2.1 Practice B. 1. If you like to eat, then you are a good cook. 2. If an animal is a bear, then it is a mammal.

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1 hapter.1 Start Thinking Sample answer: If an animal is a horse, then it is a mammal; If an animal is not a mammal, then it cannot be a horse. Any fact stated in the form of an "if-then" statement could be used, as long as it is factual and leads the reader to believe the original statement as a result..1 Warm Up 1. hexagon. right. complementary 4. straight.1 umulative Review Warm Up 1, ( 6, 1 ). ( ).1 Practice A. ( ), 4. ( 5, 1) 1. If you like the ocean, then you are a good swimmer.. If it is raining outside, then it is cold.. If you are a child, then you must attend school. 4. If are congruent, then they have equal angle measures. 5. a. conditional: If an animal is a puppy, then it is a dog; true b. If an animal is a dog, then it is a puppy; false c. If an animal is not a puppy, then it is not a dog; false d. If an animal is not a dog, then it is not a puppy; true 6. true; By definition, the sum of two complementary is false; The sides are not congruent. 8. An angle is obtuse if and only if the angle measure is greater than 90 and less than Two are supplementary if and only if the sum of their angle measures is yes; By definition, the negation of a true sentence is false, and the negation of a false sentence is true. 11. Sample answer: If two are not complementary, then the sum of their angle measures is Practice B 1. If you like to eat, then you are a good cook.. If an animal is a bear, then it is a mammal.. a. If a tree is an oak tree, then it is a deciduous tree; true b. If a tree is a deciduous tree, then it is an oak tree; false c. If a tree is not an oak tree, then it is not a deciduous tree; false d. If a tree is not a deciduous tree, then it is not an oak tree; true 4. true; Vertical share opposite rays. 5. false; The of a parallelogram are not always perpendicular. 6. A quadrilateral is a rectangle if and only if it has all perpendicular sides. 7. yes; By definition, true statements always have true contrapositives. 8. If x = 7, then x + =. 9. If m ILH = 8, then m GLH = 5 because they are complementary. If m ILH = 8, then m FLK = 8 because they are vertical. If m GLH = 5, then m KLJ = 5 because they are vertical..1 Enrichment and Extension p q p q p q p q T T T F F F T F T F T F F T T T F F F F F T T T p q ( ).1 Puzzle Time A SPELLING BEE p q p ( ) p q q F F F T F T T T F T T T A8

2 . Start Thinking yes; Sample answer: Statement 1: If I can go sledding, then there is snow on the ground. Statement : If there is snow on the ground, then it is cold outside. Statement : If I can go sledding, then it is cold outside. ; true. Warm Up ; 0.7, 0...;., ; 7, ;.9, ; 8, ; 5, 1. umulative Review Warm Up square units. 6 square units. 1 square units 4. 4 square units. Practice A 1. The next number is one more than twice the preceding number; 95, 191. The list items are letters in alphabetical order followed by letters in reverse alphabetical order; X, D. The difference of any two even integers is always even. Sample answer: 6 16 = 0 4. The product of three negative numbers is always 5 = 0 negative. Sample answer: ( )( )( ) 5. The bisector of a straight angle creates two right. 6. You got wet. 7. not possible 8. If you study, then you will pass the class. 9. If a straight angle is bisected, then each angle is a right angle. 10. Law of Syllogism 11. inductive reasoning; The conjecture is based on the assumption that the weather pattern will continue. 1. deductive reasoning; The conjecture is based on the fact that 9 14 = 188, which is even. 1. The Rocky Mountains are taller than the Appalachian Mountains. 14. P = ns. Practice B 1. The list items are letters in alphabetical order followed by numbers in decreasing numerical order starting with 6; D,. The pattern is a sequence of spider webs, each web having one more row of webs than the previous web.. The sum of two absolute values is always positive; Sample answer: + 7 = + 7 = The product of a number and its square is the number to the third power; Sample answer: ()() ( ) 5 5 = 5 5 = 15 = 5 5. If the are right, obtuse, or any acute angle other than 45, then they will not be complementary. 6. not possible 7. AOB and DOB share a common ray. 8. not possible 9. If it is Tuesday, then you water the flowers. 10. deductive reasoning; The facts of mammals and laws of logic were used to draw the conclusion. 11. inductive reasoning; The conjecture is based on the assumption that a pattern will continue. 1. no; Based on the Law of Syllogism, a series of true conditional statements will always be true. 1. Using inductive reasoning, you can make the conjecture that organic produce costs more than nonorganic produce because this was true in all of the specific cases listed in the table.. Enrichment and Extension 1. 7 guesses. lengths in ft: Stage 1: 1, Stage : 4, Stage : 16 9, Stage 4: 64 ; The expression that models the 7 n 1 pattern of the length at a given stage is 4 ( ). A9

