AMSCOGeometry Answer

Answers for Textbook Exercises

227

11. a. 1x � xb. Identity property of multiplication

12. a. 6(4 � b) � 6(4) � 6(b)b. Distributive property

13. a. 17 � 1b. Inverse property of multiplication

14. a. (24) � 24b. Commutative property of multiplication

15. a. 12(4 � 10) � 12(10 � 4)b. Commutative property of addition

16. a. � 1b. Inverse property of multiplication

17. x �

18. If a � 0, then b � 0 by the multiplicationproperty of zero. Therefore, a � b � a � 0 � a.

1-3 Definitions, Lines, and Line Segments(pages 10–11)

Writing About Mathematics1. “A hammer is a tool” is not a good definition

because it does not distinguish “hammers” fromother “tools.”

2. “A hammer is used to drive nails” is not a gooddefinition because it is not reversible. There aremany other objects that are not hammers thatcan be used to drive nails.

Developing Skills3. a. A noncollinear set of points is a set of three or

more points that do not all lie on the samestraight line.

b. The set of all pointsc. Noncollinear points do not all lie on the same

straight line.4. a. The distance between any two points on the

number line is the absolute value of thedifference of the coordinates of the points.

b. The set of real numbersc. The distance between any two points must be

a positive real number or 0.5. a. A line segment is a set of points consisting of

two points on a line, called endpoints, and allof the points on the line between theendpoints.

13

p A 1p B

A 14 B1

4

A 117 B

1-1 Undefined Terms (page 3)Writing About Mathematics

1. It may be called a plane because it is flat. It is nota mathematical plane because it does not extendendlessly in all directions. Also, when taking intoaccount the curvature of the earth or localgeographic features such as hills, it is not really aflat surface.

2. A stretched string does not go on forever. It alsohas a measurable thickness.

Developing Skills3. True 4. False 5. False6. True 7. False 8. True

Applying Skills9. Answers will vary.

10. Answers will vary.11. Answers will vary.

1-2 The Real Numbers and TheirProperties (page 6)

Writing About Mathematics1. a. The set of positive real numbers is not closed

under subtraction because, if the value of theminuend is less than or equal to the value ofthe subtrahend, then the difference is anegative real number or zero.

b. Yes. Subtracting any two real numbers resultsin another real number.

2. The set of negative real numbers is not closedunder multiplication because multiplying anytwo negative real numbers results in a positivereal number.

Developing Skills3. 0 4. 15. �a � 11 6. 07. a. 7 � 14 � 14 � 7

b. Commutative property of addition8. a. 7 � (�7) � 0

b. Inverse property of addition9. a. 3(4 � 6) � (3 � 4)6

b. Associative property of multiplication10. a. 11(9) � 9(11)

b. Commutative property of multiplication

For each proof in this Answer Key, one method is provided. Valid alternative proofs may exist and should beconsidered acceptable.

Chapter 1. Essentials of Geometry

Answers for Textbook Exercises

b. The set of all pointsc. The points in a line segment all lie on the same

line and are included between two endpoints.6. a. The measure of a line segment is the distance

between its endpoints.b. The set of all real numbersc. The measure of a line segment must be a

positive real number or 0.7. a. Congruent segments are segments that have

the same measure.b. The set of all pairs of line segmentsc. Congruent pairs of segments have the same

measure.8. AB � 2 9. BD � 3 10. CD � 2

11. FH � 4 12. GJ � 5 13. EJ � 814. Answers will vary. Example: AB � CD � DE15. P, Q, and R are noncollinear. If the points were

collinear, then PR � PQ � QR, but PR � 18while PQ � QR � 10 � 12 � 22.

Applying Skills16. a. 12 in.

b. He could either mark the board at 13 inchesor at 5 inches.

17. No. If Troy, Albany, and Schenectady were in astraight line, then the distance traveled on thereturn trip would equal the distance traveled onthe first trip.

1-4 Midpoints and Bisectors (pages 13–14)Writing About Mathematics

1. Yes. The notation is valid when and name the same line. In this case, A, B, C, and Dare collinear.

2. No. It is possible that A, M, and B arenoncollinear.

Developing Skills

3. AT � TC;4. RN � NS;5. BC � CD;6. SP � PT;7. a.

b. Let x � AB � CD and y � BC.Then AC � AB � BC � x � y and BD � BC � CD � y + x.

A B C D

SP > PTBC > CDRN > NSAT > TC

CDg

ABg

P

Q

R

228

8. 2.5 9. �1010. a. Yes; PQ = 16, QR � 2, PR � 18, and

16 � 2 � 18.b. No; �1

Applying Skills11. 30512. a. 16 in.

b. At 16 in.

Hands-On Activity1–2.

3. D bisects ;

1-5 Rays and Angles (pages 18–19)Writing About Mathematics

1. A half-line consists of the set of points that lie toone side of a point on a line. A ray consists of apoint on a line and all the points to one side ofthis endpoint. Therefore, a ray is a half-line plusan endpoint.

2. and are the same ray: they both have endpoint P, and R and S are on the same side ofP on the line.

Developing Skills

3. a. and b. E

c. and d. �y, �E, �FED, �DEF

4. a. A

b. and c. �CAD, �BAD, �CABd. �CAB

e.f. �CAB and �BADg. �BADh. No; A, B, and D are not collinear.i. No; �BAC is not the union of opposite rays.

5. a. �EAB, �CAB, �BAC, �BAEb. �DEC and �CEDc. �CBA and �ABCd. E

e. and ; and f. �DEB and �AECg. �ABE and �EBC

ECh

EAh

EBh

EDh

ABh

ABh

ADh

EDh

EFh

EDh

EFh

PSh

PRh

AD > DBAB

C

A BD

E

1-6 More Angle Definitions (pages 21–23)Writing About Mathematics

1. Disagree; a bisector is a ray. However, the angle

bisector is on .2. Acute; since an obtuse angle has a measure

greater than 90° and less than 180°, the twoangles formed have measures greater than 45°and less than 90°.

Developing Skills3. a. Acute 4. a. Obtuse

b. 12° b. 49°5. a. Obtuse 6. a. Right

b. 63° b. 45°7. a. Acute 8. a. Straight

b. 41° b. 90°9. a. Acute 10. a. Acute

b. 28.5° b. 1.5°11. 45° 12. 30° 13. 72°14. b. �ACD � �DCB

c. m�ACD � m�DCB15. b. �DAC � �CAB

c. m�DAC � m�CAB16. Two right angles17. Two acute, 45° angles18. m�LMN � m�LMP � m�NMP19. m�LMP � m�LMN � m�NMP20. m�LMN � m�LMP � m�NMP21. m�ABE � m�EBC � m�ABC22. m�BEC � m�CED � m�DEB23. m�ADC � m�CDE � m�ADE24. m�AEC � m�AEB � m�CEB25. a. �ACD, �ACF, �DCB, �BCF

b. 90° c. and d. and 26. x � 15 27. m�ABC � 6828. m�QRS = 15029. Yes. An angle bisector divides an angle into two

congruent angles, which have equal measure.Let x � m�CBD � m�PMN. Then m�ABC � 2x and m�LMN � 2x. Therefore,m�ABC � m�LMN and �ABC � �LMN.

Hands-On Activity1–2.

3. The angles have equal measure and arecongruent:m�ABD = m�ABE � m�CBD � m�CBE � 90;�ABD � �ABE � �CBD � �CBE

4. and are perpendicular.ACDE

D

A CB

E

CBh

CAh

CFh

CDh

RSTg

229

1-7 Triangles (pages 27–28)Writing About Mathematics

1. No. The statement does not specify that a triangleis a polygon. It is possible that three linesegments form a figure that is not a triangle.

2. The legs of an isosceles triangle must becongruent while the legs of a right triangle maynot be congruent.

Developing Skills

3. Legs: , Hypotenuse:4. Legs: , Hypotenuse:5. Legs: , Base:

Vertex angle: �N Base angles: �L, �M6. Legs: , Base:

Vertex angle: �R Base angles: �S, �T7–10. Answers will vary.

11. �BAC and �BCA12. �EDF and �DEF13. and

14. a. Answers will vary. Examples: is on ,

�ECB is adjacent to �BCG,b. Answers will vary. Examples: �FCE is

isosceles, AC � CB, C is the midpoint of .

Applying Skills15. 57°, 25° 16. 22 17. 21, 10, 2118. (1) With legs x � 5 and 4x � 11: 3, 7, 3

(2) With legs 3x � 13 and 4x � 11: 7, 19, 19

Review Exercises (pages 31–32)1. Set, point, line, plane2. It does not clearly define the class to which “line”

belongs.3. Collinear set of points4. Distance between two points on the real number

line5. Triangle 6. Bisector7. Opposite rays 8. Congruent angles9. Isosceles triangle 10. Line segment

11. 12. a. b. �S13. a. DE = 4, EF � 8, DF � 12

b. 3 c. 514. The midpoint of a line is a single point while a

bisector is any line that passes through themidpoint, and there are infinitely many linespassing through a point.

15. �BED and �AEC 16. �ADC17. and 18.19. �ABD and �DBC 20. 45°21. �BDA and �BDC

ACEDBE

RTLN

DE

FCG

ABg

AB

STSR

TSRTRS

LMNMNLJKLKJLCBABCA

22. No. In order for this equality to be true, A, B, andC would have to be collinear.

Exploration (pages 32–33)1. The points are the same. Each represents a

location on the surface.2. A geodesic in Euclidean geometry extends

infinitely in each direction. The geodesic on thesphere is a circle.

3. Two intersecting geodesics in Euclideangeometry form four angles. Two intersecting

230

geodesics on a sphere form eight angles.The sides of these angles do not extend infinitely.

4. The sum of the measures of the angles of atriangle in Euclidean geometry is 180°.Drawing three geodesics on a sphere forms eight triangles. The sum of the measures of the angles of a triangle on sphere is greater than 180°.

Chapter 2. Logic

2-1 Sentences, Statements, and TruthValues (pages 39–41)

Writing About Mathematics1. A sentence in grammar need only have a subject

and a predicate. A mathematical sentence used inlogic must state a fact or a complete idea that canbe judged to be true or false.

2. Answers will vary.a. Example: Today is Wednesday.b. Example: I am six feet tall.c. Example: This country is in the southern

hemisphere.Developing Skills

3. Mathematical sentence4. Mathematical sentence5. Not a mathematical sentence6. Not a mathematical sentence7. Not a mathematical sentence8. Not a mathematical sentence9. Mathematical sentence

10. Not a mathematical sentence11. She 12. We 13. y14. x 15. This 16. He17. It 18. It 19. a. True20. a. Open 21. a. False 22. a. True

b. They23. a. Open 24. a. False 25. a. True

b. x26. a. False 27. {New York}28. {Nevada, Illinois}29. {Massachusetts, New York}30. {Nevada, Illinois, Massachusetts, Alaska,

New York}31. {Alaska} 32. {triangle} 33. �34. {triangle} 35. {square, rectangle}36. {square, rhombus}37. {square, rectangle, parallelogram, rhombus}38. {trapezoid}

39. {square, rectangle, parallelogram, rhombus,trapezoid}

40. The school does not have an auditorium.41. A stop sign is not painted red.42. The measure of an obtuse angle is not greater

than 90°.43. There are not 1,760 yards in a mile.44. Michigan is a city. 45. 14 � 2 � 16 � 1246. 3 � 4 � 5 � 6 47. Today is Wednesday.48. a. p 49. a. �p 50. a. q

b. True b. False b. False51. a. �q 52. a. r 53. a. �r

b. True b. Open b. Open54. a. �q 55. a. p 56. a. q

b. True b. True b. False57. a. Summer does not follow spring. b. False58. a. August is not a summer month. b. False59. a. A year does not have 12 months. b. False60. a. She does not like spring. b. Open61. a. Summer follows spring. b. True62. a. August is a summer month. b. True63. a. A year has 12 months. b. True64. a. She likes spring. b. Open

2-2 Conjunctions (pages 46–48)Writing About Mathematics

1. No.As shown in the following truth table, �(p ∧ q)and �p ∧ �q have different truth values.

p q (p ∧ q) �(p ∧ q) �p �q �p ∧ �q

T T T F F F F

T F F T F T F

F T F T T F F

F F F T T T T

2. p, q, and r must all be true because, for aconjunction to be true, each of the conjunctsmust be true.

Developing Skills3. p ∧ q 4. p ∧ r 5. �p6. �p ∧ r 7. q ∧ �r 8. �p ∧ �q9. �r ∧ �p 10. �r ∧ p 11. �(p ∧ q)

12. �(q ∧ �p) 13. False 14. False15. True 16. True 17. False18. False 19. True 20. False21. True 22. False 23. False24. True, True 25. True, False 26. False, True27. True 28. TrueApplying Skills29. True 30. False 31. Uncertain32. False 33. True 34. False35. Uncertain 36. False37. a. True 38. a. True

b. True b. Truec. False c. False

2-3 Disjunctions (pages 51–53)Writing About Mathematics

1. The truth set of the negation of a statementcontains the elements of the replacement set thatmake the statement false. The complement of aset contains the elements of the universe that arenot members of the set.

2. When two statements are connected by theinclusive or, the disjunction is true when one ofthe statements is true or both of the statementsare true. When two statements are connected bythe exclusive or, the disjunction is true only in thecase that exactly one of the statements is true.

Developing Skills3. a. m ∨ k 4. a. c ∨ l 5. a. c ∨ m

b. True b. True b. True6. a. �k ∨ �c 7. a. l ∨ k 8. a. l ∧ c

b. False b. True b. False9. a. �(c ∨ m) 10. a. �(�k ∨ l)

b. False b. True11. a. c ∧ k 12. a. (�c ∨ �m) ∧ l

b. True b. False13. a. Spring is a season or Halloween is a season.

b. True14. a. Breakfast is a meal and spring is a season.

b. True15. a. Spring is not a season or Halloween is a season.

b. False16. a. Breakfast is a meal and Halloween is not a

season.b. True

17. a. Breakfast is not a meal or spring is not a season.b. False

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18. a. It is not the case that spring is a season andHalloween is a season.

b. True19. a. It is not the case that breakfast is a meal or

spring is not a season.b. False

20. a. Breakfast is not a meal and spring is not aseason.

b. False21. True 22. True 23. False24. False, True 25. True, False 26. False27. TrueApplying Skills28. True 29. True 30. False31. Uncertain 32. True

2-4 Conditionals (pages 58–60)Writing About Mathematics

1. a. (1) Let p be 8 is divisible by 4.(True: 8 4 � 2)Let q be 8 is divisible by 2.(True: 8 2 � 4)p → q (T → T) is true.

(2) Let p be 6 is divisible by 4.(False: 6 4 � 1.5)Let q be 6 is divisible by 2.(True: 6 2 � 3)p → q (F → T) is true.

(3) Let p be 7 is divisible by 4.(False: 7 4 � 1.75)Let q be 7 is divisible by 2.(False: 7 2 � 3.5)p → q (F → F) is true.

b. No. Every number that is divisible by 4 is alsodivisible by 2 because the prime factorizationof 4 is 2 � 2.

2. The truth value is the same when p and q areboth true or when p and q are both false.

p q p → q q → p

T T T T

T F F T

F T T F

F F T T

Developing Skills3. a. A polygon is a square.

b. A polygon has four right angles.4. a. It is noon.

b. It is time for lunch.

5. a. You want help.b. You ask a friend for help.

6. a. You are not interrupted.b. You will finish more quickly.

7. a. The length of one side of a square is s.b. The perimeter of a square is 4s.

8. a. Many people work at a task.b. The task will be completed quickly.

9. a. 2x � 7 � 11b. x � 2

10. a. You do not get enough sleep.b. You will not be alert.

11. p → r 12. q → r 13. �p → �r14. �q → �r 15. q → p 16. p → r17. a. t → l 18. a. l → t 19. a. l → b

b. True b. False b. True20. a. �b → l 21. a. �t → �l 22. a. b → t

b. True b. False b. False23. a. (�b ∧ t) → l 24. a. (b ∧ �t) → �l

b. True b. False25. True 26. True 27. True28. True 29. True 30. True31. False32. a. If July is a warm month, then I work in my

garden.b. True

33. a. If I am busy every day, then I do not work inmy garden.

b. True34. a. If I like flowers, then I work in my garden.

b. True35. a. If I do not work in my garden, then July is not

a warm month.b. True

36. a. If July is a warm month and I like flowers,then I am busy every day.

b. False37. a. If July is a warm month and I work in my

garden, then I like flowers.b. True

38. a. If July is not a warm month, then I am busyevery day and I like flowers.

b. True39. a. If I work in my garden, then July is a warm

month or I am not busy every day.b. True

40. p → q 41. q → p 42. True, False43. True 44. True 45. True

Applying Skills46. True 47. False 48. Uncertain49. Uncertain 50. True

232

2-5 Inverses, Converses, andContrapositives (pages 67–69)

Writing About Mathematics1. No. A conditional (p → q) can be true when p is

false and q is true. However, in the converse (q → p) of this case, F → T is false.

2. Yes. A conditional and its contrapositive arelogical equivalents, so they have the same truthvalues. Also, the converse of a statement and theinverse of a statement are logical equivalents, sothey have the same truth values.

Developing Skills3. a. �p → �q 4. a. �t → w

b. q → p b. �w → tc. �q → �p c. w → �t

5. a. m → �p 6. a. p → qb. p → �m b. �q → �pc. �p → m c. q → p

7. a. If 6 is not greater than 3, then �6 is notgreater than �3.

b. False c. True8. a. If a trapezoid is not a parallelogram, then a

trapezoid does not have exactly two pairs ofparallel sides.

b. True c. True9. a. If 3(3) � 9, then 3(4) � 12.

b. True c. True10. a. If 22 � 4, then 32 � 6.

b. False c. True11. If you eat Quirky oatmeal, then you lower your

cholesterol.12. If you get rich, then you enter the Grand Prize

drawing.13. If your hair curls, then you use Shiny’s hair

cream.14. If your pet grows three inches, then you feed him

Krazy Kibble.15. a. If a number is exactly divisible by 2, then the

number is even.b. True c. True

16. a. If 0.75 is rational, then it is an integer.b. True c. False

17. a. If 82 � 12 � 72, then 8 � 1 � 7.b. False c. True

18. a. If 4(5) � 6 � 14, then 4(5) � 6 � 20 � 6.b. True c. True

19. a. If Rochester is not the capital of New York,then Rochester is not a city.

b. False c. False20. a. If two angles are not supplementary, then they

do not form a linear pair.b. True c. True

21. a. If 4 � 3 � 2, then 3 � 2 � 1.b. False c. False

22. a. If a triangle is not equiangular, then all anglesof the triangle are not equal in measure.

b. True c. True23. a. If is not a counting number, then is not

greater than 0.b. False c. False

24. (4) 25. (3) 26. (2) 27. (3) 28. (2)Applying Skills29. a. If Derek lives in Nevada, then he lives in Las

Vegas; sometimes true.b. If Derek does not live in Las Vegas, then he

does not live in Nevada; sometimes true.c. If Derek does not live in Nevada, then he does

not live in Las Vegas; always true.30. a. If the probability of picking a red marble from

a bin is , then the bin contains 3 red marbles and 3 blue marbles; sometimes true.

b. If a bin does not contain 3 red marbles and 3blue marbles, then the probability of picking a red marble from the bin is not ; sometimestrue.

c. If the probability of picking a red marble from a bin is not , then the bin does not contain 3 red marbles and 3 blue marbles; always true.

31. a. If a polygon is an octagon, then it has eightsides; always true.

b. If a polygon does not have eight sides, then itis not an octagon; always true.

c. If a polygon is not an octagon, then it does nothave eight sides; always true.

32. a. If a garden grows vegetables, then it growscarrots; sometimes true.

b. If a garden does not grow carrots, then it doesnot grow vegetables; sometimes true.

c. If a garden does not grow vegetables, then itdoes not grow carrots; always true.

33. a. If the area of a rectangle is 48 square feet,then the dimensions of the rectangle are 8 feetby 6 feet; sometimes true.

b. If the dimensions of a rectangle are not 8 feetby 6 feet, then the area of the rectangle is not48 square feet; sometimes true.

c. If the area of a rectangle is not 48 square feet,then the dimensions of the rectangle are not 8feet by 6 feet; always true.

34. a. If a number is divisible by 7, then it has 7 as afactor; always true.

b. If a number does not have 7 as a factor, then itis not divisible by 7; always true.

12

12

12

12

12

233

c. If a number is not divisible by 7, then it doesnot have 7 as a factor; always true.

2-6 Biconditionals (pages 73–74)Writing About Mathematics

1. A prime number is a whole number greater than1 if and only if the prime number has exactly twofactors.

2. Yes. A biconditional is false when the truth valueof p is not the same as the truth value of q. Whenthis is the case, the conditional p → q has adifferent truth value from its converse.

Developing Skills3. True 4. False 5. True6. True 7. True 8. True9. True 10. True 11. False

12. False 13. True 14. False15. False 16. TrueApplying Skills17. a. x is divisible by 2 and 3 if and only if x is

divisible by 6.b. Let x � 2; 2 is divisible by 2 but not by 3 or 6;

(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 3; 3 is divisible by 3 but not by 2 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 5; 5 is not divisible by 2, 3, or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x = 6; 6 is divisible by 2, 3, and 6;(p ∧ q) ↔ r is T ↔ T, which is true.Let x � 8; 8 is divisible by 2 but not by 3 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 9; 9 is divisible by 3 but not by 2 or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 11; 11 is not divisible by 2, 3, or 6;(p ∧ q) ↔ r is F ↔ F, which is true.Let x � 12; 12 is divisible by 2, 3, and 6;(p ∧ q) ↔ r is T ↔ T, which is true.

c. Yes. The prime factorization of 6 is 2 � 3. If acounting number is divisible by both of thesefactors, then it is divisible by 6. If a countingnumber is not divisible by both of thesefactors, then it is not divisible by 6.

18. a. Yes b. Yes c. Yes d. Yes e. No19. A triangle is isosceles if and only if it has two

congruent sides; true.20. Two angles are both right angles if and only if

they are congruent; false.21. Today is Thursday if and only if tomorrow is not

Saturday; false.22. Today is not Friday if and only if tomorrow is not

Saturday; true.

2-7 The Laws of Logic (pages 79–80)Writing About Mathematics

1. Yes. If p → q is false, then q must be false. If q ∨ ris true, and q is false, then by the Law ofDisjunctive Inference, r must be true.

2. Yes. When �q is true, then q is false. If p ∨ q istrue, and q is false, then by the Law ofDisjunctive Inference, p must be true. Since p istrue and �q is true, p ∧ �q is also true.

Developing Skills3. True by the Law of Disjunctive Inference4. True by the Law of Detachment5. Cannot be found to be true or false because one

disjunct is known to be true6. True by the Law of Disjunctive Inference7. True by the Law of Detachment8. Cannot be found to be true or false because the

hypothesis is false9. True by the Law of Detachment

10. Cannot be found to be true or false because thehypothesis is false

11. False; the truth values of both statements in atrue biconditional must be the same.

12. Cannot be found to be true or false because theconclusion is true

13. True by applying the Law of Detachment to thecontrapositive of the conditional

14. Cannot be found to be true or false because thehypothesis is false

Applying Skills15. I take band; the Law of Detachment.

16. is irrational; the Law of Disjunctive

Inference.17. b = 4; the Law of Detachment.18. It is not 8:15 A.M.; the Law of Detachment can be

applied to the contrapositive of the conditional.19. No conclusion; a conditional with a true

conclusion may have either a true hypothesis or afalse hypothesis.

20. x is even and a prime; a biconditional with a trueconclusion has a true hypothesis.

21. It is February, and it is not summer; in a trueconjunction, both statements must be true. TheLaw of Disjunctive Inference may be applied tothe disjunction.

22. No conclusion; a conditional with a trueconclusion may have either a true hypothesis or afalse hypothesis.

23. Last Saturday we flew kites; the Law ofDisjunctive Inference.

24. I study computer science and I take welding; the

"6

234

Law of Disjunctive Inference.25. Five has exactly two factors; a biconditional with

a true hypothesis has a true conclusion.26. x is not an integer greater than 2 and prime; in a

true conjunction, both statements must be true.The Law of Detachment can be applied to thecontrapositive of the conditional.

27. Ray DF does not bisect angle CDE; thehypothesis of the contrapositive is true, so itsconclusion must be true, that is, the hypothesis ofthe conditional must be false.

2-8 Drawing Conclusions (pages 83–85)Writing About Mathematics

1. Yes. Both members of the conjunction must betrue. Since p is false, q is true by the Law ofDisjunctive Inference. Also, r is true.

2. No. Both members of the conjunction must betrue. Since the conjunct r is false, the truth valuesof p and q cannot be determined.

Developing Skills3. True 4. True 5. True6. p is true and q is true.7. p is true and q is false.8. p is false, and q cannot be determined.9. True

Applying Skills10. Laura is an investment manager. Marta is a

doctor. Shanti is a lawyer.11. Alex is a plumber. Tony is a bookkeeper. Kevin is

a teacher.12. Gamma13. Ren had chicken pot pie. Logan had pizza.

Kadoogan had a ham sandwich.14. Zach plays baseball. Steve plays soccer. David

plays basketball.15. Taylor studies Latin. Melissa studies French.

Lauren studies Spanish.16. Augustus is a truthteller. Brutus is a liar. Caesar

is a liar.

Review Exercises (pages 87–89)In 1 and 2, answers will vary. Examples are given.

1. I go to school. I play basketball. If I go to school,then I play basketball. If I play basketball, then Igo to school. If I do not play basketball, then I goto school.

2. I do not play basketball. I do not go to school. If Ido not go to school, then I play basketball. I go toschool and I do not play basketball. I do not go toschool and I play basketball.

3. No. The biconditional p ↔ q is true when p → qand q → p are both false or both true. Thecontrapositives �q → �p and �p → �q have the same truth values as the conditionals because they are logically equivalent. Therefore,�p ↔ �q has the same truth values as p ↔ q.

4. a. At first you don’t succeed.b. You should try again.

5. a. You are late one more time.b. You will get a detention.

6. True 7. False 8. True9. False 10. True 11. False

12. True 13. 5 14. p15. True 16. False, True 17. True18. False 19. False 20. False21. True 22. True23. {6, 7, 8, 9, 10} 24. {1, 2, 3, 4, 5}25. {2, 3, 5, 7} 26. {1, 4, 6, 8, 9, 10}27. {2, 3, 5, 6, 7, 8, 9, 10} 28. {7}29. {2, 3, 5} 30. {1, 2, 3, 4, 5, 6, 8, 9, 10}31. {1, 2, 3, 4, 5, 7} 32. {7}33. a. If I do not live in Oregon, then I do not live in

the Northwest.b. If I live in the Northwest, then I live in Oregon.c. If I do not live in the Northwest, then I do not

live in Oregon.d. I live in Oregon if and only if I live in the

Northwest.34. �A is not the vertex angle of isosceles �ABC.35. Janice, Sarah, Laurie36. Judy, Janice, Sue, Sarah, Laurie37. 1st Virginia, 2nd Kay, 3rd Janice38. Peter plays the violin and tennis. Carlos plays the

cello and soccer. Ralph plays the flute andbaseball.

39. Answers will vary. Examples are given.a. (1) 6 (2) Not possible

(3) 4 (4) 5b. Since there does not exist a number that is

divisible by 6 but not divisible by 2, p → q isalways true.

40. Peter

Exploration (page 90)1. a. 1 and 3 2. a. 2 and 3

b. Answers will vary. b. Answers will vary.3. a. Answers will vary. b. Answers will vary.

c. Because the hypothesis is always false, theconditional is true whether the conclusion istrue or false.

235

Cumulative Review (pages 90–92)Part I

1. 3 2. 2 3. 3 4. 3 5. 16. 3 7. 4 8. 4 9. 3 10. 1

Part II11. “I win the ring toss game” must be true and “If I

win the ring toss game, then I get a goldfish”must be true. The Law of Detachment states thatif the conditional is true and the hypothesis istrue, then the conclusion is true.

12. 3; RS is half the length of RT. Since RS � ��5 � (�1)� � 4, T is located 4 units from S.T is located at �1 � 4 � 3.

Part III

13. refers to line AB. refers to the ray that

has endpoint A and contains point B. refers

to the line segment that has endpoints A and B.

AB refers to the length of the segment .

14. By definition, the measure of straight angle ABC

is 180°. bisects �ABC, so it splits the angle

into two congruent parts, each of which measures

180 2 � 90°. Since intersects �ABC to

form 90°, or right, angles, we can say that is

perpendicular to �ABC.

Part IV15. a(2b � 1) � a(2b) � a(1) Distributive

property� (a � 2) � b � a(1) Associate

property� (2 � a) � b � a(1) Commutative

property� 2ab � a Identity property

16. a. If x is divisible by 4, then x is divisible by 12.b. If x is not divisible by 12, then x is not divisible

by 4.c. The converse can be false: 8 is divisible by 4

but not by 12.The inverse is always true. Since 4 is a factorof 12, every number that is divisible by 12 isalso divisible by 4.

d. If x is not divisible by 4, then x is not divisibleby 12.

BDh

BDh

BDh

AB

AB

ABh

ABg

3-1 Inductive Reasoning (pages 95–97)Writing About Mathematics

1. No; triangles may contain a right or obtuse angle.2. Answers will vary. Example: 3 � 5 � �2

Developing Skills3. a. Answers will vary.

b. 90°c. The sum of the measures of the acute angles

of a right triangle is 90°.4. a. Answers will vary.

b. The quadrilaterals formed are parallelograms.c. A quadrilateral with vertices at the midpoints

of the sides of another quadrilateral is aparallelogram.

5. a. Answers will vary.b. The four triangles are equilateral.c. The four triangles formed by connecting

midpoints of the sides of an equilateraltriangle are equilateral.

6. Probably true; the experiment must includelines intersecting at acute, right, and obtuseangles.

7. Probably true; the experiment must includequadrilaterals with acute, obtuse, and rightangles.

8. False; pairs of different types of numbers must beincluded: natural numbers, positive and negativeintegers, rational numbers, and irrationalnumbers.

9. Probably true; draw parallelograms of differentsizes and angle measures.

10. False; draw quadrilaterals of different shapes:trapezoids, parallelograms, rectangles, squares,rhombuses, and others.

11. Probably true; draw triangles of different shapes:acute, right, obtuse, isosceles, equilateral, andscalene.

12. False; let n � 40, then (40)2 � 40 � 41 � 1,681.This is not a prime number since 1,681 � 412.

Applying Skills13. No 14. Yes 15. No16. No 17. Yes

Hands-On Activitya. 1, 4, and 9 are perfect squares.b. 1, 4, 9, 16, and 25. The numbers are perfect

squares.c. The cards facing up will all be perfect squares.

236

3-2 Definitions as Biconditionals (pages 99–100)

Writing About Mathematics1. Doug is correct. “If a container can be used to

carry food, then it is a lunchbox” is false. Forexample, a paper bag can be used to carry food.

2. a. , but 1 � �2 is false.b. Positive real numbers

Developing Skills3. a. If a triangle is equiangular, then it has three

congruent angles.b. If a triangle has three congruent angles, then it

is equiangular.c. A triangle is equiangular if and only if it has

three congruent angles.4. a. If a line or a subset of a line is a bisector of a

line segment, then it intersects the segment atits midpoint.

b. If a line or a subset of a line intersects a linesegment at its midpoint, then it is a bisector ofthe segment.

c. A line or a subset of a line is a bisector of aline segment if and only if it intersects thesegment at its midpoint.

5. a. If an angle is acute, then its degree measure isgreater than 0 and less than 90.

b. If an angle has a degree measure greater than0 and less than 90, then it is acute.

c. An angle is acute if and only if its degreemeasure is greater than 0 and less then 90.

6. a. If a triangle is obtuse, then it has one obtuseangle.

b. If a triangle has one obtuse angle, then it isobtuse.

c. A triangle is obtuse if and only if it has oneobtuse angle.

7. a. If a set of points is noncollinear, then the setcontains three or more points that do not alllie on the same straight line.

b. If a set of points contains three or more points that do not all lie on the same straight line, then the set of points isnoncollinear.

c. A set of points is noncollinear if and only ifthe set contains three or more points that donot all lie on the same straight line.

8. a. If a part of a line is a ray, then it consists of apoint on the line, called an endpoint, and allthe points on one side of the endpoint.

122 , 1

Chapter 3. Proving Statements in Geometry

b. If a part of a line consists of a point, called anendpoint, and all the points on one side of theendpoint, then it is a ray.

c. A part of a line is a ray if and only if it consistsof a point, called an endpoint, and all thepoints on one side of the endpoint.

9. A point B is between A and C if and only if A, B, and C are distinct collinear points and AB � BC � AC.

10. Two segments are congruent if and only if theyhave the same length.

11. A point is the midpoint of a line segment if andonly if it divides the segment into two congruentsegments.

12. A triangle is right if and only if it has a rightangle.

13. An angle is straight if and only if it is the union oftwo opposite rays and its degree measure is 180.

14. Two rays are opposite rays if and only if they aretwo rays of the same line with a commonendpoint and no other point in common.

Applying Skills15. A triangle is equilateral if and only if it has three

congruent sides.16. Two angles are congruent if and only if they have

the same measure.17. Two lines are perpendicular if and only if they

intersect to form right angles.

3-3 Deductive Reasoning (pages 103–105)Writing About Mathematics

1. Yes. An equilateral triangle has three congruentsides, so it satisfies the definition for an isoscelestriangle, which is that a triangle must have twocongruent sides.

2. Yes. If B is not between A and C, then AB � BC � AC.

Developing Skills3. b. If a ray bisects an angle, then it divides the

angle into two congruent angles.4. b. If a triangle is scalene, then no two of its sides

are congruent. If two segments are notcongruent, then they do not have the samemeasure.

5. b. If two lines are perpendicular, then theyintersect to form right angles.

6. b. If A, B, and C are distinct collinear points andAB � BC � AC, then B is between A and C.

7. b. If two lines are perpendicular, then theyintersect to form right angles.

c. �LMN is a right angle.

237

8. b. If a line bisects a segment, then it divides thesegment into two congruent segments.

c. or DF � FE9. b. If points P, Q, R are collinear with

PQ � QR � PR, then Q is between P and R.c. Q is between P and R.

10. b. If two rays are opposite rays, then they form astraight angle.

c. �TSR is a straight angle.11. b. If a point is the midpoint of a line segment,

then it divides the segment into two congruentsegments.

c. or LM � MN12. b. If the degree measure of an angle is greater

than 0 and less than 90, then the angle is acute.c. �a is acute.

13. a. Statements Reasons1. M is the midpoint 1. Given.

of .2. � 2. Definition of

midpoint.3. AM � MB 3. Definition of

congruentsegments.

b. We are given that M is the midpoint of .Therefore, M divides into twocongruent segments, and . Sincecongruent segments have the same length,AM � MB.

14. a. Statements Reasons1. RS � ST 1. Given.2. 2. Definition of

congruentsegments.

3. �RST is isosceles. 3. Definition ofisosceles triangle.

b. We are given �RST with RS � ST. Segmentswith the same length are congruent, so

. An isosceles triangle is a trianglewith two congruent sides. Therefore, �RST isisosceles.

15. a. Statements Reasons

1. bisects �ACB. 1. Given.2. �ACE � �ECB 2. Definition of angle

bisector.3. m�ACE 3. Definition of

� m�ECB congruent angles.

CEh

RS > ST

RS > ST

MBAMAMB

AMB

MBAMAMB

LM > MN

DF > FE

b. We are given that bisects �ACB. If a ray bisects an angle, then it divides the angle intotwo congruent angles, so �ACE � �ECB.Congruent angles have the same measure.Therefore, m�ACE � m�ECB.

16. Statements Reasons1. DE � EF 1. Given.2. 2. Definition of

congruent segments.3. E is the midpoint 3. Definition of

of . midpoint.

17. An obtuse angle is an angle whose degreemeasure is between 90 and 180. An obtusetriangle is triangle with an obtuse angle. Sincem�A � 90 and m�B � 90, �A and �B are notobtuse. Since �ABC is an obtuse triangle, �Cmust be obtuse. Therefore, m�C 90.

18. In isosceles �ABC with �A as the vertex angle,we can only infer that legs, and , arecongruent and have equal measure. The length of

is unknown.

3-4 Direct and Indirect Proofs (pages 108–109)

Writing About Mathematics

1. Yes. �ABC is made up of the rays and ,

which intersect at vertex B. B is a point on both

lines and . Therefore, and

intersect at B.

2. Yes. The notation means that A, B, and Care collinear and that B is between A and C.

Developing Skills3. b. Statements Reasons

1. LM � MN 1. Given.2. 2. Definition of

congruentsegments.

c. Statements Reasons1. is not 1. Assumption.

congruent to .2. LM � MN 2. It two segments are

not congruent, thenthey do not havethe same measure.

3. LM � MN 3. Given.4. 4. Contradiction.LM > MN

MNLM

LM > MN

ABCg

BCg

BAg

BCg

BAg

BCh

BAh

BC

ACAB

DEF

DE > EF

CEh

238

4. b. Statements Reasons1. �PQR is a 1. Given.

straight angle.2. m�PQR � 180 2. A straight angle

measures 180°.

c. Statements Reasons1. m�PQR � 180 1. Assumption.2. �PQR is not a 2. If an angle does not

straight angle. measure 180°, thenit is not a straightangle.

3. �PQR is a straight 3. Given.angle.

4. m�PQR � 180 4. Contradiction.

5. b. Statements Reasons1. �PQR is a straight 1. Given.

angle.

2. and are 2. Definition of opposite rays. straight angle.

c. Statements Reasons

1. and are 1. Assumption.not opposite rays.

2. �PQR is not a 2. If two rays are not straight angle. opposite rays, then

their union is not astraight angle.


4. and are 4. Contradiction.two opposite rays.

6. b. Statements Reasons

1. and are 1. Given.opposite rays.

2. P, Q, and R are on 2. Definition of the same line. opposite rays.

c. Statements Reasons1. P, Q, and R are not 1. Assumption.

on the same line.

2. and are 2. If two rays are not not opposite rays. of the same line,

then they are notopposite rays.

3. and are 3. Given.opposite rays.

4. P, Q, and R are on 4. Contradiction.the same line.

QRh

QPh

QRh

QPh

QRh

QPh

QRh

QPh

QRh

QPh

QRh

QPh

7. b. Statements Reasons1. �PQR is a straight 1. Given.

angle.

2. and are 2. Definition of opposite rays. straight angle.

3. P, Q, and R are on 3. Definition of the same line. opposite rays.

c. Statements Reasons1. P, Q, and R are not 1. Assumption.

on the same line.

2. and are 2. If two rays are not not opposite rays. of the same line,

then they are notopposite rays.

3. �PQR is not a 3. If two rays are not straight angle. opposite rays, then

their union is not astraight angle.


5. P, Q, and R are on 5. Contradiction.the same line.

8. b. Statements Reasons

1. bisects �DEF. 1. Given.2. �DEG � �GEF 2. Definition of angle

bisector.3. m�DEG 3. Definition of

� m�GEF congruent angles.

c. Statements Reasons1. m�DEG 1. Assumption.

� m�GEF2. �DEG is not 2. If two angles do not

congruent to have the same �GEF. measure, then they

are not congruent.

3. does not 3. If a ray does not bisect �DEF. divide an angle into

two congruentangles, then it doesnot bisect the angle.

4. bisects �DEF. 4. Given.5. m�DEG 5. Contradiction.

� m�GEF

9. The indirect proofs were longer than the direct proofs because the statements to beproved followed directly from definitions ingeometry.

EGh

EGh

EGh

QRh

QPh

QRh

QPh

239

10. Statements Reasons1. �DEG � �GEF 1. Assumption.

2. bisects �DEF. 2. Definition of angle bisector.

3. does not bisect 3. Given.�DEF.

4. �DEG is not 4. Contradiction.congruent to �GEF.

Applying Skills

11. a. Given: is perpendicular to .

b. Prove: is the bisector of �ABD.


1. ⊥ 1. Given.2. �ABD and �CBD 2. Definition of

are right angles. perpendicular lines.3. m�ABD � 90 and 3. The measure of a

m�CBD � 90 right angle is 90°.4. �ABD � �CBD 4. Congruent angles

have the samemeasure.

5. is the bisector 5. Definition of angle of �ABD. bisector.

12. a. Given: m�EFG � 180

b. Prove: and are not opposite rays.


1. and are 1. Assumption.opposite rays.

2. �EFG is a straight 2. Definition of angle. straight angle.

3. m�EFG � 180 3. The measure of astraight angle is180°.

4. m�EFG � 180 4. Given.

5. and are 5. Contradiction.not opposite rays.

3-5 Postulates,Theorems, and Proof (pages 113–115)

Writing About Mathematics1. Yes. An angle is congruent to itself. Congruent

angles can be expressed in either order. Anglescongruent to the same angle are congruent toeach other.

2. No. A line is not perpendicular to itself.Developing Skills

3. Reflexive property of equality

FGh

FEh

FGh

FEh

FGh

FEh

BDh

ABCg

BDh

BDh

ABCg

BDh

EGh

EGh

4. Transitive property of equality5. Symmetric property of equality6. Transitive property of equality

Applying Skills7. Statements Reasons

1. y � x � 4 1. Given.2. x � 4 � y 2. Symmetric property.3. y � 7 3. Given.4. x � 4 � 7 4. Transitive property.

8. Statements Reasons1. AB � BC � AC 1. Given.2. AC � AB � BC 2. Symmetric property.3. AB � BC � 12 3. Given.4. AC � 12 4. Transitive property.

9. Statements Reasons1. M is the midpoint 1. Given.

of .2. 2. Definition of

midpoint.3. LM � MN 3. Definition of

congruent segments.4. N is the midpoint 4. Given.


midpoint.6. MN � NP 6. Definition of

congruent segments.7. LM � NP 7. Transitive property.

10. Statements Reasons1. m�FGH � m�JGK 1. Given.2. m�HGJ � m�JGK 2. Given.3. m�FGH � m�HGJ 3. Transitive property.4. �FGH � �HGJ 4. Definition of

congruent angles.

5. is the bisector of 5. Definition of angle �FGJ. bisector.

11. It is not given that �ADB and �ADC are rightangles. This may not be assumed from thediagram.

Hands-On Activitya. Definition of an equilateral triangle, transitive

property of congruence, and definition ofcongruent segments

GHh

MN > NPMP

LM > MNLN

240

b. Statements Reasons1. �ABC is 1. Given.

equilateral.2. �BCD is 2. Given.

equilateral.3. 3. Definition of

equilateral triangle.4. 4. Definition of

equilateral triangle.5. 5. Transitive property.6. AB � CD 6. Definition of

congruent segments.

3-6 The Substitution Postulate (page 117)Writing About Mathematics

1. Yes. Segments congruent to the same segmentare congruent to each other.

2. No. While and are congruent, may belocated anywhere in relation to ; that is,may not be perpendicular to .

Developing Skills3. Statements Reasons

1. MT � 1. Given.2. RM � MT 2. Given.3. RM � 3. Substitution postulate.

4. Statements Reasons1. AD � DE � AE 1. Given.2. AD � EB 2. Given.3. EB � DE � AE 3. Substitution postulate.

5. Statements Reasons1. m�a � m�b � 180 1. Given.2. m�a � m�c 2. Given.3. m�c � m�b � 180 3. Substitution postulate.

6. Statements Reasons1. y � 7 � 2x 1. Given.2. y � x � 5 2. Given.3. x � 5 � 7 � 2x 3. Substitution postulate.

7. Statements Reasons1. 12 � x � y 1. Given.2. x � 8 2. Given.3. 12 � 8 � y 3. Substitution postulate.

12RT

12RT

STPQST

PQRSPQ

AB > CD

BC > CD

AB > BC

8. Statements Reason1. BC2 � AB2 � AC2 1. Given.2. AB � DE 2. Given.3. BC2 � DE2 � AC2 3. Substitution postulate.

9. Statements Reason1. AB � 1. Given.2. = EF 2. Given.3. AB � EF 3. Transitive property.4. = EF 4. Given.5. AB � 5. Transitive property.

10. Statements Reason1. m�Q � m�R 1. Given.

� m�S � 752. m�Q � m�S 2. Given.

� m�T3. m�T � m�R � 75 3. Substitution postulate.4. m�R � m�T 4. Given.

� m�U5. m�U � 75 5. Substitution postulate.

3-7 The Addition and SubtractionPostulates (pages 122–123)

Writing About Mathematics1. Cassie is incorrect. The definition of subtraction

in terms of addition depends on the definition ofnegative numbers. However, there is no conceptof a “negative line segment,” so this definitionwould be invalid for line segments.

2. a. Yes. We are given that m�ABC � 30,m�CBD � 45, and m�DBE � 15. Then,m�ABC � m�DBE � 30 � 15 � 45 �m�CBD.

b. No. Since �ABC and �DBE are not adjacentangles, �ABC � �DBE does not representan angle.


1. and 1. Given.2. AE � ED � AD 2. Partition postulate.

BF � FC � BC3. AE � BF and 3. Given.

ED � FC4. BF � FC � AD 4. Substitution postulate.5. AD � BC 5. Transitive property.

BFCAED

12GH

12GH

"CD"CD

241

4. Statements Reasons1. �SPR � �QRP 1. Given.

and �RPQ � �PRS2. �SPR � �RPQ 2. Addition postulate.

� �QRP � �PRS3. �SPR � �RPQ 3. Partition postulate.

� �SPQ4. �QRP � �PRS 4. Partition postulate.

� �QRS5. �SPQ � �QRS 5. Substitution

postulate.

5. Statements Reasons1. and 1. Given.

2. 2. Subtraction postulate.

3. 3. Partition postulate.

4. 4. Substitution postulate.

or

6. Statements Reasons1. 1. Given.2. 2. Reflexive property.3. 3. Addition postulate.

4. 4. Given.5. 5. Partition postulate.

6. 6. Substitutionpostulate.

