Conditional Statement: Statements in if-then form are called.
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1 Monday 9/ and 2.4 Wednesday 9/ and 2.6 Conditional and Algebraic Proofs Algebraic Properties and Geometric Proofs Unit 2 Angles and Proofs Packet pages 1-3 Textbook Pg 85 (14, 17, 20, 25, 27, 35, 38) Textbook Pg 99 (10, 14, 16, 17, 18, 20, 21, 38, 39) Packet Pages 4-12 Textbook Pg 107 (23-28, 39-41) Textbook Pg 113 (6-8, 16-22) Friday 9/25 Review Packet Pages 4-12 Tuesday 9/29 Quiz Conditional Statement: in if-then form are called. Symbolic Form: means: If p then q The p portion after the if is The q portion after the then is Example 1: Write each statement in conditional form. a) An angle of 40 degrees is acute b) A student on the high honor roll has at least a 90 average. Example 2: Identify the hypothesis and the conclusion of the following statement. I will go bowling if it rains on Tuesday. Truth Value: A Statement can have a truth value of True (T) or False (F) Counterexample: A counterexample to a statement is a particular example or instance of the statement that makes the statement. Example 3: Show that this conditional statement is false by finding a counter example: a) If it is February, then there are only 28 days in the month. b) If x 2 > 0, then x > 0 Converse (of a conditional): Statement formed by switching the and the Symbolic form: Example 4: Write the converse of the statements. Is the converse true or false? a) If I study, then I get good grades. b) If two angles are complementary, they the sum of their measures is 90º. 1
2 Biconditional: A statement that is the combination of a and its. A biconditional contains the words (iff) and can only be written if *** *** Symbolic Form: (goes both ways) How to write a biconditional: Example 5: Write a biconditional for the following statement, If it is Sunday, then I am watching football. Inverse (of a conditional): statement formed by both the and the. Symbolic Form: Example 6: Write the inverse of the statement: a) If it rains, then we will get wet. b) If it does not rain then we will play ultimate frisbee. Contrapositive (of a conditional) statement formed by switching AND negating the hypothesis and the conclusion. **Note: a conditional and its contrapositive are logically equivalent;** ***they have the same truth value*** Symbolic Form: Example 7: Write the contrapositive of the following statement: If you live in Charleston, then you live in South Carolina. Example 8: Write the converse, inverse, and contrapositive of each statement. a) If you like volleyball, then you like to be at the beach. Converse: Inverse: Contrapositive: 2
3 3
4 Day 2 - Using Proofs in Algebra I. Name the property that justifies each statement. 1. If m A = m B, then m B = m A. 2. If x + 3 = 17, then x = xy = xy 4. If 7x = 42, then x = 6 5. If XY YZ = XM, then XY = XM + YZ 6. 2( x + 4) = 2x If m A + m B = 90, and m A = 30, 8. If x = y + 3 and y + 3 = 10, then x = 10. then 30 + m B = 90. II. Complete the reasons in each algebraic proof. 9. Prove that if 2(x 3) = 8, then x = 7. a) 2 (x 3) = 8 a) b) 2x 6 = 8 b) c) 2x = 14 c) d) x = 7 d) 10. Prove that if 3x 4 = a) 3x 4 = 1 x + 6, then x = x + 6 a) 2 b) 2(3x 4) = 2( 2 1 x + 6) b) c) 6x 8 = x + 12 c) d) 5x 8 = 12 d) e) 5x = 20 e) f) x = 4 f) 4
5 11. You can use the Angle Addition Postulate Given: m< AOC = 139 Solve for x and justify each step. Statement Property 1) m<aob + m< BOC = m <AOC 1) 2) x + (2x + 10) = 139 2) 3) 3x + 10 = 139 3) 4) 3x = 129 4) 5) x = 23 5) 12. You can use the definition of an angle bisector. Given: LM bisects <KLN Solve for x and justify each step. Statement Property 1) LM bisects <KLN 1) 2) m <KLM = m <MLN 2) 3) 2x + 40 = 4x 3) 4) 40 = 2x 4) 5) 20 = x 5) 12. You can use the Segment Addition Postulate. Given: AC = 21 Solve for y and justify each step. Statement 1) AB + BC = AC 1) Property 2) 2y + 3y 9 = 21 2) 3) 5y 9 = 21 3) 4) 5y = 30 4) 5) y = 6 5) 5
