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Building a System of Building a System of Geometry Knowledge 2.4Geometry Knowledge 2.4
Algebraic propertiesAlgebraic properties
Addition PropertyAddition Property If a = b , then a + c = b + cIf a = b , then a + c = b + c
Subtraction PropertySubtraction Property If a = b , then a – c = b – cIf a = b , then a – c = b – c
Multiplication PropertyMultiplication Property If a = b , then ac = bcIf a = b , then ac = bc
Division PropertyDivision Property If a = b and c If a = b and c ≠ 0 , then≠ 0 , then a/c = b/ca/c = b/c
Substitution PropertySubstitution Property
If a = b, you may replace a If a = b, you may replace a with b in any true equation with b in any true equation that has a in it, and the that has a in it, and the resulting equation will still be resulting equation will still be true.true.
Equivalence PropertiesEquivalence Properties Reflexive PropertyReflexive Property
for any number a, a = afor any number a, a = a Symmetric PropertySymmetric Property
if a = b , b = aif a = b , b = a Transitive PropertyTransitive Property
if a = b and b = c, then a = if a = b and b = c, then a = cc
Overlapping SegmentsTheoremOverlapping SegmentsTheorem
Overlapping segment theorem:Given a segment with pointsA, B, C, and D arranged as shown, the following statements are true:
If AB = CD then AC = BD
If AC = BD then AB = CD
Overlapping AnglesOverlapping AnglesTheoremTheorem
Given AED with points B and C in its interior as shown, the following statements are true:
1. If mAEB = mCED, then mAEC = mBED.
2. If mAEC = mBED, then mAEB = mCED.
Equality and CongruencyEquality and Congruency
For all these properties, you can For all these properties, you can change the equal sign to a congruent change the equal sign to a congruent sign and they are still true. sign and they are still true.
In the Reasons column of the proof In the Reasons column of the proof write Definition of Congruence.write Definition of Congruence.
Two Column ProofTwo Column Proof
Statement Reason
1. AB = CD 1. Given
2. AB + BC = BC + CD 2. Addition Property
3. AB + BC = AC 3. Segment Addition Postulate
4. BC + CD = BD 4. Segment Addition Postulate
5. AC = BD 5. Substitution Property
Given: AB = CD Prove: AC = BD
Two Column ProofTwo Column ProofGiven: a = b Prove: a + c = b + c
Statement Reason
1. a = b 1. Given
2. a + c = a + c 2. Reflexive Property of Equality
3. a + c = b + c 3. Substitution Property of Equality
Two Column ProofTwo Column Proof
Statement Reason
1. 2x − 5 = 3 1. Given
2. 2x − 5 + 5 = 3 + 5 2. Addition Property of Equality
3. 2x = 8 3. Simplify
4. 2x ÷ 2 = 8 ÷ 2 4. Division Property of Equality
5. x = 4 5. Simplify
Given: 2x − 5 = 3 Prove: x = 4
Two Column ProofTwo Column Proof
Statement Reason
1. B C 1. Given
2. mB = mC 2. Definition of Congruence
3. 5x + 12 = 47 3. Substitution Property of Equality
4. 5x = 35 4. Subtraction Property of Equality
5. x = 7 5. Division Property of Equality
Given: B C Prove: x = 7
Two Column ProofTwo Column Proof
Statement Reason
1. m1 + m3 = 180 1. Given
2. m2 + m3 = 180 2. Linear Pair Property
3. m1 + m3 = m2 + m3 3. Substitution Property of Equality
4. m1 = m2 4. Subtraction Property of Equality
5. 1 2 5. Definition of Congruence
Given: m1 + m3 = 180
Prove: 1 2
Two Column ProofTwo Column Proof
Statement Reason
1. mHGK = mJGL 1. Given
2. m∠HGK = m 1 + m 2∠ ∠ 2. Angle Addition Postulate
3. m∠JGL = m 3 + m 2∠ ∠ 3. Angle Addition Postulate
4. m 1 + m 2 = m 3 + m 2∠ ∠ ∠ ∠ 4. Substitution Property of Equality
Given: mHGK = mJGL
Prove: 1 3
5. m 1 = m 3∠ ∠ 5. Subtraction
6. 1 ∠ 3∠ 6. Definition of Congruence
Given: PQ = 2x + 5 QR = 6x – 15 PR = 46 : PQ = 2x + 5 QR = 6x – 15 PR = 46 Prove: x = 7: x = 7 P
Q
RTwo Column ProofTwo Column Proof
Statement Reason
1. PQ = 2x + 5, QR = 6x –– 15, PR = 46 1. Given
2. PQ + QR = PR 2. Segment Addition Postulate
3. 2x + 5 + 6x – 15 = 46 3. Substitution Property of Equality
4. 8x – 10 = 46 4. Simplify
5. 8x = 56 5. Addition Property of Equality
6. x = 7 6. Division Property of Equality
Two Column ProofTwo Column Proof
Statement Reason
1. Q is midpoint of PR 1. Given
2. PQ = QR 2. Definition of midpoint
3. PQ + QR = PR 3. Segment Addition Postulate
4. QR + QR = PR & PQ + PQ =PR 4. Substitution Property of Equality
5. 2QR = PR 2PQ = PR 5. Simplify
Given: Q is the midpoint of PR. Prove: PQ = and QR = Given: Q is the midpoint of PR. Prove: PQ = and QR =
6. QR = PQ = 6. Division Property of Equality
PRPRPRPR22 22
PR2
PR2
Two Column ProofTwo Column Proof
Statement Reason
1. 2(3x + 1) = 5x + 14 1. Given
2. 6x + 2 = 5x + 14 2. Distributive property
3. x + 2 = 14 3. Subtraction Property of Equality
4. x = 12 4. Subtraction Property of Equality
Given: 2(3x + 1) = 5x + 14 Prove: x = 12Given: 2(3x + 1) = 5x + 14 Prove: x = 12
Two Column ProofTwo Column Proof
Statement Reason
1. 55z – 3(9z + 12) = – 64 1. Given
2. 55z – 27z – 36 = – 64 2. Distributive Property
3. 28z – 36 = – 64 3. Simplify
4. 28z = –28 4. Addition Property of Equality
5. z= – 1 5. Division Property of Equality
Given: 55z Given: 55z – 3(9z + 12) = 3(9z + 12) = –64. Prove: z = 64. Prove: z = –1 1
Summary of PropertiesSummary of Properties
AssignmentAssignmentGeometry:Geometry:
2.4A and 2.4B2.4A and 2.4B
Section 9 - 24Section 9 - 24