7. Statements Reasons1. �LMN � �PMQ 1. Given.2. �NMQ � �NMQ 2. Reflexive property.3. �LMN � �NMQ 3. Addition postulate.

� �NMQ � �QMP4. �LMQ 4. Partition postulate.

� �LMN � �NMQ�NMP� �NMQ � �QMP

5. �LMQ � �NMP 5. Substitutionpostulate.

AC > BDBD > BC 1 CDAC > AB 1 BCABCD> BC 1 CDAB 1 BCBC > BCAB > CD

AM > BN2 NC> BN 1 NC

AM 1 MC 2 MCBC > BN1NCAC > AM1MC

> BC 2 NCAC 2 MC MC > NCAC > BC

8. Statements Reasons1. m�AEB � 180 and 1. Given.

m�CED � 1802. �AEB � �CED 2. Definition of

congruent angles.3. �AEB 3. Partition postulate.

� �AEC � �CEB�CED� �CEB � �BED

4. �AEC � �CEB 4. Substitution � �CEB � �BED postulate.

5. �CEB � �CEB 5. Reflexive property.6. �AEC � �BED 6. Subtraction postulate.7. m�AEC 7. Definition of

� m�BED congruent angles.

3-8 The Multiplication and DivisionPostulates (pages 126–127)

Writing About Mathematics1. The word “positive” is needed because only

positive quantities have square roots, and anumber has both a positive and a negative squareroot.

2. Yes. The conditions of the postulate require thatc � d. Therefore, if one of these variables is notequal to 0, then both are not equal to 0 and thefractions and are defined. If “c � 0” and “d � 0” are both removed, it is possible that c � 0 and d � 0, and that the fractions and areundefined.


1. AB � 1. Given.2. 4 � 4 2. Reflexive property.3. 4AB � BC 3. Multiplication

postulate.4. BC � CD 4. Given.5. 4AB � CD 5. Substitution

postulate.6. CD � 4AB 6. Symmetric property.

4. Statements Reasons1. m�a � 3m�b 1. Given.2. m�b � 20 2. Given.3. 3m�b � 60 3. Substitution

postulate.4. m�a � 60 4. Transitive property.

14BC

cd

ab

cd

ab

242

5. Statements Reasons1. 1. Given.2. 2MN = NP 2. Doubles of equal

quantities are equal.3. LM � 2MN 3. Given.4. LM � NP 4. Transitive property.5. 5. Definition of

congruent segments.

6. Statements Reasons1. 2(3a � 4) � 16 1. Given.2. 3a � 4 � 8 2. Halves of equal

quantities are equal.3. 4 � 4 3. Reflexive property.4. 3a � 12 4. Addition postulate.5. 3 � 3 5. Reflexive property.6. a � 4 6. Division postulate.

7. Statements Reasons1. 1. Given.2. 3 � 3 2. Reflexive property.3. 3QR � RS 3. Multiplication

postulate.4. PQ � 3QR 4. Given.5. PQ � RS 5. Transitive property.6. 6. Definition of

congruent segments.

Alternative Proof:

Statements Reasons1. PQ � 3QR 1. Given.

2. 2. Given.

3. PQ � or 3. Substitution

PQ � RS postulate.4. 4. Definition of

congruent segments.

Applying Skills8. Let m � Monday’s distance, t � Tuesday’s

distance, w � Wednesday’s distance, and f � Friday’s distance.We are given w � and w = . Then by the transitive property of equality. By themultiplication postulate, m � 2f. We are alsogiven that m � 2t, so by the transitive property ofequality, 2f � 2t. Since halves of equal quantitiesare equal, f � t or the distance Melanie walkedon Friday is equal to the distance she walked onTuesday.

13m 5 2

3f23f1

3m

PQ > RS

3 A 13RS B

QR 5 13RS

PQ > RS

QR 5 13RS

LM > NP

MN 5 12NP

9. Let L � library, P � post office, G � grocerystore, and B � bank.We are given , with LP � 4PG and GB � 3PG. By the partition postulate,PB � PG � GB. Using the substitutionpostulate, PB � PG � 3PG or PB � 4PG. By thetransitive property, LP � PB or the distancefrom the library to the post office is equal to thedistance from the post office to the bank.

10. To show that doubles of congruent segments arecongruent:

Given: B is the midpoint of , M is themidpoint of , and .

Prove:

Statements Reasons1. B is the midpoint 1. Given.

of , M is the midpoint of .

2. AC � 2AB, 2. Definition ofLN � 2LM midpoint.

3. 3. Given.4. AB � LM 4. Definition of

congruent segments.5. 2AB � 2LM 5. Doubles of equal

quantities are equal.6. AC � LM 6. Substitution postulate.7. 7. Definition of

congruent segments.

To show that doubles of congruent angles arecongruent:

Given: is the angle bisector of �ABD,is the angle bisector of �JKM, and �ABC � �JKL.

Prove: �ABD � �JKM

Statements Reasons1. is the angle 1. Given.

bisector of �ABD,

is the angle bisector of �JKM.

2. m�ABD 2. Definition of angle� 2m�ABC, bisector.m�JKM � 2m�JKL

3. �ABC � �JKL 3. Given.4. m�ABC � m�JKL 4. Definition of

congruent angles.5. 2m�ABC 5. Doubles of equal

� 2m�JKL quantities are equal.6. m�ABD � m�JKM 6. Substitution postulate.7. �ABD � �JKM 7. Definition of

congruent angles.

KLh

BCh

KLh

BCh

AC > LN

AB > LM

LNAC

AC > LNAB > LMLN

AC

LPGB

243

To show that halves of congruent segments arecongruent:

Given: B is the midpoint of , M is themidpoint of , and .

Prove:

Statements Reasons1. B is the midpoint of 1. Given.

, M is themidpoint of .

2. , 2. Definition of

midpoint.

3. 3. Given.4. AC � LN 4. Definition of

congruent segments.

5. 5. Halves of equal quantities are equal.

6. AB � LM 6. Substitutionpostulate.

7. 7. Definition ofcongruent segments.

Given: is the angle bisector of �ABD,

is the angle bisector of �JKM, and

�ABD � �JKM.Prove: �ABC � �JKL

Statements Reasons1. is the angle 1. Given.

bisector of �ABD,

is the angle

bisector of �JKM.

2. , 2. Definition of angle

bisector.3. �ABD � �JKM 3. Given.4. m�ABD � m�JKM 4. Definition of

congruent angles.

5. 5. Halves of equal quantities are equal.

6. m�ABC � m�JKL 6. Substitutionpostulate.

7. �ABC � �JKL 7. Definition ofcongruent angles.

Review Exercises (pages 129–130)1. a. If a triangle is obtuse, then it has one obtuse

angle.

5 12m/JKM

12m/ABD

m/JKL 5 12m/JKM

m/ABC 5 12m/ABD

KLh

BCh

KLh

BCh

AB > LM

12AC 5 1

2LN

AC > LN

LM 5 12LMN

AB 5 12ABC

LNAC

AB > LMAC > LNLN

AC

b. If a triangle has one obtuse angle, then it isobtuse.

c. A triangle is obtuse if and only if it has oneobtuse angle.

2. a. If two angles are congruent, then they havethe same measure.

b. If two angles have the same measure, thenthey are congruent.

c. Two angles are congruent if and only if theyhave the same measure.

3. a. If two lines are perpendicular, then theyintersect to form right angles.

b. If two lines intersect to form right angles, thenthey are perpendicular.

c. Two lines are perpendicular if and only if theyintersect to form right angles

4. We assume a postulate to be true without proof. A theorem is a statement that has beenproven.

5. Symmetric property6. No. The reflexive property does not hold true.

Example: 1 1 is false. The symmetric propertydoes not hold true. Example: If 2 1, then 1 2 is false. The transitive property holds true. Example: If 3 2 and 2 1, then 3 1 is true.

7.

Statements Reasons1. bisects 1. Given.

at M.2. M is the midpoint 2. Definition of

of . bisector.3. 3. Definition of midpoint.4. CM � MD 4. Definition of

congruent segments.

8.

Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property.RM > ST

MS > STRM > MS

R M S T

CM > MDCD

CDABg

B

A

M DC

244

9.

Statements Reasons1. 1. Given.2. 2. Partition postulate.

3. 3. Substitution postulate.

4. 4. Reflexive property.5. 5. Subtraction

postulate.

10.

Statements Reasons1. SQ � RP 1. Given.2. QR � QR 2. Reflexive property.3. SQ � QR 3. Addition postulate.

� QR � RP4. SR � SQ � QR 4. Partition postulate.

QP � QR � RP5. SR � QP 5. Substitution

postulate.

11.

Statements Reasons

1. bisects �ABD. 1. Given.2. �ABC � �CBD 2. Definition of angle

bisector.3. m�ABC � m�CBD 3. Definition of

congruent angles.4. m�CBD � m�PQR 4. Given.5. m�ABC � m�PQR 5. Transitive property.

12.

Statements Reasons

1. and bisect 1. Given.each other at E.

ABCD

A D

E

C B

BCh

A

B D

CP

QR

S Q R P

AB > CDBC > BC> BC 1 CD

AB 1 BCBD > BC 1 CDAC > AB 1 BCAC > BD

A B C D

(Cont.)

Statements Reasons2. E is the midpoint of 2. Definition of a

and of . bisector.

3. and 3. Definition of

midpoint.4. AE � EB and 4. Definition of

CE � ED congruentsegments.

5. AE � CE � EB � ED 5. Addition postulate.6. CE � BE 6. Given.7. AE � BE � CE � ED 7. Substitution

postulate.8. AB � AE � EB 8. Partition postulate.

CD � CE � ED9. AB � CD 9. Substitution

postulate.

13. A quantity may be substituted for its equal in any statement of equality. is not astatement of equality; therefore, the substitutionpostulate is not valid.

14. a. If a sequence of numbers or letters is apalindrome, then the sequence reads the samefrom left to right as from right to left.

b. If a sequence of numbers or letters reads thesame from left to right as from right to left,then the sequence is a palindrome.

c. A sequence of numbers or letters is apalindrome if and only if it reads the samefrom left to right as from right to left.

Exploration (page 130)Step 5 is not valid. Since a � b, the quantity a � b isequal to 0. Thus, we cannot apply the divisionpostulate.


1. 3 2. 4 3. 4 4. 1 5. 46. 2 7. 3 8. 2 9. 4 10. 3

Part II11. Given

� 7 � �7 Reflexive property3x � 6 Addition postulate

x � 2 Division postulate

3x 1 7 5 13

CD ' BC

CE > ED

AE > EB

CDAB

245

12. Statements Reasons1. 1. Given.2. �EDF is a right angle. 2. Definition of

perpendicular lines.3. �DEF is a right 3. Definition of right

triangle. triangle.

Part III13. Statements Reasons

1. �ABC with D a point 1. Given.on .

2. AB � AD � DB 2. Partition postulate.3. AC � AD � DB 3. Given.4. AC � AB 4. Transitive property.5. 5. Definition of

congruentsegments.

6. �ABC is isosceles. 6. Definition ofisosceles triangle.

14. Yes. By the partition postulate, PQ � QR � PR.Thus,

Therefore, PQ � 4(5) � 3 � 17 and QR � 3(5) � 2 � 17. Thus, and Q isthe midpoint of .

Part IV15. Yes. By the partition postulate,

Therefore, m�ABD � 3(24) � 18 � 90 andm�DBC � 5(24) � 30 � 90. Thus,m�ABC � 90 � 90 � 180. An angle whosedegree measure is 180 is a straight angle.

16. a. Addition postulateb. Reflexive property of equalityc. Partition postulated. Substitution postulatee. Division postulate

x 5 24

8x 2 12 5 7x 1 12

3x 1 18 1 5x 2 30 5 7x 1 12

m/ABD 1 m/DBC 5 m/ABC

PQRPQ > QR

5 5 a

7a 2 1 5 8a 2 6

(4a 2 3) 1 (3a 1 2) 5 8a 2 6

AC > AB

AB

DE ' EF

Chapter 4. Congruence of Line Segments,Angles, and Triangles

246

7. a. Given: and Prove:

b. Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property of

congruence.

8. a. Given: 2EF � DB and Prove: EF � GH

b. Statements Reasons1. 2EF � DB 1. Given.2. EF � 2. Halves of equal

quantities are equal.3. 3. Given.4. EF � GH 4. Transitive property of

equality.

9. a. Given: CE � CF, CD � 2CE, CB � 2CFProve: CD � CB

b. Statements Reasons1. CD � 2CE 1. Given.2. CE � CF 2. Given.3. CD � 2CF 3. Substitution postulate.4. CB � 2CF 4. Given.5. CD � CB 5. Transitive property of

equality.

10. a. Given: RT � RS, ,Prove: RD � RE

b. Statements Reasons1. 1. Given.2. RT � RS 2. Given.3. 3. Substitution postulate.

4. 4. Given.5. RD � RE 5. Substitution postulate.

11. a. Given: AD � BE and BC � CD Prove: AC � CE

b. Statements Reasons1. AD � BE 1. Given.2. BC � CD 2. Given.3. AD � CD 3. Subtraction postulate.

� BE – BC4. AD � AC � CD, 4. Partition postulate.

BE � BC � CE5. AC � CE 5. Substitution postulate.

RE 5 12RS

RD 5 12RS

RD 5 12RT

RE 5 12RSRD 5 1

2RT

GH 5 12DB

12DB

GH 5 12DB

AB > EFEF > CDAB > CD

AB > EFEF > CDAB > CD4-1 Postulates of Lines, Line Segments,

and Angles (pages 139–140)Writing About Mathematics

1. Points E and F must name the same point.2. No. There is only one positive real number

that is the length of a given line segment. A different number is the length of a different line segment.

Developing Skills3. a. Given: AB � AD and DC � AD

Prove: AB � DC

b. Statements Reasons1. AB � AD 1. Given.2. DC � AD 2. Given.3. AB � DC 3. Transitive property

of equality.

4. a. Given: and Prove:

b. Statements Reasons1. 1. Given.2. 2. Given.3. 3. Transitive property

of equality.

5. a. Given: m�1 � m�2 � 90, m�A � m�2Prove: m�1 � m�A � 90

b. Statements Reasons1. m�1 � m�2 � 90 1. Given.2. m�A � m�2 2. Given.3. m�1 � m�A � 90 3. Substitution

postulate.

6. a. Given: m�A � m�B, m�1 � m�B,m�2 � m�A

Prove: m�1 � m�2

b. Statements Reasons1. m�1 � m�B 1. Given.2. m�A � m�B 2. Given.3. m�1 � m�A 3. Transitive property

of equality.4. m�2 � m�A 4. Given.5. m�1 � m�2 5. Transitive property

of equality.

AD > BDBD > CDAD > CD

AD > BDBD > CDAD > CD

12. a. Given: �CDB � �CBD and �ADB � �ABD

Prove: �CDA � �CBA

b. Statements Reasons1. �CDB � �CBD 1. Given.2. �ADB � �ABD 2. Given.3. �CDB � �ADB 3. Addition postulate.

� �CBD � �ABD4. �CDA 4. Partition postulate.

� �CDB � �BDA,�CBA� �CBD � �DBA

5. �CDA � �CBA 5. Substitutionpostulate.

13. m�QSR � 37 14. m�PST � 5015. m�PST � 45Applying Skills16. a. Answers will vary.

b. Given: A triangle is equilateral.Prove: The measures of the sides are equal.

c. If triangle is equilateral then all of its sides arecongruent. Congruent sides have equalmeasures.

17. a. Answers will vary.b. Given: Points E and F are distinct and two

lines intersect at E.Prove: The lines do not intersect at F.

c. Two lines can intersect at only one point.If they intersect at both E and F, then E and F must name the same point. But we are given that E and F are distinct. This is a con-tradiction, so the lines do not intersect at F.

18. a. Answers will vary.b. Given: A line through a vertex of a triangle is

perpendicular to the opposite side.Prove: The line separates the triangle into two

right triangles.c. A line through a vertex of a triangle that is

perpendicular to the opposite side separates itinto two triangles. Perpendicular linesintersect to form right angles, so each of thetriangles has a right angle. A triangle with aright angle is a right triangle.

19. a. Answers will vary.b. Given: Two points on a circle are endpoints of

a line segment.Prove: The length of the line segment is

shorter than the length of the portionof the circle with the same endpoints.

247

c. The shortest distance between two points isthe line segment between them. Hence, itslength is shorter than the length of the arc.

20. Yes. Through two distinct points, one and onlyone line can be drawn.

4-2 Using Postulates and Definitions inProofs (pages 142–144)


1. The symbol refers to a line segment withendpoints A and C with point B lying betweenthem. The symbol refers to a line segment with endpoints A and B with point C lyingbetween them. These symbols could not refer tosegments of the same line.

2. Yes. Since m�ABD � 90, then:m�ABD � m�DBC � 180 or m�DBC � 90m�DBC � m�CBE � 180 or m�CBE � 90m�ABE � m�CBE � 180 or m�ABE � 90

Developing Skills

3. a. Given: , bisects , and bisects .

Prove:

b. Statements Reasons1. 1. Given.2. AB � CB 2. Definition of

congruent segments.3. bisects , 3. Given.

bisects .4. D is the midpoint 4. Definition of

of , E is the bisector.midpoint of .

5. , 5. Definition of midpoint.

6. AD � CE 6. Halves of equalquantities are equal.


AD > CE

CE 5 12CB

AD 5 12AB

CBAB

CBFEABFD

AB > CB

AD > CECB

FEABFDAB > CB

ACB

ABC

4. a. Given: bisects �DCB, bisects �DAB,

and �DCB � �DAB.Prove: �CAB � �DCA

b. Statements Reasons1. �DCB � �DAB 1. Given.2. m�DCB 2. Definition of

� m�DAB congruent angles.

3. bisects �DCB, 3. Given.

bisects �DAB4. m�CAB 4. Definition of angle

� , bisector.m�DCA�

5. m�CAB 5. Halves of equal � m�DCA quantities are

equal.6. �CAB � �DCA 6. Definition of

congruent angles.

5. a. Given:Prove:

b. Statements Reasons1. 1. Given.2. , 2. Partition postulate.



6. a. Given: is a segment and AB � CD.Prove: AC � BD

b. Statements Reasons1. AB � CD 1. Given.2. AC � AB � BC,

BD � BC � CD 2. Partition postulate.3. AC � BC � CD 3. Substitution

postulate.4. AC � BD 4. Substitution

postulate.

7. a. Given: is a segment, B is the midpointof , and C is the midpoint of .

Prove: AB � BC � CD

b. Statements Reasons1. B is the midpoint 1. Given.

of .AC

BDACABCD

ABCD

AE > BD

AE > BE 1 DEBD > BE 1 EDAE > AD 1 DEAD > BE

AE > BDAD > BE

12m/DCB

12m/DAB

AChCAh

ACh

CAh

248

Statements ReasonsC is the midpoint of .

2. 2. Definition of midpoint.

3. AB � BC, 3. Definition of BC � CD congruent segments.

4. AB � BC � CD 4. Transitive propertyof equality.

8. a. Given: P and T are distinct points, P is themidpoint of .

Prove: T is not the midpoint of .b. Postulate 4.9 states that a line segment has

only one midpoint. P is the midpoint of ,and T does not name P. Therefore, T is not themidpoint.

9. a. Given:Prove:

b. Statements Reasons1. 1. Given.2. 2. Symmetric property.3. 3. Subtraction

postulate.4. , 4. Partition postulate.


10. a. Given: , E is the midpoint of ,and F is the midpoint of .

Prove:

b. Statements Reasons1. 1. Given.2. AD � BC 2. Definition of

congruent segments.3. E is the midpoint 3. Given.

of .F is the midpointof .


midpoint.5. AE � FC 5. Halves of equal

quantities are equal.6. 6. Definition of

congruent segments.AE > FC

FC 5 12BC

AE 5 12AD

BC

AD

AD > BC

AE > FCBC

ADAD > BC

DE > BFBE > BF1FEDF > DE1EF> BE 2 FE

DF 2 EFEF > FEDF > BE

DE > BFDF > BE

RS

RSRS

BC > CDAB > BC,

BD

11. a. Given: , and and bisect eachother at E.

Prove:

b. Statements Reasons1. 1. Given.2. AC � DB 2. Definition of

congruent segments.3. and bisect 3. Given.

each other at E.4. E is the midpoint 4. Definition of

of and of . bisector.5. , 5. Definition of

midpoint.

6. AE � EB 6. Halves of equalquantities are equal.


12. a. Given: bisects �CDA, �3 � �1, �4 � �2Prove: �3 � �4

b. Statements Reasons

1. bisects �CDA. 1. Given.2. �1 � �2 2. Definition of angle

bisector.3. �3 � �1, 3. Given.

�4 � �24. �3 � �4 4. Substitution

postulate.

Applying Skills13. a. m�CDE � 114

b. m�FDE � 76c. m�CDG � 76d. �CDE is obtuse.

14. a. Answers will vary.b. AB � BC � 10c. The distance from A to is AB. The

distance from a point to a segment is thelength of a perpendicular from the point to thesegment. In this case, it is AB � 10.

15. a. Answers will vary.b. Given: and bisect each other at N,

, and RN � LN.

c. Statements Reasons1. and bisect 1. Given.

each other at N.2. N is the midpoint 2. Definition of

of and of . bisector.LMRS

LMRS

RS ' LMLMRS

BD

DRh

DRh

AE > EB

EB 5 12DB

AE 5 12AC

DBAC

DBAC

AC > DB

AE > EB

DBACAC > DB

249

3. RS � 2RN, 3. Definition of LM � 2LN midpoint.

4. RN � LN 4. Given.5. RS � LM 5. Doubles of equal

quantities are equal.

d. RS � LM � 48e. The distance from L to is LN. The distance

from a point to a segment is the length of aperpendicular from the point to the segment.In this case, it is LN � � 24.

4-3 Proving Theorems About Angles (pages 152–154)

Writing About Mathematics1. Yes. �AEC and �BED are supplements of

congruent angles �BEC and �AED. If twoangles are congruent, then their supplements arecongruent.

2. No. The converse is “If two angles aresupplementary, then the angles form a linearpair.” Supplementary angles do not need to beadjacent.


1. m�ACD 1. Given.� m�DCB � 90

2. �B � �DCA, 2. Given.�A � �DCB

3. m�B � m�DCB, 3. Definition of m�A � m�DCB congruent angles.

4. m�B � m�A � 90 4. Substitution postulate.5. �A and �B are 5. Definition of

complements. complementaryangles.

4. Statements Reasons1. �BFC and �BFG 1. Definition of a

form a linear pair. linear pair.�ADE and �ADCform a linear pair.

2. �BFC and �BFG 2. If two angles form aare supplementary. linear pair, then they�ADE and �ADC are supplementary.are also supple-mentary.

3. �ADC � �BFG 3. Given.4. �ADE � �BFG 4. If two angles are

congruent, then theirsupplements arecongruent.

12(48)

RS

5. Statements Reasons1. �ADB is a right 1. Given.

angle.2. 2. Given.3. �CEB is a right 3. Definition of

angle. perpendicular lines.4. �ADB � �CEB 4. If two angles are right

angles, then they arecongruent.

6. Statements Reasons1. �ABC and �BCD 1. Given.

are right angles.2. m�ABC � 90 and 2. Definition of right

m�BCD � 90 angles.3. m�EBA � m�EBC 3. Partition postulate.

� m�ABCm�ECB � m�ECD� m�BCD

4. m�EBA � m�EBC 4. Substitution postulate.� 90m�ECB � m�ECD� 90

5. �EBA and �EBC 5. Definition of are complements. complementary�ECB and �ECD angles.are complements.

6. �EBC � �ECB 6. Given.7. �EBA � �ECD 7. If two angles are

congruent, then theircomplements arecongruent.

7. Statements Reasons1. �ABC and �DBF 1. Given.

are right angles.2. m�ABC � 90 and 2. Definition of right

m�DBF � 90 angles.3. m�ABC � m�DBC 3. Partition postulate.

� m�ABCm�DBC � m�CBF� m�DBF

4. m�ABC � m�DBC 4. Substitution� 90 postulate.m�DBC � m�CBF� 90

5. �ABC and �DBC 5. Definition of are complements. complementary �DBC and m�CBF angles.are complements.

6. �DBC � �DBC 6. Reflexive property ofcongruence.

CE ' DBE

250

Statements Reasons7. �ABC � �CBF 7. If two angles are

congruent, then theircomplements arecongruent.

8. Statements Reasons

1. intersects 1. Given.at G.

2. �CGH and �DGE 2. Definition of are vertical angles. vertical angles.

3. �CGH � �DGE 3. Vertical angles arecongruent.

4. m�BHG 4. Given.� m�CGH

5. �BHG � �CGH 5. Definition ofcongruent angles.

6. �BHG � �DGE 6. Transitive property ofcongruency.

9. Statements Reasons1. �ABC and �DBC 1. Definition of a linear

form a linear pair. pair.�ACB and �ECBform a linear pair.

2. �ABC and �DBC 2. If two angles form aare supplementary. linear pair, then they�ACB and �ECB are supplementary.are supplementary.

3. �ABC � ACB 3. Given.4. �DBC � �ECB 4. If two angles are

congruent, theirsupplements arecongruent.


1. and 1. Given.intersect at E.

2. �AEC � �CEB 2. Given.

3. 3. If two angles intersect to formcongruent adjacentangles, then they areperpendicular.

11. Statements Reasons1. �ABC is a right 1. Given.

angle.2. m�ABC � 90 2. Definition of right

angles.3. m�DBA � m�CBD 3. Partition postulate.

� m�ABC

AEBg

' CEDg

CEDg

AEBg

DCg

EFg

(Cont.)

Statements Reasons4. m�DBA � m�CBD 4. Substitution postulate.

� 905. �DBA and �CBD 5. Definition of

are complementary. complementary angles.6. �BAC and �DBA 6. Given.

are complementary.7. �BAC � �CBD 7. If two angles are

complements of thesame angle, then theyare congruent.

12. m�AED � 70, m�DEB � 110, m�AEC � 11013. m�DEB � m�AEC � 120, m�AED

� m�CEB � 6014. m�BEC � m�DEA � 75,

m�AEC � m�DEB � 10515. m�CEB � m�DEA � 45,

m�BED � m�AEC � 135 16. a. y � 20, x � 30

b. m�RPL � 50, m�LPS � 130,m�MPS � 50

17. If two angles are straight angles, then they bothmeasure 180 degrees. Hence, they both have thesame measure. By definition of congruent angles,the two angles are congruent.

18. If two angles, �1 and �2, are both supplementsof the same angle, �3, then m�1 � m�3 � 180and m�2 � m�3 � 180. By the substitutionpostulate, m�1 � m�3 � m�2 � m�3. By thesubtraction postulate, m�1 � m�2 so �1 � �2.

19. If two angles, �1 and �2, are congruent and �3 and �4 are their respective supplements, thenm�1 � m�3 � 180 and m�2 � m�4 � 180.By the substitution postulate, m�1 � m�3 �m�2 � m�4. Since m�1 � m�2, by thesubtraction postulate, m�3 � m�4. Therefore,�3 � �4.

Applying Skills20. 120° 21. 40° 22. 130°23. 75° 24. 65° 25. 33°

4-4 Congruent Polygons andCorresponding Parts (page 157)

Writing About Mathematics1. No. One pair of congruent corresponding sides is

not sufficient to prove two triangles congruent.

251

2. No. �RST � �STR implies that �R � �S,�S � �T, �T � �R, , , and

. However, this is not necessarily true.

Developing Skills

3. , , , �A � �C,�ABD � �DBC, �ADB � �CDB

4. , , , �A � �C,�ADB � �CBD, �ABD � �CDB

5. , , , �A � �E,�ABD � �EBC, �D � �C

6. Reflexive property7. Symmetric property8. Transitive property9. Symmetric property

10. Symmetric property

4-5 Proving Triangles Congruent UsingSide, Angle, Side (pages 160–161)


1. No. Suppose that they are. Let t1 represent thetop of the first pole, b1 represent the point wherethe first pole meets the ground, and w1 representthe point where the wire of the first pole meetsthe ground. Define t2, b2, and w2 similarly for thesecond pole. If the poles have the same height,then . Since the poles are bothperpendicular to the ground, �t1b1w1 � �t2b2w2.Since the point where the wires meet the groundis 5 feet away from the foot of the poles,b1w1 � b2w2 � 5 feet, and so .Therefore, �t1b1w1 � �t2b2w2 by SAS. Inparticular, t1w1 � t2w2, but these represent thelengths of the wires. The assumption is false, andthe heights of the poles are unequal.

2. No. Two non-congruent triangles can have one,two, or three angles congruent. Also, two non-congruent triangles can have one side congruent,or two sides congruent if the angle between themis not congruent.

Developing Skills3. Yes 4. Yes 5. No6. Yes 7. No 8. Yes9. �D � �C

10.11. �EDA � �CDB

AD > DB

b1w1 > b2w2

t1b1 > t2b2

AD > ECBD > BCAB > EB

AB > DCDB > DBAD > CB

DA > DCBD > BDAB > CB

TR > RSST > TRRS > ST

Applying Skills12. a. Answers will vary;

Given: and bisect each other.b. Prove: �ABE � �CBD

c. Statements Reasons1. and 1. Given.

bisect each other.2. B is the midpoint 2. Definition of

of and of bisector..


4. �ABE � �DBC 4. Vertical angles arecongruent.

5. �ABE � �CBD 5. SAS.

13. a. Answers will vary;Given: ABCD is a quadrilateral; AB � CD,

BC � DA; �DAB, �ABC, �BCD, and�CDA are right angles.

b. Prove: Diagonal AC separates thequadrilateral into two congruenttriangles.

c. Statements Reasons1. AB � CD and 1. Given.

BC � DA2. and 2. Definition of

congruent segments.3. �ABC and �ADC 3. Given.

are right angles.4. �ABC � �ADC 4. Right angles are

congruent.5. �ABC � �CDA 5. SAS.

14. a. Answers will vary;Given: �PQR and �RQS form a linear pair,�PQR � �RQS, and PQ � QS.

b. Prove: �PQR � �RQS

c. Statements Reason1. PQ � QS 1. Given.2. 2. Definition of

congruent segments.3. �PQR � �RQS 3. Given.4. 4. Reflexive property.5. �PQR � �RQS 5. SAS.

RQ > RQ

PQ > QS

BC > DAAB > CD

DB > BEAB > BCDBE

ABC

DBEABC

DBEABC

252

4-6 Proving Triangles Congruent UsingAngle, Side, Angle (pages 163–164)

Writing About Mathematics1. Yes. Since the two triangles are congruent,

corresponding angles are congruent. Since one ofthe angles in the first triangle is a right angle, oneof the angles in the second triangle must also be aright angle.

2. a. No. The two triangles are not necessarilycongruent, as the figure shows.

b. �A cannot be congruent to �D. cannotbe congruent to .

Developing Skills3. Yes. Two angles and the side between them in

one triangle are congruent to two angles and theside between them in the other triangle.

4. Yes. Two angles and the side between them inone triangle are congruent to two angles and sidebetween them in the other triangle.

5. No. Only one pair of congruent angles are given.6. �ACD � �CAB7. �AED � �CEB8.


1. �E � �C, 1. Given.�EDA � �CDB

2. D is the midpoint 2. Given.of .

3. 3. Definition of midpoint.

4. �DAE � �DBC 4. ASA.

ED > DCEC

DB > DB

FECB

C

A B

D

F

E


1. bisects �ADC 1. Given.

and bisects �ABC.

2. �ADB � �BDC, 2. Definition of angle �CBD � �DBA bisector.

3. 3. Reflexive property ofcongruence.

4. �ABD � �CBD 4. ASA.

11. Statements Reasons1. 1. Given.2. �CDA � �BDA 2. If two lines are

perpendicular, thenthey intersect to formcongruent adjacentangles.

3. 3. Reflexive property ofcongruency.

4. bisects �BAC. 4. Given.5. �CAD � �BAD 5. Definition of angle

bisector.6. �ADC � �ADB 6. ASA.

12. Statements Reasons1. �DBC � �GFD 1. Given.2. �DBC and �ABD 2. Definition of a linear

form a linear pair; pair.�GFD and �DFEform a linear pair.

3. �DBC and �ABD 3. Linear pairs of anglesare supplementary; are supplementary.�GFD and �DFEare supplementary.

4. �DFE � �ABD 4. The supplements ofcongruent angles arecongruent.

5. bisects at D. 5. Given.6. D is the midpoint 6. Definition of bisector.


midpoint.8. �FDE � �ADB 8. Vertical angles are

congruent.9. �DFE � �DBA 9. ASA (steps 4, 7, 8).

4-7 Proving Triangles Congruent UsingSide, Side, Side (pages 166–167)

Writing About Mathematics1. Yes. If two triangles are congruent, then

corresponding sides are congruent. Therefore, if

FD > DBFB

FBAE

ADh

AD > AD

AD ' BC

DB > DB

BDh

DBh

253

the two sides of one triangle are congruent, thentwo sides of the other triangle must be alsocongruent and that triangle is isosceles.

2. Yes. If all corresponding sides are congruent,then by SSS, the two triangles are congruent.

Developing Skills3. Yes 4. Yes 5. No6. 7. 8.9. SAS

10. Not enough information11. ASA or SAS 12. SSS13. Not enough information14. Not enough information15. No. One triangle can be acute and the other can

be obtuse.


1. and 1. Given.

2. �ACE and �BDE 2. Perpendicular are right angles. segments meet to

form right angles.3. �ACE � �BDE 3. Right angles are

congruent.4. �AEC � �BED 4. Vertical angles are

congruent.5. bisects . 5. Given.6. E is the midpoint 6. Definition of bisector.


midpoint.8. �EAC � �EBD 8. ASA (steps 3, 7, 4).

CE > EDCED

CEDAEB

BD ' CEDAC ' CED

C

A B

D

F

E

FB > ECAD > BCBC > DC

17. Statements Reasons1. �ABC is equilateral. 1. Given.2. 2. In an equilateral

triangle, all sides arecongruent.

3. D is the midpoint 3. Given.of .

4. 4. Definition ofmidpoint.


6. �ACD � �BCD 6. SSS.

18. Statements Reasons1. PS � QS 1. Assumption.2. 2. Definition of

congruent segments.3. 3. Given.4. �PSR � �QSR 4. If two lines are



6. �PSR � �QSR 6. SAS.7. �PSR is not 7. Given.

congruent to �QSR.8. PS � QS 8. Contradiction.

19. Statements Reasons1. AC � AB � 15 1. Given.2. 2. Definition of

congruent segments.3. 3. Reflexive property of

congruence.

4. is the bisector 4. Given.

of �BAC.5. �BAD � �CAD 5. Definition of angle

bisector.6. �ABD � �ACD 6. SAS.

Review Exercises (pages 169–170)1. The measure of the angle is 125° and its

complement is 55°.2. m�PMN � 48, m�LMP � 1323. m�K � m�Q � 634. B � F; two lines cannot intersect in more than

one point.

ADh

AD > AD

AC > AB

SR > SR

RS ' PQ

PS > QS

DC > DC

AD > DBAB

AC > BC

254

5. and coincide; at a given point on a given line, one and only one perpendicular can bedrawn to the line.

6. LM � MN � LR � RN; the shortest distancebetween two points is the length of the linesegment joining these two points.

7. Yes. An angle has one and only one bisector.8. Not necessarily. Two lines cannot intersect in

more than one point. However, we are only given

that and intersect at M. If they are

distinct lines, then P does not lie on . If they coincide, then it does.

9. Yes. At a given point on a given line, only oneperpendicular can be drawn to the line.

10. Statements Reasons1. m�A � m�D � 50 1. Given.2. �A � �D 2. Definition of

congruent angles.3. AB � DE � 10 cm 3. Given.4. 4. Definition of

congruent segments.5. m�B � m�E � 45 5. Given.6. �B � �E 6. Definition of

congruent angles.7. �ABC � �DEF 7. ASA.

11. Statements Reasons1. bisects . 1. Given.2. E is the midpoint 2. Definition of bisector.


midpoint.4. m�D � m�F 4. Given.5. �D � �F 5. Definition of

congruent angles.6. �DEH � �GEF 6. Vertical angles are

congruent.7. �GFE � �HDE 7. ASA.

12. Statements Reasons1. �B � �E 1. Assumption.2. 2. Given.3. 3. Given.4. �ABC � �DEF 4. SAS.5. �ABC is not 5. Given.

congruent to �DEF.6. �B is not congruent 6. Contradiction.

to �E.

BC > EFAB > DE

DE > EFDEF

DEFGEH

AB > DE

MNg

PMg

MNg

KMg

LMg

Exploration (page 170)1.

2. We are given that STUVWXYZ is a cube. In acube, all edges are congruent. Therefore,

. In a cube, all faces are squares.All angles in a square are right angles.Therefore, �STX, �UTX, and �STU are rightangles. Right angles are congruent, so �STX � �UTX � �STU. Then �STX � �UTX � �STU by SAS.


1. 4 2. 2 3. 1 4. 45. 4 6. 2 7. 3 8. 19. 4 10. 2

Part II11. Statements Reasons

1. bisects at M. 1. Given.2. M is the midpoint 2. Definition of bisector.


midpoint.4. �R � �S 4. Given.5. �RMQ � �SMP 5. Vertical angles are

congruent.6. �RMQ � �SMP 6. ASA.

12. Statements Reasons1. DE � DG, 1. Given.

EF � GF2. , 2. Definition of

congruent segments.3. 3. Reflexive property of

congruence.4. �DEF � �DGF 4. SSS.

Part III13. Yes. Let w � “Our team wins,” c � “We

celebrate,” and p � “We practice.” Since both c ∨ p and �p are true, by the Law of DisjunctiveInference, c is true. Since �w → �c implies c → w, w is true by the Law of Detachment.Therefore, our team won.

DF > DFEF > GFDE > DG

RM > MSRS

RSPQ

ST > TX > UT

255

14. We have a system of two equations:

Solving for x in the second equation, x � 5y.Substituting for x in the first equation:

Therefore, y � 10 and x � 5(10) � 50.Part IV15. Let x � the measure of one angle, then

� the measure of the other angle.

Therefore, the angles measure 52° and 38°.16. a. �DML � �ENM � �FLN

b. We are given that �DEF is equilateral andequiangular. Thus, and

. We are also given that M, N,and L are the midpoints of the sides of thetriangle, and so ,

, and .Since halves of congruent segments are congruent,

. Therefore, �DML ��ENM � �FLN by SAS.

c. In part b, we showed that �DML � �ENM ��FLN. Since corresponding parts ofcongruent triangles are congruent,

. Therefore, �MNL isequilateral.

d. By the partition postulate,

m�DLF � m�DLM � m�MLN � m�FLNm�FNE � m�FNL � m�LNM � m�ENMm�DME � m�EMN � m�LMN � m�DML

Since all straight angles are congruent and�DLF, �FNE, and �DME are straightangles, m�DLF � m�FNE � m�DME.Thus, by the substitution postulate,

m�DLM � m�MLN � m�FLN� m�FNL � m�LNM � m�ENM� m�EMN � m�LMN � m�DML

LM > MN > LN

NF > FL > LDDM > ME > EN >

12FD 5 FL 5 LD1

2EF 5 EN 5 NF

12DE 5 DM 5 ME

DE > EF > FD/D > /E > /F

x 5 52

32x 5 78

x 1 x2 1 12 5 90

x2 1 12

y 5 10

18y 5 180

2(5y) 2 y 1 5y 1 4y 5 180

2x 2 y 5 x 1 4y

(2x 2 y) 1 (x 1 4y) 5 180

In part b, we showed that �DML � �ENM � �FLN. Sincecorresponding parts of congruent triangles arecongruent, m�DML � m�FLN � m�ENM andm�DLM � m�FNL � m�EMN. Thus, by thesubstitution postulate,

m�DLM � m�MLN � m�FLN � m�DLM � m�LNM � m�FLN� m�DLM � m�LMN � m�FLN

256

Therefore, by the subtraction postulate, we havethat m�MLN � m�LNM � m�LMN or �MLN � �LNM � �LMN, and so �NLM isequiangular.

Chapter 5. Congruence Based on Triangles

5-1 Line Segments Associated withTriangles (pages 177–178)


1. In �ABC, let �B be a right angle. The altitudedrawn to vertex B is . The altitudes drawn to vertices A and C are the sides of the triangle,

and , and therefore intersect with vertex B.

2. The altitudes intersect outside of the triangle.

Developing Skills3. a. Answers will vary.

b. �ACE � �BCEc.d. �ADC and �BDC are right angles.

4. a. Answers will vary.b. In acute triangles, the altitudes intersect inside

of the triangle. In obtuse triangles, thealtitudes intersect outside of the triangle. Inright triangles, the altitudes intersect at thevertex of the right angle.

5. a. Answers will vary.b. In all triangles, the angle bisectors intersect

inside of the triangle.6. a. Answers will vary.

b. In all triangles, the medians intersect inside ofthe triangle.

7.

Statements Reasons1. 1. Given.2. �P � �Q 2. Given.

PR > QR

R

P S Q

AF > BF

BCBA

DB

3. is a median. 3. Given.4. S is the midpoint 4. Definition of

of . median.5. 5. Definition of midpoint.6. �PSR � �QSR 6. SAS (steps 1, 2, 5).

8.

Statements Reasons1. is an angle 1. Given.

bisector of �DEF.2. bisects �DEF. 2. Definition of angle

bisector of a triangle.3. �DEG � �FEG 3. Definition of angle

bisector.4. 4. Reflexive property.5. is an altitude. 5. Given.6. 6. Definition of an

altitude of a triangle.7. �DGE � �FGE 7. If two lines are


8. �DEF � �FEG 8. ASA (steps 3, 4, 7).

9. C

A D B

EG ' DFEGEG > EG

EG

EG

E

D G F

PS > QSPQ

RS

(Cont.)

Statements Reasons1. is an altitude 1. Assumption.

of �ABC.2. 2. Definition of an

altitude of a triangle.3. �ADC � �BDC 3. If two lines are


4. 4. Reflexive property.5. is a median 5. Given.

of �ABC.6. D is the midpoint 6. Definition of a

of . median.7. 7. Definition of a

midpoint.8. �ADC � �BDC 8. SAS (steps 4, 3, 7).9. �ADC is not 9. Given.

congruent to �BDC.

10. is not an 10. Contradiction altitude of �ABC. (steps 8, 9).

Applying Skills

10. Assume that is an angle bisector of �LMO.Then bisects �LNM and �LNO � �MNO.

by the reflexive property ofcongruence. It is given that is an altitude of�LNM, so by definition, . Therefore,�LON and �MON are right angles andcongruent. �LON � �MON by SAS.

since corresponding parts of congruent triangles are congruent. However,�LMN is scalene, so this is a contradiction.Therefore, the assumption is false and is not an angle bisector.

11. Let the telephone pole and the two wires form apair of triangles with the ground, �ACD and �BCD. Side , the pole, is congruent to itself.If the pole is perpendicular to the ground, then

. This forms two congruent rightangles, �ADC and �BDC. Since D is themidpoint of , . Therefore, �ACD ��BCD by SAS. The wires, and , are corresponding parts of congruent triangles, so arecongruent and have equal length.

BCACAD > BDAB

CD ' AB

CD

NO

LN > MN

NO ' LMNO

NO > NONO

NO

CD

AD > BDAB

CDCD > CD

CD ' AB

CD

257

12. Area (�AMC) � �

Area (�BMC) � �

Since M is the midpoint of , AM � MB. By thesubstitution postulate, � .Therefore, the median divides �ABC into twotriangles with equal areas.

13. The farmer can determine the midpoint of one side of the piece of land and then build afence from that point to the opposite vertex ofthe plot.

5-2 Using Congruent Triangles to ProveLine Segments Congruent and AnglesCongruent (pages 179–181)

Writing About Mathematics1. Yes. Since �ABE � �DEF, �A � �D and

�B � �E. Congruent angles have equalmeasures, so if m�A � m�B � 90, then m�D � m�E � 90 and �D and �E are alsocomplementary.

2. Yes. The legs of an isosceles triangle arecongruent. Therefore, since the leg of oneisosceles triangle is congruent to the leg ofanother isosceles triangle, then the other legs arealso congruent. As the vertex angles arecongruent, the triangles are congruent by SAS.

Developing Skills3. a. �RPM � �SPM

b. SASc. �RPM � �SPM, �R � �S,

4. a. �ABD � �CDBb. SSSc. �ABD � �CDB, �BDA � �DBC,

�A � �C5. a. �ABE � �CDE

b. SASc. �A � �C, �B � �D,

6. a. �ABE � �CDEb. ASAc. �B � �D, ,

7. a. �PQR � �RSPb. SASc. �QRP � �SPR, �RPQ � �PRS,

8. a. �ABE � �CDEb. SSSc. �A � �C, �B � �D, �BEA � �DEC

PR > RP

BE > DEAB > CD

AB > CD

RP > SP

12MB(DC)1

2AM(DC)AB

12MB(DC)1

2bh

12AM(DC)1

2bh

9. Statements Reasons1. 1. Given.2. D is the midpoint 2. Given.

of .3. 3. Definition of midpoint.4. 4. Reflexive property.5. �CAD � �CBD 5. SSS (steps 1, 3, 4).6. �A � �B 6. Corresponding parts of

congruent triangles arecongruent.

10. Statements Reasons1. 1. Given.2. �CAB � �ACD 2. Given.3. 3. Reflexive property.4. �ABC � �CDA 4. SAS.5. 5. Corresponding parts of



bisect each other.2. E is the midpoint 2. Definition of bisector.

of and of .3. and 3. Definition of midpoint.

4. �AEC � �BED 4. Vertical angles arecongruent.

5. �AEC � �BED 5. SAS.6. �C � �D 6. Corresponding parts of

congruent angles arecongruent.

12. Statements Reasons1. KL � NM 1. Given.2. 2. Segments of equal

measure are congruent.3. �KLM and 3. Given.

�NML are right angles.

4. �KLM � �NML 4. Right angles arecongruent.