6 Name the property that justifies each statement. 1. If 3x = 120, then x = If 12 = AB, then AB = If AB = BC and BC = CD, then AB = CD. 4. If y = 75 and y = ma, then ma = If 5 = 3x 4, then 3x 4 = If 3 x 1, then 3x 5 = If m1 = 90 and m2 = 90, then m1 = m2. 8. For XY, XY = XY. 9. If EF = GH and GH = JK, then EF = JK. 10. If m = 90, then m1 = 60. Choose the number of reason in the right column that best matches each statement in the left column. A. If x 7 = 12, then x = 19 B. If MK = NJ and BG = NJ, then MK = BG 1. Distributive Property 2. Addition Property of Equality C. If y = m5 30 and m5 = 90, then y = D. If ST = UV, then UV = ST. E. If x = -3(2x 4), then x = -6x Symmetric Property 4. Substitution Property 5. Transitive Property Name the property that justifies each statement. 11. If AB + BC = DE + BC, then AB = DE. 12. m ABC = mabc 13. If XY = PQ and XY = RS, then PQ = RS. 14. If x 5, then x = If 2x = 9, then x =
7 Each statement below has reference to one of the two diagrams given. Suppose each statement is TRUE. Name the reason for each true statement. Definitions must be written out! 1. If 1 2, then. 2. m1 + m2 = mmxy 3. CD = CD 4. If AB = CD, then AB + BC = CD + BC 5. If AB = BC and BC = EF, then AB = EF 6. If D is the midpoint of. 7. If ma = mb, then mb = ma. 8. If BD = CE and CD = CD, then BD CD = CE CD. 9. AB + BE = AE 10. If bisects MXY, then If m1 = m2 and m2 = m3, then m1 = m DE + EF = DF 13. If AB + BC = AC and BC = EF, then AB = EF = AC. l If E is the midpoint of, then line l bisects. D E F Fill in the given statements or reasons 16. Given: XY = YZ Prove: Y is the midpoint of XZ 1) XY = YZ 1) 2) XY YZ 2) 3) Y is the midpoint of XZ 3) 7
8 Mini Proofs: Fill in the reasons below for each proof. A.) A B C D E Given: B is the midpoint of AC. D is the midpoint of CE. STATEMENTS Prove: AB + CD = BC + DE REASONS 1. B is the midpoint of AC. 2. AB BC CD DE D is the midpoint of CE AB = BC CD = DE AB + CD = BC + DE 4. B) P Q R S T Given: PQ = RS Prove: PQ + QR = QS STATEMENTS REASONS 1. PQ = RS QR + RS = QS QR + PQ = QS 3. C) Given: l bisects AC l bisects DF Prove: AB + DE = BC + EF E is the midpoint of DF. 2. B is the midpoint of AC. l A C B D E F 3. AB BC DE EF AB = BC 4. DE = EF 5. AB + DE = BC + EF 5. 8
9 17. Given: BD bisects AC Prove: AB = BC 1) BD bisects AC 1) 2) B is the midpoint AC 2) 3) AB BC 3) 4) AB = BC 4) ******************************************* 18. Given: M is the midpoint of PQ Prove: PM MQ Q M P 1) M is the midpoint of PQ 1) 2) PM MQ 2) 3) PM MQ 3) *********************************************** 19. Given: XP YP Prove: m bisects XY 1) XP YP 1) m X P Y 2) P is the midpoint of XY 2) 3) m bisects XY 3) ************************************** 5. Given: B is midpoint of AC D is midpoint of CE Prove: AB + CD BC DE A B C D E 1) B is midpoint of AC 1) D is midpoint of CE 2) AB BC 2) CD DE AB BC 3) CD DE 3) 4) AB + CD BC DE 4) 9
10 Practice with Segment Proofs 1. Given: RT SU Prove: RS = TU 1. RT SU RT = SU RT = RS + ST 3. SU = ST + TU 4. RS + ST = ST + TU ST = ST RS = TU 6. R S T U 2. Given: DE = EF, AD = BF Prove: AE = BE A D E F B 1. AD = BF 1. DE = EF 2. AD + DE = BF + FE AD + DE = AE 3. BE = BF + FE 4. AE = BE Given: AX = BY XC = YC Prove: AC BC 1. AX = BY 1. XC = YC 2. AX + XC = BY + YC AX + XC = AC 3. BY + YC = BC 4. AC = BC AC BC 5. A C X Y B 10
11 Fill in the blanks to complete a two-column proof of one case of the Congruent Supplements Theorem. Given: 1 and 2 are supplementary, and 2 and 3 are supplementary. Prove: 1 3 Proof: Use the given plan to write a two-column proof. Given: 1 and 2 are supplementary, and 1 3 Prove: 3 and 2 are supplementary. 11
12 Use the given plan to write a two-column proof if one case Theorem. Given: 1 and 2 are complementary, and 2 and 3 are complementary. Prove: 1 3 of Congruent Complements Given: HKJ is a straight angle. KI bisects HKJ. Prove: IKJ is a right angle. Proof: 1. a. 1. Given 2. mhkj b. 3. c. 3. Given 4. IKJ IKH mikj mikh d. 6. Add. Post. 7. 2mIKJ mikj IKJ is a right angle. 9. f. 12
13 TOUGH Proofs: If you can do these you can do ANYTHING!!!!!!!!!! Proofs Involving Definitions 1. Given: XY = YZ Prove: Y is the midpoint of XZ X Y Z 2. Given: Line m bisects XY Prove: XP = YP X P m 3. Given: B is the midpoint of AC D is the midpoint of CE Prove: AB + CD = BC + DE Y A B C D E 4. Given: PQ = MN MN = QR Prove: Q is the midpoint of PR P Q R M N 5. Given: line k bisects AC line k bisects DF Prove: AB + DE = BC + EF A B C D E F k D 6. Given: C is the midpoint of AE Prove: AB + BC = CE A B C E 13 F
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