = 6
Reasons
1. Given
2.
3.
4.
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1. 2(x – 3) + 4 = 10
2. 2x – 6 + 4 = 10
3.
4.
5. x = 6
Reasons
1. Given
2.
3.
4.
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1. 2(x – 3) + 4 = 10
2. 2x – 6 + 4 = 10
3.
4.
5. x = 6
Reasons
1. Given
2. Distributive Property
3.
4.
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1.
2.
3.
4.
5.
Reasons
2(x – 3) + 4 = 10 1. Given
2x – 6 + 4 = 10
2. Distributive Property
2x – 2 = 10
3.
4.
x=6
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1.
2.
3.
4.
5.
Reasons
2(x – 3) + 4 = 10 1. Given
2x – 6 + 4 = 10
2. Distributive Property
2x – 2 = 10
3. Substitution
4.
x=6
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1.
2.
3.
4.
5.
2(x – 3) + 4 = 10
2x – 6 + 4 = 10
2x – 2 = 10
2x = 12
x=6
Reasons
1. Given
2. Distributive Property
3. Substitution
4.
5.
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1.
2.
3.
4.
5.
2(x – 3) + 4 = 10
2x – 6 + 4 = 10
2x – 2 = 10
2x = 12
x=6
Reasons
1.
2.
3.
4.
5.
Given
Distributive Property
Substitution
Add. Prop. of Equality
Given:
2(x – 3) + 4 = 10
Prove:
x=6
Proof
Statements
1.
2.
3.
4.
5.
2(x – 3) + 4 = 10
2x – 6 + 4 = 10
2x – 2 = 10
2x = 12
x=6
Reasons
1.
2.
3.
4.
5.
Given
Distributive Property
Substitution
Add. Prop. of Equality
Division Prop. of Eq.
Given:
A-P-B on seg AB
Prove:
AP = AB - PB
Proof
Statements
1. A-P-B on seg AB
2.
·
·
·
?. AP = AB - PB
Reasons
1. Given
2.
·
·
·
?.
Given:
A-P-B on seg AB
Prove:
AP = AB - PB
Proof
Statements
1. A-P-B on seg AB
2.
·
·
·
?. AP = AB - PB
Reasons
1. Given
2. Segment-Addition
Postulate
·
·
·
?.
Given:
A-P-B on seg AB
Prove:
AP = AB - PB
Proof
Statements
1. A-P-B on seg AB
2. AP + PB = AB
·
·
·
?. AP = AB – PB
Reasons
1. Given
2. Segment-Addition
Postulate
·
·
·
?.
Given:
A-P-B on seg AB
Prove:
AP = AB - PB
Proof
Statements
1. A-P-B on seg AB
2. AP + PB = AB
3. AP = AB – PB
Reasons
1. Given
2. Segment-Addition
Postulate
3.
Given:
A-P-B on seg AB
Prove:
AP = AB - PB
Proof
Statements
1. A-P-B on seg AB
2. AP + PB = AB
3. AP = AB – PB
Reasons
1. Given
2. Segment-Addition
Postulate
3. Subtr. Prop. of Equality
Given:
MN > PQ
•
M
Prove:
•
N
MP > NQ
Proof
Statements
Reasons
•
P
•
Q
Given:
MN > PQ
•
M
Prove:
•
N
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
2.
1. Given
2.
?. MP > NQ
?.
•
P
•
Q
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
2.
1. Given
2. Add’n prop of Inequality
?. MP > NQ
?.
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
?. MP > NQ
?.
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
3. MN + NP > NP + PQ 3.
?. MP > NQ
?.
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
3. MN + NP > NP + PQ 3. Commutative prop. of
add’n
?. MP > NQ
?.
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
3. MN + NP > NP + PQ 3. Commutative prop. of
add’n
4. MN + NP = MP and 4.
NP + PQ = NQ
?. MP > NQ
?.
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
3. MN + NP > NP + PQ 3. Commutative prop. of
add’n
4. MN + NP = MP and 4. Segment Addition
Postulate
NP + PQ = NQ
?.
?. MP > NQ
Given:
MN > PQ
•
M
Prove:
•
N
•
P
•
Q
MP > NQ
Proof
Statements
Reasons
1. MN > PQ
1. Given
2. MN + NP > PQ + NP 2. Add’n prop of Inequality
3. MN + NP > NP + PQ 3. Commutative prop. of
add’n
4. MN + NP = MP and 4. Segment Addition
Postulate
NP + PQ = NQ
5. Substitution
5. MP > NQ