5. 5. Reflexive property.6. �KLM � �NML 6. SAS (steps 2, 4, 5).7. �K � �N 7. Corresponding parts of


13. a. x � 8 b. AB � 31c. BC � 32 d. EF � 32

14. a. a � 5 b. m�P � 35c. m�Q � 55 d. m�M � 55

LM > LM

KL > NM

CE > DEAE > BE

CDAB

CEDAEB

AD > CB

AC > AC

AB > CD

CD > CDAD > BD

AB

CA > CB

258

Applying Skills

15. a. Consider isosceles �ABC with vertex angle�B. The median forms �ABD and�CBD. By the definition of an isoscelestriangle, . By the reflexive propertyof congruence, . Since is amedian, . Therefore, �ABD ��CBD by SSS.

b. Since corresponding parts of congruenttriangles are congruent, �BDA � �BDC. Iftwo lines intersect to form congruent adjacentangles, then the two lines are perpendicular.Therefore, and �BDA and �BDCare right angles.

16. a. Consider quadrilateral ABCD in whichand . Since either

diagonal, or , is congruent to itself,�ABC � �CDA or �ABD � �CDB by SSS.

b. Since corresponding parts of congruenttriangles are congruent, �B � �D incongruent triangle ABC and CDA, or �A � �C in congruent triangles ABD and CDB.

17. a. Given: with ,, and

b. Statements Reasons1. , 1. Given.

,

2. �ABP � �DCP 2. SSS.3. �APB � �DPC 3. Corresponding

parts of congruenttriangles arecongruent.

c. Statements Reasons1. , 1. Given.

,

2. 2. Reflexive property.3. 3. Addition postulate.

> BC 1 CDAB 1 BCBC > BCAB > CDPB > PCPA > PD

PB > PCPA > PDAB > CD

A B C D

P

PB > PCPA > PDAB > BC > CDABCD

BDACAD > CBAB > CD

BD ' AC

AD > CDBDBD > BD

AB > CB

BD

Statements Reasons4. , 4. Partition postulate.


6. �ACP � �DBP 6. SSS (steps 1, 5).7. �APC � �DPB 7. Corresponding


18. a. Statements Reasons1. is the median 1. Given.

from P in �LNP.2. M is the midpoint 2. Definition of

of . median.3. 3. Definition of

midpoint.4. is the altitude 4. Given.

from P in �LNP.5. 5. Definition of

altitude.6. �LMP � �NMP 6. If two lines are

perpendicular,then they intersectto form congruentadjacent angles.

7. 7. Reflexive property.

8. �LMP � �NMP 8. SAS (steps 3, 6, 7).9. 9. Corresponding


10. �LNP is isosceles. 10. Definition ofisosceles triangle.

b. Statements Reasons1. �LNP is isosceles. 1. Proved in part a.2. 2. Definition of

isosceles triangle.3. is the median 3. Given.

from P to .4. 4. Definition of a

median.5. 5. Reflexive

property.6. �LMP � �NMP 6. SSS.

PM > PM

LM > MNLN

PM

LP > NP

PL > PN

PM > PM

PM ' LN

PM

LM > PMLN

PM

AC > BDBD > BC1CDAC > AB1BC

259

7. �LPM � �NPM 7. Correspondingangles of congruenttriangles arecongruent.

8. is the angle 8. Definition of an bisector from P in angle bisector.�LNP.

5-3 Isosceles and Equilateral Triangles(pages 184–185)


1. Yes. and are also corresponding sides ofcongruent triangles ACE and DBE.

2. By drawing median , Abel can only show thatand . Since �ABC is

equiangular, �A � �C. However, this is notsufficient for SSS, ASA, or SAS trianglecongruence. Therefore he cannot show that�ABD � �CBD or that �ABC is equilateral.

Developing Skills3. m�B � m�C � 304. m�S � 90, m�R � m�T � 455. x � 10, y � 30, m�D � m�E � m�F � 60

6. Statements Reasons1. � 1. Given.2. �CAB � �CBA 2. Isosceles triangle

theorem.3. �CAD � �CBE 3. If two angles are


7. Statements Reasons1. , 1. Given.

2. �BAC � �BCA, 2. Isosceles triangle �DAC � �DCA theorem.

3. �BAC � �DAC 3. Addition postulate.� �BCA � �DCA

4. �BAD � �BAC 4. Partition postulate.� �CAD,

�BCD � �BCA� �ACD

5. �BAD � �BCD 5. Substitution postulate.

AD > CDAB > CB

CBCA

BD > BDAD > CDBD

DEAE

PM

8. Statements Reasons1. , 1. Given.

2. 2. Subtraction postulate.

3. ,3. Partition postulate.

4. 4. Substitution postulate.5. �CDE � �CED 5. Isosceles triangle

theorem.

9. Statements Reasons1. 1. Given.2. �DAB � �DBA 2. Given.3. �CAB � �CBA 3. Isosceles triangle

theorem.4. �CAB � �DAB 4. Subtraction postulate.

� �CBA � �DBA5. �CAB � �CAD 5. Partition postulate.

� �DAB,�CBA � �CBD

� �DBA6. �CAD � �CBD 6. Substitution postulate.

10. a. Statements Reasons1. 1. Given.2. �A � �B 2. Isosceles triangle

theorem.3. AC � BC 3. Definition of

congruentsegments.

4. D is the midpoint 4. Given.of .E is the midpoint of .F is the midpoint of .


, midpoint.

6. 6. Halves ofcongruent segmentsare congruent.

7. �ADF � �BEF 7. SAS (steps 5, 2, 6).

AD > BEAF > BF

BE 5 12BC

AD 5 12AC

AB

BC

AC

AC > BC

AC > BC

CD > CECB > CE 1 EBCA > CD 1 DA> CB 2 EBAC 2 DADA > EBAC > CB

260

b. Statements Reasons1–7. �ADF � �BEF 1–7. Steps from

part a.8. 8. Corresponding


9. �ADF is isosceles. 9. Definition of anisosceles triangle.

11. Statements Reasons1. AB � AC 1. Given.2. 2. Definition of

congruent segments3. �B � �C 3. Isosceles triangle

theorem.4. BG � EC 4. Given.5. EG � EG 5. Reflexive property of

equality.6. BG � EG 6. Subtraction postulate.

� EC � EG7. BG � BE � EG, 7. Partition postulate.

EC � EG � GC8. BE � GC 8. Substitution postulate

(steps 6, 7).9. 9. Definition of

congruent segments.10. , 10. Given.

11. �BED � �CGF 11. If two lines areperpendicular, thenthey intersect to formcongruent adjacentangles.

12. �BED � �CGF 12. ASA (steps 3, 9, 11).13. 13. Corresponding parts

of congruent trianglesare congruent.

12. Statements Reasons1. 1. Assumption.2. �A � �D 2. Isosceles triangle

theorem.3. 3. Given.4. �AEB � �DEC 4. SAS.5. 5. Corresponding parts of


EB > EC

AB > CD

AE > DE

BD > CF

CG ' GFBE ' DE

BE > GC

AB > AC

DF > EF

(Cont.)

Statements Reasons6. is not 6. Given.

congruent to .7. is not 7. Contradiction

congruent to . (steps 5, 6).

Applying Skills

13. Let �ABC be an isosceles triangle with vertexangle B and . Draw , the anglebisector of B. By the definition of angle bisector,�ABD � �CBD. Also, by the reflexive property of congruence. Therefore,�ABD � �CBD by SAS. Since correspondingparts of congruent triangles are congruent,�A � �C.

14. Let �ABC be an equilateral triangle. Bydefinition, . Let D, E, and F bethe midpoints of , , and , respectively.The midpoint divides a segment in half. Sincehalves of congruent segments are congruent,

.Every equilateral triangle is equiangular,so �A � �B � �C. By SAS, �AEF � �BDE� �CFD. Since corresponding parts ofcongruent triangles are congruent,

.

15. We are given that BC � DC, so . Bythe isosceles triangle theorem, �CBD � �CDB.Since we are given , �CBD and �FBC are supplements and �CDB and �GDC aresupplements. Supplements of congruent anglesare congruent. Therefore, �FBC � �GDC.

16. The contrapositive of the isosceles triangletheorem states if two angles of a triangle are notcongruent, then the sides opposite these anglesare not congruent. In �PQR, m�R � m�Q.Then �R is not congruent to �Q, so is notcongruent to and PQ � PR.

5-4 Using Two Pairs of Congruent Triangles(pages 187–188)


1. No. We cannot prove that �BCE � �BDE. It isgiven that , and we know that

by the reflexive property ofcongruence, but this is not sufficient to show thatthe triangles are congruent.

2. Yes. Consider �ABC � �DEF with medians and to sides and , respectively. ThisDEABFH

CG

EB > EBCB > DB

PRPQ

‹FBDG

›

BC > DC

DE > EF > FD

BD > CD > BE > AE > AF > CF

ABCABCAB > BC > CA

BD > BD

BDAB > CB

DEAE

ECEB

261

creates triangles �ACG and �DFH. Sincecorresponding parts of congruent triangles arecongruent, , and �A � �D. Themedians cut and in half. Halves ofcongruent segments are congruent. Therefore,

and �ACG � �DFH by SAS.Therefore, since corresponding partsof congruent triangles are congruent.


1. �ABC � �DEF 1. Given.2. , 2. Corresponding parts of

�A � �D, congruent triangles are congruent.

3. AC � DF, 3. Definition of congruentAB � DE segments.

4. M is the midpoint 4. Given.of , N is the midpoint of .


6. 6. Halves of congruentsegments arecongruent.

7. �AMC � �DNF 7. SAS (steps 2, 6).

4. Statements Reasons1. �ABC � �DEF 1. Given.2. , 2. Corresponding

�A � �D, parts of congruent �ACB � �DFE triangles are

congruent.3. bisects �ACB, 3. Given.

bisects �DFE.4. m�ACG 4. Definition of angle

� , bisector.

m�DFH�

5. �ACG � �DFH 5. Halves of congruentangles are congruent.

6. �ACG � �DFH 6. ASA (steps 2, 5).7. 7. Corresponding parts


CG > FH

12m/DFE

12m/ACB

FHCG

AC > DF

AM > DN

DN 5 12DE

AM 5 12AB

DEAB

AB > DE

AC > DF

CG > FHAG > DH

DEABAC > DF


bisect each other.2. E is the midpoint 2. Definition of bisector.

of and of .


4. �AEB � �CED 4. Vertical angles arecongruent.

5. �ABE � �CED 5. SAS.6. �A � �C 6. Corresponding parts


7. �AEG � �CEF 7. Vertical angles arecongruent.

8. �AEG � �CEF 8. ASA (steps 6, 3, 7).9. 9. Corresponding parts


10. E is the midpoint 10. Definition of of . midpoint.

6. Statements Reasons1. �AME � �BMF 1. Given.2. , 2. Corresponding parts

�AEM � �BFM of congruent trianglesare congruent.

3. �AEM and �AED 3. If two angles form a are supplements, linear pair, then they�BFM and �BFC are supplementary.are supplements.

4. �AED � �BFC 4. Supplements ofcongruent angles arecongruent.

5. 5. Given.6. �AED � �BFC 6. SAS (steps 2, 4, 5).7. 7. Corresponding parts


7. Statements Reasons1. 1. Given.

2. bisects �CBA. 2. Given.3. �CBD � �ABD 3. Definition of angle

bisector.4. 4. Reflexive property.5. �CBD � �ABD 5. SAS.6. �CDB � �CDA 6. Corresponding parts


7. bisects �CDA. 7. Definition of anglebisector.

DBh

BD > BD

BDhBC > BA

AD > BC

DE > CF

AE > BF

FEG

FE > EG

DE > BEAE > CDDEB

AEC

DEBAEC

262

8. Statements Reasons1. RP � RQ 1. Given.2. 2. Definition of

congruent segments.3. �RPT � �RQT 3. Isosceles triangle

theorem.4. 4. Reflexive property.5. �RPT � �RQT 5. SAS.6. �RTP � �RTQ 6. Corresponding parts


7. 7. If two lines intersectto form congruentadjacent angles, thenthey areperpendicular.

Applying Skills

9. We are given that in quadrilateral ABCD, AB �

CD and BC � DA. It follows that and. Also, by the reflexive

property of congruence. �ABD � CDB by SSS.Corresponding parts of congruent triangles arecongruent, so �ABD � �CDB. Since M is the midpoint of , . �BME � �DMFbecause vertical angles are congruent. Therefore,�BME � �DMF by ASA. Corresponding partsof congruent triangle are congruent, so

. By definition, M is the midpoint of, so bisects at M.

10. a. We are given that l intersects at M, and Pand S are points on l on the same side of .Also, PA � PB and SA � SB, so and . by the reflexive property of congruence, so �PAS � �PBS bySSS. Corresponding parts of congruenttriangles are congruent, so �APM � �BPM.

by the reflexive property of congruence, so �PAM � �PBM by SAS.Corresponding parts of congruent trianglesare congruent, so �PMA � �PMB. If twolines intersect to form congruent adjacent angles, then they are congruent, so l ⊥ .

b. Same as part a.c. Both methods require proving two pairs of

triangles congruent and then usingcorresponding angles to showperpendicularity.

AB

PM > PM

PS > PSSA > SBPA > PB

ABAB

EMFBMDEMFEM > MF

BM > DMBD

BD > BDBC > DAAB > CD

RT ' PQ

RT > RT

RP > RQ

5-5 Proving Overlapping TrianglesCongruent (pages 189–190)


1. Yes. By the reflexive property of congruence,�B � �B, , and �C � �C, so �DBC � �ECB by SAS. Then the medians tothe legs, and , are corresponding parts ofcongruent triangles and, therefore, congruent.

2. The triangle must be equilateral.Developing Skills

3. Statements Reasons1. , , 1. Given.

and �A and �B are right angles.

2. �A � �B 2. Right angles arecongruent.

3. 3. Reflexive property.4. 4. Addition property.5. , 5. Partition postulate.


7. �DAF � �CBE 7. SAS (steps 1, 2, 6).8. 8. Corresponding


4. Statements Reasons1. 1. Given.2. �S � �S 2. Reflexive property.3. 3. Given.4. 4. Subtraction

postulate.5. 5. Partition postulate.


7. �SRQ � �STP 7. SAS (steps 1, 2, 6).8. �R � �T 8. Corresponding



2. �DAB and �CBA 2. Perpendicular lines are right angles. intersect to form

right angles.

CB ' ABDA ' ABDA > CB

SP > SQSQT > SQ 1 QTSPR > SP 1 PR> SQT 2 QTSPR 2 PRPR > QT

SPR > SQT

DF > CE

AF > BEEB > EF 1 FBAF > AE 1 EFAE 1 EF > EF 1 FBEF > EF

AE > FBDA > CB

BECD

BC > BC

263

Statements Reasons3. �DAB � �CBA 3. Right angles are

congruent.4. 4. Reflexive property.5. �DAB � �CBA 5. SAS (steps 1, 3, 4).6. 6. Corresponding


6. Statements Reasons1. �BAE � �CBF, 1. Given.

�BCE � �CDF,

2. 2. Reflexive property.3. 3. Addition postulate.4. 4. Partition postulate.


6. �ACE � �BDF 6. ASA (steps 1, 5).7. , �E � �F 7. Corresponding


7. Statements Reasons1. M is the midpoint of 1. Given.

, N is the midpoint of .

2. , 2. Definition ofmidpoint.

3. 3. Given.4. 4. Substitution

postulate.5. �R � �S 5. Isosceles triangle

theorem.6. 6. Reflexive property.7. �RSM � �SRN 7. SAS (steps 4, 5, 6).8. 8. Corresponding


RN > SM

RS > RS

MR > NSTM > TN

TN > NSTM > MRTS

TR

AE > BF

AC > BDBD > BC 1 CDAC > AB 1 BCAB 1 BC > BC 1 CDBC > BCAB > CD

AC > BD

AB > AB

8. Statements Reasons1. , 1. Given.2. 2. Addition postulate.



5. �B � �B 5. Reflexive property.6. �ABE � �CBD 6. SAS (steps 1, 5, 4).7. �BDC � �BEA 7. Corresponding


8. �ADC � �CEA 8. Supplements ofcongruent anglesare congruent.

Applying Skills

9. Let �ABC be an isosceles triangle with vertexangle B and . Let the angle bisectors ofthe base angles be and . By the isosceles triangle theorem, �BAC � �BCA. Since anangle bisector splits an angle in half, and halvesof congruent angles are congruent,�BAE � �BCD. By the reflexive property ofcongruence, �B � �B. Therefore,�ABE � �CBD by ASA. Becausecorresponding parts of congruent triangles are congruent, the angle bisectors, and , arecongruent.

10. Let �ABC be an isosceles triangle with vertex B,with D, E, and F the midpoints of , , and

, respectively. By the definition of midpoint,. Since �ABC is isosceles, .

The midpoints D and E split these sides in half.Since halves of congruent segments arecongruent, so . By the isosceles triangletheorem, �A � �C. Then �ADF � �CEF bySAS. Corresponding parts of congruent trianglesare congruent, so . Since and are congruent sides of �DEF, the triangle isisosceles.

11. We will use an indirect proof. Let be themedian and the altitude to side in scalenetriangle �ABC. Then D is the midpoint of , so

. Also, �ADB and �ADC are rightangles and therefore congruent. By the reflexiveproperty of congruence, . Therefore,�ABD � �ACD by SAS. Corresponding parts

AD > AD

BD > CDBC

BCAD

EFDFDF > EF

AD > CE

AB > BCAF > CFCA

BCAB

CDAE

CDAEAB > CB

AB > CBCB > CE 1 EBAB > AD 1 DB5 CE 1 EBAD 1 DB

DB > EBAD > CE

264

of congruent triangles are congruent, so �B � �C. However, that contradicts thestatement that �ABC is scalene, so is not an altitude.

5-6 Perpendicular Bisector of a LineSegment (pages 195–196)

Writing About Mathematics1. Perpendicular lines are two lines that intersect to

form right angles. If two lines intersect to formcongruent adjacent angles, then they areperpendicular. If two points are each equidistantfrom the endpoints of a line segment, then thetwo points determine the perpendicular bisectorof a line.

2. Corollary 5.1b and Example 1 are converses ofeach other. Corollary 5.1b can be rewritten as, “Ifa triangle is isosceles, then the vertex lies on theperpendicular bisector of the base.” Example 1can be rewritten as, “If a point lies on theperpendicular bisector of a line segment, then thepoint is the vertex of an isosceles triangle formedby the endpoints of the line segment and thepoint.”


1. is the ⊥ bisector 1. Given.of .

2. �RSA and �RSB 2. Definition of are right angles. perpendicular lines.

3. �RSA � �RSB 3. Right angles arecongruent.

4. S is the midpoint of 4. Definition of . bisector.

5. 5. Definition ofmidpoint.

6. 6. Reflexive property.7. �RSA � �RSB 7. SAS (steps 5, 3, 6).8. �ARS � �BRS 8. Corresponding parts

of congruent anglesare congruent.

4. Statements Reasons1. PR � PS, QR � QS 1. Given.2. is the ⊥ bisector 2. If two points are

of . each equidistant from the endpointsof a line segment,then the pointsdetermine the ⊥bisector of the linesegment.

RSPQ

RS > RS

AS > BSASB

ASBRS

AD

(Cont.)

Statements Reasons3. RT � ST 3. If a point is on the

perpendicular bisectorof a line segment, thenit is equidistant fromthe endpoints of theline segment.

5. Statements Reasons1. AB � BC 1. Given.

� CD � DA2. is the ⊥ 2. If two points are

bisector of , each equidistant is the ⊥ from the endpoints

bisector of . of a line segment, thenthe points determinethe ⊥ bisector of theline segment.

6. Statements Reasons1. �ACE � �BCE 1. Given.2. 2. Reflexive property.3. �AED � �BED 3. Given.4. �AEC � �BEC 4. Supplements of


5. �AEC � �BEC 5. ASA (steps 1, 2, 4).6. , 6. Corresponding parts


7. AC � BC, 7. Definition of AE � BE congruent segments.

8. is the ⊥ 8. If two points are bisector of . each equidistant

from the endpoints ofa line segment, thenthe points determinethe ⊥ bisector of theline segment.

Applying Skills

7. Let intersect at M. We are given that PA � PB and TA � TB, so and

. By the reflexive property ofcongruence, .Therefore, �APT � �BPTby SSS. Corresponding parts of congruenttriangles are congruent, so �APM � �BPM.

by the reflexive property ofcongruence, so �APM � �BPM by SAS.Therefore, since corresponding partsof congruent triangles are congruent. By

definition, M is the midpoint of , so is a PTg

AB

AM > MB

PM > PM

PT > PTTA > TB

PA > PBABPT

g

ADBCED

AE > BEAC > BC

CE > CE

ACBD

BDAC

265

bisector.Also, adjacent angles �AMP and �BMPare congruent because they are correspondingparts of congruent triangles. Therefore,

.

8. Let bisect �B in �ABC. Then �ABD � �CBD. Also, let . Then �ADB and �CDB are right angles andcongruent. By the reflexive property of congruence, . �ADB � �CDB byASA. Corresponding parts of congruent trianglesare congruent, so . Therefore, �ABCis isosceles.

9. Since the line forms a linear pair of angles withthe side of the triangle, the sum of the anglemeasures should be 180.

However, � 27 � 69 and � 15 � 111.Therefore, these two angles are not right angles,and so the line through the vertex and theopposite side are not perpendicular.

5-7 Basic Constructions (page 202–203)Writing About Mathematics

1. The perpendicular bisector to a side of a triangleis an altitude if it goes through the oppositevertex. The altitude from a vertex is aperpendicular bisector if it goes through themidpoint of the opposite side. A perpendicularbisector need not contain the opposite vertex andan altitude need not contain the midpoint of theopposite side.

2. In both constructions, a line is drawn through twopoints equidistant from the endpoints of asegment. However, in Construction 3, theendpoints of the segment are given and thepoints equidistant from them are constructed,while in Construction 6, a point is given and theendpoints of the segment and a second pointequidistant from those endpoints are constructed.

Developing Skills3. a. Use Construction 1.

b. Use Construction 1. Then, on the ray whoseendpoint is newly constructed, repeatConstruction 1 to create a second, non-overlapping segment.

c. Use Construction 3.d. Follow part b. Then use Construction 3 to

bisect the second line segment constructed.

32(84)1

2(84)a 5 84

A 12a 1 27 B 1 A 3

2a 2 15 B 5 180

AB > CB

BD > BD

BD ' ABBD

AB ' PTg

4. a. Use Construction 2.b. Use Construction 2. Then, on the newly

constructed side, repeat Construction 2 tocreate a second, adjacent angle.

c. Use Construction 4.d. Follow part b twice. Then, use Construction 4

to bisect the most recently constructed angle.5. a. Use Construction 1 with .

b. Use Construction 1 with . Open the compass to a radius equal to BC. With thepoint on an endpoint of the newly constructedsegment, draw an arc above the segment.Open the compass to a radius equal to CD.With the point on the other endpoint of thenewly constructed segment, draw a second arcabove the segment intersecting the first arc.Draw a segment from each endpoint to thisintersection.

c. Follow part b, substituting BC for CD.d. Follow part b using only .

6. a. Use Construction 4.b. Use Construction 4. Then, use Construction 4

to bisect one of the congruent halves.c. Same as part b.

7. a. Use Construction 3 with . Draw a segmentfrom C to the newly constructed midpoint.

b. Use Construction 6 with and point C.c. Use Construction 6 with and point A.d. Use Construction 4 with �A.

8. a. Answers will vary.b. Yes. Place the point of the compass at the

point at which the perpendicular bisectorsintersect. Place the pencil on any vertex of thetriangle and draw a circle. Since the point ofintersection is equidistant from each of thevertices, all three vertices lie on the circle.

Hands-On Activity

a. Construct a segment congruent to .Construct lines perpendicular to this segmentat each of the endpoints. Construct a segment congruent to at each endpoint on the perpendiculars, in the same direction. Draw asegment connecting the endpoints not on thefirst segment.

b. Construct a segment congruent to .Construct intersecting arcs with radius ABcentered at each of the endpoints. Draw asegment from each endpoint to thisintersection.

c. Construct a segment congruent to .Construct a line perpendicular to this segmentat an endpoint. Bisect the angle formed.

AB

AB

AB

AB

BCAB

AB

CD

ABBC

266

d. Follow part b. Bisect one of the angles of thetriangle.

e. Construct a segment congruent to . Constructa point C not on the line containing the segment.Connect the endpoints of the segment to point Cto form a triangle. Construct the perpendicularbisector of each side of the triangle. Construct acircle whose center is the intersection of theperpendicular bisectors and the radius is thedistance to any of the vertices of the triangle.

Review Exercises (pages 204–205)1. x � 60, y � 30 2. m�PQR � 50 3. 464. m�PQS � 65, m�SQR � 115

5. Statements Reasons1. �ABC is equilateral. 1. Given.2. 2. Definition of

equilateral triangle.3. D lies on the 3. Given.

perpendicular bisectors of , ,and .

4. DA � DB = DC 4.A point on theperpendicular bisectorof a segment is equi-distant from the end-points of that segment.

5. 5. Segments with equalmeasure arecongruent.

6. �ADC � �BDC 6. SSS (steps 2, 5).� �ADB

7. �ADC, �BDC, and 7. Definition of �ADB are isosceles. isosceles triangle.

6. Statements Reasons1. 1. Assumption.2. is a median of 2. Given.

�ABC.3. 3. The median from the

vertex of an isoscelestriangle is perpen-dicular to the base.

4. is the altitude 4. Definition of altitude.to side .

5. is not the 5. Given.altitude to side .

6. is not 6. Contradiction congruent to . (steps 1, 5).BCAC

ABCD

ABCD

CD ' AB

CDAC > BC

DA > DB > DC

CABCAB

AB > BC > CD

AB

7. Statements Reasons1. �ABC and �ABD 1. Given.

are isosceles with as base.

2. , 2. An isosceles triangle has two congruent sides.

3. AC � BC, 3. Definition of AD � BD congruent segments.

4. is the perpen- 4. If two points are dicular bisector each equidistant of . from the endpoints of a

line segment, then thetwo points determinethe ⊥ bisector of theline segment.

8. Statements Reasons1. is the median 1. Given.

to .2. D is the midpoint 2. Definition of median.

of .3. 3. Definition of midpoint.4. 4. Given.5. 5. Transitive property.6. �A � �ACD, 6. Isosceles triangle

�B � �BCD theorem.7. m�A � m�ACD, 7. Definition of

m�B � m�BCD congruent angles.8. m�ACB 8. Partition postulate.

� m�ACD� m�BCD

9. m�ACB 9. Substitution postulate.� m�A � m�B

9. a. Use Construction 5.b. Use Construction 4 to bisect �ADC.c. m�EDC � 45d. m�EDB � 135

10. a. Results will vary.

b. Draw opposite . Use Construction 6 to

construct a line through P perpendicular to

.

Exploration (page 205)1. Statements Reasons

1. 1. Given.

2. 2. Addition postulate.> MN 1 NBAL 1 LM> MN > NBAL > LM

SQRg

QRh

QSh

AD > CDCD > DBAD > DB

AB

ABCD

AB

CDg

AD > BDAC > BCAB

267



5. is an altitude 5. Given.of �CLM.

6. 6. Definition of altitude.7. �CMA � �CMB 7. If two lines are


8. 8. Reflexive property.9. �ACM � �BCM 9. SAS (steps 4, 7, 8).

10. 10. Corresponding partsof congruent trianglesare congruent.

2–5. Results will vary.


1. 3 2. 4 3. 2 4. 15. 2 6. 1 7. 3 8. 49. 3 10. 4

Part II11. m�PQS � 67.2, m�SQR � 112.8

m�PQS � 3(16.4) � 18 � 67.2m�SQR � 7(16.4) � 2 � 112.8

12. 5(4 � x) � 32 � x Given.20 � 5x � 32 � x Distributive

property ofmultiplication.

20 � 5x � x � 32 � x � x Addition postulate.20 � 6x � 32 � 0 Additive inverse.20 � 6x � 32 Additive identity.

6x � 12 Subtractionpostulate.

x � 2 Division postulate.

a 5 16.4

10a 5 164

10a 1 16 5 180

3a 1 18 1 7a 2 2 5 180

CA > CB

CM > CM

CM ' AB

CM

AM > MBMB > MN 1 NBAM > AL 1 LM


1. 1. Given.2. �ABF is the 2. Given.

supplement of �A.3. �FBD is the 3. If two angles form a

supplement of linear pair, then they �ABF. are supplementary.

4. �A � �FBD 4. If two angles aresupplements of thesame angle, then theyare congruent.

5. 5. Given.6. 6. Addition postulate.



9. �AEC � �BFD 9. SAS (steps 1, 4, 9).

14. Statements Reasons1. 1. Given.2. is the bisector 2. Given.

of �ABC.3. �CBD � �ABD 3. Definition of angle

bisector.4. 4. Reflexive property.5. �CBE � �ABE 5. SAS.6. 6. Corresponding parts


Part IV15. Statements Reasons

1. Point P is not on 1. Given..

2. PB � PC 2. Given.

‹ABCD

›

AE > CE

BE > BE

BDhAB > CB

AC > BDBD > BC 1 CDAC > AB 1 BC> BC 1 CD

AB 1 BCAB > CD

AE > BF

268

Statements Reasons3. 3. Segments with equal

measures arecongruent.

4. �CBP � �BCP 4. Isosceles triangletheorem.

5. �ABP and �CBP 5. If two angles form aare supplements. linear pair, then they�DCP and �BCP are supplementary.are supplements.

6. �ABP � �DCP 6. If two angles arecongruent, then theirsupplements arecongruent.

16. Statements Reasons1. and are ⊥ 1. Given.

bisectors of each other.

2. Let M be the 2. Construction.intersection of and .

3. M is the midpoint of 3. Definition of bisector.and of .


5. �AMB � �CMB 5. If two lines are � �CMD � �AMD perpendicular, then

they intersect to formcongruent adjacentangles.

6. �AMB � �CMB 6. SAS.� �CMD � �AMD

7. 7. Corresponding parts of congruent trianglesare congruent.

8. AB � BC 8. Definition of � CD � DA congruent segments.

> CD > DAAB > BC

BM > DMAM > CM

BDAC

BDAC

BDAC

PB > PC

Chapter 6.Transformations and the Coordinate Plane

6-1 The Coordinates of a Point in a Plane(pages 213–214)


1. Since is a horizontal segment, must be a vertical segment with BC � AC � 7. Therefore,the coordinates of B are (5, 3) or (5, �11).

2. The area of the polygon is equal to the sum ofthe area of two triangles EFD and GFD. If O isthe origin, then the area of the triangle EFD is

BCAC

= . The area of a

triangle GFD is .Therefore, the area of a polygon FEDG equals 22.5 square units.

Developing SkillsIn 3–12, part a, answers will be graphs.

3. b. 14 sq units 4. b. 20 sq units5. b. 20 sq units 6. b. 16 sq units7. b. 21 sq units 8. b. 24 sq units

12FD 3 GO 5 1

2 3 5 3 4 5 10

12 3 5 3 5 5 12.51

2FD 3 EO

9. b. 20 sq units 10. b. 6 sq units11. b. 10 sq units 12. b. 16 sq units13. D(1, 4)14. R(2, �7), S(�1, �7) or R(2, �1), S(�1, �1)15. a. Graph 16. a. Graph

b. 24 sq units b. 6 sq units

6-2 Line Reflections (pages 220–221)Writing About Mathematics

1. Every real number corresponds to exactly onepoint on the number line. Every point on the realnumber line corresponds to exactly one number.Hence, there is a one-to-one correspondencebetween points on the number line and the realnumbers.

2. No. Two different points can have the sameimage. For example, points (1, 3) and (2, 3) bothmap to (2, 3).

Developing Skills3. Since distance is preserved under a line

reflection and PQ � QR, the images of thesesegments also have equal length, that is,P�Q� � Q�R� and . Therefore,�P�Q�R� is isosceles.

4. Since angle measure is preserved under a linereflection and m�N � 90, m�N� � 90, and�L�M�N� is a right triangle.

5. Yes. Since distance is preserved under a linereflection, AB � A�B�, BC � B�C�, and CA � C�A� so , , and

. By SSS, �ABC � �A�B�C�.6. Since , �ACB is a right angle, and

m�ACB � 90. Since angle measure is preservedunder a line reflection, m�A�C�B� � 90.Therefore, and intersect to form rightangles, and .

7. Yes. Midpoint is preserved under a linereflection.

8. Yes. Collinearity is preserved under linereflection.

9. Point D; a point is on the line reflection if it is itsown image.

10. Since the triangle has line symmetry with respectto the altitude from S, RS � ST and ,and the triangle is isosceles. However, since itdoes not have line symmetry with respect to the altitude from R, RS � RT and is notcongruent to , and the triangle is not equilateral. Therefore, �RST is an isoscelestriangle with exactly two congruent sides.

RTRS

RS > ST

ArCr ' BrCrBrCrArCr

AC ' BCCA > CrAr

BC > BrCrAB > ArBr

PrQr > QrRr

269

Applying Skills

11. (1) Let O be the point where intersects k.(2) By the definition of a line reflection, k is the

perpendicular bisector of and . LetM be the point where and k intersectand N be the point where and kintersect.

(3) Then, and at M and N, respectively. Therefore,m�AMO � m�A�MO � 90 and m�BMO � m�B�MO � 90. Also,

and by the reflexive property of congruence.

(4) Thus, by SAS, �AMO � �A�MO and�BNO � �B�NO.

(5) Since corresponding parts of congruenttriangles are congruent, and

, and so AO � A�O and OB � OB�.

(6) By the partition postulate,A�B� � A�O � OB� and AB � AO � OB.By the substitution postulate,A�B� � AO � OB � AB. Therefore,

.

12. The lines of symmetry of a rhombus are itsdiagonals. Therefore, the lines of symmetry of thebaseball diamond would be from home plate andsecond base and from first base and third base.

13. E, D, C14.

6-3 Line Reflections in the CoordinatePlane (pages 225–227)

Writing About Mathematics1. The distance is the absolute value of the

difference of the y-coordinates.

�b – a� � 1�b � a� � ��1� � �b � a�

� �(�1)(b � a)� � �a � b�Thus, �b � a� and �a – b� have the same value.

2. Second quadrantDeveloping SkillsIn 3–17, part a is a graph of the point indicated in b.

3. b. (2, �5) 4. b. (1, �3) 5. b. (�2, �3)6. b. (2, 4) 7. b. (0, �2) 8. b. (�3, 5)9. b. (�1, 4) 10. b. (�2, �3) 11. b. (2, 3)

H O X

AB > ArBr

OB > OBrAO > ArO

BN > BrNAM > ArM

BBr ' kAAr ' k

BBrAAr

BBrAAr

AB

12. b. (1, 0) 13. b. (5, 3) 14. b. (5, �3)15. b. (�2, 4) 16. b. (�5, �1) 17. b. (2, 2)Applying Skills18. (1) The line of reflection, the x-axis, is a

horizontal line. P(a, b) and P�(a, �b) have the same x-coordinates, so is a segment of a vertical line. Every horizontal isperpendicular to every vertical line.Therefore, the x-axis is perpendicular to .

(2) Let Q be the point at which intersects the x-axis, so the y-coordinate of Q is 0.The length of a vertical line segment is theabsolute value of the difference of the y-coordinates of the endpoints. Therefore,

and . Since PQ � P�Q, Q is the midpoint

of and the x-axis bisects .(3) Steps 1 and 2 prove that the x-axis is the

perpendicular bisector of . By the definition of a line reflection, P�(a, �b) is theimage of P(a, b).

19. a. A line is a line of symmetry of a figure if thefigure is its own image under a reflection inthat line. Since collinearity is preserved undera line reflection, it suffices to show that eachvertex of ABCD is the image of anothervertex under .

Therefore, y � x is a line of symmetry.b. Under :

Therefore, the y-axis is a line symmetry.20. Under :

Since F and G do not map to one of the verticesof EFGH, the y-axis is not a line of symmetry.

21. y � 0 and x � 3

H(0, 3) S H(0, 3)

G(5, 3) S Gr(25, 3)

F(5, 23) S Fr(25, 23)

E(0, 23) S E(0, 23)

ry-axis

D(0, 4) S Dr(0, 4) 5 D

C(4, 0) S Cr(24, 0) 5 A

B(0, 24) S Br(0, 24) 5 B

A(24, 0) S Ar(4, 0) 5 C

ry-axis

D(0, 4) S Dr(4, 0) 5 C

C(4, 0) S Cr(0, 4) 5 D

B(0, 24) S Br(24, 0) 5 A

A(24, 0) S Ar(0, 24) 5 B

ry5x

PPr

PPrPPr5 Zb Z

PrQ 5 Z2b 2 0 Z PQ 5 Zb 2 0 Z 5 Zb Z

PPrPPr

PPr

270

Hands-On Activity 1a.

,

b.

, ,

c.

, ,

d.

, , , Sr(9, 22)Rr(3, 24)Qr(5, 0)Pr(7, 26)

Q�

S

Q

O

R�

P

1

1R

P�

S�

U

T

q

Lr(2, 25)Kr(23, 25)Jr(0, 24)

y

x

J�

p

JK

M

N

L�K�

L

O 11

Gr(22, 22)Fr(22, 3)Er(1, 0)

y

O

1 x

1

E�

l F G

E

I

HF�

G�

Br(2, 6)Ar(0, 0)

y

O

1 x1A�

A

C

BDB� k

Hands-On Activity 2a.

b.

c.

d.

6-4 Point Reflections in the CoordinatePlane (pages 231–232)

Writing About Mathematics1. Yes. Since a point reflection is a transformation

and a transformation is a one-to-onecorrespondence, the endpoints of the segmentare in a one-to-one correspondence with theendpoints of the image of the segment. Thus,

, , , and. Since collinearity is preserved, the

image of is and the image of is .AB

ArBrArBrABRP(Br) 5 B

RP(B) 5 BrRP(Ar) 5 ARP(A) 5 Ar

A�

x

A

B

O

C�

1

1

C

B�

y

C�

x

C

A

O 1

1

BA�

y

B�

A

B B�

A�

A

A�

271

2. Not necessarily. For example, let P be the origin, l be the line y � x, and m be the x-axis. Then while

and . The two images are not

the same.Developing Skills

3. (�1, �5) 4. (2, �4) 5. (1, 0)6. (0, �3) 7. (�6, �6) 8. (1, 5)

Applying Skills9. Suppose that A � P. Then A is its own image

under , that is, A� � A. Let B� be the image ofB under . By the definition of a point reflection, P is the midpoint of , and so PB � PB�. By the substitution postulate,AB � A�B�. Therefore, distance is preservedunder a point reflection.

10. Given: �ABC and �A�B�C�, the image of�ABC under .

Prove: m�ABC � m�A�B�C�Proof: Consider �A�B�C�, the image of �ABCunder a reflection in point P. Since distance ispreserved, AB � A�B�, BC � B�C�, and CA � C�A�, so , , and

. Thus, �A�B�C� � �ABC by SSS.Since corresponding parts of congruent trianglesare congruent, �ABC � �A�B�C�. Therefore,m�ABC � m�A�B�C�, and angle measure ispreserved.

11. Given: and A�, D�, and B�, the images of A,D, and B, respectively, under .

Prove: D� is a point on .Proof: Since A, D, and C are collinear, AD � DB� AB. Since distance is preserved, AD � A�D�,DB � D�B�, and AB � A�B�. By substitution,A�D� � D�C� � A�B�. If D� were not on ,then A�D� � D�B� A�B� because a straight lineis the shortest distance between two points. ButA�D� � D�C� � A�B�. Therefore, A�, D�, and B�

are collinear, and collinearity is preserved undera point reflection.

12. Given: M, the midpoint of , and A�, M�, andB�, the images of A, M, and B,respectively, under .

Prove: M� is the midpoint of .Proof: Since distance is preserved under a pointreflection, A�M� � AM, and M�B� � MC. Since

ArBrRP

AB

ArBr

ArBrRP

ADB

CA > CrArBC > BrCrAB > ArBr

RP

BBrRP

RP

rx-axis(2b, 2a) 5 (2b, a)ry5x(2a, 2b) 5 (2b, 2a)

RO(a,b) 5 (2a,2b)

M is the midpoint of , AM � MB, and, by thesubstitution postulate, A�M� � M�B�. Therefore,M� is the midpoint of , and midpoint ispreserved under a point reflection.

13. H, I, O, X, Z14. Under :

Each vertex maps to another vertex, so the figureis its own image under a reflection in the origin.Therefore, it has point symmetry.

15. a. (2, �6) b. (1, �6)c. No. For example, while

. The two images are notthe same.

16. a. (4, �8) b. (7, �2)c. Yes. Point reflection preserves collinearity.

6-5 Translations in the Coordinate Plane(pages 236–238)

Writing About Mathematics1. A point P(x, y) is fixed if its coordinates are the

same before and after the translation. However,under a translation , the coordinates of its image P� are (x � a, y � b). Therefore, P � P� ifand only if a � b � 0. A translation other than

does not have any fixed points.2. Let A have the coordinates (x, y). Then A� is

ry-axis(x, y) � (�x, y). A is a reflection in the linex � 3, which is the same as translating �3 units inthe x-direction, reflecting in the y-axis, and thentranslating 3 units back in the x-direction.

T–3,0(�x, y) � (�x � 3, y)

ry-axis(�x � 3, y) � (x � 3, y)

T3,0(x � 3, y) � (x � 6, y)

The coordinates of A are the same as those obtained under the translation

.a. Yes b. Yes c. Yes

Developing Skills3. a. P b. H c. O

d. L e. K4. a. Graph b. A�(6, �1), B�(11, 0), C�(5, 3)

c. Graph5. a. Graph b. Graph

c. A�(�6,5), B�(�2, 7), C�(�1, 4)

(x, y) S (x 1 6, y)

T0,0

Ta,b

r(2, 0)(3, 6) 5 (1, 26)rx-axis(3, 6) 5 (3, 26)

S(0, 5) S Q(0, 25)

R(5, 0) S P(25, 0)

Q(0, 25) S S(0, 5)

P(25, 0) S R(5, 0)

RO

ArBr

AB

272

6. a � 6, b � 07.8. a. Graph b.

c. B�(7, 5), C�(8, 9) d. Graph9. a. Graph b. A�(�1, �5), B(1, �5)

c. Graph d. none

Applying Skills10. a. Given: �ABC and , the image of

�ABC under .Prove: m�ABC �

Proof: Consider �A�B�C�, the image of�ABC under a translation . Since distanceis preserved, AB � A�B�, BC � B�C�, and CA � C�A� or , , and

. Thus, �A�B�C� � �ABC by SSS.Since corresponding parts of congruenttriangles are congruent, �ABC � �A�B�C�.Therefore, m�ABC � m�A�B�C�, and anglemeasure is preserved.

b. Given: and A�, D�, and B�, the images ofA, D, and B, respectively, under .

Prove: D� is a point on .Proof: Since A, D, and C are collinear,AD � DB � AB. Since distance is preserved,

, , and . Bysubstitution, A�D� � D�C� � A�B�. If D� werenot on , then because a straight line is the shortest distancebetween two points. But, A�D� � D�C� � A�B�.Therefore, A�, D�, and B� are collinear, andcollinearity is preserved under a translation.

c. Given: M, the midpoint of , and A�, M�, andB�, the images of A, M, and B,respectively, under .

Prove: M� is the midpoint of .Proof: Since distance is preserved,A�M� � AM, and M�B� � MB. Since M is the midpoint of , AM � MB, and, by thesubstitution postulate, A�M� � M�B�.Therefore, M� is the midpoint of , andmidpoint is preserved under a translation.

In 11–13, part a, answers will be graphs.

11. b. L�(4, 0), M�(0, 0), N�(0, 2)c. L(4, 0), M(8, 0), N(8, 2)d. P(4, 0), Q(8, 0), R(8, 2)e. The two triangles coincide.

ArBr

AB

ArBr

Ta,b

AB

ArDr1DrBr . ArBrArBr

AB 5 ArBrDB 5 DrBrAD 5 ArDr

ArBrTa,b

ADB

CA > CrArBC > BrCrAB > ArBr

Ta,b

m/ArBrCrTa,b

/ArBrCr

(x, y) S (x 1 3, y 1 4)(x, y) S (x, y 2 4)

12. b. D�(�2, 3), E�(1, 3), F�(0, 0)c. D(�2, 3), E(1, 3), F(6, 0)d. R(�2, 3), S(1, 3), T(6, 0)e. The two triangles coincide.

13. b. A�(1, �2), B�(5, �2), C�(4, �5)c. A(1, �4), B(5, �4), C(4, �1)d. Yes. a � 0, b � �6

14. Let d be the distance between the two lines. Forvertical lines, x � 2d and y � 0. For horizontallines, x � 0 and y � 2d.

15. a. Graphb. A�(1, 0), B�(5, 0), C�(4, �3)c. A(1, �4), B(5, �4), C(4, �1)d. ; yes

6-6 Rotations in the Coordinate Plane(pages 242–243)

Writing About Mathematics1. a. The image of P(x, y) under this rotation is

P�(�x, �y).b. Reflection in the originc. Yes, the two transformations are equivalent.

2. Reflection in the origin and 180° rotation aboutthe origin.

Developing Skills3. a. D 4. a. J

b. E b. Kc. F c. Ld. J d. De. K e. Ef. A f. Gg. B g. Hh. C h. I

5. a. A�(1, 2), B�(1, 5), C�(�4, 5), D�(�4, 2)b. Yesc. The midpoint of is (3.5, �1), the midpoint

of is (1, 3.5). Yes, the image of themidpoint is the midpoint of .

6. a. Q(�3, 1) b. R(�1, �3)c. S(3, �1) d. P�(3, �1)e. S and P� are the same since rotating a point

90° clockwise is equivalent to rotating thepoint 270° counterclockwise.