3 . areas in ft : Stage 1: 1, Stage : 8, 64 Stage : 9 81, Stage 4: 51 ; The expression that models that 79 pattern of the shaded area at a given stage is 8 ( ) 9 n Sample answer: U V S T R. Puzzle Time LADY BUGS 9. Sample answer:. Start Thinking Sample answer: doors, windows, scale, stairs, water lines. Warm Up umulative Review Warm Up 1. m ABD =, m BD =. m ABD = 85, m BD = 85. m ABD =, m BD = 4. m ABD = 64, m BD = 64. Practice A 1. Sample answer: There is exactly one line through points and H.. Sample answer: Line contains points G and D.. Sample answer: H and GE intersect at point D. 4. Sample answer: Points B, H, and E are noncollinear and define plane M. 5. Sample answer: Plane M contains the noncollinear points B, H, and E. 6. Sample answer: Points G and E lie in Plane M so, GE lies in plane M. 7. Sample answer: X G A H Y Q A 10. yes 11. no 1. yes 1. no 14. yes 15. If three points are noncollinear, then there exists exactly one plane that contains them; converse: If there exists exactly one plane that contains three points, then the three points are noncollinear. inverse: If three points are collinear, then there are multiple planes that contain all three points. contrapositive: If there are multiple planes that contain three points, then the three points are collinear; The converse, inverse, and contrapositive are true. 16. no; Three lines must intersect each other at three points.. Practice B 1. Sample answer: There is exactly one line through points and G.. Sample answer: EF contains points E and F.. Sample answer: G and EF intersect at point J. 4. Sample answer: Plane A contains the noncollinear points D, H, and I. 5. Sample answer: Points E and F lie in plane B. So, EF lies in plane B. 6. Sample answer: Planes A and B intersect at G. 7. Sample answer: E D B B D A A10

4 8. Sample answer: A Q A T B S B E D B X A P s F 9. no 10. yes 11. yes 1. no 1. converse: If two planes share a line, then the two planes intersect. inverse: If two planes do not intersect, then their intersection is not a line. contrapositive: If the intersection of two planes is not a line, then the two planes do not intersect; The converse, inverse, and contrapositive are true. 14. yes; Because three noncollinear points define a plane, two points on a line define an infinite number of planes. 15. no; The line that passes through a point on a plane does not lie in the plane unless there is another point on the line that is also in the plane. 16. no; yes; Because of the Plane Line Postulate (Post..6), EF only lies in plane Z when it contains two points in plane Z.. Enrichment and Extension 1. There exists exactly one plane that contains both lines m and n.. Line-Point Postulate (Post..); A line contains at least two points.. Line Intersection Postulate (Post..); If two lines intersect, then their intersection is exactly one point. 4. Three Point Postulate (Post..4); Through any three non-collinear points, there exists exactly one plane. 5. Plane-Line Postulate (Post..6); If two points lie in a plane, then the line containing them lies in the plane a b c D s r M 10. X; If two planes intersect, their intersection is a line; Plane Intersection Postulate (Post..7). 11. no; Points, D, E, and X would all be collinear, but line EX and X intersect only at point X, so this is impossible. 1. no; This does not follow the Plane-Line Postulate (Post..6) because D lies in plane P but not in plane Q, and B lies in plane Q but not in plane P.. Puzzle Time BEAUSE IT WOULD TAKE THE GEESE FOREVER TO WALK.4 Start Thinking Sample answer: One instance when it is necessary is when there is a quotient containing the variable and addition or subtraction in the numerator and a real number in the denominator; x + 4 = Warm Up 1. Each side of the equation was subtracted by rather than added; f = 17; f + = 17 + ; f = 6. Each side of the equation was divided by 8, 8r 4 1 rather than 8; 8r = 4; = ; r = 8 8. The right side of the equation was multiplied by 4 7 rather than 7 4 ; 4m = ( ) m = ( ) m = 8.5 ; ; Each side of the equation was multiplied by 6, n 6 n rather than 6; = ; = 6 ; n = 18 A11

