Math 2 Geometry Based on Elementary Geometry, 3 rd ed, by Alexander & Koeberlein 1.5 Introduction to Geometric Proof.

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Presentation transcript:

Math 2 Geometry Based on Elementary Geometry, 3 rd 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

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x = 6 1.Given

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x = 6 1.Given

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x = 6 1.Given 2.Distributive Property

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x – 2 = x = 6 1.Given 2.Distributive Property

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x – 2 = x = 6 1.Given 2.Distributive Property 3.Substitution 4. 5.

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x – 2 = x = 12 5.x = 6 1.Given 2.Distributive Property 3.Substitution 4. 5.

Proof Given:2(x – 3) + 4 = 10 Prove: x = 6 StatementsReasons 1.2(x – 3) + 4 = x – = x – 2 = x = 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 StatementsReasons 1.2(x – 3) + 4 = x – = x – 2 = x = 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons M N P Q

Proof Given:MN > PQ Prove: MP > NQ StatementsReasons 1. MN > PQ 2. ?. MP > NQ 1. Given 2. ?. M N P Q

Proof Given:MN > PQ Prove: MP > NQ StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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 StatementsReasons 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