6-7 Glide Reflections (page 246)Writing About Mathematics

1. No. Under a glide reflection, a translation isperformed in the direction of the line ofreflection. A translation has no fixed points.Since the translation is along the line ofreflection, the fixed points of the line reflection

ArBrArBr

AB

T0,26

273

are moved by the translation. Therefore, theresulting transformation has no fixed points.

2. No. For example, consider the translation that is in a direction perpendicular to the x-axis.Then, the image of (x, y) under followed by

is (x, �y � 1). However, the image of (x, y)under followed by is (x, �y � 1). Thetwo images do not coincide.

Developing SkillsIn 3–8, part b, answers will be graphs.

3. a. A�(5, �1), B�( 9, �4), C�(7, �5)c. Yes, since the translation is in the direction of

the line of reflection.d.

4. a. A�(�1, �5), B�(�5, �2), C�(�3, �1)c. Yes, since the translation is in the direction of


5. a. A�(6, �1), B�(10, �4), C�(8, �5)c. Yes, since the translation is in the direction of


6. a. A�(�4, 4) , B�(�8, 7), C�(�6, 8)c. No, since the translation is not in the direction

of the line of reflection.d.

7. a. A�(4, 4), B�(7, 8), C�(8, 6)c. Yes, since the translation is in the direction of


8. a. A�(0, 3), B�(�3, 7), C�(�4, 5)c. No, since a rotation is not equivalent to a line

reflection.d.

Applying Skills9. A(�2, �5)

10. a. The line of reflection is x � 0. The translationis .

b. C�(0, �4)11. a. The line of reflection is the x-axis or y � 0.

The translation is .b. C�(0, �2)

12. a. Translations and glide reflectionsb. Rotations and point reflectionsc. Reflections

6-8 Dilations in the Coordinate Plane(pages 249–250)

Writing About Mathematics1. The length of each side of the image of the

triangle is k times the length of its corresponding

T24,0

T0,211

(a, b) S (2b 1 1, a 1 2)

(a, b) S (b 1 3, a 1 3)

(a, b) S (2a 2 3, b 1 3)

(a, b) S (a 1 5, 2b)

(a, b) S (2a, b 2 6)

(a, b) S (a 1 4, 2b)

rx-axisT0,1

T0,1

rx-axis

T0,1

side in the original triangle. A dilation is not anisometry since distance is not preserved.

2. The length of each side of the image of thetriangle is times the length of its correspondingside in the original triangle.

Developing Skills3. (12, 28) 4. (�16, 8)5. (8, 0) 6. (�4, 36)7. (10, 10) 8. (5, 50)9. (�15, 25) 10. (0, 20)

11. (3, �1) 12. (2, 0)13. 14.15. (�4, 20) 16. (4, 6)17. 18. (15, 3)19. (6, �2) 20. (6, 12)21. D2 followed by rx-axis 22. D7 followed by ry-axis23. followed by rx-axis 24. followed by ry-axis

Applying Skills25. a. A�(4, �4), B�(12, �4), C�(12, 12), D�(4, 12)

b. Since A� and B� have the same y-coordinate,is a horizontal segment. Similarly, is

a horizontal segment. Since A� and D� havethe same x-coordinate, is a verticalsegment. Similarly, is a vertical segment.Vertical and horizontal lines are perpendicular. Thus, the sides of meet to form right angles, and is arectangle.

26. Dk(a, b) � (ka, kb), RO(ka, kb) � (�ka, �kb)D

�k(a, b) � (�ka, �kb)The two images are the same. Therefore, Dkfollowed by a point reflection in the origin givesthe same result as D–k.

27. a, b. C�

x

C

A

O 22 B

A�

y

B�

ArBrCrDrArBrCrDr

BrCrArDr

CrDrArBr

D14

D12

(3, 212)

(5, 312)(23, 221

2)

1k

274

c. As we can see from the graph, AB is not equalto A�B�. Therefore, distance is not preservedunder the dilation.

Hands-on Activitya. Yes b. Yes c. Yes

6-9 Transformations As Functions (pages254–255)

Writing About Mathematics1. No. For example, consider a reflection in y � x

and the y-axis. while . The two images are

not the same.2. Tyler is correct. Every opposite isometry changes

the order or orientation from clockwise tocounterclockwise or vice versa. Therefore, everypair (even number) of opposite isometries resultsin the original order or orientation. Every non-pair (odd number) of opposite isometries resultsin the reverse order or orientation.

Developing Skills3. A�(�3, 2) 4. A�(�3, 2)5. A�(�3, �2) 6. A�(�2, �3)7. A�(�3, 7) 8. A�(�3, 2)9. A�(2, 3) 10. A�(�6, 4)

11. A�(�2, �4)12. A reflection in the y-axis13. RO14. a. D�(0, �2), E�(2, �5), F�(1, 1)

b. Glide reflectionc. Opposite isometry

Applying Skills

15. A(a, b) and A�(�b, �a) are points. Let B(a, �a)and C(�b, b) be points on the line of reflection y � �x. The distance BA � �b � (�a)� � �b � a�and the distance BA� � ��b � a� � �b � a�.Therefore, B is equidistant from A and A�.Similarly, the distance CA � �a � (�b)� � �a � b�and the distance CA� � ��a � b� � �a � b�.Therefore, C is equidistant from A and A�. If twopoints are each equidistant from the endpoints ofa line segment, then they lie on the perpendicular bisector of the line segment. Therefore, ,which is a subset of the line y � �x, is theperpendicular bisector of . By the definitionof a line reflection, the image of is

.Ar(2b,2a)A(a,b)

AAr

CB

ry5x + ry-axis(a, b) 5 (b, 2a)ry-axis + ry5x(a, b) 5 (2b, a)

16.

The images are the same. Therefore,.

17.

The images are the same. Therefore,.

Review Exercises (pages 257–258)1. x � �2 2. y � �43. a. Graph

b. 16.5 sq units4. a. P�(5, 3)

b. Every point on the x-axis5. a. P�(�5, 3) 6. a. P�(3, 5)

b. (0, 0) b. (0, 0)7. a. P�(5, 0) 8. a. P�(7, 1)

b. No fixed point b. No fixed point9. a. P�(�3, �5) 10. a. P�(3, 5)

b. (0, 0) b. (0, 0)11. a. P�(�6, �2)

b. No fixed point12.

13.

14. S 15. Z16. Line and glide reflections 17. Dilation18. a. Graph

b. A�(�3, �2) , B�(�3, �7) , C�(2, �7)c. A(3, �2), B(3, �7), C(�2, �7)d. Reflection in the x-axis

19. a. Graphb. R�(4, 1), S�(1, 1), T�(1, �2)c. R(4, 7) , S(1, 7), T(1, 10)

ry52x + ry5x 5 RO

RO(a, b) 5 (2a, 2b)ry52x + ry5x(a, b) 5 ry52x(b, a) 5 (2a, 2b)

ry5x + ry52x 5 RO

RO(a, b) 5 (2a, 2b)ry5x + ry52x(a, b) 5 ry5x(2b, 2a) 5 (2a, 2b)

275

20. a. (3, 0), (3, 0), (�1, 0), (�1, 0)b. Two different points have the same image.

Hence, the correspondence is not one-to-one.21. a. M�(3, �9), A�(6, �6), T�(�6, �6)

b. M(3, �9), A(6, �6), T(�6, �6)c. Yes

Therefore, two transformations are the same.

Exploration (page 259)Answers will vary.


1. 2 2. 1 3. 2 4. 4 5. 46. 3 7. 4 8. 3 9. 1 10. 2

Part II11. b � �2. Since A is a fixed point under a

reflection in y � x, A is on the line y � x, and so b � �2.

12. (2, �1)


1. is perpendicular 1. Given.bisector of .

2. E is the midpoint of 2. Definition of and of . bisector.

3. and 3. Definition of midpoint.

4. �AEB � �BEC 4. Perpendicular lines � �CED � �AED intersect to form

congruent adjacentangles.

5. �AEB � �AED 5. SAS.� �CED � �CEB

6. �ABD � �ADB and 6. Corresponding �DBC � �BDC parts of congruent

triangles arecongruent.

7. �ABD � �DBC 7. Addition postulate.� �ADB � �BDC

8. �ABC 8. Partition postulate.� �ABD � �DBC�ADC� �ADB � �BDC

9. �ABC � �ADC 9. Substitutionpostulate.

BE > EDAE > EC

BEDAEC

BEDAEC

D3 + rx-axis(a, b) 5 D3(a, 2b) 5 (3a, 23b)rx-axis + D3(a, b) 5 rx-axis(3a, 3b) 5 (3a, 23b)

14. Let x � m�S, then 3x � 12 � m�R. Since theangles are supplementary:

Therefore, m�S � 42 and m�R � 138.Part IV15. Statements Reasons

1. �ABC is not scalene. 1. Assumption.2. �ABC has at least 2. Definition of a

one pair of congruent scalene triangle.sides.

3. �A � �B or 3. Isosceles triangle �B � �C or theorem.�A � �C

4. m�A � m�B or 4. Definition of m�B � m�C or congruent angles.m�A � m�C

5. m�A � m�B, 5. Givenm�B � m�C,m�A � m�C

6. �ABC is scalene. 6. Contradiction.

x 5 424x 5 168

x 1 3x 1 12 5 180

276

16. Statements Reasons1. T3,–3(�ABC) 1. Given.

� �A�B�C�2. A�B� � AB, 2. In a translation,

B�C� � BC distance ispreserved.

3. XY � A�B�, 3. Given.YZ � B�C�

4. XY � AB, 4. Transitive property YZ � BC of equality.

5. , 5. Definition of congruent segments.

6. �Y � �B 6. Given.7. �ABC � �XYZ 7. SAS.

YZ > BCXY > AB

Chapter 7. Geometric Inequalities

7-1 Basic Inequality Postulates (pages 266–267)

Writing About Mathematics1. No. The reflexive and symmetric properties do

not hold for inequality.2. No. By the trichotomy postulate, it is possible

that AB � BC.Developing Skills

3. a. Trueb. A whole is greater than any of its parts.

4. a. Falseb. The shortest distance between two points is

the length of the line segment joining thesetwo points.


6. a. Trueb. Definition of a linear pair of angles

7. a. Falseb. Definition of straight angle

8. a. Trueb. Trichotomy postulate

9. a. Trueb. Transitive property of inequality

10. a. Trueb. Transitive property of inequality

11. a. Trueb. Substitution postulate

12. a. Trueb. Substitution postulate


1. 1. Given.2. �A � �CBA 2. Isosceles triangle

theorem.3. m�A � m�CBA 3. Definition of

congruent angles.4. m�CBD m�CBA 4. Given.5. m�CBD m�A 5. Substitution

postulate.

14. Statements Reasons1. 1. Given.2. PR PQ 2. A whole is greater

than any of its parts.3. PQ � RS 3. Given.4. PR RS 4. Substitution

postulate.

PQRS

AC > BC

15. Statements Reasons1. 1. Given.2. LM � NM 2. Definition of congruent

segments.3. 3. Given.4. KM LM 4. A whole is greater than

any of its parts.5. KM NM 5. Substitution postulate.

16. Statements Reasons1. KM KN, 1. Given.

KN NM2. KM NM 2. Transitive property of

inequality.3. NM � NL 3. Given.4. KM NL 4. Substitution postulate.

7-2 Inequality Postulates InvolvingAddition and Subtraction (page 269)

Writing About Mathematics1. No. It is possible for the differences to be

unequal in the same order or equal.2. Subtracting unequal quantities is the same as

adding the opposites of the unequal quantities.To find the opposites, we multiply the inequalityby �1, which changes the order of the inequality.Then, if unequal quantities are added to equalquantities, the sums are unequal in the sameorder.

Developing Skills3. If equal quantities (8 � 8) are added to unequal

quantities (10 7), then the sums are unequal inthe same order (18 15).

4. If equal quantities (11 � 11) are added tounequal quantities (4 � 14), then the sums areunequal in the same order (15 � 25).

5. If equal quantities (3 � 3) are subtracted fromunequal quantities (x + 3 12), then thedifferences are unequal in the same order (x 9).

6. If equal quantities (5 � 5) are added to unequalquantities (y � 5 � 5), then the sums are unequalin the same order (y � 10).

7. If unequal quantities (8 6) are added tounequal quantities in the same order (5 3),then the sums are unequal in the same order (13 9).

8. If equal quantities (2 � 2) are subtracted fromunequal quantities (7 � 12), then the differencesare unequal in the same order (5 � 10).

KLM

LM > NM

277

9. If equal quantities (1 � 1) are subtracted fromunequal quantities (y 8), then the differencesare unequal in the same order (y � 1 7).

10. If equal quantities (a � b) are subtracted fromunequal quantities (180 90), then thedifferences are unequal in the same order (180 � a 90 � b).


1. AB � AD, 1. Given.BC � DE

2. AB � BC � AD � DE 2. If equal quantities are added tounequal quantities,then the sums areunequal in thesame order.

3. AC � AB + BC, 3. Partition postulate.AE � AD � DE

4. AC � AE 4. Substitutionpostulate.

12. Statements Reasons1. AE BD, 1. Given.

AF � BF2. AE � AF � BD � BF 2. If equal quantities

are subtractedfrom unequalquantities, then thedifferences areunequal in thesame order.

3. AE � AF � FE, 3. Partition postulate.BD � BF � FD

4. FE FD 4. Substitutionpostulate.

13. Statements Reasons1. AE � BE 1. Given.2. 2. Definition of

congruentsegments.

3. �EAB � �EBA 3. Isosceles triangletheorem.

4. m�EAB � m�EBA 4. Definition ofcongruent angles.

5. m�DAC m�DBC 5. Given.6. m�DAC � m�EAB 6. If equal quantities

m�DBC � m�EBA are added tounequal quantities,then their sums areunequal in thesame order.

AE > BE

(Cont.)

Statements Reasons7. m�DAB � m�DAC 7. Partition

� m�EAB, postulate.m�CBA � m�DBC

� m�EBA8. m�DAB m�CBA 8. Substitution

postulate.

14. Yes. If equal weights are added to unequalweights (Blake weighs more than Caleb), thentheir new weights are unequal in the same order(Blake still weighs more than Caleb).

15. Not necessarily. If unequal weights aresubtracted from unequal weights (Blake weighsmore than Andre), then their new weights maybe unequal in the same order (Blake weighsmore than Andre), equal in different orders(Andre weighs more than Blake), or equal(Blake and Andre weigh the same).

7-3 Inequality Postulates InvolvingMultiplication and Division (page 272)

Writing About Mathematics1. No. It is only true for positive values of a.

Otherwise, if unequal quantities (1 � 2) aremultiplied by negative equal quantities, then theproducts are unequal in the opposite order (a 2a). If a � 0, then a � 2a.

2. No. It is possible that ac � 0 and bd � 0. Then a � b. It is also possible that ac � 0 and bd � 0 or ac � 0 and bd 0. Then ac � bc.

Developing Skills3. If unequal quantities (8 7) are multiplied by

positive equal quantities (3 � 3), then theproducts are unequal in the same order (24 21).

4. If unequal quantities (30 � 35) are divided bynegative equal quantities (�5 � �5), then thequotients are unequal in the opposite order (�7 � �6).

5. If unequal quantities (8 6) are divided bypositive equal quantities (2 � 2), then thequotients are unequal in the same order (4 3).

6. If unequal quantities (3x 15) are divided bypositive equal quantities (3 � 3), then thequotients are unequal in the same order (x 5).

7. If unequal quantities are multiplied bynegative equal quantities (�2 � � 2), then the products are unequal in the opposite order (8 �x).

A x2 . 24 B

278

8. If unequal quantities are multiplied bypositive equal quantities (6 � 6), then the products are unequal in the same order (y � 18).

9. Always true; if unequal quantities are multipliedby positive equal quantities, then the productsare unequal in the same order.

10. Always true; if equal quantities are added tounequal quantities, then the sums are unequal inthe same order.

11. Never true; let a � 5, b � 4, and c � 1. Then 1 � 5 � 1 � 4 or �4 � �3.

12. Always true; if equal quantities are subtractedfrom unequal quantities, then the differences areunequal in the same order.

13. Sometimes true; let a � 5, b � 3, and c � 2. Then5 � 3 3 � 2 or 2 1. Let a � 5, b � 4, and c � 2. Then 5 � 4 � 4 � 2 or 1 � 2.

14. Never true; let a � 5, b � 3, and c � 2. Then .

15. Always true; if unequal quantities are divided bypositive equal quantities, then the quotients areunequal in the same order.

16. Never true; let a � 3, b � 2, and c � 1. Then �3 � �2.

17. Always true; transitive property of inequality.Applying Skills18. Statements Reasons

1. BD � BE 1. Given.2. D is the midpoint 2. Given.

of , E is the midpoint of .

3. BA � 2BD, 3. Definition of BC � 2BE midpoint.

4. 2BD � 2BE 4. If unequal quantitiesare multiplied bypositive equalquantities, then theproducts are unequal inthe same order.

5. BA � BC 5. Substitution postulate.

19. Statements Reasons1. m�DBA 1. Given

m�CAB2. 2m�DBA 2. If unequal quantities

2m�CAB are multiplied bypositive equalquantities, then theproducts are unequal inthe same order.

BCBA

25 , 2

3

A y6 , 3 B

Statements Reasons3. m�CBA � 2m�DBA, 3. Given.

m�DAB � m�CAB4. m�CBA m�DAB 4. Substitution

postulate.

20. Statements Reasons1. AB AD 1. Given.2. 2. If unequal

quantities aremultiplied bypositive equalquantities, then theproducts areunequal in thesame order.

3. , 3. Given.4. AE AF 4. Substitution

postulate.

21. Statements Reasons1. m�CAB � m�CBA 1. Given.2. m�CAB � m�CAD � 2. Partition postulate.

m�DAB, m�CBA �m�CBE � m�EBA

3. m�CAD � m�DAB, 3. Definition of anglem�CBE � m�EBA bisector.

4. m�CAB � 2m�DAB, 4. Substitution m�CBA � 2m�EBA postulate.

5. 2m�DAB � 2m�EBA 5. Substitutionpostulate.

6. m�DAB � m�EBA 6. If unequalquantities aredivided by positiveequal quantities,then the quotientsare unequal in thesame order.

7-4 An Inequality Involving the Lengths ofthe Sides of a Triangle (pages 275–276)

Writing About Mathematics1. If s is not the longest side, then 12 is the longest

side. By the triangle inequality theorem, 12 � s +7, so s 5.

2. a. No. This may not be true for negative values.Example: �5 � �4 � �3, but �5 �3 �(�4) or �5 �7.

b. Yes. If a, b, and c are the lengths of the sides ofa triangle, then they are positive. By thetriangle inequality theorem, a � b � c.

AF 5 12ADAE 5 1

2AB

12AB . 1

2AD

279

Developing Skills3. Yes 4. No 5. Yes6. No 7. Yes 8. No9. Yes 10. Yes

11. 2 � s � 6 12. 19 � s � 4313. 0 � s � 13 14. 2.9 � s � 22.115. Since 3x = x + 2x, by the triangle inequality

theorem, x, 2x, and 3x cannot be the lengths ofthe sides of a triangle.

16. a 4; solve the inequalities given by the triangleinequality theorem:

17. Statements Reasons1. ABCD is a 1. Given.

quadrilateral.2. AD � AC � CD, 2. Triangle inequality

AC � AB � BC theorem.3. AC � CD 3. If equal quantities

� AB � BC � CD are added tounequal quantities,then the sums areunequal in thesame order.

4. AD � AB � BC � CD 4. Transitive propertyof inequality.

18. Statements Reasons1. AB � AD �BD 1. Triangle inequality

theorem.2. BC � BD � DC 2. Partition postulate.3. AD � DC 3. Given.4. BC � BD � AD 4. Substitution

postulate.5. AB � BC 5. Substitution

postulate.

19. Statements Reasons1. PZ � PQ � QZ 1. Triangle inequality

theorem.2. PY � PZ 2. If equal quantities

� PY � PQ � QZ are added tounequal quantities,then the sums areunequal in thesame order.

24 , a

a 2 2 , a 1 a 1 2

4 , a0 , a

a 1 2 , a 1 a 2 2a , a 1 2 1 a 2 2

(Cont.)

Statements Reasons3. YQ � XY � XQ 3. Triangle inequality

theorem.4. YQ � PY � PQ 4. Partition postulate.5. PY � PQ � XY � XQ 5. Substitution

postulate.6. PY � PQ � QZ 6. If equal quantities

� XY � XQ � QZ are added tounequal quantities,then the sums areunequal in thesame order.

7. XZ � XQ � QZ 7. Partition postulate.8. PY + PQ � QZ 8. Substitution

� XY � XZ postulate.9. PY � PZ � XY � XZ 9. Transitive property

of inequality.

Hands-On Activitya. 1, 6 in.; 2, 5 in.; 2, 6 in.; 3, 4 in.; 3, 5 in.; 3, 6 in.;

4, 4 in.; 4, 5 in.; 5, 5 in.; 5, 6 in.; 6, 6 in.b. {1, 6, 6}, {2, 5, 6}, {2, 6, 6}, {3, 4, 6}, {3, 5, 6}, {3, 6, 6},

{4, 4, 6}, {4, 5, 6}, {5, 5, 6}, {5, 6, 6}, {6, 6, 6}c. {1, 1, 6}, {1, 2, 6}, {1, 3, 6}, {1, 4, 6}, {1, 5, 6}, {2, 2, 6},

{2, 3, 6}, {2, 4, 6}, {3, 3, 6}d. In part b, the sum of the two shortest sides is

greater than 6. In part c, the sum is less than orequal to 6.

7-5 An Inequality Involving an ExteriorAngle of a Triangle (pages 280–281)

Writing About Mathematics1. Yes. By the exterior angle inequality theorem, the

exterior angles that are not adjacent to the rightangle have measures greater than the right angle,that is, greater than 90°.Therefore, they are obtuse.

2. Yes.The exterior angle adjacent to the right angleforms a linear pair with the right angle.Therefore,it also measures 90° and is a right angle.

Developing Skills3. a. �TRP

b. �T, �S4. a. True

b. Definition of median5. a. True

b. A whole is greater than any of its parts.6. a. True

b. Exterior angle inequality theorem7. a. True

b. A whole is greater than any of its parts.8. a. True

b. Exterior angle inequality theorem

280

9. a. Trueb. Exterior angle inequality theorem


11. False12. a. True

b. A whole is greater than any of its parts.13. FalseApplying Skills14. Statements Reasons

1. Let N be the midpoint 1. Every line segment of . has one and only

one midpoint.2. Draw , extending 2. Two points

the ray through N to determine a line. A point G so that line segment can

. be extended to anylength.

3. Draw . 3. Two pointsdetermine a line.

4. Extend through C 4. A line segment canto point F. be extended to any

length.5. m�ACF � m�ACG 5. Partition postulate.

� m�FCG6. 6. Definition of

midpoint.7. �ANB � �GNC 7. Vertical angles are

congruent.8. �ANB � �CNG 8. SAS (steps 2, 6, 7)9. �A � �ACG 9. Corresponding


10. m�ACF m�ACG 10. A whole is greaterthan any of itsparts.

11. m�ACF m�A 11. Substitutionpostulate.

12. �ACF � �BCD 12. Vertical angles arecongruent.

13. m�BCD m�A 13. Substitutionpostulate.

15. Statements Reasons1. �ABD � �DBE 1. Given.

� �ABE,�ABE � �EBC� �ABC

2. m�ABD � m�ABE, 2. A whole is greater m�ABE � m�ABC than any of its parts.

3. m�ABD � m�ABC 3. Transitive propertyof inequality.

AN > NC

BC

GC

BN > NG

BNh

AC

16. Statements Reasons1. DE � EF 1. Given.2. 2. Definition of

congruent segments.3. �EFD � �EDF 3. Isosceles triangle

theorem.4. m�EFD � m�EDF 4. Definition of

congruent angles.5. m�EFG m�EDF 5. Exterior angle

inequality theorem.6. m�EFG m�EFD 6. Substitution

postulate.

17. Statements Reasons1. �ABC with 1. Given.

m�C � 90.2. Extend through 2. A line segment may

C to point D, so that be extended to any is formed. length.

3. Exterior angle �BCD 3. Definition of an and interior angle exterior angle.�BCA form a linear pair.

4. m�BCD � m�MCA 4. If two angles form a� 180 linear pair, then

they aresupplementary.

5. m�BCD � 90 � 180 5. Substitutionpostulate.

6. m�BCD � 90 6. Subtractionpostulate.

7. 0 � m�A � 90 7. Exterior angleinequality theorem.

8. �A is acute. 8. Definition of anacute angle.

18. Statements Reasons1. m�RMP m�RTM, 1. Exterior angle

m�RTM m�SRT inequality theorem.2. m�RMP m�SRT 2. Transitive property

of inequality.

19. Statements Reasons1. FC � FD 1. Given.2. 2. Definition of

congruent segments.3. �DCF � �CDF 3. Isosceles triangle

theorem.4. �EDF � �BCF 4. If two angles are

congruent, thentheir supplementsare congruent.

FC > FD

ACD

AC

DE > EF

281

5. m�EDF � m�BCF 5. Definition ofcongruent angles.

6. m�ABF m�BCF 6. Exterior angleinequality theorem.

7. m�ABF m�EDF 7. Substitutionpostulate.

7-6 Inequalities Involving Sides and Anglesof a Triangles (pages 284–285)

Writing About Mathematics1. a. If the measures of two angles of a triangle are

equal, then the lengths of sides opposite theseangles are equal.

b. Yes2. a. If two angles of a triangle are congruent, then

sides opposite these angles are congruent.b. The converse statement and the

contrapositive statement both handle thesame relationship between the angles and thesides opposite the angle in a triangle.However, the converse deals with congruenceand the contrapositive deals with equality.

Developing Skills3. �B 4. 5.6. 7. �B 8. �C9. 10.

11. 12. RT RS ST


1. m�ABC m�CBD 1. Given.2. m�CBD m�CAB 2. Exterior angle

inequality theorem.3. m�ABC m�CAB 3. Transitive property

of inequality.4. AC BC 4. If the measures of

two angles of atriangle are unequal,then the lengths ofthe sides of oppositethese angles areunequal and thelonger side isopposite the largerangle.

14. a. The measure of the exterior angle adjacent to�C measures 90 degrees. By the exteriorangle inequality theorem, the nonadjacentinterior angles, �A and �B, have measuresless than 90 degrees. By definition, �A and�B are acute.

GH , FG , FHCD . BC . BDRS

ABACDF

b. �C is the largest angle, so hypotenuse ,which is opposite �C, is the longest side.

15. The exterior angle adjacent to the obtuse angleforms a linear pair with the obtuse angle. Sincethe measure of the obtuse angle is greater than90 degrees, the measure of the exterior angle isless than 90 degrees. By the exterior angleinequality theorem, the measures of the remoteinterior angles are also less than 90 degrees.Therefore, the two angles are acute.

Review Exercises (pages 286–287)1. A whole is greater than any of its parts.2. Transitive property of inequality3. Exterior angle inequality theorem4. If the measures of two angles of a triangle are

unequal, then the lengths of the sides oppositethese angles are unequal and the longer side liesopposite the larger angle.

5. If equal quantities are added to unequalquantities, then the sums are unequal in the sameorder.

6. Triangle inequality postulate7. If the measures of two angles of a triangle are

unequal, then the lengths of the sides oppositethese angles are unequal and the longer side liesopposite the larger angle.

8. A whole is greater than any of its parts.

9. Statements Reasons1. AE BD, EC DC 1. Given.2. AE + EC BD + DC 2. If unequal quantities

are added to unequalquantities in thesame order, the sumsare unequal in thesame order.

3. AC = AE + EC, 3. Partition postulate.BC = BD + DC

4. AC BC 4. Substitutionpostulate.

5. �B �A 5. If the lengths of twosides of a triangle areunequal, then themeasures of theangles opposite thesesides are unequaland the larger anglelies opposite thelonger side.

AB

282

10. Statements Reasons1. AD DC 1. Given.2. m�ACD m�CAD 2. If the lengths of two

sides of a triangle areunequal, then themeasures of theangles opposite thesesides are unequaland the larger anglelies opposite thelonger side.

3. �ABC � �CDA 3. Given.4. �ACD � �BAC 4. Corresponding parts

of congruent tri-angles are congruent.

5. m�ACD � m�BAC 5. Definition ofcongruent angles.

6. m�BAC m�CAD 6. SubstitutionPostulate.

7. does not 7. Definition of angle bisect �A. bisector.

11. Statements Reasons1. �ABC is isosceles 1. Given.

with .2. m�CAB � m�CBA 2. Base angles of an

isosceles triangle areequal in measure.

3. m�CBA m�DBA 3. A whole is greaterthan any of its parts.

4. m�CAB m�DBA 4. Substitutionpostulate.

5. DB DA 5. If the measures oftwo angles of a tri-angle are unequal,then the lengths ofthe sides of oppositethese angles are un-equal and the longerside is opposite thelarger angle.

12. Statements Reasons1. �RST is isosceles 1. Given.

with RS � ST.2. m�SRT � m�STR 2. Base angles of an

isosceles triangle areequal in measure.

3. �SRT and �SRP 3. Definition of anform a linear pair. exterior angle of a�STR and �STQ triangle.form a linear pair.

CA > CB

AC

(Cont.)

Statements Reasons4. m�SRP � m�STQ 4. If two angles are


13. ADB � 7 units; AEFGHIB � 7 units. The lengthof one side of a triangle is less than the sum ofthe lengths of the other two sides. Therefore,CB � AD � 1 and AC � CB � AC � AD � 1.We know that AC � 2 and AD � 4, so AC � CB � 7. Thus, ACB � ADB.a. ACB is the shortest path.b. ADB and AEFGHB are the longest paths.

Exploration (page 287)1. a. Proofs will vary.

Let ABC and DEF be two triangles with, , and m�B m�E.

Draw a point G on such that �ABG � �E. Then �ABG � �DEF by SAS.Corresponding parts of congruent trianglesare congruent, so and AG � DF.Since a whole is greater than any of its parts,AC AG. By the substitution postulate,AC DF.

b. Answers will vary.2. a. Proofs will vary.

Let ABC and DEF be two triangles with , , and AC DF. Draw a

point G on such AG � DF. Then �ABG � �DEF by SSS. Corresponding parts of congruent triangles are congruent, so�ABG � �E and m�ABG � m�E. Since awhole is greater than any of its parts,m�ABC m�ABG. By the substitutionpostulate, m�ABC m�E.

b. Answers will vary.


1. 3 2. 4 3. 3 4. 15. 4 6. 3 7. 1 8. 19. 4 10. 2

Part II11. Yes. A statement and its contrapositive are

logically equivalent. Therefore, “If our meeting isnot cancelled, then the snow does not continue tofall” is true. By the law of detachment, since it istrue that “our meeting is not cancelled,” we canconclude that “snow does not continue to fall.”

ACBC > EFAB > DE

AG > DF

ACBC > EFAB > DE

283

12. A�(�2, �4), B�(�2, 0), C�(0, �1)T

�4,�5(0, 3) � (�4, �2), ry=x(�4, �2) � (�2, �4)T

�4,�5(4, 3) � (0, �2), ry=x(0, �2) � (�2, 0)T

�4,�5(3, 5) � (�1, 0), ry=x(�1, 0) � (0, �1)Part III13. Statements Reasons

1. AP � BP 1. Assumption.2. 2. Definition of con-

gruent segments.3. bisects . 3. Given.4. R is the midpoint of 4. Definition of

. bisector.5. 5. Definition of

midpoint.6. 6. Reflexive property.7. �APR � �BPR 7. SSS.8. �ARP � �BRP 8. Corresponding


9. 9. If two lines intersectto form congruentadjacent angles,then they areperpendicular.

10. is not ⊥ to . 10. Given.11. AP � BP 11. Contradiction.

14. Statements Reasons1. bisectors �DAB, 1. Given.

bisects �DCB.2. �DAC � �BAC, 2. Definition of angle

�DCA � �BCA bisector.3. 3. Reflexive property.4. �DAC � �BAC 4. ASA.5. �B � �D 5. Corresponding parts

of congruenttriangles arecongruent.

Part IV15. m�PTR � m�QTS because they are vertical

angles, so x � y.m�PTR � m�RTQ � 180 since they aresupplements, so

x � 2x � y � 180

x � 2x � x � 180

4x � 180

x � 45

m�PTR � m�QTS � 45,m�RTQ � m�PTS � 2(45) � 45 � 135

AC > AC

CAhACh

ARBPR

PR ' ARB

PR > PR

AR > BRARB

ARBPR

AP > BP

16. a. m�BCA � m�DCB � 180, so

6x � 8 � 4x � 12 � 180

10x � 20 � 180

10x � 160

x � 16

m�BCA � 6(16) � 8 � 104,m�DCB � 4(16) � 12 � 76

284

b. m�A � m�B � m�BCA; we are given thatm�A � m�B. Since m�B � m�DCB andm�DCB � m�BCA, by the transitiveproperty of inequality, m�B � m�BCA.

c. BC � AC � AB; if the measures of two anglesof a triangle are unequal, then the lengths ofthe sides opposite these angles are unequal andthe longer side lies opposite the larger angle.

Chapter 8. Slopes and Equations of Lines

8-1 The Slope of a Line (page 295) Writing About Mathematics

1. “Delta y” means change of position with respectto the y-axis.

2. No. “No slope” is not the same as “zero slope.”“Zero slope” is a well-defined number. “Noslope” occurs when the formula for the slope of a line, , is undefined.

Developing Skills3. b. m � 1 c. Slant upward4. b. m � 3 c. Slant upward5. b. m � �2 c. Slant downward6. b. m � 1 c. Slant upward7. b. m � c. Slant downward8. b. No slope c. Vertical9. b. m � �1 c. Slant downward

10. b. m � 0 c. Horizontal11. b. m � 2 c. Slant upward12.

13. y

O1x

(�1, 3)

1

y

O1 x

(0, 1)

212

m 5y2

2 y1

x2 2 x1

14.

15.

16.

17. y

O1

x

(�4, 7)

�1

y

O1x

(�3, 2)

1

y

O1x

(�4, 5)

1

y

O1x

(2, 5)

1

18.

19.

20.

21.

22.

23. y

O1 x(�2, 0)

�1

y

O1x

(0, �2)

�1

y

O1 x

(�1, 0)

1

y

O1x

(�1, 5)

1

y

O1 x

(�2, 3)

�1

y

O1 x

(1, 3)

�1

285

Applying Skills24. a–c.

d. C(5, 10)e. , the altitude from C, is a vertical segment,

and by the reflexive property ofcongruence. Since and are horizontalsegments, AD � �5 � 2� � 3 and BD � �8 � 5� � 3. Thus, . Also,�ADC and �BDC are right angles sincevertical and horizontal lines areperpendicular. Therefore, �ADC � �BDCby SAS, and since corresponding parts of congruent triangles are congruent.

25. C(3, 2) and D(9, 2); C(3, �6) and D(9, �6)26. m �27. 7.5 feet

8-2 The Equation of a Line (pages 299–300)Writing About Mathematics

1. No. If the slope of is equal to the slope of ,

then and could be two distinct parallel

lines.2. a. Yes. The point-slope form of the equation of a

line requires that be set equal to the slope m. Since m does not exist, the equalitycannot be used.

b. A line with no slope is vertical. The equationof a vertical line is x � a, where a is the x-coordinate of any point on the line.

Developing Skills3. 4.5. 6.7. 8.9. 10.

11. 12.13. 14. x � 215. a. Yes. Slope of � slope of �

b. y 5 12x 1 32

12QRPQ

y 5 22x 1 4x 5 1y 5 5y 5 x 2 4y 5 23x 1 5y 5 21

2x 1 32y 5 24x 1 5y 5 2xy 5 24x 1 6y 5 x 2 1x 5 0

y 2 bx 2 a

CDg

ABg

CDAB

34

AC > BC

AD > BD

BDADCD > CD

CD

x

C

A

O1

1

B

y

D

16. a. No. Slope of � slope of

b. ; ;

Applying Skills17. a. y � 40x � 20 b. $140

c. The increase in cost of repair per hour of workd. The initial charge to repair a TV not including

any labor18. a. y � 18x � 8 b. $152

c. The price per cartridged. The shipping cost

19. To find the x-intercept, substitute y � 0:

Therefore, the x-intercept is (a, 0).To find the y-intercept, substitute x � 0:

Therefore, the y-intercept is (0, b).

8-3 Midpoint of a Line Segment (pages 306–307)


1.

✔

2.

✔

Developing Skills3. (3, 4) 4. (6, 6) 5. (5, 4)6. (2, 2) 7. (0, 0) 8. (4, 5.5)9. (0.5, 4) 10. (1, 8) 11. (�2, �3)

12. (3, 3.5) 13. (0.75, 2.75) 14. (0.5, 6)15. B(0, 5) 16. B(3, 15) 17. A(6, 3)18. A(�1, 2) 19. M(2, 6.5) 20. B(0, 0)Applying Skills21. a. B(9, 1), D(1, 7)

b. Midpoint of � � (5, 4)

Midpoint of � � (5, 4)A 9 1 12 , 7 1 1

2 BBDA 1 1 9

2 , 1 1 72 BAC

a 1 b 5 a 1 b2b 2 b 1 a 5? a 1 b

2 Ab 2 b 2 a2 B 5? 2 A a 1 b

2 Bb 2 b 2 a

2 5? a 1 b2

a 1 b 5 a 1 b2a 1 b 2 a 5? a 1 b

2 Aa 1 b 2 a2 B 5? 2 A a 1 b

2 Ba 1 b 2 a

2 5? a 1 b2

y 5 b

0a 1

yb 5 1

xa 1

yb 5 1

x 5 a

xa 1 0b 5 1

xa 1

yb 5 1

LNg

:y 5 35x 1 12

5

MNg

:y 5 23x 1 83LM

g:y 5 3

4x 1 94

MNLM

286

22. a. Q(x2, y1), S(x1, y2)

b. Midpoint of �

Midpoint of

�

23. a. M (2, 3) b. y � x � 1 c. N(2, 5)d. y � �x � 7 e. (4, 4) f. P(5, 4)g. y � 4 h. Yes i. Yes; (4, 4)

8-4 The Slopes of Perpendicular Lines(pages 310–312)

Writing About Mathematics1. The line x � 5 has no slope. Therefore, there can

be no reciprocal.2. Both are correct. The negative reciprocal of

.

Developing Skills3. a. m � 4 b. m �4. a. m � 1 b. m � �15. a. m � �1 b. m � 16. a. m � 2 b. m �

7. a. m � b. m �

8. a. m � b. m � �2

9. a. m � �2 b. m �

10. a. m � b. m � �211. a. No slope b. m � 012. a. m � 0 b. No slope

13. 14. y � �2x

15. y � 3x � 18 16. x � 217. No 18. y � �2x � 119. y � �x � 3 20. y � 2x � 121. y � 3x 22.

23. y � �3 24.Applying Skills25. a. y � �x � 1

b. Yes. It intersects M(1, 0), which is the midpoint of .

c. In �DEF, the altitude from D is also the median. Call this segment . By the definitionof altitude, , so �DGE � �DGF.By the definition of median, G is the midpointof , so . by the reflexive property of congruence. Therefore,�DGE � �DGF by SAS. Then since corresponding parts of congruenttriangles are congruent, which makes�DEF isosceles.

DE > DF

DG > DGEG > FGEF

DG ' EFDG

EF

y 5 72x 2 14

y 5 12x 1 74

y 5 715x 1 21

2

12

12

12

2323

2

212

14

213 5 2 5 3

A x2 1 x12 ,

y2 1 y12 B 5 A x1 1 x2

2 , y1 1 y2

2 BQS

A x1 1 x22 ,

y1 1 y22 BPR

1�1

3

26. The slopes of and are both . The slopesof and are both . Any two consecutive sides have slopes that are negative reciprocals.Therefore, any two consecutive sides areperpendicular and all angles are right angles. Bydefinition, ABCD is a rectangle.

27. a. 1 b. y � �x � 8c. No slope d. y � 3e. �0.5 f. y � 2x � 7g. The point (5, 3) makes the equations of the

perpendicular bisector true:

✔ ✔

✔

28. a. Altitude at A: y � x � 2;altitude at B: y � �0.5 x � 2;altitude at C: x � 2

b. (0, 2)29. a. The measure of the exterior angle adjacent to

�M is 90°. By the exterior angle inequalitytheorem, the nonadjacent interior angles, �Land �M, both measure less than 90°.Therefore, �L and �M are acute.

b. , ,c. �L, �N, �M

Hands-On ActivityIn a–c, answers will vary. Examples are given.

a. l1: y � �x � 7 b. l1: y �

l2: y � �x � 11 l2: y �

c. l1: y �

l2: y �

8-5 Coordinate Proof (pages 315–317)Writing About Mathematics

1. Yes. The slope of the segment connecting (a, 0)and (0, b) is . The slope of the segment

connecting (0, b) and (c, 0) is . If the segments are perpendicular, then the slopes are negative

reciprocals and would equal . Since

, the two sides of the triangle are

perpendicular, and the triangle is a right triangle.2. Yes. The point (0, b) lies on the y-axis and (a, 0)

and (c, 0) are on the x-axis. Therefore, the

5 ab 5 2b

c

b�a�1

2bc

b�a�1

2bc

2ba

213x 1 5

213x 1 20

3

32x 2 4

32x 2 4

LNLMMN

3 5 3y 5 3

3 5 33 5 33 5

? 2(5)273 5? 25 1 8

y 5 2x27y 5 2x 1 8

223BCAD

32DCAB

287

altitude from (0, b) is on the y-axis. Since a and chave the same sign, this altitude lies outside thetriangle. Thus, the triangle is obtuse.

Developing Skills

3. Midpoint of � (2, 2)Midpoint of � (2, 2)The two segments have the same midpoint.Therefore, the two segments bisect each other.

4. Midpoint of �

Midpoint of �The two segments have the same midpoint.Therefore, the two segments bisect each other.Slope of �

Slope of � 7The slopes are negative reciprocals.Therefore, thetwo segments are perpendicular to each other.

5. a. K � (3, 2)b. Slope of � �3

Slope of �

The slopes are negative reciprocals. Therefore,is perpendicular to .

c. Since the altitude of the triangle is also itsmedian, the triangle is isosceles.

6. a. Slope of � �3Slope of � 3The slopes are negative reciprocals. Therefore,the two sides are perpendicular.

b. �C c. d. (5, 5)e. y � 2x � 5 f. y � 2x � 5g. Yes. The altitude and the median are the same

line segment.

In 7–8, part a, answers will be graphs.7. b. Midpoint of � (�2, 4)

Midpoint of � (6, 4)Midpoint of � (4, 0)

c. Slope of � 2Slope of �

Slope of � 0d. : ; : ; : x � 4e. (4, 1)

8. a. Midpoint of � (6, �1)Midpoint of � (6, �1)The two segments have the same midpoint.Therefore, the segments bisect each other.

b. Slope of �

Slope of � �2The slopes are negative reciprocals.Therefore, the two segments areperpendicular to each other.

BD

12AC

BDAC

ACy 5 232x 2 5BCy 5 21

2x 1 3ABAC

223BC

ABACBCAB

AB

BCAC

LNMK

13LN

MK

BD

217AC

A512, 11

2 BBD

A512, 11

2 BAC

CDAB

Applying Skills9. Since and are vertical segments,

SP � �b � 0� � �b� and RQ � �b � 0� � �b�.Therefore, . is a horizontal segment.Since horizontal and vertical segments areperpendicular, and .�SPQ and �RQP are both right angles, so�SPQ � �RQP. By the reflexive property,

. Then �SPQ � �RQP by SAS.Corresponding parts of congruent triangles arecongruent, so .

10. (1) The midpoint of is .

The midpoint of is .

Therefore, and intersect each other attheir midpoints so, by definition, bisect eachother.

(2) The slope of � 1 and the slope of � �1, so the slopes are negative

reciprocals. Therefore, and areperpendicular to each other.

11. Let A � (0, 0), B � (2a, b), and C � (a, b). Sinceis a horizontal segment, the altitude from C is

a vertical line intersecting in the point D(a, 0). Then, AD � �0 � a� � �a� and DB � �a � 2a� � �a�, so . Since and

are vertical and horizontal segments,respectively, they are perpendicular to eachother. Thus, �ADC and �CDB are both right angles and �ADC � �CDB. Since bythe reflexive property, �ADC � �BDC by SAS.Therefore, because corresponding parts of congruent triangles are congruent. Bydefinition �ABC is isosceles.

12. Let A � (�a, 0), B � (a, 0), and C � (0, b).Under the translation , the images of thevertices of the triangle are:

By Exercise 11, �A�B�C� is an isosceles triangle.Since distance is preserved under translation, theimage of a triangle under a translation iscongruent to the given triangle by SSS.Therefore, �ABC is also an isosceles triangle.

13. a. E(a, b); F(c, d)b. Slope of �

Slope of � 2b 2 2d2a 2 2c 5 b 2 d

a 2 cBC

b 2 da 2 cEF

C(0, b) S Cr(a, b)B(a, b) S Br(2a, b)

A(2a, 0) S Ar(0, 0)

Ta,0

AC > CB

CD > CD

ABCDAD > DB

ABAB

FHEGFH

EG

FHEG

A 0 1 a2 , a 1 0

2 B 5 A a2, a2 BFH

A 0 1 a2 , 0 1 a

2 B 5 A a2, a2 BEG

SQ > PR

PQ > PQ

RQ ' PQSP ' PQ

PQSP > RQ

RQSP

288

14. a. We are given a segment with endpoints A(�a, 0) and B(a, 0), and points P(0, b) andQ(0, c). Consider �AOP and �BOP, with O the origin. AO � �0 � (�a)� � a and BO � �0 � a� � a, so . Since and

are vertical and horizontal segments,respectively, they are perpendicular to eachother. Thus, �AOP and �BOP are both rightangles, and �AOP � �BOP. By the reflexive property, , and �AOP � �BOP bySAS. Therefore, because corre-sponding parts of congruent triangles arecongruent, and P is equidistant from the endpoints of . Similarly, and�AOQ � �BOQ by SAS. Therefore,

and Q is equidistant from theendpoints of .

b. If two points are each equidistant from theendpoints of a line segment, then the pointsdetermine the perpendicular bisector of the

line segment. Therefore, by part a, is the

perpendicular bisector of .