5 .4 umulative Review Warm Up 1. AD, ADB, BD. EHG, EHF, GHF.4 Practice A 1. Equation Explanation and Reason x + 4 = 1 Write the equation; Given x = 7 Subtract 4 from each side; Subtraction Property x = 9 Divide each side by ; Division Property. Equation Explanation and Reason ( x + 1) = 15 Write the equation; Given 6x + = 15 Multiply; Distributive Property 6x = 1 Subtract from each side; Subtraction Property x = Divide each side by 6; Division Property. Equation Explanation and Reason 1 ( 16x 18 ) = ( x + 16) Write the equation; Given 8x 4 = x + Multiply; Distributive Property 6x = 6 Add 4 to each side and subtract x from each side; Addition and Subtraction Properties x = 6 Divide each side by 6; Division 4. Equation Explanation and Reason p p v = v Write the equation; Given Divide each side by ; = v Division Property p Rewrite the equation; = Symmetric Property 5. Equation Explanation and Reason V = πr h Write the equation; Given V Divide each side by πr ; = h πr Division Property V Rewrite the equation; h = πr Symmetric Property 6. Equation Explanation and Reason S = πrs + πr Write the equation; Given S πr = πrs Subtract πr from each side; Subtraction Property S πr Divide each side by πr; = s πr Division s = S πr Rewrite the equation; πr Symmetric Property 7. Multiplication Property 8. Transitive Property m K 10. GH 11. ; x 1 1. Equation Explanation and Reason A = w + h Write the equation; + hw Given A h = w + hw Subtract h from A h = w( + h) each side; Subtraction Property Factor w; Distributive Property A h Divide each side by = w + h + h; Division Property A h Rewrite the equation; w = + h Symmetric Property w = 1. units; P = ommutative and Addition Properties A1

6 .4 Practice B 1. Equation Explanation and Reason ( x 4) + = x Write the equation; Given x 1 + = x Multiply; Distributive Property x = 7 Add 9 to each side and subtract x from each side; Addition and Subtraction Properties x =.5 Divide each side by ; Division. Equation Explanation and Reason 1( x + 5) = x + ( x 1) Write the equation; Given x 5 = ( x + x 1) Multiply; Distributive Property x 5 = x + 6x Multiply; Distributive Property 10x = Add 5 to each side and subtract 9 x from each side; Addition and Subtraction Properties of x = 0. Divide each side by 10; Division. Equation Explanation and Reason I I I Divide each side by r ; = m r m = 1 Write the equation; mr Given = mr Multiply each side by ; Multiplication Property Division Property I Rewrite the equation; = r Symmetric Property 4. Equation Explanation and E = Reason 1 Write the mv + 9.8mh equation; Given 1 1 E mv = 9.8mh Subtract mv from each side; Subtraction 1 E mv Divide each side = h by 9.8 m; Division 9.8m h 1 E mv = 9.8m Rewrite the equation; Symmetric 5. Multiplication and Subtraction Properties of 6. Transitive and Addition Properties x + y = 5x y 9. Equation Explanation and Reason 1 Write the equation; V = bh Given V Multiply each side by and divide = b h each side by hl; Multiplication and Division Properties V Rewrite the equation; Symmetric b = h Property b = 8m 10. Sample answer: m BD = 65 so m GF = 65 and m FD = 115, so m FE = m DE = 57.5 A1

7 .4 Enrichment and Extension = + ( 1) 4.. n S a n d. = + ( 1). a. Given b. Addition Property c. Multiplication Property d. Simplify. e. Distributive Property f. Subtraction Property g. Distributive Property 4. a. n = c( 1 + r) b. n c( 1 r) = + Given S a n d n. Multiplication Property S n 1 d a n =. Subtraction Property. ( ) ( n 1) S d = a 4. Division Property n 1 = 1. V πh ( r h). V πh ( r h) V πh V πh V πh =. Multiplication Property = r h. Division Property + h = r 4. Addition Property + h = r 5. Division Property n = 1 + r Division Property c n 1 = r Subtraction Property c c. r = % d. c = ; Solve the formula n n = c( 1 + r), for r to yield 1 c = r. Substitute n = 10.4 and r = 0.04 into the equation to yield = Using the c Addition Property, add 1 to each side to obtain 10.4 = Next, multiply by c and c 10.4 divide by 1.04 to obtain c = Puzzle Time MAKE APPLESAUE.5 Start Thinking The formula for the area of a triangle is derived directly from the formula for the area of a rectangle. By drawing a diagonal, the rectangle is now split into two congruent tri. So, each triangle is half the area of the rectangle, and the formula for the area of a triangle 1 is A = bh..5 Warm Up 1. complement: 1, supplement: 11. complement: 70, supplement: 160. complement: 7, supplement: complement: 67.4, supplement: complement: 6, supplement: complement: 16, supplement: umulative Review Warm Up 1. BE, DE Practice A 1. Symmetric Segment ongruence. Reflexive Angle ongruence A14

