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