8-6 Concurrence of the Altitudes of aTriangle (pages 321–322)

Writing About Mathematics1. Yes. Using the formula for the coordinates of the

orthocenter from the proof, the orthocenter is . Since A is to the left of the origin, a is

negative. Since C is to the right of the origin, c ispositive. Thus, ac is a negative number, and so �ac is a positive number. Since b is positive,is a positive number. Therefore, the y-coordinateof the orthocenter is positive, and so theorthocenter is located above the origin.

2. No. Using the formula for the coordinates of theorthocenter from the proof, the orthocenter is

. Since both A and C are located to the right of the origin, both a and c are positive. Thus,ac is positive, and so �ac is negative. Since b is positive, is negative. Therefore, the y-coordinate of the orthocenter is negative, andso the orthocenter is located below the origin.

Developing Skills

3. a. Slope of �

Slope of EF 5 234

43DE

2acb

A0, 2acb B

2acb

A0, 2acb B

AB

PQg

ABAQ > BQ

QO > QOAB

AP > BPPO > PO

ABPOAO > BO

The slopes are negative reciprocals. Therefore,and are perpendicular, and so �DEF

is a right angle.b. Since �DEF is a right angle, is the

altitude from D to and is the altitudefrom F to . The two altitudes intersect at E.Therefore, E is the orthocenter.

4. (0, 1) 5. (0, 8) 6. (2, �2.5)7. (1, 6) 8. (3.5, 3.2) 9. (12.5, �5.5)

Applying Skills10. a. y � 2x � 6 b. y � �x � 6

c. (0, 6) d. x � 0e. Point P has coordinates (0, 3) and so it lies on

the line x � 0.11. a. y � �x b. y � x �

c. d. y �

e. Substitute x � 1 and y � 1 into the equationfrom part d:

✔

The coordinates of P make the equation true.Therefore, P lies on the line.

Hands-On Activity(1) b. (0, 6) (2) b. (6, �2) (3) b. (0, �4)

Review Exercises (pages 323–324)1. a. (3, �2) b. m � c. y �

2. a. (2, �1) b. m � � c. y �

3. a. (3, �3) b. m � �1 c. y � �x4. a. m � �0.5 b. y � �0.5x � 4

c. (2, 3) d. y � 2x � 15. a. Slope of � 2

Slope of �

Therefore, the two segments areperpendicular, and the triangle is a righttriangle.

b. c. x � 1 d. y � �3x � 7

e. The equation of the line that contains is. This line is perpendicular to the

median from S. Therefore, the median is alsothe altitude.

6. (6, 7)

y 5 13x 2 11

3

RT

A212, 21

2 B

212ST

RS

23x5 1 15

35

223x22

3

1 5 1

1 5? 77

1 5? 237(1) 1 10

7

237x 1 10

7A2212, 21

2 B103

13

DEEFEF

DE

EFDE

289

7. a. Midpoint of � (3, 3)Midpoint of � (2, 5)Midpoint of � (0, 3)

b. Slope of � 1Slope of � 0Slope of � �2

c. Slope of the altitude from F � �1Slope of the altitude from D � undefinedSlope of the altitude from E �

d. Perpendicular bisector of : x � 2Perpendicular bisector of : y � �x � 6Perpendicular bisector of : y � 0.5x � 3

e. (2, 4)8. a. Slope of � 1

Slope of � �1The two sides are perpendicular and form aright angle. Therefore, the triangle is a righttriangle.

b. The slope of is �1. Therefore, isperpendicular to since the slope of is1, and �ANM � �CNM since they are bothright angles. because N is themidpoint of . by the reflexiveproperty of congruence. Therefore, �ANM ��CNM by SAS. Since corresponding parts ofcongruent triangles are congruent,and AM � CM. Since M is the midpoint of

, AM � MB. By the transitive property ofequality, AM � BM � CM, and M is equi-distant from the vertices of A, B, and C.

9. a. Midpoint of � (�2, 0)

b. Since is a vertical line and is a horizontal line, the two lines are perpendicular. Since A is the midpoint of ,

is a perpendicular bisector. Since E is onthe perpendicular bisector of , it is equidistant from points D and F. Therefore,DE � FE, and .

10. Slope of � 3

Slope of �

Therefore, the two sides are perpendicular, and�D is a right angle.

213DC

g

DFg

DE > FE

DFAE

DF

AEg

DFg

DF

AB

AM > CM

NM > NMACAN > NC

ACACNMNM

CBAC

DFDEEF

12

DFEFDE

DFEFDE

Exploration (page 324)a.

Area of the enclosing rectangle � 8(8) � 64Area of Triangle 1 � (1)(8) � 4

Area of Triangle 2 � (5)(7) � 17.5

Area of Triangle 3 � (3)(8) � 12Area of enclosed triangle � 64 � (4 � 17.5 � 12)

� 30.5 sq unitsb.


Area of Triangle 2 � (5)(5) � 12.5

Area of Triangle 3 � (3)(10) � 15Area of enclosed triangle � 80 � (20 � 12.5 � 15)

� 32.5 sq unitsc.


Area of Triangle 2 � (2)(10) � 1012

12

x

A

O�1

By

C

1

12

12

12

x

A

O1�1

By

C

12

12

12

x

C

A O11

By

290

Area of Triangle 3 � (1)(4) � 2Area of remaining rectangle � 2Area of enclosed triangle � 66 � (33 � 10 � 2 � 2) � 19 sq units

Yes. The area of the quadrilateral is 19.


1. 3 2. 1 3. 2 4. 15. 2 6. 4 7. 2 8. 19. 2 10. 3

Part II11. a.

b. See graph; B� � (5, �7), C� � (9, 0)c. a � 4, b � �2

12. Yes. Lines and have slopes of �2 and ,

respectively. Since these slopes are negative

reciprocals, the lines are perpendicular and �C is

a right angle. Therefore, �ABC is a right triangle.Part III13. Statements Reasons

1. bisects �RTS. 1. Given.2. �RTQ � �STQ 2. Definition of angle

bisector.3. 3. Given.4. �TQR and �SQT are 4. Definition of

right angles. perpendicular lines.5. �TQR � �SQT 5. Right angles are

congruent.6. 6. Reflexive property.7. �TQR � �TQS 7. ASA.8. 8. Corresponding

sides of congruenttriangles arecongruent.

9. �RST is isosceles. 9. Definition ofisosceles triangle.

TR > TS

TQ > TQ

TQ ' RS

TQh

12BC

gACg

x

AO

�1

B

yC

�1C�

A�

B�

12

14. Let e � “Evanston is the capital of Illinois.”Let c � “Chicago is the capital of Illinois.”Let s � “Springfield is the capital of Illinois.”Then in symbols, the given statements are:

�e → �c

s ∨ c

�e

Since �e is true, �c is true by the Law ofDetachment.Since �c is true, c is false.Since c is false and s ∨ c is true, s is true by theLaw of Disjunctive Inference.Therefore, Springfield is the capital of Illinois.

Part IV15. Substitute the first equation for y in the second

equation, and solve for x:

x 5 1

7x 5 7

x 1 6x 2 2 5 5

x 1 2(3x 2 1) 5 5

291

Substitute x � 1 into either equation to find y:

Therefore, the coordinates of the intersectionpoint are (1, 2).

16. Statements Reasons1. and bisect 1. Given.

each other at G.2. G is the midpoint of 2. Definition of

and of . bisector.3. and 3. Definition of

midpoint.4. �CGD � �EGF 4. Vertical angles are

congruent.5. �CGD � �EGF 5. SAS.6. 6. Corresponding

sides of congruenttriangles arecongruent.

CD > EF

DG > GECG > GF

DGECGF

DGECGF

y 5 2

y 5 3(1) 2 1

y 5 3x 2 1

Chapter 9. Parallel Lines

9-1 Proving Lines Parallel (page 334)Writing About Mathematics

1. Corresponding angles2. Yes. This is the contrapositive of Theorem 9.1a: If

two coplanar lines cut by a transversal are notparallel, then the alternate interior angles formedare not congruent.

Developing Skills3. If two coplanar lines are cut by a transversal so

that the alternate interior angles formed arecongruent, then the two lines are parallel.

4. If two coplanar lines are cut by a transversal sothat the corresponding angles are congruent,then the two lines are parallel.

5. If two coplanar lines are cut by a transversal sothat the interior angles on the same side of thetransversal are supplementary, then the lines areparallel.

6. Vertical angles (�2 and �4) are congruent. If twocoplanar lines are cut by a transversal so that theinterior angles on the same side of the transversalare supplementary, then the lines are parallel.

7. Vertical angles (�2 and �4) are congruent. Iftwo coplanar lines are cut by a transversal so thatthe corresponding angles are congruent, then thelines are parallel.

8. Vertical angles (�7 and �5) are congruent. Iftwo coplanar lines are cut by a transversal so thatthe interior angles on the same side of thetransversal are supplementary, then the lines areparallel.

Applying Skills

9. Given: intersects and ; �1 � �5.

Prove:

Statements Reasons

1. is not parallel 1. Assumption.

to .2. �1 � �3 2. Vertical angles are

congruent.3. �3 is not congruent 3. If two coplanar

to �5. lines cut by atransversal are notparallel, then thealternate interiorangles formed arenot congruent.

4. �1 is not congruent 4. Transitive property.to �5.

5. �1 � �5 5. Given.

6. 6. Contradiction.ABg

7CDg

CDg

ABg

ABg

7CDg

CDg

ABg

EFg

10. Given: and

Prove:Statements Reasons

1. and 1. Given.

2. �1 and �2 are right 2. Definition of angles. perpendicular lines.

3. �1 � �2 3. Right angles arecongruent.

4. 4. If two coplanar lines are cut by atransversal so thatthe alternate interiorangles are congruent,then the two lines areparallel.

11. m�A � m�B � 3x � (180 � 3x) � 180. Sinceinterior angles on the same side of the transversal are supplementary, .

12. We are given that , so �BCD is a right angle and m�BCD � 90. Therefore,m�BCD � m�ADC � 180. Since interior same angles on the same side of the transversal aresupplementary, .

13. a. Statements Reasons1. and bisect 1. Given.

each other at E.2. E is the midpoint 2. Definition of

of and of . bisector.3. , 3. Definition of

midpoint.4. �CEA � �DEB 4. Vertical angles are

congruent.5. �CEA � �DEB 5. SAS.

b. Statements Reasons1. �CEA � �DEB 1. Part a.2. �ECA � �EDB 2. Corresponding parts


c. Statements Reasons1. �ECA � �EDB 1. Part b.2. 2. If two coplanar lines

are cut by a trans-versal so that thealternate interiorangles formed arecongruent, then thetwo lines areparallel.

CA y DB

CE > EDAE > EB

CDAB

CDAB

AD y BCCD

DC ' BCAD y BCAB

ABg

y CDg

CDg

' EFg

ABg

' EFg

ABg

7CDg

CDg

' EFg

ABg

' EFg

292

14. Given: intersects and ; �1 � �2.

Prove:

Statements Reasons

1. intersects 1. Given.

and ;�1 � �2.

2. �1 � �3 2.Vertical angles arecongruent.

3. �2 � �3 3. Transitive property.

4. 4. If two coplanar lines are cut by a transversalso that thecorresponding anglesare congruent, then thetwo lines are parallel.

9-2 Properties of Parallel Lines (pages 340–341)

Writing About Mathematics1. a. Yes. The inverse of Theorem 9.1a states: If two

coplanar lines are cut by a transversal so thatthe alternate interior angles formed are notcongruent, then the lines are not parallel. Thecontrapositive of this statement, which has thesame truth value, is: If two coplanar lines cutby a transversal are parallel, then the alternateinterior angles formed are congruent. This isthe same as Theorem 9.1b.

b. Yes. The inverse of Theorem 9.6 states: If atransversal is not perpendicular to one of twoparallel lines, then it is not perpendicular tothe other. Alternate interior angles of parallellines are congruent, so if the transversal doesnot form a right angle with one of the parallellines, it will not form a right angle with theother.

2. The measures of the angles formed by theparallel lines and the transversal are all 90degrees. Alternate interior angles of parallel linesare congruent, so their measures are equal. If theangles are also supplementary, then they mustboth be right angles.

Developing Skills3. 80 4. 150 5. 1206. 75 7. 115 8. 509. 42 10. 80 11. 65

12. 60 13. 44 and 136 14. 21 and 21

ABg

y CDg

CDg

ABg

EFg

ABg

y CDg

CDg

ABg

EFg

15. a. m�1 � 70, m�2 � 110, m�3 � 70, m�4 � 50,m�5 � 130, m�6 � 50, m�7 � 50, m�8 � 60,m�9 � 70

b. 120 c. Yes d. 120e. 110 f. 130 g. 180

16. m�A � m�C � 75; m�ABC � m�D � 105Applying Skills17. Statements Reasons

1. and 1. Given.

transversal 2. �3 � �5 2. If two parallel lines

are cut by atransversal, then thealternate interiorangles formed arecongruent.

3. �4 is the 3. If two angles form supplement of �3. a linear pair, then they

are supplementary.4. �4 is the 4. If two angles are

supplement of �5. congruent, then theirsupplements arecongruent.


1. , 1. Given.2. �AEF � �EFD 2. If two parallel lines are

cut by a transversal,then the alternateinterior angles formedare congruent.

3. �AEF is a right 3. Perpendicular lines angle. intersect to form right

angles.4. �EFD is a right 4. Definition of

angle. congruent angles.

5. 5. Perpendicular lines intersect to form rightangles.

19. When two parallel lines are cut by a transversal,alternate interior angles are congruent. Theseangles are also congruent to their respectivevertical angles. Since these vertical angles arealternate exterior angles, alternate exteriorangles are congruent.

EFg

' CDg

EFg

' ABg

ABg

y CDg

EFg

ABg

y CDg

293


1. bisects exterior 1. Given.�BCD

2. �BCE � �DCE 2. Definition of anglebisector.

3. 3. Given.4. �DCE � �A 4. If two parallel lines

are cut by a trans-versal, then corre-sponding angles arecongruent.

5. �BCE � �B 5. If two parallel lines arecut by a transversal,then the alternateinterior angles formedare congruent.

6. �A � �B 6. Substitution postulate.

21. a. Statements Reasons1. �CAB � �DCA 1. Given.

2. 2. If two coplanar lines are cut by atransversal so thatthe alternate inte-rior angles formedare congruent, thenthe two lines areparallel.

b. Statements Reasons

1. 1. Part a.2. �CAB � �DCA, 2. Given.

�DCA � �ECB3. �CAB � �ECB 3. Transitive property.4. �ECB � �ABC 4. If two parallel lines

are cut by a trans-versal, then thealternate interiorangles formed arecongruent.

5. �CAB � �ABC 5. Transitive property.6. m�ABC � 6. Angles forming a

m�CBG � 180 linear pair aresupplementary.

7. m�CAB � 7. Substitution m�CBG � 180 postulate.

8. �CAB is the sup- 8. Definition of sup-plement of �CBG. plementary angles.

AB y DCEg

AB y DCEg

CEh

y AB

CEh

22. If two parallel lines are cut by a transversal, thenthe two interior angles on the same side of thetransversal are supplementary. Therefore, since

and �P is a right angle, its supplement,�S, is also a right angle. Similarly, since and �P and �S are right angles, their respectivesupplements, �Q and �R, are also right angles.

23. If two parallel lines are cut by a transversal, thenthe two interior angles on the same side of thetransversal are supplementary. Therefore, since

and �K is an acute angle measuring less than 90°, its supplement, �N, must measuremore than 90° and be obtuse. Similarly, since

and �K is acute, its supplement, �L,must be obtuse. Since �N is an obtuse anglemeasuring more than 90°, its supplement, �M,must measure less than 90° and be acute.

9-3 Parallel Lines in the Coordinate Plane(pages 345–347)

Writing About Mathematics1. They are the same line.2. Vertical lines are parallel, but they do not have

the same slope since vertical lines do not haveslope.

Developing Skills3. Perpendicular 4. Parallel 5. Parallel6. Neither parallel nor perpendicular7. Parallel 8. Perpendicular9. y � �3x � 4 10.

11. 12. y � �3

Applying Skills

13. a. b. c.d. 3 e. 3 f. y � 3x � 1g. (2, 5)

14. a. 1 b. �1

c. �1; since and , .d. 1; since and , .e. y � x � 2 f. y � �x g. (�1, 1)

15. a. Slope of , slope of

Slope of , slope of

Therefore, and .

b. The slopes of the adjacent sides, and , are not negative reciprocals. Therefore, no sidesare perpendicular and PQRS does not have aright angle.

232

14

QR y SPPQ y RS

SP 5 232QR 5 23

2

RS 5 14PQ 5 1

4

AB y DCDA ' DCDA ' ABBC y DADA ' ABBC ' AB

y 5 212x 1 621

2212

y 5 12x 1 3

y 5 13x 1 4

LM y NK

KL y MN

QR y SPPQ y RS

294

16. a. Slope of , slope of

Slope of � �2, slope of � 3Only one pair of sides has equal slopes.Therefore, KLMN has only one pair ofparallel sides.

b. The slopes of the adjacent sides, and 3, are negative reciprocals, so the respective sidesare perpendicular and form right angles.Therefore, �K and �N are right angles.

Hands-On Activity 1a. If two coplanar lines are each perpendicular to

the same line, then they are parallel.b. (1)

(2)(3)

Hands-On Activity 21. Answers will vary.

a. A(a, b); B(c, d); C(e, f)

b. Midpoint of

Midpoint of

c. Slope of

Slope of

The slope of the midsegment is equal to theslope of the third side of the triangle.Therefore, they are parallel.

2. Results will vary.

9-4 The Sum of the Measures of theAngles of a Triangle (pages 351–352)

Writing About Mathematics1. Yes. The sum of the angles of a triangle is 180

degrees. Therefore, if one angle has a measuregreater than 90 degrees, then the sum of theother two angles must be less than 90 degrees,and they must both be acute.

2. No. By definition, exactly two angles can besupplementary. Since a triangle has three angles,and no angle of a triangle can measure 0 degrees,the sum of the measure of any two angles mustbe less than 180 degrees.

Developing Skills3. Yes 4. No 5. No6. Yes 7. 40 8. 509. 54 10. 50 11. 80

12. 45 13. 52 14. 35

5 d 2 bc 2 aDE 5

AB 5 d 2 bc 2 a

BC 5 E A c 1 e2 ,

d 1 f2 B

AC 5 D A a 1 ec ,

b 1 f2 B

y 5 219x 1 4

y 5 12x 2 148y 5 1

4x 1 194

213

NKLM

MN 5 213KL 5 21

3

c 1 e2 2 a 1 e

2

d 1 f2 2

b 1 f2

15. 20 16. 140 17. 9018. 54 19. 12020. m�ACD � 60; m�ACB � 12021. m�ACD � 90; m�ACB � 9022. m�ACD � 60; m�B � 2023. m�B � 95; m�ACD � 135; m�ACB � 45Applying Skills24. 46°, 67°, 67° 25. 70°, 55°, 55° 26. 20°, 80°, 80°27. m�N � 58; measure of exterior angle � 12228. a. m�A � 72; m�B � 67; m�C � 41 b.29. m�A � m�C � 30; m�B � 120; measure of

exterior angle � 6030. Let �ABC be a triangle with �BCA the exterior

angle at C. Then m�A � m�B � m�C � 180since the sum of the measures of the angles of atriangle is 180. Also, �BCA and �C form a linearpair, so m�BCA � m�C � 180. By thesubstitution postulate, m�A � m�B � m�C �m�BCA � m�C. By the subtraction postulate,m�A � m�B � m�BCA, or the measure of anexterior angle of a triangle is equal to the sum ofthe measures of the nonadjacent interior angles.

31. a. Graph b. Graphc. is a vertical line, so BC � |�1 � 2| � 3.

is a horizontal line, so CD � |�1 � 2| � 3.Since BC � CD, and �BDC isisosceles. Horizontal and vertical lines areperpendicular, so �BCD is a right angle.

d. m�BDC � 45; each acute angle of anisosceles right triangle measures 45°.

e. is a horizontal line and, therefore,perpendicular to . By the partition postulate, m�DBA � m�DBC � m�ABC.Since m�DBC � 45 and m�ABC � 90,m�DBA � 135.

32. In hexagon ABCDEF, draw . This divides the hexagon into two quadrilaterals, ABCDand ADEF. The sum of the measures of theangles in each quadrilateral is 360°. Therefore,the sum of the measures of the angles in ABCDEF � 360 � 360 � 720°.

33. bisects �ABC, so �ABD � �CBD.bisects �ADC, so �ADB � �CDB.by the reflexive property of congruence, so�ABD � �CBD by SAS. Corresponding partsof congruent triangles are congruent, so �A � �C.

BD > BDDBh

BDh

AD

BCABg

BC > CD

CDBC

BC

295

9-5 Proving Triangles Congruent by Angle,Angle, Side (pages 356–357)

Writing About Mathematics1. No, it simply means that SSA is insufficient to

prove the triangles congruent. The triangles mayor may not be congruent.

2. Yes. and are perpendicular, so �C is a right angle and measures 90°. Since the sum ofthe measures of the angles of a triangle is 180°,the measures of �CAB and �CBA must be 90°,which makes them complementary.

Developing Skills3. Yes, by AAS. 4. Yes, by AAS.5. No. AAA is insufficient to prove congruence.6. Yes, by AAS.7. No. SSA is insufficient to prove congruence.8. Yes, by SAS.

Applying Skills

9. Let �ABC � �DEF, so �A � �D and. Draw , the altitude from �A in

�ABC, and , the altitude from �F in �DEF,to form right triangles �ABG and �DFH. Tworight triangles are congruent if the hypotenuseand an acute angle of one right triangle arecongruent to the hypotenuse and an acute angleof the other right triangle, so �ACG � �DFH.Corresponding parts of congruent triangles are congruent, so the altitudes, and , arecongruent.

10. Let �ABC � �DEF, so �A � �D, ,and . Draw , the median from �A in �ABC, and , the median from �F in�DEF, to form triangles �ABG and �DFH.Since G is the midpoint of and H is the midpoint of , AG � and DH � .Halves of congruent segments are congruent, so

, and �ACG � �DFH by SAS.Corresponding parts of congruent triangles arecongruent, so the altitudes, and , arecongruent.

11. Let �ABC � �DEF, so �A � �D, ,and �C � �F. Draw , the bisector of �A in �ABC, and , the bisector of �F in �DEF, to form triangles �ABG and �DFH. By the definition of angle bisector, m�ACG �

and m�DFH = . Halves of congruent angles are congruent, so �ACG � �DFH, and

12m/F

12m/C

FHCG

AC > DF

FHCG

AG > DH

12DE1

2ABDEAB

FHCGAB > DE

AC > DF

FHCG

FHCGAC > DF

BCAC

�ACG � �DFH by AAS. Corresponding partsof congruent triangles are congruent, so thealtitudes, and , are congruent.


1. �A � �C and is 1. Given.the bisector of �ABC.

2. �ABD � �CBD 2. Definition of anglebisector.

3. 3. Reflexive property.4. �ABD � �CBD 4. AAS.5. �ADB � �CDB 5. Corresponding parts

of congruent tri-angles are congruent.

6. bisects �ADC. 6. Definition of angle bisector.


and .2. 2. If a transversal is

perpendicular to oneof two parallel lines,then it is perpendic-ular to the other.

3. �B and �C are right 3. Definition of angles. perpendicular lines.

4. �B � �C 4. Right angles arecongruent.

5. �AEB � �DEC 5. Vertical angles arecongruent.

6. �ABE � �DCE 6. AAS.7. , 7. Corresponding parts


8. and bisect 8. Definition of each other. bisector.

14. a. Under the translation T9,0, A(�6, 0) → D(3, 0),B(�1, 0) → E(8, 0), and C(�5, 2) → F(4, 2).Distance is preserved under translation, so�ABC � �DEF by SSS.

b. Answers will vary. Under rx=1, A(�6, 0) → (8, 0),B(�1, 0) → (3, 0), and C(�5, 2) → (7, 2).Under rx=5.5, (8, 0) → D(3, 0), (3, 0) → E(8, 0),and (7, 2) → F(4, 2).Distance is preserved under reflection, so�ABC � �DEF by SSS.

15. Let �ABC and �DEF be right triangles with �Band �E the right angles, , and AC > DF

BECAED

AE > DEBE > CE

CD ' BECAB ' BEC

AB > CDAB y CD

DBh

BD > BD

BDh

FHCG

296

�A � �D. Since right angles are congruent,�B � �E. Therefore, �ABC � �DEF by AAS.

16. Let P be a point on the bisector of �ABC.By the definition of angle bisector, �ABP ��CBP. By the reflexive property of congruence,

. Draw and . Then �BAP and �BCP are right angles andcongruent. Therefore, �ABP � �CBP by AAS.Corresponding parts of congruent triangles are congruent, so and PA � PC. By defi-nition, P is equidistant from and from .

17. Through a point E on side , draw parallelto . If two parallel lines are cut by a transversal, then the corresponding angles arecongruent. Therefore, �A � �BED and �C � �BDE. By the reflexive property ofcongruence, �B � �B. If AAA were sufficientto establish congruence, then �EBD would becongruent to �ABC. But AB � DB and CB � DB, so �EBD is not congruent to �ABC.

9-6 The Converse of the Isosceles TriangleTheorem (pages 360–362)

Writing About Mathematics1. Yes. The contrapositive of Theorem 7.3, which is

also true, states: If the measures of the anglesopposite two sides of a triangle are equal, thenthe lengths of the sides opposite these angles areequal.

2. Yes. The sum of the measures of the angles of atriangle is 180°, so if two angles measure 45° and90°, then the third must measure 45°. Since twoangles of the triangle have equal measures, theyare congruent, so the sides opposite these anglesmust be congruent. By definition, the righttriangle is isosceles.

Developing Skills3. a. 70 b. Isosceles4. a. 30 b. Isosceles5. a. 65 b. Isosceles6. a. 60 b. Not isosceles7. x � 4 8. PR � RQ � 219. m�R � 72; m�N � 36

10. AB � AC � 36; BC � 4611. x � 10 � 2x � 2x � 30 � 180, so x � 40. The

three angles then measure 50°, 80°, and 50°. Sincetwo of the angles of the triangle measure 50°, thetriangle has two congruent angles and, therefore,

ACDEBC

BCh

BAh

PA > PC

PC ' BCh

PA ' BAh

BP > BP

BPh

two congruent sides. By definition, the triangle isisosceles.

12. x � 35 � 2x � 10 � 3x � 15 � 180, so x � 25. Thethree angles all measure 60°. Therefore, all of theangles are congruent and the triangle isequiangular. If a triangle is equiangular, then it isequilateral.

13. 3x � 18 � 4x � 9 � 10x � 180, so x � 9. Themeasures of the three angles are 45°, 45°, and 90°.Since the triangle has a right angle, it is a righttriangle. Since two of the angles of the trianglemeasure 40°, the triangle has two congruentangles and, therefore, two congruent sides. Bydefinition, the triangle is isosceles.

14. 120 15. 360Applying Skills16. Statements Reasons

1. �ABP � �PCD 1. Given.2. �ABP and �PBC are 2. If two angles form a

supplements. �PCD linear pair, then and �PCB are they are supple-supplements. mentary.

3. �PBC � �PCB 3. If two angles arecongruent, thentheir supplementsare congruent.

4. 4. If two angles of a triangle are congru-ent, then the sidesopposite these an-gles are congruent.

5. �BPC is isosceles. 5. Definition ofisosceles triangle.

17. Statements Reasons1. �PAB � �PBA 1. Given.2. 2. If two angles of a

triangle are congru-ent, then the sidesopposite these an-gles are congruent.

3. PA � PB 3. Definition of con-gruent segments.

4. P is on the perpen- 4. A point is on the dicular bisector of . perpendicular

bisector of a linesegment if and onlyif it is equidistantfrom the endpointsof the line segment.

AB

PA > PB

PB > PC

297


1. bisects �DBC. 1. Given.2. m�DBC � 2m�DBE 2. Definition of angle

bisector.

3. 3. Given.4. �A � �DBE 4. If two parallel lines

are cut by a trans-versal, then thecorresponding an-gles are congruent.

5. m�A � m�DBE 5. Definition ofcongruent angles.

6. m�DBC 6. Exterior angle � m�A � m�C theorem.

7. 2m�A 7. Substitution � m�A � m�C postulate.

8. m�A � m�C 8. Subtractionpostulate.

9. �A � �C 9. Definition ofcongruent angles.

10. 10. If two angles of a triangle arecongruent, then thesides oppositethese angles arecongruent.

19. Statements Reasons1. �PBC � �PCB, 1. Given.

�APB � �DPC2. , 2. If two angles of a

triangle are congru-ent, then the sidesopposite these an-gles are congruent.

3. �ABP and �PBC 3. If two angles form a are supplements. linear pair, then �DCP and �PCB are they are supple-supplements. mentary.

4. �ABP � �DCP 4. If two angles arecongruent, thentheir supplementsare congruent.

5. �ABP � �DCP 5. AAS.6. 6. Corresponding parts

of congruent trian-gles are congruent.

20. In �ABC, �A � �B. Draw , the altitude from �C. By definition, , so �ADCand �BDC are right angles and congruent.

CD ' ABCD

AP > DP

AB > CDPB > PC

AB > CB

BEh

y AC

BEh

by the reflexive property of congruence. �ACD � �BCD by AAS.Therefore, because correspondingparts of congruent triangles are congruent.

9-7 Proving Right Triangles Congruent byHypotenuse, Leg (pages 365–367)

Writing About Mathematics1. In two right triangles, the hypotenuse and one

leg, both legs, or one leg and its adjacent acuteangle must be congruent for the triangles to beproved congruent.

2. PD � PQ. The distance from a point to a line isthe length of the perpendicular from the point tothe line. Therefore, �PDQ is a right angle and

can be considered a leg of right triangle.Then, is the hypotenuse of the triangle, and the hypotenuse is longer than either leg.

Developing Skills3. a. m�BCA � 80

b. m�PAN � 20; m�PBN � 30; m�APB � 130c. m�PCL � 40; m�PBL � 30; m�BPC � 110d. m�PAM � 20; m�PCM � 40;

m�APC � 120e. No. �APB and �BPL form a linear pair so

are supplementary. Since m�APB � 130,m�BPL � 50. Similarly, �APC and �CPLform a linear pair so are supplementary.Since m�APC � 120, m�CPL � 60.m�BPL � m�CPL, so they are not congruent.

Therefore, does not bisect �CPB.4. a. m�A � 45; m�B � 45

b. m�PAB � 22.5; m�PBA � 22.5;m�APB � 135

c. m�PBC � 22.5; m�PCB � 45;m�BPC � 112.5

d. m�PAC � 22.5; m�PCA � 45;m�APC � 112.5

e. Yes. Extend through P to a point M.Since angles forming a linear pair are supple-mentary and m�APC � m�BPC � 112.5, wehave that m�APM � m�BPM � 67.5. Thus,

�APM � �BPM and bisects �APB.5. a. m�A � 20; m�B � 20

b. m�PAB � 10; m�PBA � 10; m�APB � 160c. m�PBC � 10; m�PCB � 70; m�BPC � 100d. m�PAC � 10; m�PCA � 70; m�APC � 100

CPh

CPh

ALh

PQPD

CA > CB

CD > CD

298

e. Yes. Extend through P to a point M.Since angles forming a linear pair are supple-mentary and m�APC � m�BPC � 100, wehave that m�APM � m�BPM � 80. Thus,

�APM � �BPM and bisects �APB.6. m�TRS � 70; m�RST � 60; m�SPT � 1257. a–c. Graph

d. No. Since �A is not congruent to �B,�PAB and �PBA (halves of these angles) are also not congruent. If the measures of two angles of a triangle are unequal, thelengths of the sides opposite these angles are unequal. Therefore, in �APB, sincem�PAB � m�PBA, AP � BP. Similarly,AP � CP and BP � CP.

e. Equilateral


1. at D, 1. Given.

at F.2. �PDB and �PFB 2. Definition of

are right angles. perpendicular lines.3. PD � PF 3. Given.4. 4. Definition of con-

gruent segments.5. 5. Reflexive property.6. �ABP � �PFB 6. HL.7. �ABP � �CBP 7. Corresponding parts


9. When �ADC is an isosceles right triangle,� and �A and �C are congruent

complementary angles. Since ⊥ ,�BDC and �C are also complementary. If twoangles are complements of the same angle,then they are congruent. Thus, �BDC ��A � �C, and so � by the converse of the Isosceles Triangle Theorem. By HL,we can conclude that �ABD � �DBC. Thus,�ABD � �DBC when �ADC is an isoscelesright triangle.

10. If and are parallel, then �PAB and�PBA are supplementary. By the partitionpostulate, and

. Since a whole isgreater than any of its parts, m�PAB � m�Aand m�PBA � m�B. But in �ABC, the sum of

m/B 5 m/CBP 1 m/PBAm/A 5 m/CAP 1 m/PAB

BMAL

BCDB

ACDBDCAD

BP > BP

PD > PF

PF ' BCh

PD ' BAh

CPh

CPh

the measures of �A, �B, and �C is 180°. Thus,m�A � m�B � 180, and so:

Since �PAB and �PBA are supplementary,m�PAB � m�PBA � 180, and we arrive at

, a contradiction. Thus,the assumption is false, and and are notparallel.

11. Statements Reasons1. , 1. Given.2. �ABD and �CDB 2. Definition of

are right angles. perpendicular lines.3. 3. Given.4. 4. Reflexive property.5. �ABD � �CDB 5. HL.6. �A � �C, 6. Corresponding

�ADB � �CBD parts of congruenttriangles arecongruent.

7. 7. If two coplanar lines are cut by atransversal so thatthe alternate inte-rior angles formedare congruent, thetwo lines areparallel.

12. a. Statements Reasons1. In �QRS, the 1. Given.

bisector of �QRSis perpendicular to

at P.2. �QRP � �SRP 2. Definition of angle

bisector.3. 3. Reflexive property.4. �QPR and �PSR 4. Definition of

are right angles. perpendicular lines.5. �QPR � �PSR 5. Right angles are

congruent.6. �QPR � �PSR 6. ASA.7. 7. Corresponding


8. �QRS is isosceles. 8. Definition ofisosceles triangle.

QR > SR

RP > RP

QS

AD y CB

BD > BDAD > CB

BD ' DCAB ' BD

BMAL180 , m/A1m/B , 180

m/PAB1m/PBA , m/A1m/B , 180

299

b. Statements Reasons1. �QPR � �PSR 1. Part a.2. 2. Corresponding parts

of congruent trian-gles are congruent.

3. P is the midpoint 3. Definition of of . midpoint.

13. Given: Isosceles triangle �ABC with vertex �B,D the midpoint of , , and

.Prove:Proof: By the isosceles triangle theorem, since

in �ABC, �A � �C. Since D is themidpoint of , . Since and

, �CED and �AFD are right anglesand congruent. Therefore, �CED � �AFD byAAS. Then because correspondingparts of congruent triangles are congruent.

14. a. In ABCD, �A and �C are right angles, so �A� �C. Since AB � CD, . By thereflexive property of congruence, ,so �ABD � �CDB by HL. Sincecorresponding parts of congruent triangles are congruent, and AD � BC.

b. From part a, �ABD � �CDB. Correspondingparts of congruent triangles are congruent, so�ABD � �CDB.

c. From parts a and b, �ABD � �CDB. Sincecorresponding parts of congruent triangles are congruent, �ADB � �CBD and m�ADB � m�CBD. The acute angles of aright triangle are complementary, so m�CDB � m�CBD � 90. By the substitutionpostulate, m�CDB � m�ADB � 90.Since m�CDB � m�ADB � m�ADC,m�ADC � 90, and �ADC is a right angle.

15. In ABCD, �ABC and �BCD are right anglesand therefore congruent. Since AC � BD,

. By the reflexive property ofcongruence, , and �ABC � �DCB byHL. Corresponding parts of congruent trianglesare congruent, so and AB � CD.AB > CD

BC > BCAC > BD

AD > BC

BD > BDAB > CD

DE > DF

DF ' ABDE ' BCAD > CDAC

AB > BC

DE > DFDF ' AB

DE ' BCAC

QS

QP > SP

16. Since and �ABP and �ADP are right angles and congruent. We are

given that PB � PD, so . by the reflexive property of congruence. Therefore,�ABP � �ADP by HL. Corresponding parts ofcongruent triangles are congruent, so �BAP ��DAP or �CAS � �EAS. By the definition of

angle bisector, bisects �CAE.

9-8 Interior and Exterior Angles ofPolygons (pages 371–372)

Writing About Mathematics1. Yes. A diagonal may be drawn from every vertex

of a polygon to every other vertex except foritself and the two adjacent vertices. Therefore,each vertex of an n-sided polygon is an endpointof (n � 3) diagonals.

2. Yes. Each vertex is an endpoint of (n – 3)diagonals, so n vertices are the endpoints of n(n � 3) diagonals. A diagonal has two endpoints, so there are diagonals.

Developing Skills3. a. 180 b. 900 c. 1,2604. a. 360 b. 1,080 c. 540 d. 1,4405. a. 360 b. 360 c. 360 d. 3606. a. 90 b. 907. a. 72 b. 1088. a. 60 b. 1209. a. 45 b. 135

10. a. 40 b. 14011. a. 30 b. 15012. a. 18 b. 16213. a. 10 b. 17014. a. b.15. a. 12 b. 8 c. 6 d. 316. a. 4 b. 6 c. 9 d. 1817. a. 3 b. 4 c. 5 d. 7

e. 10 f. 17 g. 12 h. 22Applying Skills18. 8 sides 19. 9 sides20. No. If two angles were greater than 180 degrees,

then the sum of the interior angles would alreadyexceed 360 degrees, which is the sum of themeasures of the angles of a quadrilateral.

21. a. From �A in ABCDE, draw diagonals andto form �ABC and �AED. Since

ABCDE is regular, it is both equilateral and equiangular. Therefore,

, �B � �E, and �ABC � �AED by SAS.AB > BC > AE > ED

ADAC

1713784

7

n2(n 2 3)

APS›

AP > APPB > PD

PD ' ADE›,

PB ' ABC›

300

b. Since in �ABC and in�ABD, the triangles are isosceles.

c. Since �ABC � �AED, and correspondingparts of congruent triangles are congruent,

and �DAC is isosceles.

22. a. From �L in LNMRST, draw diagonals ,, and . Since LMNRST is regular, it is

both equilateral and equiangular. Therefore,, �M � �T, and

�LMN � �LTS by SAS.

b. �LMN � �LTS and corresponding parts of congruent triangles are congruent, so

. Because LMNRST is regular,

. By the reflexive property ofcongruence, so �LNR � �LSR by SSS.

c. m�M � m�T � 120;m�MLN � m�MNL � m�TLS

� m�TSL � m�NLR � m�SLR � 30;m�LSR � m�LNR � 90;m�LRS � m�LRN � 60

23. a. Slope of � slope of � �1Slope of � slope of � 1The slopes of adjacent sides are negativereciprocals, therefore , ,

, and �A, �B, �C, and �D areright angles.

b. The x-axis is horizontal and the y-axis isvertical, so they intersect to form right angles, which are congruent, so �AOB � �BOC � �COD � �DOA.Also, AO � CO � BO � DO � 2, so

and �AOB � �BOC � �COD � �DOAby SAS.

c. Quadrilateral ABCD is equiangular because it contains four right angles. It is equilateralbecause each of its sides is the hypotenuse of a congruent right triangle. Therefore, it isregular.

Hands-On Activitya. No b. Yesc. The angle bisectors of a polygon are concurrent

if and only if the polygon is regular.

Review Exercises (pages 375–376)1. 15 2. 30 3. 24 4. 555. a. 18 b. 72 c. 72 d. 108 e. 72

AO > CO > BO > DO

CD ' DABC ' CDAB ' BC

DABCCDAB

LR > LRNR > RSLN > LS

LM > MN > LT > TS

LSLRLN

AC > AD

AE > EDAB > BC

6. 30 7. 6 8. 369. 42, 48, 90 10. 120 11. 80

12. 40 13. 154 14. 6015. 28 16. 20, 80, 80 17. 9018. 30 19. 12 20. 1,26021. In �ABC, �C is a right angle so m�C � 90.

Therefore, m�B � 90 since �B arecomplementary so �B must be acute. If themeasures of two angles of a triangle are unequal,the lengths of the sides opposite these angles areunequal and the longer side lies opposite thelarger angle. Therefore, AB � AC.

22. Since and bisect each other,and . Vertical angles are

congruent, so �AEC � �BED. Therefore,�AEC � �BED by SAS. Corresponding parts of congruent triangles are congruent, so �EAC � �EBD. If two coplanar lines are cut bya transversal so that the alternate interior anglesformed are congruent, then the two lines are parallel. Therefore, .

23. In �BPA, , so by the isosceles triangle theorem �PBC � �PCB. �PBA and �PBCform a linear pair, so they are supplementary.Also, �PCD and �PCB form a linear pair, sothey are supplementary. If two angles arecongruent, then their supplements are congruent, so �PBA � �PCD. It is given that�APB � �DPC, so �ABP � �DCP by ASA.Corresponding parts of congruent triangles are congruent, so .

24. In �BPC, �PBC � �PCB. If two angles of atriangle are congruent, then the sides opposite these angles are congruent, so . It is given that �PBC � �PCB. �PBA and �PBCform a linear pair, so they are supplementary.Also, �PCD and �PCB form a linear pair, sothey are supplementary. If two angles arecongruent, then their supplements are congruent,so �PBA � �PCD. Since , �ABP ��DCP by SAS. Corresponding parts of congruent triangles are congruent, so .

25. No. The sum of the measures of the interior an-gles of a pentagon is 540 degrees, and angles A, B,C, and D have measures which already sum to 540.Since m�E � 0, Herbie’s pentagon is not possible.

Exploration (page 376)Results will vary.

PA > PD

AB > DC

PB > PC

PA > PD

BP > CP

AC y BD

CE > EDAE > EBCEDAEB

301


1. 4 2. 2 3. 1 4. 3 5. 26. 3 7. 1 8. 1 9. 1 10. 2

Part II

11. ; m�C � 45 and m�B � 90. If the measures of two angles of a triangle are unequal, thelengths of the sides opposite these angles areunequal and the longer side lies opposite the larger angle. Therefore, is the longest side.

12. Since a point on the perpendicular bisector of asegment is equidistant from the endpoints of the segment, AP � BP. Therefore, and�ABP is isosceles.

Part III

13. Since ABCD is equilateral,. By the reflexive

property of congruence, , so �ABC � �ADC by SSS. Corresponding parts of congruent triangles are congruent, so �BAC � �DAC and �BCA � �DCA. By thedefinition of angle bisector, bisects �DAB and �DCB.

14. We are given that . If two parallel lines are cut by a transversal, then the alternateinterior angles formed are congruent. Therefore,�ECB � �EDA and m�ECB � m�EDA.By the exterior angle theorem,

m�DEB � m�EBC � m�ECB.

By the substitution postulate,

m�DEB � m�EBC � m�EDA.Part IV15.

3x � 36, 4x � 48, 8x � 96The smallest exterior angle is the supplement of the largest interior angle, so its measure is 180 � 96 � 84.

16. The midpoint of

The slope of

The perpendicular bisector has a slope of �1 andgoes through (3, 2):

The equation of the perpendicular bisector is y � �x � 5.

b 5 52 5 23 1 by 5 2x 1 b

AB 56 2 (22)7 2 (21) 5 1

AB 5 A21 1 72 , 22 1 6

2 B 5 (3, 2)

x 5 1215x 5 180

3x 1 4x 1 8x 5 180

ADg

y CBg

AC

AC > ACAB > BC > CD > DE

AB > BP

AC

AC

10-2 The Parallelogram (pages 383–385)Writing About Mathematics

1. No. If opposite sides are supplementary, thequadrilateral is not necessarily a parallelogram.For example, consider a quadrilateral with anglemeasures 30°, 40°, 150°, and 140°. Oppositeangles are supplementary but not congruent, andso the quadrilateral is not a parallelogram.

2. No. Consider parallelogram ABCD with thediagonals intersecting at the point E. The fourtriangles formed are �ABE, �BCE, �CDE, and �ADE. Since and , in orderfor �ABE to be congruent to �BCE, wouldneed to be congruent to adjacent side .However, this is not necessarily true.

Developing Skills3. a. 70, 110, 110 b. 65, 115, 115

c. 90, 90, 90 d. 50, 130, 130e. 25, 155, 155 f. 12, 168, 168

4. x � 50; m�A � 80; m�B � 1005. x � 34; m�A � 78; m�B � 1026. x � 70; m�A � m�C � 110; m�B � m�D � 707. x � 22; m�A � m�C � 66; m�B � m�D � 1148. x � 15; m�A � m�C � 75; m�B � m�D � 1059. x � 120; m�A � m�C � 60; m�B � m�D � 120

10. x � 5; AB � CD � 2711. x � 2; y � 2; AB = CD � 12; BC � DA � 812. x � 3; AC � 1213. y � 2; DB � 9

Applying Skills

14. In parallelogram ABCD, diagonal divides the parallelogram into two congruent triangleswith �ABC � �CDA. Since corresponding parts of congruent triangles are congruent,and . Therefore, the opposite sides of a parallelogram are congruent.

15. In a parallelogram ABCD, diagonal divides the parallelogram into two congruent triangleswith �ABC � �CDA. Since corresponding partsof congruent triangles are congruent, �B � �D.Similarly, diagonal divides the parallelogram into two congruent triangles with �ABD ��CDB, and �A � �C. Therefore, the oppositeangles of a parallelogram are congruent.