8 AB = AB 1. Reflexive Property. AB AB. Definition of 1. BF bisects AF 1. A B. m A = m B. Definition of congruent. m B = m A. Symmetric Property 4. B A 4. Definition of congruent. AFB BF. Definition of angle bisector. FD BF. Given 4. BF FD 4. Symmetric Property of Angle ongruence (Thm..) 5. AFB FD 5. Transitive Property of Angle ongruence (Thm..) 1. AG bisects D IJ bisects E BH bisects ED. E = ED K = KE EF = FD. KE = E FD = ED. Definition of segment bisector. Definition of segment bisector 4. KE = FD 4. Transitive Property 5. KE = FD 5. Division Property 6. KE FD 6. Definition of.5 Practice B AB D... AB = D. Definition of. D = AB. Symmetric Property 4. D AB 4. Definition of 1. A B B. m A = m B m B = m. Definition of congruent. m A = m. Transitive Property 4. A 4. Definition of congruent 1. E bisects AI B bisects AE FH bisects EI. AE EI AD DE EG GI. AE = EI AD = DE EG = GI 4. AD + AD + AE EG + EG = EI. Definition of midpoint. Definition of 4. Segment Addition Postulate (Post. 1.) 5. AD = AE 5. Properties of Addition EG = EI 6. AD = EG 6. Substitution Property 7. AD = EG 7. Division Property 8. AD EG 8. Definition of A15

9 m KMN = 8. m KMN + m JMK = m JMK = 90. Definition of complementary. Substitution Property 4. m JMK = 6 4. Subtraction Property 5. m PTS = Given 6. m PTS + m STR m STR 6. Definition of supplementary 7. Substitution Property 8. m STR = 6 8. Subtraction Property 9. m JMK = m STR 9. Transitive Property 10. JMK STR 10. Definition of congruent 1. AD BDE. m AD = m BDE. Definition of congruent. m AD + m ADE. Definition of supplementary 4. m BD + m BDE 4. Definition of supplementary 5. m AD + m ADE 5. Transitive Property = m BD + m BDE 6. m AD + m ADE = m BD + m AD 6. Substitution Property 7. m ADE = m BD 7. Subtraction Property 8. ADE BD 8. Definition of congruent.5 Enrichment and Extension 1. RT = z; 1. T is the midpoint of RS.. z = RS;. RT TS. Definition of midpoint. RT = TS. Definition of 4. TS = z 4. Given 5. RT = z 5. Substitution Property 1. T is the midpoint of RS.. RT TS. Definition of midpoint. RT = TS. Definition of 4. TS = z 4. Given 5. RT = z 5. Transitive Property 6. RT + TS = RS 6. Segment Addition Postulate (Post. 1.) 7. z + z = RS 7. Substitution Property 8. z = RS 8. Simplify. A16

10 z. RW = ; 1. T is the midpoint of RS.. RT TS. Definition of midpoint. RT = TS. Definition of 4. TS = z 4. Given 5. RT = z 5. Transitive Property 6. W is the midpoint of RT. 6. Given 7. RW WT 7. Definition of midpoint 8. RW = WT 8. Definition of 9. RW + WT = RT 9. Segment Addition Postulate (Post. 1.) 10. RW + RW = RT 10. Substitution Property 11. RW = RT 11. Simplify RT RW = 1. Division Property z RW = 1. Substitution Property 6..5 Puzzle Time THEY ALL DO.6 Start Thinking 1. m ZYQ = 45. m ZQP = 45. Given. m ZYQ = m ZQP. Substitution 4. ZYQ ZQP 4. Definition of congruent 5. m XYQ + m ZYQ 5. Definition of linear pair 6. m ZQP + m ZQR 7. m XYQ + m ZYQ = m ZQP + m ZQR 8. m XYQ + m ZQP = m ZQP + m ZQR 6. Definition of linear pair 7. Substitution 8. Substitution 9. m XYQ = m ZQR 9. Subtraction 10. XYQ ZQR 10. Definition of congruent a + b 4. coordinate of point P: ; 4 point Q: 5 a + b a. x = 10, y = b. x = 18, y = 8 coordinate of Straight Angle Sample answer: factoring polynomials.6 Warm Up Obtuse Angle Right Angle Acute Angle 1. x = 9. y = 5. x = 5 4. y = 9 5. x = 7 6. x = 7 A17

