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