BD

AC

AD > BCAB > CD

AC

BCAB

BE > EDAE > EC

302

16. Statement Reason1. Parallelograms EBFD 1. Given.

and ABCD2. 2. Opposite sides of a

parallelogram arecongruent.

3. �E � �F, 3. Opposite angles of �BAD � �DCB a parallelogram are

congruent.4. �BAD and �EAD 4. If two angles form a

are supplements, linear pair, then �DCB and �FCB are they are supple-supplements. mentary.

5. �EAD � �FCB 5. If two angles arecongruent, thentheir supplementsare congruent.

6. �EAD � �FCB 6. AAS.

17. In a parallelogram, consecutive angles aresupplementary. Thus, the corners adjacent to thecorner forming a right angle are also right angles.In a parallelogram, opposite angles arecongruent. Thus, the corner opposite the cornerforming a right angle is also a right angle.Therefore, Petrina’s floor has four right angles.

18. Suppose that ABCD is a parallelogram. Thenopposite angles �A and �C are congruent andm�A � m�C. However, we are given that m�Aand m�C are not equal, which is a contradiction.Therefore, the assumption is false, and ABCD isnot a parallelogram.

19.

20. Since , , and �ABC � �PQR by SAS. Since a diagonaldivides a parallelogram into two congruenttriangles, �ABC � �CDA and �PQR � �RSP.By the transitive property of congruence,

/B > /Q,QR > BCPQ > AB

B C

A D

Q R

P S

ED > BF

Chapter 10. Quadrilaterals

�ABC � �CDA � �PQR � �RSP. Sincecorresponding parts of congruent triangles are congruent, ,

, and �D � �B � �S � �Q. Since �C and �A are supplementary to �B, and �Rand �P are supplementary to �Q, �B � �Q,�C � �A � �R � �P because the supplementsof congruent angles are congruent. Thus, all sidesand angles of one parallelogram are congruent tothe corresponding sides and angles of the otherparallelogram, and so the parallelograms arecongruent.

10-3 Proving That a Quadrilateral Is aParallelogram (pages 387–388)


1. because opposite sides of aparallelogram (EBFD) are congruent.

2. because opposite angles in aparallelogram (EBFD) are congruent.

Developing Skills3. A pair of opposite sides is both parallel and

congruent.4. Two pairs of opposite sides are congruent.5. Both pairs of opposite angles are congruent.6. Both pairs of opposite sides are parallel.7. The diagonals bisect each other.

8. Since , �D is the supplement of �A, and �B is the supplement of �C. Since thesupplements of congruent angles are congruentand �A � �C, . If both pairs ofopposite angles are congruent, then thequadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.

9. Since is supplement of , . is a

transversal that cuts and . Thus, �S is the supplement of �R. Since the supplements ofcongruent angles are congruent, . Ifboth pairs of opposite angles are congruent, thenthe quadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.

10. Since �GDF � �DFE, �EDF � �GFD, andby the reflexive property, �EDF �

�GFD by ASA. Therefore, andsince corresponding parts of

congruent triangles are congruent. If oppositesides of a quadrilateral are congruent, then thequadrilateral is a parallelogram. Therefore,DEFG is a parallelogram.

GD > FEGF > DE

DF > DF

/Q > /S

QRg

SPg

SRPSg

y QRg

/Q/P

/B > /D

ABg

y CDg

/DEB > /BFD

DE > FB

SR > PQDC > AB >DA > BC > SP > QR

303

11. In a parallelogram, opposite sides are congruent.Thus, . Since E and F are the midpointsof and , respectively, and

. Thus, since congruenthalves of congruent segments are congruent.

Since ABCD is a parallelogram, , and so

and are also parallel. Therefore, AEFD is a parallelogram because one pair of sides is bothcongruent and parallel.

12. EFGH is a parallelogram, so and. Segments of parallel lines are

parallel, so . Since F is the midpoint of, . By the transitive property of

congruence, . If one pair of opposite sides of a quadrilateral is both congruent andparallel, the quadrilateral is a parallelogram.Therefore, FJGH is a parallelogram.

13. a. Since P, Q, R, and S are midpoints of , ,, and , respectively, ,

, , and . SinceABCD is a parallelogram, opposite sides arecongruent, and so and .Since congruent halves of congruent segmentsare congruent, , ,

, and . Since ABCD is a parallelogram, opposite angles are congruent,and so �A � �C and �D � �B. Therefore,�APS � �CRQ and �BQP � �DSRby SAS.

b. In part a we showed that �APS � �CRQ and�BQP � �DSR. Since corresponding partsof congruent triangles are congruent,

and . Therefore, PQRS is a parallelogram since both pairs of oppositesides are congruent.

14. Yes. Since sum of the measures of the angles in aquadrilateral is 360°, the fourth angle also has tomeasure 90°. Therefore, the quadrilateral is aparallelogram since both pairs of opposite anglesare congruent.

Applying Skills

15. Since the diagonals and bisect eachother, and . Since vertical angles are congruent, �AEB � �DEC.Therefore, �ABE � �CDE by SAS. Therefore,

and �CDE � �EBA because corresponding parts of congruent triangles are AB > DC

BE > EDAE > ECDBAC

SP > RQSR > PQ

DS > QBSA > CQRD > PBAP > RC

BC > DAAB > DC

DS > SACR > RDBQ > QCAP > PBDACD

BCAB

FJ > GHEF > FJEJ

FJ y GHEF > GH

EF y GH

AEg

DFg

ABg

y DEg

AE > DFAE > EBDF > FEDEAB

AB > DE

congruent. is a transversal that cuts and

with congruent interior angles �CDE and �EBA. If two lines are cut by a transversal andthe alternate interior angles formed arecongruent, then the two lines are parallel. Thus,

, and so ABCD is a parallelogram since a pair of sides is both congruent and parallel.

16. If the diagonals of a quadrilateral bisect eachother, then the quadrilateral is a parallelogram.Therefore, a quadrilateral drawn by joining theendpoints of two line segments that bisect eachother is a parallelogram.

17. Slope of

Slope of Slope of

Slope of Both pairs of opposite sides are parallel.Therefore, ABCD is a parallelogram.

18. Suppose that the diagonal paths bisect eachother. Then the quadrilateral is a parallelogram.However, we are given that the quadrilateral isnot a parallelogram, a contradiction. Therefore,the assumption is false, and the diagonal paths donot bisect each other.

19. Yes. Since �ABC � �A�B�C�, �C � �C�,, and .

Thus, opposite angles �C and �C� are congruentin the quadrilateral, ,and . Since A coin-cides with B� and B coincides with A�,

andby

the partition postulate. Substitutingand

, we arrive at . Thus, m�CAC� � m�CBC� by the

transitive property, and �CAC� � �CBC�.Therefore, both pairs of opposite angles arecongruent in the quadrilateral, and so thequadrilateral is a parallelogram.

20. a. In a parallelogram, opposite sides arecongruent. Thus, . Since M and Nare midpoints, and .Since congruent halves of congruent segmentsare congruent, and .

Since ABCD is a parallelogram, and

. Therefore, AMND and MBCN are MBg

y NCg

AMg

y DNg

MB > CNAM > DN

CN > NDAM > MBAB > CD

1 m/CABm/CBCr 5 m/CrBrArm/CrArBr

m/CAB 5m/CBA 5 m/CrBrAr

m/CACr 5 m/CAB 1 m/CrBrArm/CBCr 5 m/CBA 1 m/CrArBr

m/CAB 5 m/CrArBrm/CBA 5 m/CrBrAr

/CAB > /CrArBr/CBA > /CrBrAr

AD 5 1

CD 5 212

BC 5 1

AB 5 212

ABg

y CDg

DCg

ABg

DBg

304

parallelograms because, in both quadrilaterals,one pair of sides is both congruent andparallel.

b. Since opposite angles of parallelogramsABCD, AMND, and MBCN are congruent,�A � �C, �D � �B, , �D ��AMN, �NMB � �C, and �MNC � �B. Bythe transitive property, �A � �NMB ��DNM � �C and �D � �MNC � �AMN� �B. Thus, corresponding angles arecongruent.

Since opposite sides of a parallelogramsAMND and MBCN are congruent,

and . By transitivity,. Since M and N are

midpoints, and . Thus,corresponding sides are congruent.

Therefore, the two parallelograms arecongruent since corresponding sides andangles are congruent.

10-4 The Rectangle (pages 392–393)Writing About Mathematics

1. Yes. If the parallelogram were a rectangle, thenall of its angles would be right angles, but thiscontradicts the given condition.

2. Yes. By Exercise 16 on page 388, thequadrilateral is a parallelogram. If the diagonalsof a parallelogram are congruent, theparallelogram is a rectangle. The given segmentsform the diagonals of this quadrilateral. Since thesegments are congruent, the quadrilateral is alsoa rectangle.

Developing Skills

3. Since diagonals of a rectangle are congruent andalso bisect each other, . Therefore,

is isosceles.

4. x � 8; AC � BD � 38; AE � BE � 145. y � 10; AE � DE � 22; AC � BD � 446. a � 4; BE � ED � 13; BD � AC � 267. x � 12; AE � 12; BD � AC � 24; BE � 128. ; m�AEB � 110;

m�AED � 709. x � 40; m�AEB � m�DEC � 120;

m�CAB � 30; m�CAD � 6010. y � 40; m�AED � 50; m�AEB � 130;

m�CAB � 25; m�CAD � 65Applying Skills

11. Sides and are horizontal segments sincethe endpoints of the segments have the same

DCAB

m/ACB 5 m/CAD 5 55

nAEBAE > EB

DN > NCAB > MBAD > NM > CB

NM > CBAD > NM

/A > /DNM

y-coordinates. Similarly, and are verticalsegments. Thus, , , ,and . Since perpendicular lines meet toform right angles, �A, �B, �C, and �D are rightangles.

12. Since rectangle ABCD is a parallelogram,opposite sides and are congruent. Since �DAB and �ABC are right angles, �DAB ��ABC. Since by the reflexive property, �DAB � �CBA by SAS. Therefore,

because corresponding sides of congruent triangles are congruent.

13. Since opposites sides of a parallelogram arecongruent, . Since the diagonals arecongruent, . Thus, since by the reflexive property, �DAB � �CBA by SSS.Consecutive angles of a parallelogram aresupplementary, and so �DAB and �CBA aresupplementary. However, they are alsocorresponding angles of congruent triangles�DAB and �CBA. Thus, �DAB � �CBA, andthese two angles are both congruent andsupplementary. Since congruent andsupplementary angles are right angles, �DABand �CBA are both right angles. Therefore,parallelogram ABCD is also a rectangle.

14. Since M is the midpoint of , . Sinceopposite sides of a rectangle are congruent,

. Since the angles of a rectangle are allright angles, �S � �R. Thus,by SAS. Since corresponding sides of congruenttriangles are congruent, .

15. a. Slope of Slope of

Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel, andthe quadrilateral is a parallelogram. The slopes of adjacent sides and are negative reciprocals. Therefore, these twosides are perpendicular and form a right angle.Since a parallelogram with one right angle is arectangle, ABCD is a rectangle.

b.

c. 90° clockwise rotation about the origin.d. Under a rotation, angle measure is preserved.

Thus, , , , and, and the angles of areArBrCrDr/D > /Dr

/C > /Cr/B > /Br/A > /Ar

A 32, 2 B

ABAD

BC 5 2AD 5 2

DC 5 212AB 5 21

2

PM > QM

nQRM > nPSMSP > RQ

SM > MRRS

AB > ABDB > ACDA > CB

AC > BD

AB > AB

CBDA

CD ' DABC ' CDAB ' BCDA ' AB

DABC

305

all right angles. Therefore, is arectangle.


Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel andthe quadrilateral is a parallelogram.

b. Suppose that PQRS were a rectangle. Then allof its angles would be right angles. However,since the slopes of adjacent sides and are not negative reciprocals, and are not perpendicular, and �Q is not a right angle,a contradiction. Therefore, the assumption isfalse, and PQRS is not a rectangle.

c. P�(�2, �5), Q�(�6, �2), R�(�2, 2), S�(2, �1)d. Since reflections and translations preserve

distance, , , ,and . Thus, corresponding sides arecongruent. Since reflections and translationspreserve angle measure, ,

, , and . Thus,corresponding angles are congruent, and

.

17. Suppose that ABCD were a parallelogram.Since �A is a right angle, ABCD is a rectangle.However, we are given that ABCD is not arectangle, a contradiction. Therefore, theassumption is false, and ABCD is not aparallelogram.

18. Since she can only measure distance, she can first find the midpoint of the segment formed by the stakes she has already placed in theground. Through this midpoint, she can drawanother line. She can then draw two other points on each side of the midpoint at an equaldistance away. The two remaining stakes can be placed on these two points to form arectangle.

19. The diagonals of a rectangle are congruent andbisect each other. Archie should locate themidpoint of each string by folding it in half. Then,with one end of each string at a different vertex,he should cross the strings at their midpoints. Thetwo remaining ends of the stings are located atthe other two vertices.

PQRS > PrQrRrSr

/S > /Sr/R > /Rr/Q > /Qr/P > /Pr

SP > PrSrRS > RrSrQR > QrRrPQ > PrQr

PQQRPQQR

SR 5 243PS 5 1

PQ 5 243QR 5 1

ArBrCrDr

10-5 The Rhombus (pages 396–398)Writing About Mathematics

1. Yes. In rhombus ABCD with diagonals andintersecting at E, the sides are all congruent,

and the diagonals bisect each other and areperpendicular. Thus, � ,

, , and �AED, �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles).Thus, �AED, �CED, �CEB, and �AEB are allright triangles. Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.

2. Yes. The diagonals of a rhombus separate therhombus into four congruent right triangles.

Thus, each triangle has a base of length and a

height of . The area of each triangle is .

Therefore, the area of the rhombus is

.

Developing Skills3. AB, BC, CD, DA4. AE and EC; DE and EB5. AC, DB 6.7. No, since the diagonals are not congruent.8. No 9. C, D, A, B, E

10. Yes11. a. The diagonal bisects �P into two 60°

angles and �R into two 60° angles. Since thesum of the measures of the angles of a triangleis 180°, �S and �Q both measure 60°. Thus,diagonal separates the rhombus into two equilateral triangles �PRS and �PRQ.

b. 24 cm12. Yes13, 14. No 15. Yes

Applying Skills

16. In rhombus ABCD with diagonals and intersecting at E, the sides are all congruent, andthe diagonals bisect each other and are perpendicular. Thus, � ,

, , and �AED , �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles). Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.Since corresponding sides of congruent triangles

DE > EBAE > ECAD > BCAB > DC

DBAC

30°20°

120°40° 90°

PR

PR

/AEB, /BEC, /CED, AED

(4) A d1d28 B 5

d1d22

d1d28

d22

d12


DBAC

306

are congruent, , ,, and .

Therefore, bisects �DAB and �DCB, andbisects �CDA and �CBA.

17. In parallelogram ABCD with diagonals andintersecting at E, since the

diagonals bisect each other. Since the diagonalsare perpendicular, �AED and �AEB are bothright angles. Thus, �AED � �AEB. Since

by the reflexive property,�AED � �AEB by SAS. Therefore,because corresponding parts of congruenttriangles are congruent, and ABCD is a rhombus.


Slope of Slope of The slopes of opposite sides are equal.Therefore, the opposite sides are parallel,and ABCD is a parallelogram.

b. Slope of � 1 Slope of The slopes are negative reciprocals. Therefore,the two diagonals are perpendicular to eachother.

c. Yes. If the diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Since ABCD is a parallelogram withperpendicular diagonals, ABCD is a rhombus.

19. �KLM is an equilateral triangle. Each angle of an equilateral triangle measures 60°. Thus,�KLM is a 60° angle.

20. In rhombus ABCD with diagonals and intersecting at E, the sides are all congruent, andthe diagonals bisect each other and are perpendicular. Thus, � ,

, , and �AED, �DEC,�AEB, and �BEC are all right angles (sinceperpendicular lines meet to form right angles).Thus, �AED, �CED, �CEB, and �AEB are allright triangles. Since right angles are congruent,�AED � �CED � �CEB � �AEB by SAS.

21. Since the diagonals bisect each other, thequadrilateral is a parallelogram. Since thediagonals are perpendicular to each other, theparallelogram is a rhombus.

22. In parallelogram ABCD, opposite angles arecongruent, so . If bisects�DAB and �DCB, then and/DAC > /CAB

AC/DAB > /DCB


DBAC

DB 5 21AC

BC 5 5CD 5 15

DA 5 5AB 5 15

AD > ABAE > AE

DE > EBDBAC

DBAC

/ABE > /CBE/ADE > /CDE/DCE > BCE/DAE > /EAB

. Since congruent halves ofcongruent angles are congruent. �DAC hascongruent base angles, and so �DAC is isosceles.Since �DAC is isosceles, , and the parallelogram has two congruent consecutivesides. Therefore, ABCD is a rhombus.

23. Since a rhombus is a parallelogram, the diagonalsof ABCD bisect each other. Since the diagonalsof a rhombus are perpendicular to each other,the diagonals of ABCD are the perpendicular bisectors of each other. Therefore, is theperpendicular bisector of . Thus, P is on theperpendicular bisector of . If a point is on the perpendicular bisector of a segment, then it isequidistant from the endpoints of the segment.Therefore, AP � CP, and so .

24. a. Slope of

Slope of

The slopes of opposite sides and arethe same. Therefore, and are parallel.Since and are both vertical lines, they are parallel. Therefore, oppositesides of quadrilateral ABCD are parallel, andthe quadrilateral is a parallelogram.

b. (0, 0)c. Slope of

Slope of The slopes are negative reciprocals.Therefore, the diagonals are perpendicular toeach other.

d. Yes. If the diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Since ABCD is a parallelogram withperpendicular diagonals, ABCD is a rhombus.

25. a, b. In Exercise 20 on page 388, we proved thatAMND and MBCN are both parallelograms. Since , AMND isa rhombus since it is a parallelogram withtwo congruent consecutive sides. Since M isthe midpoint of , . By thetransitive property, . Sinceopposite sides are congruent inparallelogram ABCD, . Thus,

by transitivity, and MBCN isalso a rhombus.

c. Since opposite angles of parallelogramsABCD, AMND, and MBCN are congruent,

MB > BCAD > BC

MB > ADAM > MBAB

AD > AM

DB 5 22

AC 5 12

CBADABDC

ABDC

DC 5 234

AB 5 234

AP > CP

ACAC

DB

AD > DC

/DCA > /ACB

307

�A � �C, �D � �B, ,�D � �AMN, �NMB � �C, and �MNC � �B. By the transitive property,�A � �NMB � �DNM � �C and �D � �MNC � �AMN � �B. Thus,corresponding angles are congruent.

Since the sides of rhombuses AMND andMBCN are all congruent,

and .Thus, corresponding sides are congruent.

Therefore, the two parallelograms arecongruent since corresponding sides andangles are congruent.

Hands-On Activity1–2. Answers will vary.3–4. The two endpoints of the diagonal constructed

in step 1 will be two opposite vertices of therhombus. Since any point on the perpendicularbisector of a segment is equidistant from theendpoints of the segment, the third vertex willbe equidistant from the two opposite vertices.Thus, the two consecutive sides formed will becongruent. The fourth vertex will be on theperpendicular bisector at the same distanceaway from the segment as the third vertex.Since the diagonals formed bisect each other,the resulting quadrilateral is a parallelogram.Since the parallelogram has two congruentconsecutive sides, the parallelogram is arhombus.

5. Answers will vary.

10-6 The Square (pages 401–402)Writing About Mathematics

1. Yes. Since a square is a rhombus, its diagonalsseparate it into four congruent right triangles andbisect the angles of the square. Since each angleof a square measures 90°, the resulting angles areall 45°. Thus, the base angles of the trianglesformed are congruent, and the triangles arecongruent isosceles right triangles.

2. Yes. If a quadrilateral is equilateral, thenconsecutive sides are congruent. If aquadrilateral is equiangular, then it is a rectangle.If a quadrilateral is a rectangle with twocongruent consecutive sides, then it is a square.Therefore, an equilateral and equiangularquadrilateral is a square. Conversely, since asquare is a rhombus, it is equilateral, and since allof its angles are right angles, it is equiangular.

NM > MB > BC > CNNM > DNAD > AM >

/A > /DNM

Developing Skills3. Since a square is a rhombus, its diagonals are the

perpendicular bisectors of each other.

4. x � 3; AC � BD � 11; AM � MB �

5. a � b � 5; AB � BC � CD � DA � 106. x � 54; y � 18 7. Yes8. Yes9. No 10. Yes

11. See the counterexample for Exercise 9.Applying Skills12. A rhombus is a parallelogram. Thus, ABCD is a

parallelogram. Since a parallelogram with oneright angle is a rectangle, ABCD is a rectangle. Arectangle with two congruent consecutive sides isa square. Therefore, since ABCD is also a rhombus, , and so ABCD is a square.

13. Since a square is a rhombus, its diagonals are theperpendicular bisectors of each other.

14. See the answer to Exercise 1.

15. Since the diagonals bisect each other, thequadrilateral formed is a parallelogram. Sincethe diagonals are congruent, the parallelogram isa rectangle. Since the diagonals areperpendicular to each other, the parallelogram isa rhombus. Thus, the parallelogram is a rhombusand a rectangle. In particular, it is a rhombus witha right angle. Therefore, since a rhombus with aright angle is a square, the quadrilateral formedis a square.

16. Let ABCD be a square with A(0, 0), B(2a, 0),C(2a, 2a), and D(0, 2a). Then P, the midpoint of

, is at (0, a); Q, the midpoint of , is at (2a, a); R, the midpoint of , is at (a, 2a); and S the midpoint of , is at (0, a). The slope of

� the slope of � 1, so . The slopeof � the slope of � �1, so .Therefore, PQRS is a parallelogram. Since isvertical and is horizontal, the diagonals areperpendicular to each other and PQRS is arhombus. Since the slopes of and areQRPQ

SQPR

QR y SPSPQRPQ y SRSR PQ

DACD

BCAB

AB > BC

512

308

negative reciprocals, �PQR is a right angle andPQRS is a square.

17. a. Midpoint of � (4, 1)Midpoint of � (4, 1)The midpoints are the same. Therefore, thediagonals bisect each other.

b. is a horizontal segment. is vertical segment. Since vertical and horizontal linesare perpendicular, the two segments areperpendicular to each other.

c. Since the diagonals bisect each other, PQRS isa parallelogram. Since PR � SQ � 6, thediagonals are congruent, and so parallelogramPQRS is a rectangle. Thus, PQRS has a rightangle. Since the diagonals are perpendicular toeach other, parallelogram PQRS is also arhombus. A rhombus with a right angle is asquare. Therefore, PQRS is a square.

d. 90° rotation about the origin.18. a. Midpoint of � (1, 2)

Midpoint of � (1, 2)The midpoints are the same. Therefore, thediagonals bisect each other.

b. is a horizontal segment. is a vertical segment. Since vertical and horizontal linesare perpendicular, the two segments areperpendicular to each other.

c. Since the diagonals bisect each other, ABCDis a parallelogram. Since AC � DB � 8, thediagonals are congruent, and so parallelogramABCD is a rectangle. Thus, ABCD has a rightangle. Since the diagonals are perpendicular toeach other, parallelogram ABCD is also arhombus. A rhombus with a right angle is asquare. Therefore, ABCD is a square.

d.

10-7 The Trapezoid (pages 408–409)Writing About Mathematics

1. No. If ABCD is a trapezoid with , then �A and �D are supplementary. Thus, if �A is aright angle, �D must also be a right angle.

2. No. Suppose that trapezoid ABCD has threeobtuse angles �A, �B, and �C. Then, �A and�B cannot be supplementary, and �B and �Ccannot be supplementary (since each angle measures more than 90°). Thus, cannot beparallel to , and cannot be parallel to .DCABCB

AD

ABg

y CDg

T23,1

DBAC

BDAC

SQPR

QSPR

The trapezoid has no parallel sides, acontradiction.

Developing Skills3. a. b.

c.4. x � 6 5. y � 3; AD � 66. x � 20; m�A � m�B � 75; m�D � m�C � 1057. x � 10; ;

8. AD � DC � BC � 11, AB � 229. True. An isosceles trapezoid is a trapezoid in

which nonparallel sides are congruent.10. False. Two nonparallel sides can be congruent to

one of the bases.

11. False. Base angles are congruent only in anisosceles trapezoid.

12. True. The diagonals of a trapezoid are congruentif and only if the trapezoid is isosceles.

13. True. The trapezoid is a quadrilateral. The sum ofthe measures of the angles of a quadrilateral is360 degrees.

14. True. The trapezoid consists of two parallel linescut by two transversals. The angles formed by thetwo sides and the two parallel bases form twopairs of supplementary angles.

Applying Skills

15. Since , �T is the supplement of �Q, and�S is the supplement of �R. Since thesupplements of congruent angles are congruentand �Q � �R, �T � �R.

16. Draw P on such that . Then QPST is a parallelogram since both pairs of sides areparallel. In a parallelogram, consecutive anglesare supplementary. Thus, �Q is the supplementof �QPS. Since �QPS and �SPR form a linearpair, �SPR is the supplement of �QPS. Thesupplements of congruent angles are congruent.Thus, �Q � �SPR since �QPS is congruent toitself. By the transitive property, �SPR � �R,and so �SPR is an isosceles triangle by theconverse of the isosceles triangle theorem. In particular, . But in parallelogram QPST,opposite sides are congruent. Thus, . Bythe transitive property, .QT > SR

QT > PSPS > SR

QT y PSQR

QR y ST

m/BAD 5 m/ABC 5 120m/ADC 5 m/BCD 5 60

m/DAB 5 70m/ABC 5 70m/BCD 5 110

309

17. a. Draw and . Then, �DEBand �CFA are both right angles. Since

, �DEB � �CFA by HL. Thus,�CAB � �DBA because corresponding partsof congruent triangles are congruent. Since

and (by the reflexiveproperty), �ACB � �BDA by SAS.Therefore, because correspondingparts of congruent triangles are congruent.

b. Because we cannot assume that the two baseangles are congruent.

18. Since translations preserve distance, we canassume that B, C, and D are to the right of theorigin and that C is to the right of D.

AB � �0 � b� � b

DC � �d � c� � c � d

Thus, .

M � and

Thus, , and

MN = .

19. a. Slope of Slope of Slope of Slope of

b. Since the slopes of only two sides are equal,only two sides are parallel. Therefore, ABCDis a trapezoid.

c. E(3,5), F(0, �1) d. Slope of e. Since the slope of the median is the same as

the slope of the bases, it is parallel to them.

20. Suppose that the diagonals of the trapezoidbisected each other. A trapezoid is aquadrilateral. Since a quadrilateral withdiagonals that bisect each other is aparallelogram, the trapezoid is a parallelogram.Therefore, it has two pairs of parallel sides, acontradiction.

21. A trapezoid is isosceles if and only if the diag-onals are congruent. Therefore, if the diagonalsare not congruent, the trapezoid is not isosceles.

22. If a quadrilateral does not have pair ofconsecutive angles that are supplementary, thenno pair of opposite sides is parallel. Therefore,the quadrilateral is not a trapezoid.

EF 5 2

DA 5 22.5CD 5 2BC 5 21AB 5 2

12(AB1CD)

MN 5 Pd2 2 c 1 b2 P 5 1

2(b 1 c 2 d)

N 5 A c 1 b2 , e2 BA d

2, e2 B

12(AB 1 CD) 5 1

2(b 1 c 2 d)

AD > BC

AB > ABDB > AC

AC > DB

CF ' ABDE ' AB

10-8 Areas of Polygons (pages 410–412)Writing About Mathematics

1. Yes. The area of ABCD is and thearea of PQRS is . Since AB � PQ andBC � QR, by substitution, the area of PQRS is

. Therefore, the two areas are equal.2. No. The area of a parallelogram is equal to the

product of the length of a base and the height,where the height is the distance from the base toits parallel side. If we are only given that the sidesof the parallelogram are congruent, the height ofeach parallelogram may be different, and so eachparallelogram can have a different area.

Developing Skills3. 40 sq units4. a. Graph

b. and because alternate interior angles formed by atransversal cutting two parallel lines are congruent. Since by the reflexiveproperty, �ABC � �DBC by ASA.

c. ABCD is a parallelogram since both pairs ofopposite sides are parallel. From Example 1,the area of the parallelogram is (AB)(CD) � bh. The area of congruentfigures are congruent, so

(area of �ABC) � (area of �DBC).Therefore, 2(area of �ABC) � bh, or

(area of �ABC) � .

5. 24 sq units6. a. Answers will vary.

b. Since the lines perpendicular to the same lineare parallel, . Since segments ofparallel lines are parallel, . Thus,FECD is parallelogram, and CE � DF sinceopposite sides of a parallelogram arecongruent.

c. Area of ABCD � area of �AFD � area ofFECD � area of �CEB

Let AF � x and EB � y.Then .

Area of �AFD

7. 16 sq units5 1

2h(b1 1 b2)

5 hS12(b1 2 b2) 1 b2T

5 hS12(x 1 y) 1 b2T

5 12xh 1 b2h 1 12yh

x 1 y 5 b1 2 b2

FE y DCCE y DF

12bh

BC > BC

/ACB > /CBD/ABC > /BCD

(AB)(BC)

(PQ)(QR)(AB)(BC)

310

8. a. Answers will vary.b. By Exercise 20 on page 398, the diagonals of a

rhombus separate the rhombus into fourcongruent right triangles. Therefore,�ABE � �CBE � �CDE � �ADE.

c. The diagonals of a rhombus separate therhombus into four congruent right triangles.Thus, each diagonal is a height of one of thetriangles, and the area of �ABE is

.

d. The area of the rhombus is the sum of eachtriangle formed by the intersecting diagonals.Since the triangles are all congruent, the area

of the rhombus is .

9. 12 sq unitsApplying Skills

10. a. Midpoint of Midpoint of

The midpoints are the same. Therefore, thetwo diagonals of ABCD bisect each other.

is a horizontal segment. is avertical segment. Horizontal and vertical linesare perpendicular. Therefore, the twodiagonals of ABCD are perpendicular.

If the diagonals of a quadrilateral bisecteach other, then the quadrilateral is a paral-lelogram. Thus, ABCD is a parallelogram. Ifthe diagonals of a parallelogram areperpendicular, then the parallelogram is arhombus. Thus, ABCD is a rhombus.

b. 2411. a. Midpoint of � (1, 2)

Midpoint of � (1, 2)The midpoints are the same. Thus, the

diagonals bisect each other, and ABCD is a parallelogram. Since the diagonals arevertical and horizontal segments, they areperpendicular to each other. Thus,parallelogram ABCD is a rhombus. Since AC � DB � 6, the diagonals are congruent,and parallelogram ABCD is a rectangle. Inparticular, ABCD has a right angle. Arhombus with a right angle is a square.Therefore, ABCD is a square.

b. 18 sq unitsc. A�(2, 2), B�(�1, �1), C�(�4, 2), D�(�1, 5)d. 18

DBAC

DBAC

DB 5 (2, 1)AC 5 (2, 1)

(4) A d1d28 B 5

d1d22

12 A

d12 ?

d22 B 5

d1d28

e. The diagonals of AEA�F are congruent andare the perpendicular bisectors of each other.Therefore, AEA�F is a square.

f. 8 sq units12. a. 189 sq units b. 9 units13. The two triangles have the same base . Also,

, and so SM � TN where and are the altitudes from S and T to . Let h � SM � TN. Thus, the areas of both trianglesare .

14. a. 16 sq units b. 16 sq units c. 32 sq units15. The length of the median is equal to one-half

the sum of the bases. Thus, EF = .Therefore, area of the trapezoid is

or bysubstitution.

16. a. 24 sq unitsb. Area of sq units

Area of sq unitsArea of sq units

c. 16.5 sq units17. 35 sq unitsHands-On Activity

1. BD = d,

2. Area of �ACD � area of �ACB �

3. Area of ABCD �4. Area of ABCD � s2

5. 6.

7. Length of a hypotenuse � (length of a leg)

Review Exercises (pages 414–415)1. No. If a parallelogram has one right angle, then

the parallelogram is a rectangle by the definitionof a rectangle. But in a rectangle, all the anglesare right angles.

2. x � 24; 72°, 108°, 72°, 108° 3.4. If a transversal is perpendicular to one of two

parallel lines, then it is perpendicular to the other. Thus, since transversal isperpendicular to , it is also perpendicular to

. If two lines are each perpendicular to thesame line, they are parallel. Thus, since

and , . Sincesegments of parallel lines are parallel, .Thus, MBND is a parallelogram. Since oppositesides of a parallelogram are congruent,

DN y MBDM y NBNB ' DCDM ' DC

DCAB

DM

m/APC 5 135

!2

d 5 s 3 !2s 5 d!2

d2

2

5 d2

4

12 3 d 3 d

2

AE 5 d2, EC 5 d

2, BE 5 d2, ED 5 d

2

nSMP 5 2nQLR 5 3

2

nPNQ 5 4

(DH)(EF)12(DH)(AB 1 CD)

12(AB 1 CD)

12(AB)(h)

ABTNSMAB y DC

AB

311

and . Since and , �AMD and �CNB are both right

angles, and so �AMD and �CNB are both righttriangles. Therefore, �AMD � �CNB by HL.

5. ANCM is a parallelogram since and(segments of parallel lines are

parallel). Thus, since opposite sides of aparallelogram are congruent, ,

, , and ,and so DC � AB and NC � AM. By the partition postulate, DC � DN � NCand AB � AM � MB. Thus,DN = DC � NC � AB � AM � MB, and so

. Therefore, bySSS.

6. a.b.

7. If �PTQ is isosceles with �PTQ the vertexangle, then . Thus, since doubles ofcongruent segments are congruent, ,and PQRS is a parallelogram with congruentdiagonals. Therefore, since a parallelogram withcongruent diagonals is a rectangle, PQRS is arectangle.

8. Since a rectangle is a parallelogram, its oppositesides are congruent. Thus, . Since P isthe midpoint of , . Since �C and �Bare both right angles, �C � �B. Thus,

by SAS. Therefore, sincecorresponding parts of congruent triangles arecongruent, .

9. a. Yes. Every equilateral triangle is equiangular.b. No. An equilateral quadrilateral does not

have to be equiangular. For example, arhombus is equilateral but consecutive anglesare not necessarily congruent.

10. Since ABEF and BCDE are rhombuses, theirdiagonals are perpendicular, so �EGB and�EHB are right angles. Opposite sides of a

rhombus are parallel, so . If two parallel lines are cut by a transversal, then the alternateinterior angles formed are congruent. Therefore,�ABE � �DEB. Since the sides of a rhombus

are congruent, and , and so

. By the reflexive property, ,so �ABE � �DEB by SAS. Correspondingparts congruent triangles are congruent, so�GEB � �HBE and m�GEB � m�HBE.

BE > BEAB > DE

BE > DEAB > BE

FEg

y ABg

AP > DP

nABP > nDCP

BP > PCBCAB > DC

PR > SQPT > TQ

/EBA, /EBC, /EDC, /EDA/EAD, /ECB, /ECD

nAND > nCMBDN > MB

NC > AMAN > MCDC > ABAD > BC

AM y CNAN y CM

NB ' DCDM ' ABAD > BCDM > NB

Since �EGB and �EHB are right triangles,their acute angles are complementary.Then, m�GBE � m�GEB � 90 and m�HBE � m�HEB � 90. By the substitutionpostulate, m�GBE � m�HBE � 90 andm�GEB � m�HEB � 90. Therefore, �GEHand �GBH are right angles and BHEG is arectangle.

11. Since a square is a rhombus, the result fromExercise 10 applies. Thus, BHEG is a rectangle,and BHEG has a right angle. Since ABEF is a square, . Thus, BHEG is also a rhombus. A rhombus with a right angle is asquare. Therefore, BHEG is a square.

12. a. Graphb. Parallelogram; slope of � slope of � 0;

slope of � slope of � 2.5. The slopes of opposite sides are equal. Therefore,opposite sides are parallel, and ABCD is aparallelogram.

c. 30 sq units13. a. Midpoint of � midpoint of � (2, 2)

Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, thequadrilateral is a parallelogram. Therefore,DEFG is a parallelogram.

b. The diagonals are perpendicular since theyare horizontal and vertical segments. If thediagonals of a parallelogram areperpendicular to each other, then theparallelogram is a rhombus. Therefore, DEFGis a rhombus.

c. Slope of and slope of . The slopes are not negative reciprocals. Thus, thetwo consecutive sides are not perpendicular,and �FGD is not a right angle. The angles in asquare are all right angles. Therefore, DEFG isnot a square.

d. Since DEFG is a rhombus, the area of the

quadrilateral is by Exercise 8 on page 411.

14. a. Midpoint of

� (b � d, a)

Midpoint of

� (b � d, a)

Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, the

DB 5 A 2b 1 2d2 , 0 1 2a

2 B

AC 5 A 2b 1 2d 1 02 , 2a 1 0

2 B

(DF)(GE)2

GF 5 23GD 5 22

3

FDGE

BCDADCAB

EG > GB

312

quadrilateral is a parallelogram. Therefore,ABCD is a parallelogram.

b. P(b, 0), Q(2b � d, a), R(b � 2d, 2a), S(d, a)c. Midpoint of

� (b � d, a)

Midpoint of

� (b � d, a)

Since the midpoints are the same, the twodiagonals bisect each other. If the diagonals ofa quadrilateral bisect each other, thequadrilateral is a parallelogram. Therefore,PQRS is a parallelogram.

15. a. 12 � xy b. 16 � 2(x � y)c. Solving for x in the second equation gives:

Substituting this value of x into the firstequation and solving for y gives:

Therefore, y � 6 or y � 2. When y � 6,x � 2, and when y � 2, x � 6. Thus, thedimensions of the rectangle are 6 centimetersby 12 centimeters. (The length and width canbe either of these two dimensions.)

16. In �ABD and �PQS, �A � �P, , and. Thus, �ABD � �PQS by SAS. Since

corresponding parts are congruent, .Since we are given that and ,�BCD � �QRS by SSS. Thus, �C � �Rbecause corresponding parts are congruent. Bythe partition postulate:

But since corresponding parts are congruent incongruent triangles:

m/BDC 5 m/QSR

m/DBC 5 m/SQR

m/ADB 5 m/PSQ

m/ABD 5 m/PQS

m/S 5 m/PSQ 1 m/QSR

m/Q 5 m/PQS 1 m/SQR

m/D 5 m/ADB 1 m/BDC

m/B 5 m/ABD 1 m/DBC

CD > RSBC > QRBD > QS

DA > SPAB > PQ

0 5 (y 2 6)(y 2 2)

0 5 2y2 1 8y 2 12

12 5 8y 2 y2

12 5 (8 2 y)y

12 5 xy

x 5 82y

SQ 5 A d 1 2b 1 d2 , a 1 a

2 B

PR 5 A b 1 b 1 2d2 , 0 1 2a

2 B

By the substitution postulate:

Thus, in ABCD and PQRS, all correspondingangles and sides are congruent, and so the twoquadrilaterals are congruent.

Exploration (page 416)a.

m/D 5 m/PSQ 1 m/QSR 5 m/S

m/B 5 m/PQS 1 m/SQR 5 m/Q

313

b. Rectangle, square, isosceles trapezoidc. Circumcenter: square, rectangle, isosceles

trapezoidNo circumcenter: rhombus, parallelogram(without a right angle), non-isosceles trapezoid

d. See graph of part a.e. Answers will vary. For example: a circumcircle is

a circle such that all of the vertices of a polygonare on the circle and the center is thecircumcenter of the polygon.


1. 4 2. 2 3. 1 4. 1 5. 16. 4 7. 3 8. 2 9. 1 10. 2

Part II11. Since the sum of the angles of a triangle is 180°:

Therefore, the measures of the angles of thetriangle are 3(16) � 48°, 4(16) � 5 � 69°, and5(16) � 17 � 63°.


1. 1. Given.2. Through G, draw 2. Through a given

. point not on a given line, thereexists one and onlyone line parallel tothe given line.

3. 3. If two of three lines in the same planeare each parallel tothe third line, thenthey are parallel toeach other.

4. m�EGJ � m�FGH 4. Partition postulate.� m�z

HJg

y CDg

HJg

y ABg

ABg

y CDg

x 5 1612x 5 192

12x 2 12 5 180

3x 1 4x 1 5 1 5x 2 17 5 180

(Cont.)

Statements Reasons5. �x � �EGJ, 5. If two coplanar

�y � �FGH lines are cut by atransversal, thenthe alternateinterior anglesformed arecongruent.

6. m�x � m�EGJ, 6. Definition ofm�y � m�FGH congruent angles.

7. m�x � m�y � m�z 7. Substitutionpostulate.

Part III

13. Since the triangles are isosceles with bases and , and

by the isosceles triangletheorem. Since �CAB � �FDE,�CAB � �FDE � �CBA � �FED by thetransitive property. Since ,�ABC � �DEF by ASA.

14. since radii of a circle arecongruent. Since , by the transitive property and �OCD is equilateral.

DC > CO > DOAO > DCAO > DO > CO > BO

AB > DE

/FDE > /FED/CAB > /CBADEAB

314

Since and segments of parallel lines areparallel, . Therefore, AOCD contains a pair of opposite sides that are both congruentand parallel, and AOCD is a parallelogram. In aparallelogram, opposite sides are congruent.Thus, , and �AOC and �BOD areboth equilateral.

Part IV

15. Slope of Slope of

Slope of Slope of The slopes of only two pairs of sides are equal.Therefore, KLMN has only one pair of parallelsides, and so it is a trapezoid.

16. Under :

Under :

Cs(2, 4) S Cr(22, 4)

Bs(0, 4) S Br(0, 4)

As(2, 2) S Ar(22, 2)

ry-axis

C(4, 2) S Cs(2, 4)

B(4, 0) S Bs(0, 4)

A(2, 2) S As(2, 2)

ry5x

NK 5 1NM 5 212

LM 5 1KL 5 215

AD > OC

AO y DCAB y CD

Chapter 11.The Geometry of Three Dimensions

11-1 Points, Lines, and Planes (pages 422–423)

Writing About Mathematics1. Yes. If two lines intersect, then they determine a

plane. If two lines are parallel, then they lie in thesame plane. Since skew lines are neither parallelnor intersecting, skew lines can be defined as twolines that do not lie in the same plane.

2. Yes. Two intersecting lines determine a plane, soA, B, C, and D all lie in the same plane. Sincethey are distinct points in the same plane, theycan be the vertices of a quadrilateral.

Developing Skills

3. Since , the lines are coplanar and have no points in common, so A, B, C, and D are fourpoints that lie in the same plane. Since AB � CD,ABCD is not a parallelogram, so is notparallel to . A quadrilateral in which two and only two sides are parallel is a trapezoid.Therefore, ABCD is a trapezoid.

BCAD

CDg

ABg

y

4. Since , the lines are coplanar and have no points in common, so A, B, C, and D are fourpoints that lie in the same plane. A parallelogramis a quadrilateral in which two pairs of opposite

sides are parallel. Since , ABCD is a

parallelogram.

5. Two intersecting lines determine a plane, so and are coplanar. Therefore, A, B, C, and D are four points that lie in the same plane. Sincethe diagonals are perpendicular bisectors of eachother, �AEB and �AED are congruent right angles and . By the reflexive propertyof congruence, . Then �AEB ��AED by SAS, and . A quadrilateral with diagonals that are perpendicular is arhombus, and a rhombus with two congruentconsecutive sides is a square. Therefore, ABCD is a square.

AB > ADAE > AE

BE > DE

BEDAEC

ADg

y BCg

y CDg

ABg

In 6–9, answers will vary. Examples are given.6. and , and 7. and and 8. and , and 9. The skew lines

10. a. p ↔ (q ∧ r)b. p states that two lines in space are parallel. By

definition, parallel lines in space are coplanar,so q is true.

Applying Skills11. He should choose a base with three legs.

Consider the bottom of each leg of the camerabase. There is one and only one plane containingthree non-collinear points, so if the base hasthree legs, the bottom of each leg will touch theground at the same time and the base will besturdy. Four points can be in different planes, sothe bottom of each leg is not guaranteed to touchthe ground at the same time and it can wobble.

12. Two intersecting lines determine a plane, so ifeach pair of strings intersect, then they all lie inthe same plane and the stakes to which they areconnected lie in the same plane.

13. The four triangles are equilateral, so, ,, and . By the

transitive property of congruence,and

triangles are congruent by SSS.

11-2 Perpendicular Lines and Planes (pages 432–433)

Writing About Mathematics1. Yes. Since the four dihedral angles have equal

measures, they each measure 90 degrees and areright dihedral angles. Perpendicular planes aretwo planes that intersect to form a right dihedralangle.

2. a. Yes. If a line is perpendicular to a plane, it isperpendicular to each line in the planethrough the point of intersection of the lineand the plane.

b. CubeDeveloping Skills

3. False. At a given point on a given line, any planeperpendicular to the line at that point containsinfinitely many lines perpendicular to the line atthe point.

4. True. If a line not in a plane intersects the plane,then it intersects the plane in exactly one point.

AB > BC > AC > AD > CD > BD

AB > BD > ADAC > CD > ADBC > CD > BDAB > BC > AC

CGAEBFAEHGAEDC,AEHEDHADAE

315

5. False. If a line is perpendicular to a plane at B, and C and D are two other points on the

plane, then is not perpendicular to .6. False. The line and the plane can intersect the

given line in the same point.7. False; p ⊥ r and q ⊥ r, but p is not perpendicular

to q.

8. True. If a line is perpendicular to a plane, thenevery plane containing the line is perpendicularto the given plane.

9. False; p ⊥ r at A and q ⊥ r at A.

10. False. Let plane p be perpendicular to at A,

and let intersect . Through a given point on a plane, there is only one line perpendicular toa given plane. Therefore, p is not perpendicular

to .11. False. Let the two perpendicular planes, p and q,

intersect in line . Let line be perpendic-

ular to plane p in q. Then cannot be perpen-dicular to plane q because it lies in the plane.