11 .6 umulative Review Warm Up F P M L N.6 Practice A N Q 1. A BD, BD EDF, A EDF, DF BDE; A BD by definition because they have the same measure. BD EDF by the Vertical Angles ongruence Theorem (Thm..6). A EDF by the Transitive Property. DF BDE by the Vertical Angles ongruence Theorem (Thm..6).. 1 4, 5, 6,, 6, 5, 5 6; 1 4, 5, and 6 by the Vertical Angles ongruence Theorem (Thm..6). by definition because they have the same measure. 6 and 5 by the Transitive Angle ongruence (Thm..). 5 6 by substitution.. x = 1, y = 8 4. x = 5, y = 19 P M D D N Q E and are supplementary. 1 and are supplementary.. m 1 + m m 1 + m. m 1 + m = m 1 + m Proof: Because 1 and are supplementary and 1and are supplementary, m 1 + m and m 1 + m by the definition of supplementary. By the Transitive Property of Angle ongruence (Thm..), m 1 + m = m 1 + m. By the Subtraction Property, m = m. So, by the definition of congruent..6 Practice B. Definition of supplementary. Transitive Angle ongruence (Thm..) 4. m = m 4. Subtraction Definition of congruent 1. D B, DA AB, BA AD, BAD BD, D BA, B BAD, D BD, and B BD; D B by the Right Angles ongruence Theorem (Thm..). DA AB and BA AD by definition because they have the same measures. Because m DA + m BA = 90 and by the Angle Addition Postulate (Post. 1.4), DA + BA DAB and m DAB = 90. By the same reasoning, m BD = 90. So, BAD BD by the Right Angles ongruence Theorem (Thm..). D BA, B BAD, D BD, and B BD by the Transitive Property. A18

12 . 1, 4, 1 5, 6, 4 6; 1, and 4 by the Vertical Angles ongruence Theorem (Thm..6). 1 5 by definition because they have the same angle measure. Because 1 and form a linear pair and 5 and 6 form a linear pair, m 1 + m = 180 and m 5 + m 6 by the Linear Pair Postulate (Post..8). So, by the ongruent Supplements Theorem (Thm..4), 6. Because 1 and 4 form a linear pair and 5 and 6 form a linear pair, m 1 + m 4 and m 5 + m 6 by the Linear Pair Postulate (Post..8). So, by the ongruent Supplements Theorem (Thm..4), x = 8, y = x = 4, y = is a right angle. Given is a right angle. Given are supplementary. Given Vertical Angles ongruence Theorem (Thm..6) Definition of a right angle is a right angle. Right Angle ongruence Theorem (Thm..) Definition of supplementary Subtraction Because 1 is a right angle and 1 by the Vertical Angles ongruence Theorem (Thm..6), 1 by the Right Angle ongruence Theorem (Thm..). Because 5 is a right angle, m 5 = and 8 are supplementary so, m 5 + m 8 by the definition of supplementary. By the Subtraction Property, m 8 = 90. So, 8 is a right angle by the definition of a right angle. Because and 8 are right, 8 by the Right Angles ongruence Theorem (Thm..)..6 Enrichment and Extension 1. false. true. false 4. false 5. true 6. true Right Angle ongruence Theorem (Thm..) is a right angle. Definition of a right angle AB BD ED BD. ABD EDB is a right angle. is a right angle.. m ABD = 90 m EDB = m AB + m BD = 90 m ED + m DB = 90. Definition of perpendicular lines. Definition of right angle 4. Angle Addition Postulate (Post. 1.4) 5. AB ED 5. Given 6. m AB = m ED 6. Definition of congruent 7. m AB + m BD 7. Substitution = m ED + m DB 8. m ED + m BD 8. Substitution = m ED + m DB 9. m BD = m DB 9. Subtraction 10. BD DB 10. Definition of congruent A19