Applying Skills

12. , so �RPA and �SPA are right anglesand congruent. P is the midpoint of , so

. by the reflexive property of congruence, so �RPA � �SPA by SAS.Corresponding parts of congruent triangles are congruent, so .

13. From steps 1 and 2, and .by the reflexive property of

congruence, so �RAB � �SAB by SSS.Corresponding parts of congruent triangles are congruent, so �RAB � �SAB.

14. From step 1, . From step 4, .by the reflexive property of

congruence, so �RPQ � �SPQ by SSS.Corresponding parts of congruent triangles arecongruent, so �RPQ � �SPQ.

PQ > PQRQ > SQRP > SP

AB > ABBR > BSAR > AS

AR > AS

AP > APRP > SPRS

l ' AP

BCg

BCg

ABg

ACg

ACg

ABg

ABg

r

q

A

p

r

q

p

CDg

ABg

ABg

15. We are given that ⊥ plane q at M, themidpoint of . A line is perpendicular to aplane if and only if it is perpendicular to eachline in the plane through the intersection of theline and the plane. Since R is a point in p,

and �AMR and �BMR are rightangles and congruent. by thedefinition of midpoint and by the reflexive property of congruence. Therefore,�AMR � �BMR by SAS. Corresponding partsof congruent triangles are congruent, so

and AR � BR.

16. We are given that M is the midpoint of , so. We are also given that

and . and by the reflexive property of congruence. Therefore,�AMS � �BMS and �AMR � �BMR by SSS. Corresponding parts of congruent trianglesare congruent, so �AMS � �BMS and �AMR � �BMR.If two lines intersect to form congruent adjacentangles, then they are perpendicular, so and . If a line is perpendicular to each of two intersecting lines at their point ofintersection, then the line is perpendicular to the plane determined by these lines. Therefore, is perpendicular to the plane determined by M, R,and S.

17. We are given equilateral �ABC in plane p and

perpendicular to plane p. A line is perpendicular to a plane if and only if it isperpendicular to each line in the plane throughthe intersection of the line and the plane.Therefore, and , so �BADand �CAD are right angles and congruent. Since�ABC is equilateral, . by the reflexive property of congruence. Therefore,�ABD � �ACD by SAS. Corresponding partsof congruent triangles are congruent, so

.

18. We are given that and intersect at A and

determine plane p, and is perpendicular to plane p at A. A line is perpendicular to a plane ifand only if it is perpendicular to each line in theplane through the intersection of the line and the

plane. Therefore, and , so �BAD and �CAD are right angles andcongruent. We are given that AB � AC, so

, and by the reflexive AD > ADAB > AC

ADg

' ACADg

' AB

ADg

ACg

ABg

BD > CD

AD > ADAB > AC

AD ' ACAD ' AB

ADg

AB

AB ' MRAB ' MS

MS > MSMR > MRSA > SBRA > RBAM > BM

AB

AR > BR

MR > MRAM > BM

AB ' MR

ABAB

316

property of congruence. Therefore,�ABD � �ACD by SAS.

19. We are given �QRS in plane p with perpendicular to plane p. A line is perpendicularto a plane if and only if it is perpendicular toeach line in the plane through the intersection of the line and the plane. Therefore,and , so �QST and �RST are rightangles and congruent. We are given that �QTS � �RTS, and by the reflexiveproperty of congruence. Therefore, �QST ��RST by ASA. Corresponding parts of con-gruent triangles are congruent, so .

20. If a line is perpendicular to two intersecting linesat their point of intersection, then the line isperpendicular to the plane determined by theselines. Therefore, the pole is perpendicular to theground.

21. A line is perpendicular to a plane if and only if itis perpendicular to each line in the plane throughthe intersection of the line and the plane.Therefore, the pole and the ground between thepole and each wire form right angles. The pole iscongruent to itself. The wires are of equal lengthsand are, therefore, congruent. The two trianglesformed by the pole, a wire, and the groundbetween the pole and wire are congruent by HL.Corresponding parts of congruent triangles arecongruent, so the other legs of the triangles arecongruent, and the wires are fastened to theground at equal distances from the pole.

11-3 Parallel Lines and Planes (pages 439–440)

Writing About Mathematics1. No. For example, consider the floor and any two

adjacent walls of a room. The walls are bothperpendicular to the floor but intersect in thecorner of the room.

2. Yes. Draw a line l perpendicular to the givenplane. If two planes are parallel, then a lineperpendicular to one plane is perpendicular tothe other plane. Thus, since l is perpendicular tothe given plane and since both planes are parallelto the given plane, both planes are perpendicularto l. Two planes perpendicular to the same lineare parallel. Therefore, the two planes are alsoparallel.

TQ > TR

ST > ST

ST ' RSST ' QS

ST

Developing SkillsIn 3–9, there may be more than one answer.

3. a. Plane EFGH ⊥ plane BCFE , , and

.

b. and

, but .

4. a. is parallel to planes ABEH and CDGF,

and .

b. is parallel to planes DCFG and BCFE,

but .

5. a. is perpendicular to and , and

.

b. is perpendicular to and , but

and intersect.

6. a. and do not intersect and are parallel.

b. and do not intersect but are skew

lines.7. a. Plane ABEH is perpendicular to planes

AHGD and BEFC, and .b. Plane ABEH is perpendicular to AHGD and

EFGH, but .

8. a. Plane EFGH is parallel to and , and

.

b. Plane EFGH is parallel to and , but

.

9. a. and is a skew line of both lines.

b. , is a skew line of , but

intersects .Applying Skills

10. We are given that p || q. �ABC determines aplane. If a plane intersects two parallel planes,then the intersection is two parallel lines.Therefore, . If two coplanar parallel linesare cut by a transversal, then the correspondingangles are congruent. Therefore, �ABC ��ADE and �ACB � �AED. Since �ABC isisosceles, �ABC � �ACB. By the transitiveproperty of congruence, �ADE � �AED. If twoangles of a triangle are congruent, then the sidesopposite these angles are congruent. Therefore,

and �ADE is isosceles.AD > AE

DE y BC

BEg

FEg

AHg

FEg

AHg

y BEg

DFg

AHg

y BEg

ABg

' ADg

ADg

ABg

ABg

y IJg

IJg

ABg

AHGD ' EFGH

AHGD y BEFC

DCg

AHg

BEg

AHg

DFg

DCg

DFg

DCg

ADg

AHg

y BEg

BEg

AHg

ABg

DCFG ' BCFE

AHg

ABEH y CDGF

IJg

AHg

y DCFGAHg

y DCFG

Plane DCFG ' plane BCFE

AHg

' EFGH

AHg

y BCFE

317

11. Consider the two triangles and .Because , and , so

. Since we are givenand by the reflexive

property, �PQA � �PQB by HL. Hence,.

12. Yes. Two lines perpendicular to the same planeare parallel. Thus, each pair of posts is parallel toeach other.

13. No. Parallel planes are everywhere equidistant.He has to make the poles the same length.

11-4 Surface Area of a Prism (pages 444–445)

Writing About Mathematics1. a. Equilateral triangle. An equilateral triangle is

equiangular. The base cannot change shapebecause the angles must always measure 60degrees.

b. Rhombus. The base can change shape becausewhile opposite angles must be congruent,these angle measures can change.

c. Yes. The lateral sides are rectangles and theyare perpendicular to the bases.

2. a. No. The rectangular faces are notperpendicular to the bases, so the height of theprism will be less than the length of thealtitude of one of the rectangular faces.

b. Yes. The faces that are parallelograms areperpendicular to the bases, so the height ofthe altitude will be equal to the length of thealtitude of one of the faces that areparallelograms.

Developing Skills3. 158 cm2

4. 1,500 in.2 or 125 ft2

5. 292 ft2

6. 9,605.62 cm2 or 96.0562 m2

7. a. Rectangleb. 9.00 cm 14.5 cm, 40.0 cm 14.5 cm,

41.0 cm 14.5 cmc. 1,665 cm2

8. a. Rectanglesb. 5 cm 12 cm, 5 cm 12 cm, 6 cm 12 cmc. 216 cm2

9. 6 faces. There are 2 bases, and each base has 4sides, so there are 4 lateral faces.

10. No. A parallelepiped must have parallelogramsas bases.

11. 162 in.2

AQ > BQ

PQ > PQPA > PBm/PQA 5 m/PQB 5 90

PQ ' QBPQ ' QAPQ ' pnPQBnPQA

Applying Skills12. We are given that the bases of a parallelepiped

are ABCD and EFGH with ABCD � EFGH.A parallelepiped is a prism, and the lateral edges of a prism are congruent and parallel.Therefore, or AE � BF � CG � DH, and .

13. a. We are given a prism with congruent bases�ABC and �DEF. The bases of a prism areparallel, so a line perpendicular to one of theplanes is perpendicular to the other.Therefore, and ;

and ; and and. If a plane contains a line

perpendicular to another plane, then theplanes are perpendicular. Therefore, thelateral faces are all perpendicular to the bases.By definition, the prism is a right prism andthe lateral faces are rectangles.

b. The lateral faces are congruent polygonswhen �ABC and �DEF are equilateral. Then

. Also,since parallel planes are

everywhere equidistant. The sides are allrectangles, so all angles are right angles.

14. 4 ft15. Let the bases of the parallelepiped be ABCD and

DEFG. A parallelepiped is a prism, and thelateral edges of a prism are parallel, so

. ABCD and DEFG arecongruent parallelograms, so and

. The lateral faces of the parallelepipedare parallelograms, so and .Similarly, .

16. Rhombuses. The lateral edges of a prism are con-gruent. Since the lateral faces are squares, thesides are all congruent to the lateral edges.Therefore, the edges of the bases are congruent.The bases of a parallelepiped are parallelograms,and a parallelogram with all sides congruent is arhombus.

17. By Exercise 16, a parallelepiped with squarelateral faces has bases that are rhombuses. Sinceone of the bases has a right angle, that base is asquare. A parallelepiped is a prism, so the basesare congruent. Therefore, both bases are squaresand the parallelepiped is rectangular. Since theedges are all congruent, all of the faces of theparallelepiped are congruent, and it is a cube.

BC y AD y EF y DGCD y FGAB y DE

DE y FGAB y CD

AD y BE y CF y DG

AD > BE > CFAB > BC > AC > DE > EF > FD

BE ' EFBE ' BCAD ' DEAD ' AB

AD ' DFAD ' AC

AE y BF y CG y DHAE > BF > CG > DH

318

18. a. 512 ft2 b. 8 ftHands-On Activity

1–3. Demonstrated with manipulatives.

11-5 Volume of a Prism (page 448)Writing About Mathematics

1. No. The bases need only have equal area.Polygons with equal area can be different shapes,and therefore, not congruent.

2. Yes. The height of the prism is equal to the heightof each of the lateral sides if the sides are allperpendicular to the base. By definition, such aprism is a right prism. The lateral sides of a rightprism are rectangles.

Developing Skills3. 72 ft3 4. 27.2 ft3 5. 157.5 in.3

6. 1,080 cm3 7. 50,008 cm3

Applying Skills8. 25 cm 9. 18 cm

10. AB : CD � 1 : 2;

11. Let the bases of the prism be �ABC and�A�B�C�. Consider lateral side ABB�A and aplane p parallel to the bases and intersecting thelateral sides. Then plane p will intersect lateral

side ABB�A in a line that is parallel to

and . Since ABB�A� is a parallelogram, we have �A � �B� and �A� � �B�. Since

, corresponding angles are congruent,so and . By thetransitive property, and

. Thus, both pairs of oppositeangles are congruent in , so is

a parallelogram. Similarly, since , we have and .Thus, is also a parallelogram because

and by thetransitive property. Since opposite sides of a parallelogram are congruent, and

. Therefore, plane p intersects lateral side in a segment that is congruent tothe base edges of the lateral side. Since was an arbitrary side, the result is true for theother two sides. Therefore, the figure formed byplane p is congruent to the bases of the prism.

ABBrArABBrAr

AB > AsBsArBr > AsBs

/Br > /ArAsBs/Ar > /AsBsBrAsBsBrAr

/B > /AsBsBr/A > /ArAsBsAsBsg

y ABg

ABBsAsABBsAs/AsBsB > /A

/AAsBs > /B/Br > /AsBsB/Ar > /AAsBs

AsBsg

y ArBrg

ArBrg

ABg

AsBsg

12 5 AB

CD

12CD 5 AB

12(AB)(CD) 5 (AB)2

12(AB)(CD) 5 (PQ)2

12. Let the sides of a right prism be s1, s2, . . . , snand the height be h. All of the lateral sides of a right prism are rectangles. The lateral area �s1h � s2h � . . . � snh � (s1 � s2 � . . . � sn)h.Since the perimeter of base � s1 � s2 � . . . � sn,the lateral area is equal to the perimeter of thebase times the height of the prism.

11-6 Pyramids (pages 452–453)Writing About Mathematics

1. They are both correct. In an equilateral triangle,the perpendicular bisectors intersect at the samepoint at which the medians intersect.

2. No. The lateral faces of a pyramid are congruentif the pyramid is regular.

Developing Skills3. 240 cm2 4. 12 ft2 5. 4,608 cm2

6. 576 cm3 7. 46 in.3 8. 2 ft3

9. cm3 10. 96 in.2

Applying Skills11. a. 49.5945 cm2 b. 148.7835 cm2

c. 198.378 cm2 d. 144.48531 cm3

12. 216 ft3

13. Let the base of the regular pyramid be �ABC,and the vertex, D. The lateral faces of a regular pyramid are isosceles triangles, so and

. By the transitive property ofcongruence, , so the lateral edges are congruent.

14. We are given a pyramid with square base ABCDand vertex F with . Since ABCD is a square, its diagonals are congruent and bisect each other at a point E, so

. By the reflexiveproperty of congruence, . Therefore,�AEF � �CEF by SSS, so �AEF � �CEF. Iftwo lines intersect to form congruent adjacent angles, then they are perpendicular, so

. Since ABCD is a square, .Since and we are given ,�ABF � �CBF by SSS. Thus, we also have�AEF � �BEF by SSS. Since correspondingangles are congruent, �AEF � �BEF, and �FEB is a right angle. Thus, is alsoperpendicular to . A line perpendicular to two lines in a given plane is perpendicular to the plane. Thus, is the altitude of the pyramid and intersects the base in its center E. Since aregular pyramid is a pyramid whose base is aregular polygon and whose altitude is

FE

DBFE

AF > FCFB > FBAB > BCAC ' EF

EF > EFAE > CE > DE > EB

AF > CF

AD > BD > CDBD > CD

AD > BD

2,25813

319

perpendicular to the base at its center, thepyramid is regular.

15. Let the base of the regular pyramid be �ABC,and the vertex, D. The lateral faces of a regularpyramid are congruent isosceles triangles, so �ABD � �BCD � �ACD with �

and �DAB � �DBC � �DCA. Draw thealtitudes such that , , and

. This forms congruent right angles�DEA, �DFB, �DGA. Therefore, �DEA ��DFB � �DGA by AAS, so the altitudes

, , and are congruent.16. Let the sides of the base of a regular pyramid

be s1, s2, . . . , sn . Then the perimeter p � s1 � s2 � . . . � sn. The lateral faces of aregular pyramid are triangles, so the lateral

area � �

� .

17. a. When both perimeter and slant height aredoubled, the lateral area increases by amultiple of 4. When tripled, the lateral areaincreases by a multiple of 9.

b. When both the sides of the base and theheight are doubled, the volume increases by amultiple of 8. When tripled, the volumeincreases by a multiple of 27.

11-7 Cylinders (pages 455–456)Writing About Mathematics

1. No. When the radius is doubled and the height is halved, the volume is .Therefore, the volume is doubled.

2. Yes. When the radius is doubled and the height is halved, the lateral surface area is

. Therefore, the lateral surface area remains unchanged.

Developing Skills3. a. 4,080p cm2 b. 6,392p cm2 c. 69,360p cm3

4. a. 96p in.2 b. 128p in.2 c. 192p in.35. a. 864p in.2 or 6p ft2

b. 1,512p in.2 or 10.5p ft2

c. 7,776p in.3 or 4.5p ft3

6. a. 15,000p cm2 or 1.5p m2

b. 35,000p cm2 or 3.5p m2

c. 750,000p cm3 or 0.75p m3

7. 6.2 cm 8. 6.4 cm9. 9 in. 10. 16.3 cm

Applying Skills11. 27.7 in.3 12. 12,117 gal

2p(2r) A 12h B 5 2prh

p(2r)2 A 12h B 5 2pr2h

12phs

12hs(s1 1 s21c1 sn)

12s1hs 1 12s2h21c1 12snhs

DGDFDE

DG ' ACDF ' BCDE ' AB

CDAD > BD

13. a. 47.7 cm3 b. 158.3 cm3

c. Volume of the lateral surface of the vase(without the base) � 158.3 cm3

But this is the height of the vase not includingthe base. The height of the vase is 14.65 � 0.75 � 15.4 centimeters.

d. cm2

14. 2.2 in., 5.5 in.

11-8 Cones (pages 458–459)Writing About Mathematics

1. No. For the lateral areas to be equal, the pyramidand the cone must have equal slant heights.

2. No. The volume of the larger cone is 4 times thevolume of the smaller since .

Developing Skills

3. a. b. c.4. a. b. c.5. a. b. c.6. a. b. c.7. 4.5 in. 8. 3.5 cm 9. 6 ft

Applying Skills10. 377 ft3 11. 4 gal 12. 112.32p in.3

13. Let the cone cut by the plane create �ABC withvertex B. since they are both the slantheight, hs, of the cone. Let hc, the altitude of thetriangle, intersect at D. Then �BDA and�BDC are right angles. The altitude of anisosceles triangle is also the median, so

. Since hc � r, and�ADC and �BDC are isosceles right triangles.The acute angle of an isosceles right trianglemeasures 45°, so m�ABD � m�CBD �m�ABC � 90 and �ABC is a right triangle.

11-9 Spheres (pages 463–464)Writing About Mathematics

1. Yes. The surface area is .2. No. The volume of the cone is one-fourth the

volume of the sphere since the volume of the cone is and the volume of the sphere is .4

3pr313pr3

4pr2 5 4p A 12d B 2

5 pd2

AD > BD > CDAD > CD

AC

AB > BC

128p cm3144p cm280p cm21,344p cm31,176p cm2600p cm2100p cm390p cm265p cm212p cm324p cm215p cm2

13p(2r)2hc 5 4

3pr2hc

2p(4.5)(15.4) < 435.4

< 14.65 cmh 5 158.3pf (4.5)2

2 (4.1)2g

158.3 5 phf(4.5)2 2 (4.1)2g

158.3 5 p(4.5)2h 2 p(4.1)2h

320

Developing Skills3. S � 225p in.2, V � 563p in.34. S � 697p cm2, V � 3,067p cm3

5. S � 16p ft2, V � 11p ft3

6. S � 1,989p cm2, V � 14,786p cm3

7. 5 ft 8. 2.6 cm 9. 6 in.10. in.3

Applying Skills11. 7 cups 12. 13.1 oz13. 6,3361,600p mi2 14. 4,665,600p mi2

15. LA � SA � 1,024p cm2 16. LA � SA � 4pr2

Hands-On Activitya. 9; each face of the prism has four symmetry

planes perpendicular to that face: two thatcontain the midpoints of opposite sides and twothat contain the opposite vertices.

b. 2; one symmetry plane is the plane parallel to thebases at the midpoint of the altitude of the prism;the other is the plane containing the altitudes ofthe bases.

c. 4; two symmetry planes are the planesperpendicular to the base containing thediagonals of the base, the other two are theplanes perpendicular to the base containing theperpendicular bisectors of the sides of the base.

d. 9; each face of the cube has four symmetry planesperpendicular to that face: two that contain themidpoints of opposite sides and two that containthe opposite vertices.

e. Infinitely many; a symmetry plane is any planecontaining a radius of the sphere.

f. Infinitely many; one symmetry plane is the planeparallel to the bases at the midpoint of thealtitude of the cylinder; the others are any planescontaining a radius of the base.

Review Exercises (pages 468–469)1. Yes. If a line is perpendicular to each of two

intersecting lines at their point of intersection,then the line is perpendicular to the planedetermined by these lines.

2. No. Through a given point on a line, there can beonly one plane perpendicular to the given line.

3. Only if and coincide. Through a given point on a plane, there is only one line to thegiven plane.

4. Yes. Two lines perpendicular to the same planeare coplanar.

5. Yes. Two planes are perpendicular if and only ifone plane contains a line perpendicular to theother.

RSg

RTg

45713p

6. Yes. Two planes are perpendicular if and only ifone plane contains a line perpendicular to theother.

7. Yes. If a line is perpendicular to a plane, then anyline perpendicular to the given line at its point ofintersection with the given plane is in the plane.

8. Yes. If a line is perpendicular to a plane, thenevery plane containing the line is perpendicularto the given plane.

9. Yes. If a plane intersects two parallel planes, thenthe intersection is two parallel lines. Therefore,

intersects if planes p and q are not parallel.

10. Yes. Two planes are perpendicular to the sameline if and only if they are parallel.

11. No. The lateral edges of a prism are parallel.12. Yes. The lateral edges of a prism are congruent.13. Yes. The volume of a prism is equal to the area of

the base times the height of the prism.14. No. The volume of the pyramid is equal to one-

third the area of the base times the height of thepyramid.

15. Yes. If two planes are equidistant from the centerof a sphere and intersect the sphere, then theintersections are congruent circles.

16. No. A great circle of a sphere is the intersectionof the sphere and a plane through the center ofthe sphere. The intersection of plane p and thesphere is not a great circle and the intersection ofplane q and the circle is a great circle. Therefore,the circles are not congruent.

17. L � 384 cm2, S � 512 cm2, V � 768 cm3

18. L � 180 in.2, S � 288 in.2, V � 270 in.319. L � 128 ft2, S � 254 ft2, V � 252 ft3

20. L � 60 in.2, S � 96 in.2, V � 48 in.321. L � 204 ft2, S � 283 ft2, V � 314 ft3

22. L � 396 cm2, S � 704 cm2, V � 1,385 cm3

23. 2.8 m24. cm25. a. 117.8 in.3 b. 7.2 in.3 c. 16 scoops

Exploration (pages 470–471)b.

No. No. No.vertices faces edges

Tetrahedron 4 4 6

Cube 8 6 12

Octahedron 6 8 12

Dodecahedron 20 12 30

Icosahedron 12 20 30

7!2

CDg

ABg

321

Vertices + Faces � 2 � Edgesc. Results will vary.


1. 1 2. 2 3. 4 4. 35. 3 6. 4 7. 4 8. 49. 4 10. 3


1. Isosceles �ABC 1. Given.and �DEF with

2. 2. Definition of isosceles triangle.

3. �B � �E 3. Given.4. �ABC � �DEF 4. SAS.

12. Statements Reasons1. Altitude bisects 1. Given.

�C in �ABC.2. �ADC � �BDC 2. Definition of altitude.3. �ACD � �BCD 3. Definition of angle

bisector.4. �A � �B 4. If two angles of one

triangle arecongruent to twoangles of anothertriangle, then thethird angles arecongruent.

5. 5. If two angles of a triangle arecongruent, then thesides opposite theseangles are congruent.

6. �ABC is isosceles. 6. Definition ofisosceles triangle.


1. BCDE is a 1. Given.parallelogram.

2. 2. Definition of parallelogram.

3. 3. Opposite sides of a parallelogram arecongruent.

4. B is the midpoint of 4. Given..ABC

BC > DE

BC y ED

AC > BC

CD

BC > EFAB > DE

(Cont.)

Statements Reasons5. ABDE is a parallel- 5. If one pair of

ogram. opposite sides of aquadrilateral is bothcongruent andparallel, then thequadrilateral is aparallelogram.

14. is perpendicular to plane p at B, themidpoint of . By the definition of midpoint,

. Let D be a point on plane p. If a lineis perpendicular to a plane, then any lineperpendicular to the given line at its point ofintersection with the given plane is in the line.Therefore, and �ABD and �CBDare right angles and congruent. by the reflexive property of congruence, so

BD > BDAD ' ABC

AB > CBABC

ABC

322

�ABD � �CBD by SAS. Corresponding partsof congruent triangles are congruent, so

and AD � CD. Therefore, any point on p is equidistant from A and C.

Part IV

15. y � �3x � 1; the median is the line throughB(�1, 4) and the midpoint of . The midpoint

is at or (1, �2).

m � � �3

y � �3x � b

�2 � �3(1) � b

b � 1

16. A�(�3, �1), B�(1, 5); R90° (1, 3) � (�3, 1),rx-axis (�3, 1) � (�3, �1); R90° (5, �1) � (1, �5),rx-axis (1, �5) � (1, 5)

4 2 (22)21 2 1

A23 1 52 , 22 1 22

2 BAC

AD > CD

Chapter 12. Ratio, Proportion, and Similarity

12-1 Ratio and Proportion (pages 479–480)Writing About Mathematics

1. Yes. Consider the proportion . If the means are interchanged with the extremes, and theextremes are interchanged with the means, the resulting proportion is . This proportion is equivalent to the original.

2. Yes. If the extremes are perfect squares, then theproportional equation can be written as

or with a and b integers. Thus, ,an integer.

Writing About Mathematics3. Yes 4. Yes 5. No6. Yes 7. No 8. Yes9. , 10. ,

11. , 12. 20

13. 14. 6 15. 40.516. 2 17. 18.19. �6 or 5Applying Skills20. AB � 12, BC � 21 21. 8 cm, 40 cm22. 18 cm, 30 cm 23. 20 in., 24 in., 28 in.24. If the measures of the sides of the triangle are in

the ratio 2 : 3 : 7, then the measures of the sides ofthe triangle are 2x, 3x, and 7x for some x. In atriangle, the length of any side of the trianglemust be less than the sum of the lengths of the

6106306!3

315 5 2

1032 5 15

10

186 5 12

41812 5 6

4305 5 6

1306 5 5

1

x 5 aba2b2 5 x2

a2

x 5 xb2

ba 5 d

c

ab 5 c

d

other two sides. Since 7x � 2x � 3x � 5x, themeasures of the sides of the triangle cannot be inthe ratio 2 : 3 : 7.

25. Length � 30, width � 48 26. 40°, 140°

12-2 Proportions Involving Line Segments(pages 484–485)

Writing About Mathematics1. In any segment, the midpoint divides the

segment into two congruent parts. Thus, the ratioof the lengths of the two parts formed by themidpoint is always equal to 1. Therefore, themidpoints of any two line segments divide thetwo segments proportionally.

2. No. Let x � AB � BC � CD � DE � EF.Then BF � BC � CD � DE � EF � 4x, and AB : BF � x : 4x � 1 : 4.

Developing Skills3. 18 4. 8.5 5. 126. 5 7. 1 : 2 8. 3 : 19. 76 10. 138 11. 6 cm

12. 21 13. 12 14. 3015. AB � 6, BC � 1516. a. 2 : 9 b. 7 : 3 c. 4 : 7 d. 6 : 717. Since AB : BC � 2 : 3, AB � 2x and BC � 3x for

some x. Since DE : EF � 2 : 3, DE � 2y and EF � 3y for some y. Thus, AC � AB � BC � 5x

and DF � DE � EF � 5y. and ABAC 5 2x

5x 5 25

. Therefore, and

AB : AC � DE : DF.Applying Skills18. a. A line segment joining the midpoints of two

sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, , joining the midpoints ofsides and , is parallel to and

. Since is a segment of and, . Since M is the midpoint of

, , so CM � NL. Thus,

and LMCN is a parallelogram because a pair of opposite sides is bothcongruent and parallel.

b. 18 units19. By Exercise 18, PMQC is a parallelogram. Since

�C is a right angle, PMQC is a parallelogramwith a right angle. Therefore, PMQC is arectangle.

20. a. A line segment joining the midpoints of twosides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpoints

of sides and , and .

Since is a segment on , is also parallel to . In particular, . Since

Q is the midpoint , , so AQ � MP by the transitive property. We aregiven PM � MD. Thus, AQ � DM, and

. Therefore, QADM is a paral-lelogram since it has a pair of opposite sidesthat are congruent and parallel. Exercises 18and 19 show that PMQC is a rectangle. Thus,�CQM is a right angle. Since �AQM and�CQM together form a linear pair, �AQM isalso a right angle. Thus, QADM is a parallelogram with a right angle, so it is arectangle.

b. Since Q is the midpoint of , .Since both QADM and PMQC are rectangles,�AQM and �CQM are both right angles, so�AQM � �CQM. Since by thereflexive property, �AQM � �CQM by SAS.Therefore, since correspondingparts of congruent segments are congruent.

AM > CM

QM > QM

AQ > QCAC

AQ > DM

12AC 5 AQAC

AQ y DMACDMPM

hDM

12AC 5 MPMP y ACCBAB

MP

CM 5 NL

12AC 5 CMAC

CM y NLAC y NLACCM1

2AC 5 NLACBCAB

NL

ABAC 5 DE

DFDEDF 5

2y5y 5 2

5

323

c. Since P is the midpoint of , .Since PMQC is a rectangle, �CPM is a rightangle. Since �CPM and �BPM together forma linear pair, �BPM is also a right angle, so �CPM � �BPM. Since by thereflexive property, �CPM � �BPM by SAS.Therefore, because correspondingparts of congruent segments are congruent. Bypart b, . By the transitive property,

, so M is equidistant fromthe vertices of �ABC.

21. Given: and with

Prove:

Statements Reasons1. 1. Given.2. 2. The product of

the means equalsthe product of the extremes.

3. (BC)(DF) � (BC)(EF) 3. Subtraction � (AC)(EF) � (EF)(BC) postulate.

4. 4. Distributive � property.


6. 6. If the products of two pairs of fac-tors are equal,one pair of factorscan be the meansand the other theextremes of aproportion.

22. Let M be the midpoint of , N be the midpointof , P be the midpoint of , and Q be themidpoint of . Then, diagonal divides the quadrilateral into two triangles �ACD and�ACB. A line segment joining the midpoints oftwo sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpoints of thesides of �ACD and joins the midpoints ofthe sides of �ACB, we have , ,

, and . Two lines parallelto the same line are parallel. Thus, .

Since and , we have 12AC 5 MN1

2AC 5 QP

QP y MN

12AC 5 MN1

2AC 5 QP

MN y ACQP y ACMN

QP

ACDACDBC

AB

BCAB 5 EF

DE

(BC)(DE) 5 (EF)(AB)(EF)(AC2BC)

(BC)(DF2EF)

(AC)(EF)(BC)(DF) 5

BCAC 5 EF

DF

BCAB 5 EF

DE

BCAC 5 EF

DFDEFABC

AM > CM > BMCM > AM

BM > CM

MP > MP

CP > PBCB

QP � MN, and and are congruent.Therefore, MNPQ is a parallelogram since it has a pair of opposite sides that are congruentand parallel.

12-3 Similar Polygons (pages 488–489)Writing About Mathematics

1. Yes. Two polygons are similar if and only if allpairs of corresponding angles are congruent andthe ratios of the lengths of all pairs ofcorresponding sides are equal. Given any twosquares, all pairs of corresponding angles arecongruent since all right angles are congruent.The ratios of the lengths of any two pairs ofcorresponding sides are equal because the sidesof a square are all congruent. Therefore, any twosquares are similar.

2. No. For example, a square is not similar to anequilateral triangle since it does not have thesame number of sides.

Developing Skills3. 14. Yes. Two polygons are congruent if and only if all

pairs of corresponding parts are congruent. Thus,given any pair of corresponding sides of twocongruent polygons, the ratio of the lengths ofthe sides is equal to 1. Therefore, the twopolygons are similar.

5. No. Similar polygons are not congruent unlessthe ratio of the lengths of all pairs ofcorresponding sides is equal to 1. In that case,corresponding sides have equal lengths and arecongruent, and corresponding angles arecongruent.

6. 1 7. 27, 338. 3, , 4 9. 3a, 3b, 3c

Applying Skills10. Since the first triangle is equilateral, each side

has the same length, a. Since the second triangleis equilateral, each side has the same length, b.The ratio of the length of any side in the firsttriangle to the length of any side in the second triangle is . The angles of both triangles are all congruent since they all measure 60°. Therefore,the triangles are similar.

11. In any two regular polygons with the same num-ber of sides, the measure of any interior angle is

. Thus, all pairs of corresponding angles are congruent. If the length of one side of one ofthe regular polygons is a, then all of its sides have

180(n 2 2)n

ab

412

MNQP

324

the same measure. Similarly, if the length of oneside of the other regular polygon is b, then all ofits sides have the same measure. Thus, the ratio of the lengths of any pair of corresponding sidesis a : b, so the two regular polygons are similar.

12. a. A line segment joining the midpoints of twosides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpointsof the sides of �ABC, and

. Since M and N are midpoints of the sides of �ABC, we have and

, and the ratios of the lengths ofall pairs of corresponding sides are equal.Since , �CMN � �CAB and �CNM � �CBA because correspondingangles are congruent. Since �C � �C, allpairs of corresponding angles are congruent.Therefore, �ABC � �MNC.

b. 2 : 113. a. A line segment joining the midpoints of two

sides of a triangle is parallel to the third sideand its length is one-half the length of the third side. Thus, since joins the midpointsof the sides of �MNC, and

. Since M and N are midpoints of

the sides of �MNC, we have and

, and the ratios of the lengths of all pairs of corresponding sides are equal.Since , �CPQ � �CMN and �CQP� �CNM because corresponding angles arecongruent. Since �C � �C, all pairs ofcorresponding angles are congruent.Therefore, �PQC � �MNC. In Exercise 12awe showed that �MNC � �ABC. Therefore,since similarity is transitive, �PQC � �ABC.

b. 4 : 114. The angles of rectangles ABCD and EFGH are

all congruent since all right angles are congruent.Thus, all pairs of corresponding angles arecongruent. Since opposite sides are congruent ina rectangle, AB � DC, DA � BC, EF � GH, andHE � FG. Since , by substitution. Therefore, the ratios of the lengthsof all pairs of corresponding sides are equal, andthe two rectangles are similar.

DCHG 5 DA

HEABEF 5 BC

FG

PQ y MN

12NC 5 QC

12MC 5 PC

12MN 5 PQ

PQ y MNPQ

MN y AB

12CB 5 CN

12AC 5 MC

12AB 5 MN

MN y ABCM

15. Since and opposite sides of a

parallelogram are congruent, by substitution. Thus, the ratios of the lengths of allpairs of corresponding sides are equal. Since �K � �P, �M ��R because opposite sides of aparallelogram are congruent. In parallelogramKLMN, �N and �L are both the supplement of�K. In parallelogram PQRS, �S and �Q areboth the supplement of �P. Since thesupplements of two congruent angles arecongruent, �N � �L � �S � �Q. Therefore, allpairs of corresponding angles are congruent, andthe two parallelograms are similar.

12-4 Proving Triangles Similar (pages 494–495)

Writing About Mathematics1. Yes. The two right angles are congruent since all

right angles are congruent. Thus, if one acuteangle in one right triangle is congruent to anacute angle in the other triangle, the twotriangles are similar by AA�.

2. No. If the ratios of the lengths of two pairs ofcorresponding sides are equal, correspondingangles are not necessarily congruent. Thus, thetwo triangles do not need to be similar.

Developing Skills3. A line segment joining the midpoints of two sides

of a triangle is parallel to the third side and itslength is one-half the length of the third side.Thus, since joins the midpoints of the sides of �ABC, and . Since D and E

are midpoints of the sides of �ABC,

and . Thus, the ratios of the lengths of all pairs of corresponding sides are equal. Since

, �CDE � �CAB and �CED � �CBAbecause corresponding angles are congruent.Since �C � �C, all pairs of corresponding an-gles are congruent. Therefore, �ABC � �DEC.

4. 5 5. 6 6. 97. 12 8. 6 9. 15

10. 6 11. 10 12. 1613. 30 14. 9 15. DE � 4, DA � 3Applying Skills

16. In the proof of Theorem 12.5, we showed that CB � C�E and �A�B�C� � �DEC�. If twopolygons are similar, then their corresponding sides are in proportion, so .DE

ArBr 5 CrECrBr

DE y AB

12CB 5 CE

12AC 5 CD

12AB 5 DEDE y AB

DE

MNRS 5 NK

SP

KLPQ 5 LM

QR

325

Substituting CB � C�E into the proportion gives

. We are given that so by

the transitive property, . Therefore,

(DE)(A�B�) � (AB)(A�B�) or DE � AB.

17. Choose point D on such that B�D � BC.Choose point E on such that .Corresponding angles of parallel lines arecongruent, so and

. Therefore,by AA�. If two polygons are similar,

then their corresponding sides are in propor-tion, so . Substituting B�D � BC

into the proportion gives . We are

given that so by the transitivity

property, , and EB� � AB.Therefore, by SAS and

. Then by the transitiveproperty of similarity, .

18. Since �ABC is an isosceles right triangle,

and �A � �B. Since is the angle

bisector of �ACB, �ACD � �DCB.by the reflexive property of congruence. Then�ACD � �BCD by ASA. Thus, the angles�ADC and �BDC are congruent. If the anglesof a linear pair are congruent, then they are bothright angles. In particular, �CDA is a right angle.Therefore, �CDA � �C and �A � �A, so�ABC � �ACD by AA�.

19. Since and alternate interior angles in parallel lines are congruent, �DCA � �BAC.Since vertical angles are congruent,�DEC � �AEB. Therefore, by AA�,�ABE � �CDE.

20. We are given that �DAE � �BCE. Sincevertical angles are congruent, �AED � �CEB,so �ADE � �CBE by AA�.

21. Since perpendicular lines meet to form rightangles, �DFA and �BEA are both right angles.Thus, �DFA � �BEA since right angles arecongruent. Since ABCD is a parallelogram,opposite angles are congruent, so �D � �B.Therefore, �ABE � �ADF by AA�.

22. Since and are vertical and horizontal segments, respectively, they are perpendicular, so�EDC is a right angle. Similarly, �ABC is a rightangle. Thus, �EDC � �ABC. In �DEC, DE � 2

DCDE

AB y CD

CD > CD

CDh

AC > CB

nArBrCr , nABCnEBrD , nABC

nEBrD > nABC

EBrArBr 5 AB

ArBr

ABrArBr 5 BC

BrCr

EBrArBr 5 BC

BrCr

EBrArBr 5 BrD

BrCr

nEBrDnArBrCr , /CrArBr > /DEBr

/ArCrBr > /EDBr

ArCr y DEArBrBrCr

DEArBr 5 AB

ArBr

BCBrCr 5 AB

ArBrDE

ArBr 5 CBCrBr

and DC � 1. In �ABC, AB � 2 and CB � 4.Thus, , and the ratios of two pairs of corresponding sides are equal. Therefore,�ABC � �CDE by SAS�.

23. SP � 4 and SR � 2. Since :

, , and

. Therefore, the ratios of all

three pairs of corresponding sides are equal, so�PQS � QRS by SSS�.

24. Since and are vertical and horizontalsegments, respectively, , so �BOA is a right angle. Similarly, �DCA is a right angle.Thus, �BOA � �DCA. In �OAB, BO � 6 and OA � 4. In �CDA, CA � 3 and DC � 2. Thus,

. Therefore, since the ratios of two pairs of corresponding sides are equal and thecorresponding angles included between them arecongruent, �OAB � �CDA by SAS�.

25. Let the base of the pyramid be �ABC and let Pbe the vertex of the pyramid. If a plane p parallelto the base cuts the lateral faces of the pyramid,triangle �A�B�C� is formed (with A� on , B�

on , and C� on .) Consider the lateral face�ABP. Since p is parallel to the base, .A line is parallel to one side of a triangle andintersects the other two sides if and only if thepoints of intersection divide the sides propor-tionally. Thus, . Two line segments are divided proportionally if and only if the ratio ofthe length of a part of one segment to the lengthof the whole is equal to the ratio of thecorresponding lengths of the other segment.Thus, . Since �P � �P, �PA�B� ��PAB by SAS�. Therefore,

since corresponding sides are in proportion. Asimilar argument applied to the other two

lateral faces shows that and

that . By the transitive property,

. Therefore,�ABC � �A�B�C� by SSS�.

12-5 Dilations (pages 499–501)Writing About Mathematics

1. Yes. Since �ABC � �A�B�C�,because corresponding sides are in proportion.

2. (2, 1)

ABArBr 5 BC

BrCr 5 ACArCr

CrArCA 5 ArBr

AB 5 BrCrBC

PArPA 5 CrAr

CA

PBrPB 5 BrCr

BC

ArBrAB 5 PAr

PA 5 PBrPB

PArPA 5 PBr

PB

PArArA 5 PBr

BrB

ArBr y ABCPBP

AP

BOCA 5 OA

DC 5 2

BO ' OAOABO

SQSR 5 2!2

2 5 !2

PQQR 5 2!2

2 5 !2SPQS 5 4

2!2 5 !2

PQ 5 QS 5 2!2

CBDE 5 AB

DC 5 2

326

Developing Skills3. (3, 2) 4. 5. (6, 1)

6. 7. (24, 24) 8. (6, 39)

9. (�12, 21) 10. 11. (�2, 1)12. (�3, �4) 13. (1.5, 1) 14. (�10, �5.5)15. (6, �9) 16. (10, 7.5) 17.18. (�3, 6) 19. (2, 3) 20. (10, 6)21. 22.23. 24.

25.Applying Skills26. Under D

�3, A�(0, �15) and B�(�15, 0). The slopeof AB � the slope of A�B�, and the slope of A�B � the slope of AB�. AA� � BB� � 20, so the diagonals are congruent. Since , thequadrilateral is an isosceles trapezoid.

27. Since is a transversal cutting parallel

lines and , corresponding angles �HBAand �BED are congruent. Similarly, since

is a transversal cutting parallel lines

and , corresponding angles �EBC and �GEFare congruent. Thus, m�HBA � m�BED andm�EBC � m�GEF. By the partition postulate,180 � m�HBA � m�ABC � m�EBC and 180 � m�BED � m�DEF � m�GEF. Thus,m�ABC � 180 � m�HBA � m�EBC andm�DEF � 180 � m�BED � m�GEF. Bysubstituting m�HBA � m�BED and m�EBC� m�GEF, m�ABC � 180 � m�BED �m�GEF. Therefore, m�ABC � m�DEF and �ABC � �DEF.

28. a. A�(10, �15), B�(20, �15), C�(20, 5), D�(10, 5)b. Since and are both horizontal

segments, they are parallel. Since andare both vertical segments, they are

parallel. Therefore, A�B�C�D� is aparallelogram.

c. Since dilations preserve angle measure, allpairs of corresponding angles are congruent.

and

. Therefore, the ratios of the lengths of all pairs of corresponding sides are equal,and ABCD � A�B�C�D�.

d. Since dilations preserve angle measure,�B � �B�. In part c, we showed that

210 5 1

5

ABArBr 5 CD

CrDr 5DADrAr 5 BC

BrCr 5 420 5 1

5

BrCrDrAr

CrDrArBr

EFg

BCg‹

HBEG›

DEg

ABg

‹HBEG

›

AB y ArBr

RO, R1808, D21, or ry-axis + rx-axis

ry-axis + D15

ry-axis + D14

rx-axis + D4rx-axis + D1.5

A10, 203 B

A 1, 158 B

A213, 27

3 BA25

3, 0 B

. Therefore, by SAS�,

�ABC � �A�B�C�.

29. b. A�(6, 3), B�(3, 6), C�(�3, 6), D�(�6, 3),E�(�6, �3), F�(�3, �6), G�(3, �6), H�(6, �3)

c. HA � BC � DE � FG � 2d. H�A� � B�C� � D�E� � F�G� � 6e. A�B� � C�D� � E�F� � G�H�; draw diagonals

and . Under a dilation, angle measureis preserved, so �ABC � �A�B�C� by AA�.Since and AB � ,A�B� � .

f. Since dilations preserve angle measure, allpairs of corresponding angles are congruent.By parts c and d, the ratios of the lengths of allpairs of corresponding sides are equal to 1 : 3.Therefore, the two polygons are similar.

30. a. 10 sq units b. 90 sq unitsc. 160 sq units d. 250 sq unitse. The area of the image of a figure is equal to

the product of the area of the figure and k2,the square of the constant of dilation.

31. a. A(a, b) → A�(ka, kb)

B(a, b � d) → B�(ka, kb � kd)

C(c, b) → C�(kc, kb)

D(c, b � d) → D�(kc, kb � kd)

Since A� and B� have the same x-coordinate,is a vertical segment. Similarly, since C�

and D� have the same x-coordinate, isalso a vertical segment. Vertical segments areparallel. Therefore, the images and are parallel.

b. A(a, b) → A�(a, b)

B(c, d) → B�(kc, kd)

C(a � e, b) → C�(ka � ke, kb)

D(c � e, d) → D�(kc � ke, kd)

The slope of

and the slope of

.

The two slopes are the same. Therefore, theimages and are parallel.

12-6 Proportional Relations AmongSegments Related to Triangles (pages 505–506)


1. Both are correct. The two ratios are equal: .25 5

CrDrArBr

5 kd 2 kbkc 1 ke 2 ka 2 ke 5 kd 2 kb

kc 2 ka 5 d 2 bc 2 a

DrCr5 d 2 bc 2 a

5kd 2 kbkc 2 ka 5

k(d 2 b)k(c 2 a)BrAr

CrDrArBr

CrDrArBr

3!2!2AB

ArBr 5 BCBrCr 5 1

3

ArCrAC

ABArBr 5 BC

BrCr

327

2. No. Consider two similar triangles �ABC and�A�B�C�. Then, since the ratio of any pair ofcorresponding sides is equal to k, AB � kA�B�,BC � kB�C�, and CA � kC�A�. Thus, theperimeter of the first triangle is AB � BC � CA� k(A�B� � B�C� � C�B�), and the ratio of the

two perimeters is .

Developing Skills3. 6 cm 4. 4 : 3 5. 9 in.6. 12 in. 7. 6 : 7 8. 16 mm

9. a. 7 : 6 b. m, 14 mc. 44 m, m d. Yes


1. �ABC � �A�B�C� 1. Given.with ratio of similitude k : 1.