13 m WYZ = m TWZ = 45. TWZ and SWZ are a linear pair. WYZ and XYW are a linear pair.. TWZ and SWZ are supplementary. WYZ and XYW are supplementary. 4. m TWZ + m SWZ m WYZ + m XYW 5. m TWZ + m SWZ = m WYZ + m XYW m SWZ = 45 + m XYW. Definition of linear pair. Linear Pair Postulate (Post..8) 4. Definition of supplementary 5. Transitive 6. Substitution 7. m SWZ = m XYW 7. Subtraction 8. SWZ XYW 8. Definition of congruent 1. The hexagon is regular.. All interior of the hexagon are congruent.. 1 is congruent to an interior angle of the hexagon. 4. is supplementary to an interior angle of the hexagon. 5. is supplementary to 1.. Definition of regular hexagon. Vertical Angles ongruence Theorem (Thm..6) 4. Linear Pair Postulate (Post..8) 5. Substitution 6. m 1 + m 6. Definition of supplementary.6 Puzzle Time MAKE SURE ONE OF THEM IS A MATH umulative Review 1. Each additional figure has an additional inscribed equilateral triangle in the previous triangle.. Each additional figure has a bottom row containing one more solid circle than the previous figure.. Each additional figure is the same rectangle with one more equal division. A0

14 4. Add starting at ; 10, 1 5. Add starting at 1; 1, Add 1 starting at 4; 56, Add 9 starting at 7; 4, 5 8. Multiply by 9. Multiply by starting at ;, starting at ;, Increase what you add by 1 each term. Start at 1 and start by adding ; 15, Increase what you add by 1 each term. Start at and start by adding ; 0, 7 1. a. 88 ft b. 468 ft 1. a. 54 in. b. 16 in s c g x m r j 6. 6a f 8. 6y b 40. 8k w 4. g p b = m = k = p = a = x = ,000, x = x = 96. x = x = x = 99. x = x = x = 10. x = x = 104. x = 105. x = 106. x = x = 108. x = 109. x = x = x = a. The chair is not wood. b. The rug is brown. 11. a. The photograph is in color. b. Your homework is not finished a. It is not cold outside. b. The bicycle is green. 50. r = 51. h = 9 5. w = x + 8x x 4x t = 54. c = 55. e = x 6x x x a. 5 lunches b. 10 lunches c. $ x + 18x x 6x d. $9.5 e. $7 57. a. 7 push-ups b. 9 push-ups c. day 8 d. day x 16x x 60x x 6x + 1 5x 10x x 48x x 16x + 16 A1

15 17. ( x + )( x 7) 18. ( x 1)( x + 11) 19. ( x 7)( x + 4) 10. ( x 5)( x ) 11. ( x )( x ) 1. ( x + 4)( x 9) hapter.1 Start Thinking B 1. ( x + 1)( x 1) 14. ( x + )( x ) 15. ( x + 5)( x 5) 16. ( x )( x + 4) A 17. ( x 5)( x 7) 18. ( 5x + )( x + 4) 19. x = 8 and x = 140. x = 6 and x = 141. x = 1 and x = x = 9 and x = x = 5 and x = x = 10 and x = x = and x = x = 5 and x = 147. x = 11 and x = and = and = x = and x = 151. a. 7 words per min b words c. 105 words d words right triangle; no; no; Because points B and connect perpendicular lines, you cannot plot either point to make a perpendicular segment or a parallel segment..1 Warm Up 1. Sample answer: B. GE. G 4. AB, B, BD 5. Sample answer: FE and FG 6. Sample answer: D.1 umulative Review Warm Up 1. K ( 4, 11). J ( 7, 18). K ( 1, ).1 Practice A 1. AB and D. A and D. no; AB D and by the Parallel Postulate (Post..1), there is exactly one line parallel to AB through point. 4. no; They are intersecting lines. 5. and 8, and and 7, 4 and and 5, and 6, and 7, 4 and 8 8. and 5, and 8 9. no; By definition, skew lines are not coplaner. 10. pairs; 4 pairs; ( n ) pairs 11. a. AB and D, A and BD b. A and D, BD and D c. and 5, and 8 A

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