2. 2. The ratio of the lengths ofcorresponding sidesof similar polygonsis equal to the ratioof similitude.

3. M is the midpoint of 3. Given.and M� is the mid-

point of .4. AC � 2CM and 4. Definition of

A�C� � 2C�M� midpoint.5. or 5. Substitution (steps

2, 4).

6. �C � �C� 6. In similar triangles,correspondingangles arecongruent.

7. �BCM � �B�C�M� 7. SAS�.

11. Statements Reasons1. �ABC � �A�B�C� 1. Given.2. �C � �C� and 2. In similar triangles,

�B � �B� corresponding an-gles are congruent.

3. is the bisector of 3. Given.

�B and is the bisector of �B�.

4. m�CBE � m�B and 4. Definition of angle

m�C�B�E� � m�B� bisector.12

12

BrErg

BEh

CMCrMr 5 CB

CrBr 5 k1

2CM2CrMr 5 CB

CrBr 5 k1

ArCrAC

ACArCr 5 CB

CrBr 5 k1

5113

1613

k(ArBr 1 BrCr 1 CrBr)ArBr 1 BrCr 1 CrBr 5 k

1

251

(Cont.)

Statements Reasons5. �CBE � �C�E�B� 5. Halves of con-

gruent angles arecongruent.

6. �BCE � �B�C�E� 6. AA�.

12. Let ABCD and A�B�C�D� be two similar parallelograms. Since and �B � �B�, �ABC � �A�B�C� by SAS�.Therefore, because the ratios of the lengths of corresponding sides in similartriangles �ABC and �A�B�C� are equal.

13. Let �ABC � �DEF with ratio of similitude k : 1. Then AC : DF � k : 1, or AC � k(DF). If twotriangles are similar, then the lengths ofcorresponding altitudes have the same ratio asthe lengths of any two corresponding sides. Let

be the altitude to and be the altitudeto . Then BG : EH � k : 1 or BG � k(EH).The area of �ABC � (AC)(BG) and the areaof �DEF � (DF)(EH). By the substitutionpostulate, (AC)(BG) � [k(DF)][k(EH)] � k2(DF)(EH). Therefore, the ratio of the areas is k2 : 1.

14. a. Let ABCD be a trapezoid with parallel bases and . Let E be the intersection point of

the diagonals. Since , alternate interior angles are congruent, so �ABE ��CDE. Since vertical angles are congruent,�DEC � �AEB. Therefore, �ABE � �CDE

b. In similar triangles, the ratio of the lengths ofcorresponding sides is equal to the ratio ofsimilitude, k. Since �ABE � �CDE,

. Therefore, the ratio of similitude is equal to the ratio of the length ofthe parallel sides.

12-7 Concurrence of the Medians of aTriangle (pages 509–510)


1. Yes. Since the point P divides in the ratio 2 : 1, that is, AP : PM � 2 : 1, is twice thelength of . Therefore, can be divided intotwo congruent segments that are equal to thelength of .

2. Yes. In an isosceles triangle, the altitude to thevertex is also the median. Therefore, in anisosceles triangle, the altitude to the vertex is alsothe perpendicular bisector.

PM

APPMAP

AM

EBED 5 EA

EC 5 ABDC 5 k

AB y CDCDAB

DFEHACBC

ACArCr 5 AB

ArBr 5 BCBrCr

ABArBr 5 BC

BrCr

328

Developing Skills3. (�1, 2) 4. (�1, 1) 5. (1, 3)6. (3, 0) 7. (1, 3) 8. (�2, 4)9. (�4, �1) 10. (7, 1)

Applying Skills11. a. B(�2, 0) and C(2, 0)

b. M(�1, 3) and N (1, 3)c. y � �x � 2d. y � x � 2e. (0, 2)

12. A(3, 3), B(3, �3), C(�6, 0)

12-8 Proportions in a Right Triangle (pages 514–515)

Writing About Mathematics1. �ACD � �BCD when �ABC is an isosceles

right triangle. When �ABC is isosceles, andare the two congruent hypotenuses of right

triangles �ACD and �BCD, respectively. Since � , �ACD � �BCD by HL.

2. �RST is an isosceles right triangle. The altitude iscongruent to itself. Also, it is perpendicular to thehypotenuse, forming congruent right angles. Thealtitude also separates the hypotenuse into twocongruent segments, so the two smaller trianglesare congruent by SAS. Corresponding parts of congruent triangles are congruent, so �and �RST is an isosceles right triangle.

Developing Skills3. 12 4. 2 5. 20 6. 47. 6 8. 25 9. 9 10. 6

11. 12 12. 6Applying Skills13. 10 in. 14. 32 ft. 15. 2 m, 18 m16. 17. 4, 16 18. 6,19. 18 cm 20. 12 in.21. a. 7 and 28 b. and

12-9 Pythagorean Theorem (pages 519–521)Writing About Mathematics

1. Yes. Let �ABC be obtuse with �C the obtuseangle, AB � c, BC � a, and AC � b. The altitudelies outside the triangle and �ACD is a righttriangle. Let DC � x and AD � y. Then,b2 � x2 � y2. �ABD is also a right triangle, so c2 � (a � x)2 � y2 or c2 � a2 � 2ax � x2 � y2. Bythe substitution postulate, c2 � a2 � 2ax � b2.Since a2 � 2ax � b2 � a2 � b2, c2 � a2 � b2.

2. Yes. The diagonals of a parallelogram bisect eachother and separate the parallelogram into four

7!514!5

12!22!5

STRT

CDCD

CBCA

triangles. If the lengths of the diagonals are 6 and8 and one side of the parallelogram is 5, then a 3-4-5 right triangle is formed. Thus, all of the fourtriangles are right triangles and the diagonals areperpendicular. A parallelogram withperpendicular diagonals is a rhombus.

Developing Skills3. Yes 4. No 5. No6. Yes 7. Yes 8. Yes9. 11.3 cm 10. 52 cm 11. 160 mm

12. 24 ft 13. ft14. a. ft b. 36 ft15. 13 cm 16. 27 ft, 36 ft17. 20 ft, 21 ft, 29 ftApplying Skills18. The diagonals of the rectangle are 26 feet long.

Thus, if he marks the corners of two triangleswith sides that are 10, 24, and 26 feet in length,the two triangles formed are right triangles(because 102 � 242 � 262) and the resultingquadrilateral is a rectangle.

19. 2,448 ft2 20. in.21. If the adjacent sides of a parallelogram are 21

and 28 feet long and the diagonal is 35 feet, then212 � 282 � 352. By the converse of thePythagorean Theorem, a right triangle is formedwith the diagonal and the two adjacent sides, andthe parallelogram has a right angle. Aparallelogram with a right angle is a rectangle.Therefore, the parallelogram is a rectangle.

22. The diagonals of a parallelogram bisect eachother and separate the parallelogram into fourtriangles. If the diagonals of the parallelogramare 48 and 140 centimeters long, then one of thetriangles formed has sides that are 24, 70, and 74 centimeters long. By the converse of thePythagorean Theorem, since 242 � 702 � 742, oneof the triangles formed by the diagonals is a righttriangle, so all of the four triangles formed areright triangles. Therefore, the diagonals areperpendicular, and the parallelogram is arhombus.

23. 7.5 ft 24. ft

12-10 The Distance Formula (pages 524–526)

Writing About Mathematics1. Yes. Since vertical segments have the same x-

coordinate, x1 � x2 � 0 in the distance formula.Similarly, since horizontal segments have thesame y-coordinate, y1 � y2 � 0. In such cases, the

2!3

5!3

!64!61

329

distance formula reduces to �y1 � y2� and �x1 � x2�,respectively, or the length of a vertical and ahorizontal segment.

2. Let x2 � x1 be equal to some number a. If a � 0,then �a� � a, so �a�2 � a2. If a � 0, then �a� � a,so �a�2 � a2 � 0. If a � 0, then �a� � �a, so �a�2 � (�a)2 � a2.

Developing Skills3. 5 4. 13 5.6. 7. 8. 109. 10. 11. 16 or �8

12. 5 or �113. AB � BC � CD � DA � 5. The lengths of the

sides of the quadrilateral are all the same.Therefore, the quadrilateral is a rhombus.

14. a. . Thus, and�PQR is an isosceles triangle.

b. , so and . By the

converse of the Pythagorean Theorem, �PQRis a right triangle.

c. The midpoint, D, of is (4, 0), and. Therefore, the

midpoint of the hypotenuse is equidistantfrom the vertices of �PQR.

15. a. , , and .Therefore, the lengths of the sides of thetriangle are unequal, and the triangle isscalene.

b.and . By the converse of the Pythagorean Theorem, �LMN is a right triangle.

c. The midpoint, D, of is , and

. Therefore, the midpoint of the hypotenuse is equidistantfrom the vertices of �LMN.

16. ; therefore, .

17. a. y � 3x � 13 b. D(�4, 1) c.18. a. ED � DF = FE � 6

b. Yes. The sides all have the same length, so thetriangle is equilateral.

19. a. A�(2, �2), B�(8, 2), C�(4, 8)b. and . Thus, the

length of is not equal to the length of itsimage, , so distance is not preserved under the dilation.

ArBrAB

ArBr 5 2!13AB 5 !13

2!10

DE > FEDE 5 FE 5 !58

DL 5 DN 5 DM 5 52!2

A412, 21

2 BNM

(MN)2 5 (5!2)2 5 50(LM)2 1 (NL)2 5 (2!10)2 1 (!10)2 5 50

NL 5 !10MN 5 5!2LM 5 2!10

DP 5 DQ 5 DR 5 !10PQ

(PQ)2 5 (2!10)2 5 405 40(PR)2 1 (QR)2 5 2(2!5)2PQ 5 2!10

PR > QRPR 5 QR 5 2!5

6!25!22!23!2

2!10

c. , ,, and .

Therefore, , and corresponding sides are in proportion. BySSS�, �ABC � �A�B�C�.

d. In part c, we showed that �ABC � �A�B�C�.Since corresponding angles are congruent insimilar triangles, �A � �A�, �B � �B�, and�C � �C�. Therefore, angle measure ispreserved under this dilation.

20. a. and .Opposite sides are congruent, so ABCD is aparallelogram.

b. A�(6, 0), B�(9, �3), C�(12, 3), D�(9, 6)c. and

. Opposite sides are congruent, soA�B�C�D� is a parallelogram.

d. Yes. Since A�B�C�D� is also a parallelogram,, and the images of the two parallel

lines and are also parallel.

21. and .Opposite sides are congruent, so ABCD is aparallelogram. However, consecutive sides arenot congruent, so ABCD is not a rhombus.

22. , so �ABC is isosceles., so (AB)2 � 80 and � 80. By the converse of the

Pythagorean Theorem, �ABC is a right triangle.

Applying Skills

23. Let , and .

Then, . The lengths of the two sides are the same, so �ABC is an isoscelestriangle.

24. , so the diagonals arecongruent. The slope of , and the slope of

. Since the slopes are negativereciprocals, . Therefore, the diagonalsof a square are congruent and perpendicular.

25. Since and are both horizontal segments,

and BC � DA � a. Thus, ABCD is a

parallelogram. The length of is . If

a2 � b2 � c2, then AB � DA � a, and the two

consecutive sides and are congruent.DAAB

"b2 1 c2AB

BC y DA

DABC

EG ' FHFH 5 21

EG 5 1EG 5 FH 5 a!2

AB 5 BC 5 "a2 1 b2

C 5 (a, 0)A 5 (2a, 0), B 5 (0, b)

5 2(2!10)2(CA)2 1 (CB)2AB 5 4!5

CA 5 CB 5 2!10

BC 5 DA 5 !10AB 5 CD 5 2!5

CDABArBr y CrDr

5 3!5BrCr 5 DrArArBr 5 CrDr 5 3!2

BC 5 DA 5 !5AB 5 CD 5 !2

ABArBr 5 BC

BrCr 5 CACrAr 5 1

2

CrAr 5 2!26ArBr 5 BrCr 5 2!13CA 5 !26AB 5 BC 5 !13

330

A parallelogram with two consecutive congruentsides is a rhombus. Therefore, if a2 � b2 � c2,then ABCD is a rhombus.

26. a.

b.

c.

On the other hand:

Therefore, PM � MQ.

27. a. W�(kw, ky) and X�(kx, kz)

b.

Review Exercises (pages 529–531)1. , , 152. Yes. In �ABC, the measure of the other

acute angle is 90 � 67 � 23°. Therefore,�ABC � �LMN by AA�.

3. 94. and �C � �C. Therefore, since

the ratios of two corresponding sides are inproportion, and the angles included between the two pairs of sides are congruent,�EFC � �ABC by SAS�.

5. 86. a. 16 b. 32 c. d.7. 15.6 cm 8. 12.2 cm9. a. P�(4, �14), Q�(16, 2) b. 1 : 2

c. M(5, 3) d. M�(10, �6)e. N(10, �6) f. Yes, since M� � N

16!58!5

ECAC 5 CF

CB 5 13

2212121

2

5 k ? WX5 k"(x 2 k)2 1 k2(z 2 y)2

5 "k2(x 2 k)2 1 k2(z 2 y)2

WrXr 5 "(kx 2 kw)2 1 (kz 2 ky)2

5 $A x2 2 x12 B 2

1 A y2 2 y12 B 2

5 $A 2x2 2 x1

2 x22 B 2

1 A 2y2 2 y1

2 y22 B 2

5 $A 2x22 2

x1 1 x22 B 2

1 A 2y22 2

y1 1 y22 B 2

MQ 5 $Ax2 2 x1

1 x22 B 2

1 Ay2 2 y1

1 y22 B 2

5 $A x2 2 x12 B 2

1 A y2 2 y12 B 2

5 $A x1 1 x2

2 2x12 B 2

1 A y1 1 y2

2 2y12 B 2

5 $A x1 1 x22 2

2x12 B 2

1 A y1 1 y22

_

2y12 B 2

PM 5 $A x1 1 x22 2 x1 B

2 1 A y1

1 y22 2 y1 B

2

MQ 5 $Ax2 2 x1

1 x22 B 2

1 Ay2 2 y1

1 y22 B 2

PM 5 $A x1 1 x22 2 x1 B

2 1 A y1

1 y22 2 y1 B

2


1. 1. Given.2. �ABE � �DCE 2. If two parallel lines are

cut by a transversal, thealternate interior anglesare congruent.

3. �CED � �AEB 3.Vertical angles arecongruent.

4. �ABE � �DCE 4. AA�.


1. �ABE � �DCE 1. Given.2. 2. Definition of

similar polygons.3. 3. If the points at which a

line intersects two sidesof a triangle dividethose sides proportion-ally, then the line isparallel to the thirdside.

12. 12 cm 13. 25 in. 14. 43.8 cm15. a. , ,

b. and.

By the converse of the Pythagorean Theorem,�RST is a right triangle.

16. a. AC � AB � 5, so �ABC is isosceles since ithas two sides with the same length.

b. D(2, 4)c. and d. and

(AB)2 � 52 � 25. By the converse of thePythagorean Theorem, �ADB is a righttriangle. Therefore, is perpendicular to

, so is the altitude to .

17. a. A�(�6, 3), B�(6, �3), C�(0, 9)b. ,

,c. , so corresponding

sides are in proportion. By SSS�,�ABC � �A�B�C�.

d. P(0, 1)e. CP � 2 and PM � 1. Thus, .

f. P�(0, 3) g. Yesh. C�P� � 6 and P�M� � 3. Thus, .

18. 26 cm

CrPrPrMr 5 2

1

CPPM 5 2

1

ABArBr 5 BC

BrCr 5 CACrAr 5 1

3

CrAr 5 6!2ArBr 5 BrCr 5 3!5CA 5 2!2AB 5 BC 5 2!5

BCADBCAD

(DB)2 1 (AD)2 5 (2!5)2 1 (!5)2 5 25DB 5 !5AD 5 2!5

(ST)2 1 (TR)2 5 (!10)2 1 (2!10)2 5 50(RS)2 5 50

TR 5 2!10ST 5 !10RS 5 5!2

y ABg

EFg

CECA 5 CF

CB

y CDg

ABg

331

Exploration (pages 531–532)a. , ,

b.

4 � 4 ✔

Since AFGD and FGCB are both rectangles,corresponding angles are congruent.Therefore, AFGD � FGCB, so AFGD andFGCB are both golden rectangles.

c. Yes d. Results will vary.


1. 2 2. 4 3. 2 4. 25. 1 6. 4 7. 2 8. 39. 2 10. 2


1. 1. Given.2. �EBC � �ECB 2. Isosceles Triangle

Theorem3. �ABE is the supple- 3. If two angles form a

ment of �EBC and linear pair, then �DCE is the supple- they are supple-ment of �ECB. mentary.

4. �ABE � �DCE 4. Supplements ofcongruent angles arecongruent.

5. 5. Given.6. �ABE � �DCE 6. SAS.7. 7. Corresponding parts


12. Statements Reasons1. at R. 1. Given.2. Points A and B in 2. Given.

plane p3. and 3. If a line is perpen-

dicular to a plane,then it is perpendic-ular to each line inthe plane through theintersection of theline and the plane.

SR ' RBSR ' AR

SR ' plane p

AE > DE

AB > CD

BE > CE

4 0 5 2 1(2)(2) 0 (1 1 !5)(!5 2 1)

1 1 !52 0 2

!5 2 1

AFAD 0 FG

BF

BF 5 !5 2 1AF 5 1 1 !5EC 5 EF 5 !5

(Cont.)

Statements Reasons4. �ARS � �BRS 4. Perpendicular lines

meet to form rightangles and all rightangles are congruent.

5. 5. Reflexive property.6. �ARS � �BRS 6. SAS.7. 7. Corresponding parts


Part III13. a. Since the radius is one-half the slant height,

hc � 5.00 ft.

Lateral area �

�b. The radius, slant height, and height of the

cone form a right triangle. Thus,

Volume �

14. a. Since is a horizontal segment, the altitudefrom C to is a vertical line with the same x-coordinate as C. Thus, the equation of thealtitude is x � 2.

b. The slope of . Thus, the altitude fromB to will be a line with a slope of passing through (4, 0). Using the point-slopeformula, the equation of the altitude is:

c. Substitute x � 2 into the equation:

Therefore, the coordinates of D are (2, 1).

d. passes through the points (�1, 0) and

(2, 1). Therefore, the slope of .

The equation of can be found by using the

point-slope formula:

y 5 13x 1 13

y 2 0 5 13(x 1 1)

ADg

ADg

5 1 2 02 1 1 5 1

3

ADg

5 15 21 1 2

y 5 212(2) 1 2

y 5 212x 1 2

y 5 212x 1 2

y 2 0 5 212(x 2 4)

212AC

AC 5 2

ABAB

13pr2hc 5 1

3p(2.50)2(4.33) < 28.3 ft3

hc 5 "h2s 2 r2 5 "5.002 2 2.502 5 4.3301c

49.1 ft2

12pr2hs 5 1

2p(2.50)2(5.00)

SA > SB

SR > SR

332

e. The slope of is �3. Since the slopes of

and are negative reciprocals, they are

perpendicular. Therefore, contains the

altitude from A to .Part IV15. Statements Reasons

1. �ADE is an isosceles 1. Given.triangle.

2. is the median 2. Given.to .

3. 3. In an isosceles triangle, the medianfrom the vertexangle is perpen-dicular to the base.

4. 4. If a transversal is perpendicular toone of two parallellines, it is perpen-dicular to the other.

5. �DFG, �AFG, 5. Definition of per-�CGF, and �BGF pendicular lines.are all right angles.

6. ABCD is a rectangle. 6. Given.7. �FDC, �DCG, 7. The angles of a

�FAB, �GBA are rectangle are all all right angles. right angles.

8. FGCD and FGAB 8. A quadrilateral are both rectangles. with four right an-

gles is a rectangle.9. and 9. In a rectangle,

opposite sides are congruent.

10. 10. Definition of amedian.

11. 11. Transitive property.12. G is the midpoint 12. Definition of

of . midpoint.

16. a.

b.

The two images are the same for bothtransformations. Therefore, thetransformations are equivalent.

R908(x, y) 5 (2y, x)ry5x + rx-axis(x, y) 5 ry5x(x, 2 y) 5 (2y, x)

Cr 5 ry5x + rx-axis(6, 4) 5 ry5x(6, 24) 5 (24, 6)Br 5 ry5x + rx-axis(3, 1) 5 ry5x(3, 21) 5 (21, 3)Ar 5 ry5x + rx-axis(2, 5) 5 ry5x(2, 25) 5 (25, 2)

CB

GC > GB

DF > FA

FA > GBDF > GC

EFGg

' BC

EF ' ADAD

EF

BC

ADg

ADg

BCBC

13-1 Arcs and Angles (pages 541–542)Writing About Mathematics

1. Yes. If two lines intersect at the center of a circle,they form four central angles that are alsovertical angles. Since vertical angles arecongruent, the arcs intercepted by each pair ofvertical central angles are also congruent.

2. Yes. Since the degree measure of a circle is 360,the four arcs each measure 90 degrees, and thecentral angles that intercept the arcs each

measure 90 degrees. Therefore, and are perpendicular.


11. 120 12. 170 13. 91 14. 8915. 91 16. 138 17. 138 18. 13319. 133 20. 227 21. 180 22. 18023. 100 24. 110 25. 35 26. 11527. 115 28. 210 29. 145 30. 14531. 150 32. 215Applying Skills33. In a circle, central angles are congruent if their

intercepted arcs are congruent. Since ,�AOB � �COD. All radii of the same circle are

congruent, so . Therefore,�AOB � �COD by SAS.

34. �AOC and �BOD are vertical angles, so theyare congruent. Radii of the same circle are congruent, so . Then �AOC � �BOD by SAS. Since correspondingparts of congruent triangles are congruent,

.

35. � 90, � 27036. Since and , the diagonals of ABCD, are

perpendicular, ABCD is a rhombus. �AOC and�BOC are right angles since perpendicular linesintersect to form right angles. Radii of the same circle are congruent, so .Therefore, �AOB and �BOC are isosceles righttriangles and m�ABO � m�CBO � 45. Thenm�ABC � m�ABO � m�CBO � 90. Arhombus with a right angle is a square, so ABCDis a square.

Hands-On ActivityIn 1–2, results will vary.

3. By the definition of the degree measure of an arc, since � 60, m�AOB � 60 and sincemABC

OA > OB > OC

BDACmADCCmACC

AC > BD

OA > OB > OC > OD

OA > OB > OC > OD

ABC > CDC

BDg

ACg

333

, m�A�O�B� � 60. Therefore,�AOB � �A�O�B�. Radii of the same circle arecongruent, so and . By definition, �AOB and �A�O�B� are isoscelestriangles with m�OAB � m�OBA � 60 andm�O�A�B� � m�O�B�A� � 60. Then �OAB � �OBA � �O�A�B� � �O�B�A�and �AOB � �A�O�B� by AA�.

13-2 Arcs and Chords (pages 551–552)Writing About Mathematics

1. The apothem is perpendicular to the chord, sothe triangle formed is a right triangle. Also, theapothem is the distance from the center to thechord, so the length of one leg is 3 inches. Thehypotenuse of the triangle is the radius andmeasures 5 inches. By the Pythagorean Theorem,the length of the other leg is 4 inches, so a 3-4-5right triangle is formed. However, the apothemintersects the chord at its midpoint, so the lengthof the chord is 8 inches.

2. No. If the arcs belong to two circles with differentradii, then the arcs are not congruent.

Developing Skills3. 3 in. 4. 4.5 cm 5. 1.5 ft6. 12 mm 7. cm 8. 10 in.9. 24 ft 10. 14 cm 11. 12.4 mm

12. yd 13. radius � 11, diameter � 2214. 5 15. 25 16. 3017. 18. 12 19.20. OB � 30, DE � 6021. OB � 30, OC � 22.Applying Skills

23. We are given and .Since congruent circles are circles with congruentradii, ,and �COD � �AOB � �A�O�B� by SSS.Corresponding parts of congruent triangles arecongruent, so �COD � �AOB � �A�O�B�.

24. We are given and .Since congruent circles are circles with congruent radii, ,and �COD � �AOB � �A�O�B� by SSS. Cor-responding parts of congruent triangles are con-gruent, so �COD � �AOB � �A�O�B�. In acircle or in congruent circles, if central angles arecongruent, then their intercepted arcs arecongruent. Therefore, .CDC > ABC > ArBrC

OA > OB > OC > OD > OrAr > OrBr

CD > AB > ArBr(O > (Or

OA > OB > OC > OD > OrAr > OrBr

CD > AB > ArBr(O > (Or

LM15!3

5!512!5

2!5

!6

OrAr > OrBrOA > OB

mArBrC 5 60

Chapter 13. Geometry of the Circle

25. We are given that diameter intersects chord

at E and bisects at B. Then, .In a circle, congruent chords have congruentcentral angles, so �COE � �DOE. All radii of a circle are congruent, so , and

by the reflexive property of congruence. Therefore, �OEC � �OED bySAS. Corresponding parts of congruent trianglesare congruent, so �DEO � �CEO. If two linesintersect to form congruent adjacent angles, then they are perpendicular, so . Adiameter perpendicular to a chord bisects thechord, so E is the midpoint of and .

26. 5 cm27. In a circle, if the lengths of two chords are

unequal, then the shorter chord is farther from the center. is farther from the center than

, so AB < BC. is farther from the centerthan , so BC < AC. By the transitive property of inequality, AB < BC < AC. If the lengths oftwo sides of a triangle are unequal, then themeasures of the angles opposite these sides areunequal and the larger angle lies opposite thelonger side. Therefore, �B is the largest angle of�ABC.

13-3 Inscribed Angles and Their Measures(pages 555–558)


1. Draw diameter . Choose any point C on thecircle and draw and . Inscribed �ACBintercepts a semicircle, so it is a right angle.Triangle ACB is a right triangle with hypotenuse

and legs and .2. Yes. The measure of an inscribed angle of a circle

is equal to one-half the measure of itsintercepted arc. Therefore, m�ABC � 25 andm�PQR � 25, so �ABC � �PQR.


11. 190 12. 25013. a. 160 14. a. 50 15. a. 34

b. 44 b. 100 b. 68c. 56 c. 50 c. 84d. 112 d. 80 d. 34e. 200 e. 160 e. 42

CBCAAOB

CBCAAOB

ACBCBC

AB

DE > CECD

AOB ' CD

OE > OEOC > OD

CBC > BDCCDCCD

AOB

334

16. a. 40 17. a. 54b. 80 b. 48c. 80 c. 50d. 40 d. 55e. 100 e. 50

18. a. 60 19. a. 120b. 90 b. 120c. 210 c. 120d. 45 d. 60e. 105 e. 60f. 30 f. 60

Applying Skills20. a. Since vertical angles are congruent,

�LPR � �SPM. If two inscribed angles of acircle intercept the same arc, then they arecongruent. Therefore, �RLP � �MSP and�LPR � �SPM by AA�.

b. 8 cm

21. If and , .

Therefore, . In a circle, congruent arcs have congruent chords, so and �ABCis isosceles.

22. a. In a parallelogram, opposite angle arecongruent. Therefore, �ABC � �ADC orm�ABC � m�ADC. The measure of aninscribed angle of a circle is equal to one-halfthe measure of its intercepted arc, so

m�ABC � and m�ADC � .

By the substitution postulate,

or .

b.c. By the reasoning in part a, m�BAD �

m�BCD � 90 and ABCD is equiangular. Anequiangular parallelogram is a rectangle, soABCD is a rectangle.

23. It is given that ,

so . Since

� 360, � 180. The measure of an inscribed angle of a circle is equal to one-half themeasure of its intercepted arc, so m�DEF � 90and �DEF is a right triangle.

24. It is given that . Since and are diameters, and �ABC and �BCD areinscribed in semicircles. Therefore, �ABC and�BCD are right angles and congruent. If twoinscribed angles of a circle intercept the samearc, then they are congruent, so �BAC � �CDB.Thus, �ABC � �DCB by ASA.

BODAOCAB > CD

mDEFCmDEFC 1 mFGDCmDEFC 5 mFGDC

mDEC 1 mEFC 5 mFGDC

mABCC 5 mADCC 5 180

mABCC 5 mADCC5 12mADCC12mABCC

12mADCC1

2mABCC

AB > CAABC > CAC

mCAC 5 130mBCC 5 130mABC 5 100

25. In a circle, congruent chords have congruent arcs.

Since , . Vertical angles are congruent, so �AEB � �DEC. If two inscribedangles of a circle intercept the same arc, thenthey are congruent, so �BAE � �CDE.Therefore, �ABC � �DCB by AAS.

26. Let ABCD be inscribed in circle O with .

Draw . If parallel lines are cut by a transversal, then alternate interior angles arecongruent, so �ACD � �BAC or m�ACD �m�BAC. The measure of an inscribed angle of acircle is equal to one-half the measure of its

intercepted arc, so m�ACD � and

m�BDC � . By the substitution postulate,

or . Therefore,

. In a circle, congruent arcs have congruent chords, so and ABCD is anisosceles trapezoid.

27. We are given that . Draw . If parallel lines are cut by a transversal, then alternateinterior angles are congruent, so �ACD � �BAC or m�ACD � m�BAC. Themeasure of an inscribed angle of a circle is equalto one-half the measure of its intercepted arc, so

m�ACD � and m�BDC � . By

the substitution postulate, or

. Therefore, .

28. Since , . The measure of an inscribed angle of a circle is equal to one-halfthe measure of its intercepted arc, so

m�ACD � and m�BDC � .Halves of equal quantities are equal, so m�ACD� m�BDC and �ACD � �BDC. If two linesare cut by a transversal so that the oppositeinterior angles formed are congruent, then they are parallel. Therefore, .

29. In a circle, all radii are congruent, so . Vertical angles are

congruent, so �AOB � �COD. Then �AOB � �COD by SAS. Corresponding partsof congruent triangles are congruent, so �BAC � �DCA. If two lines are cut by atransversal so that the opposite interior anglesformed are congruent, then they are parallel.Therefore, .AB y CD

OA > OB > OC > OD

AB y CD

12mBCC1

2mADC

mADC 5 mBCCADC > BCCADC > BCCmADC 5 mBCC12mADC 5 1

2mBCC12mBCC1

2mADC

ACAB y CD

AD > BCADC > BCC

mADC 5 mBCC12mADC 5 1

2mBCC12mBCC

12mADC

AC

AB y CD

AB > CDABC > CDC

335

30. In Exercise 26, we proved that a trapezoidinscribed in a circle must be isosceles. Since

, ABCD is a trapezoid and thus isosceles. It follows that . Two chords are congruent if and only if their arcs are

congruent. Thus, , so

. Similarly,and DBCE is an isosceles trapezoid. Again,two chords are congruent if and only if their arcs are congruent. Thus, .Since , DCFE is also an isosceles

trapezoid with . Thus,

, and

.

31. In a circle, two chords are congruent if and only

if their arcs are congruent. Since is not

congruent to , is not congruent to .Since opposite sides of ABCD are not congruent,ABCD is not a parallelogram.

13-4 Tangents and Secants (pages 564–567)Writing About Mathematics

1. No. Since l is tangent to circle O at A, .Since m is tangent to circle O at B, .Both and are segments of . If two lines are each perpendicular to the same line,then they are parallel. Therefore, l || m and thelines do not intersect.

2. A polygon inscribed in a circle intersects thecircle at each of its vertices, whereas a circleinscribed in a polygon intersects each of its sidesat exactly one point.

Developing Skills

3. AB � 10, BC � 15, CA � 15; since BC � CA,so �ABC is isosceles.

4. AB � 40, BC � 50, CA � 30; since 302 � 402 �502, �ABC is a right triangle.

5. a. 25 6. a. 7 7. a. 24b. 20 b. 14 b. 16c. 40 c. 10 c. 36

8. a. 7 9. a. 41 10. a. 12b. b. 40 b. 12c. c. 40 c. 5

d. 1311. a. 3 12. a. 4 d. 8

b. 6 b. 3 e. 11c. 8 c. 3 f. 12d. 10

!13!13

BC > CA

AOBOBOAm ' OB

l ' OA

CDABCDCABC

CAC > DBC > DFC > ECCDFC > DEC 1 EFC > FCC 1 EFC > ECC

ED > FC

EF y DCECC > DBC > CAC

CB y EDBAC > DAC 1 BAC > BDCCAC > CBC 1 DAC > BCC

DA > BCAB y DC

13. a. cm b. 1.73 cm14. a. 80 b. 70 c. 50 d. 160

e. 80 f. 10 g. 20 h. 100i. 110 j. 130 k. 360

Applying Skills

15. and are tangent to circle O at Q and R,respectively. Tangent segments drawn to a circlefrom an external point are congruent, so

. by the reflexive property ofcongruence. Radii of a circle are congruent, so

. Therefore, �OPQ � �OPR by SSS.Corresponding parts of congruent triangles arecongruent, so �OPQ � �OPR. By definition,

bisects �RPQ.

16. and are tangent to circle O at Q and R,respectively. Tangent segments drawn to a circlefrom an external point are congruent, so

. by the reflexive property ofcongruence. Radii of a circle are congruent, so

. Therefore, �OPQ � �OPR by SSS.Corresponding parts of congruent triangles arecongruent, so �QOP � �ROP. By definition,

bisects �QOR.

17. a. Tangent segments drawn to a circle from anexternal point are congruent, so .By the isosceles triangle theorem,�PQR � �PRQ.

b. From part a, and �PQR � �PRQ.by the reflexive property of

congruence, so �QPS � �RPS by SAS.Corresponding parts of congruent triangles are congruent, so and QS � RS.Also, �QSP � �RSP. If two lines intersect toform congruent adjacent angles, then they are perpendicular. Therefore, .

c. The length of the altitude to the hypotenuse ofa right triangle is the mean proportionalbetween the lengths of the segments of the hypotenuse. Thus, if x � OS, then or

. Solving for x gives x � 2 or 8.Since OS < SP, OS � 2 and SP � 8.

18. a. We are given perpendicular tangents. Since , m�ACB � 90.

If two tangents are drawn to a circle from anexternal point, then the line segment from thecenter of the circle to the external pointbisects the angle formed by the tangents.Therefore, m�ACO � 45. A tangent line is

AC ' BCAC and BC

2x2110x 5 16

x4 5 4

10 2 x

OP ' QR

QS > RS

SP > SPPQ > PR

PQ > PR

OPh

OQ > OR

PO > POPQ > PR

PRg

PQg

POh

OQ > OR

PO > POPQ > PR

PRg

PQg

!3

336

perpendicular to a radius at a point of tangency, so and m�OAC � 90.The sum of the measures of the angles in atriangle is 180, so m�AOC � 45. Since �OAC � �AOC, .

b. Since �AOC is a 45-45-degree right triangle,

OC � .

c. From part a, m�ACB � 90 and .Radii of a circle are congruent, so .By the same reasoning in part a, . Bythe transitive property of congruence,

. Since AOBC isequilateral and has a right angle, it is a square.

19. Isosceles �ABC with vertex A is circumscribedabout circle O, and D, E, and F are the points of tangency of , , and , respectively. By the isosceles triangle theorem, �DBE � �FCE. Iftwo tangents are drawn to a circle from anexternal point, then the line segment from thecenter of the circle to the external point bisectsthe angle formed by the tangents. Therefore,m�OBE � and m�OCE � ,so �OBE � �OCE. Since is tangent to circle O at E, , forming congruent rightangles �BEO and �CEO. by thereflexive property of congruence, so �BEO ��CEO by AAS. Corresponding parts ofcongruent triangles are congruent, so and E is the midpoint of .

20. a. If a line is tangent to a circle, then it isperpendicular to a radius at a point on the

circle. Therefore, and . If two lines are each perpendicular to the sameline, then they are parallel, so . We are given that OA < O�B. Since one pair ofopposite sides of OABO� is not bothcongruent and parallel, OABO� is not a

parallelogram. Therefore, cannot be

parallel to .

b. From part a, and .Therefore, �OAC and �O�BC are rightangles and congruent. By the reflexiveproperty of congruence, �C � �C. Therefore,�OAC � �O�BC by AA�.

c. AC � 8, AB � 4, OC � , O�C � ,

OO� � "17

3!172!17

ABg

' OrBABg

' OA

ABg

OOrg

OA y OrB

ABg

' OrBABg

' OA

BCBE > CE

OE > OEBC ' OE

BC

12m/FCE1

2m/DBE

ACBCAB

AO > OB > BC > AC

BC > BOOA > OB

AC > AO

"2OA

AC > AO

AC ' OA

21. a. We are given that OA � O�B, so .If a line is tangent to a circle, then it isperpendicular to a radius at that point on the

circle. Therefore, and ,forming congruent right angles �OAC and�O�BC. Vertical angles are congruent, so�OCA � �O�CB. �OCA � �O�CB by AAS.Corresponding parts of congruent triangles are congruent, so and OC � O�C.

b. From part a, �OCA � �O�CB.Corresponding parts of congruent trianglesare congruent, so and AC � BC.

Hands-On Activitya. Results will vary.b. (1) By definition, the center, P, is the

intersection of the bisectors of the angles of aregular polygon. If a point lies on thebisector of an angle, then it is equidistantfrom the sides of the angle. Thus, the center Pis equidistant from the sides of the regularpolygon. The distance from a point to a lineis equal to the length of the perpendicularfrom the foot. Thus, the apothems are allcongruent.

(2) Each circle constructed in part a used one ofthe apothems as the radius. Since theapothems are all congruent, every apothemis also a radius of the circle, so the foot ofeach apothem is on the circle.

(3) By construction, each side of the regularpolygon is perpendicular to one of theapothems at the foot of the apothem. If a lineis perpendicular to a radius at a point on acircle, then it is tangent to the circle. Sincethe foot of each apothem is on the circle andeach apothem is a radius of the circle, thesides of the regular polygon are tangent tothe circle.

c.

13-5 Angles Formed By Tangents, Chords,and Secants (pages 572–574)


1. Yes. Let be tangent to the circle O at A. Then

and m�AOB � 90. Let be a secant that intersects circle O at A and C. Thenm�OAB � m�OAC � m�CAB. Therefore,m�OAC � 90 and the radius is not

perpendicular to the secant .ACg

OA

ACg

ABg

' OA

ABg

r2 5 a2 1 A s2 B

2

AC > BC

OC > OrC

ABg

' OrBABg

' OA

OA > OrB

337

2. Yes. Vertical angles are congruent, and eachcentral angle is equal to the measure of itsintercepted arc. One-half of the sum of two equalintercepted arcs is the intercepted arc.


12. 130 13. 130 14. 18015. 20 16. 100 17. 8018. 30 19. 60

20. ,21. 55 22. 140 23. 3524. 140 25. 70 26. 22027. a. 125 b. 55 c. 55

d. 90 e. 90 f. 125g. m�AED � . Therefore,

. A line through the center of acircle that is perpendicular to a chord bisectsthe chord, so bisects .

28. a. 150 b. 150 c. 30d. 30 e. 75 f. 105

Applying Skills

29. a. We are given that . If two parallel lines are cut by a transversal, then thealternate interior angles formed arecongruent. Therefore, �OCP � �OEA and�OPC � �OAE, and �OPC � �OAE byAA�.

b. , , ,

, , m�P � 3030. Vertical central angles are congruent and

intercept congruent arcs. Therefore, �FOE ��GOD and . We are given that

. By the partition postulate,

and

. Thus, by substitution,

. The measure

of an angle formed by a tangent and a secant is

equal to one-half the difference of the inter-

cepted arcs, so

and . By substitu-

tion,

, so �A � �C.Therefore, and �AOC is isosceles.OA > OC

12(mEFBC 2 mGBC) 5 m/C

m/A 5 12(mDGBC 2 mFBC) 5

m/C 5 12(mEFBC 2 mGBC)

m/A 5 12(mDGBC 2 mFBC)

5 mDGBC 2 mDGC 5 mDGCmEFCmFBC 5 mEFBC 25 mGBC

mDGBC 2 mDGCmEFBC 2 mEFC 5 mFBCmEFBC 5 mDGBC

mEFC 5 mDGC

mACC 5 105mBDC 5 75

mFBC 5 60mCFC 5 45mADC 5 75

AB y CPg

ACBD

BD ' AC

12(55 1 125) 5 90

mRSQC 5 225mRQC 5 135

31. Let D be a point on major . Since

m�AOB � 120, � 120 and � 240.The measure of an angle formed by two tangentsintersecting outside the circle is equal to one-halfthe difference of the measures of the intercepted

arcs, so m�P � . Tangent segments drawn to a circle from an externalpoint are congruent, so �PAB � �PBA. Sincethe sum of the measures of the angles of atriangle is 180, m�PAB � m�PBA � 60.Therefore, �ABP is equiangular. If a triangle isequiangular, then it is equilateral, so �ABP isequilateral.

32. We are given secant intersecting circle O at

A and B and chord . Assume that m�CBD �

. �ABD is an inscribed angle, so m�ABD

� . Since �ABD and �CBD form a linear

pair, m�CBD � m�ABD � 180. By the

substitution postulate,

or . However,

since .

Our assumption is false and m�CBD � .

13-6 Measures of Tangent Segments,Chords, and Secant Segments (pages 579–581)

Writing About Mathematics1. Yes. If two chords intersect, the product of the

measures of the segments of one chord is equalto the product of the segments of the other. Since AB � 24, and M is the midpoint of , AM �BM � 24. Because 12 12 � 144, any chord ofcircle O that intersects at M is separated by M into two segments such that the product of thelengths is also 144.

2. No. If two secant segments are drawn to a circlefrom an external point, then the product of thelengths of one secant segment and its externalsegment is equal to the product of the lengths ofthe other secant segment and its externalsegment. Since (AP)(BP) � (CP)(DP) and AP > CP, BP must be less than CP.


AB

AB

12mBDC

mBDC 1 mADC 1 mABC 5 360mADC 2 360

mBDC 1 mBDC 1 mADC 5 360

12mBDC 1 12mADC 5 180

12mADC

12mBDC

BD

ABCg

12(240 2 120) 5 60

mADBCmABCADBC

338

12. 27 13. 20 and 6 14. 14 and 215. 16 16. 6 17. 418. 819. AC � 24, AB � 6, BC � 1820. AC � 20, AB � 5, BC � 1521. AC � 16, AB � 4, BC � 1222. AC � 25, AB � 9, BC � 1623. 20 24. 12 25. 30

26. ,

27. 8 28.29. AD � 3, AE � 6 30. AD � 7, AE � 1231.

13-7 Circles in the Coordinate Plane (pages 584–587)

Writing About Mathematics1. Yes. Consider a circle with center (h, k) and

radius r. Two points on the circle are (h � r, k)and (h � r, k). The chord connecting these pointsgoes through the center, so it is a diameter. Thesepoints have the same y-coordinate, so thisdiameter is horizontal. Two other points on thecircle are (h, k � r) and (h, k � r). The chordconnecting these points goes through the center,so it is a diameter. These points have the same y-coordinate, so this diameter is vertical.

2. Yes. Dividing each side by 3 gives .This is the equation of a circle centered at theorigin with a radius of 2.

Developing Skills

3.4.5.6.7.8.9.

10.11.12.13.14.15.16.17.18.19.20.21.22. x2 1 (y 1 2)2 5 36

(x 1 1)2 1 y2 5 1(x 1 4)2 1 (y 1 4)2 5 25(x 2 5)2 1 (y 1 5)2 5 9(x 2 4)2 1 (y 2 3)2 5 16(x 1 3)2 1 (y 2 2)2 5 4(x 2 10)2 1 (y 1 1)2 5 73(x 2 1)2 1 (y 2 6.5)2 5 76.25(x 1 6)2 1 (y 2 1)2 5 9x2 1 (y 2 12)2 5 25(x 1 1)2 1 (y 2 3)2 5 16(x 2 2)2 1 (y 2 9)2 5 16x2 1 y2 5 16x2 1 y2 5 4(x 1 3)2 1 (y 1 3)2 5 4(x 2 6)2 1 y2 5 81(x 2 4)2 1 (y 1 2)2 5 100(x 1 2)2 1 y2 5 36(x 2 1)2 1 (y 2 3)2 5 9x2 1 y2 5 9

x2 1 y2 5 4

152

8!2

AC 5 15$215AB 5 3$21

5

23.

24.

25.

26.

27.

xO1

y

1(0, 0)

xO 1

y

�1 ( , )�34

52

1

y

�1

( , 1)�32

O x

xO1

y

1

(�4, 4)

xO1

1y

(2, �5)

339

28.

29.30. Yes. The equation of the circle is x2 � y2 � 32.

Substituting (4, 4) into the equation gives 42 � 42

� 32. Therefore, (4, 4) is a point on the circle.31. Yes. It is equivalent to ,

which is the equation for a circle centered at (�2, 1) with a radius of 5.

Applying Skills32. a. y � �x � 2 b. x � 0

c. d. (0, 2)

e. The perpendicular bisector of a side isequidistant from the endpoints of each side, sothe circumcenter, which is the intersection ofthe three perpendicular bisectors, isequidistant from the vertices of �ABC.

f. Yes. Let (0, 2) be the center of the circle andA(2, 6) be a point on the circle. Then the radiusis and the equation of the circle is x2 � (y � 2)2 � 20. Substituting B(�4, 0) and into the equation gives

, so B is on the circle.Substituting C(4, 0) into the equation gives 42

� (0 � 2)2 � 16 � 4 � 20, so C is on the circle.33. a. Yes. The x-coordinate of the incenter is 3.

6(3) � 8y � 10

18 � 8y � 10

y � �1

The incenter is at (3, �1).b. The incenter is on each of the angle bisectors,

and a point on the bisector of an angle isequidistant from the sides of the angle.

c. Yes. The equation of is y � 2, so (3, 2) is on. Since the incenter of the triangle isPR

PR

5 16 1 4 5 20242 1 (0 2 2)2

!20

y 5 13x 1 2

(x 1 2)2 1 (y 2 1)2 5 25

(x 2 2)2 1 (y 2 3)2 5 169

xO 1

y

1(1, 1)

equidistant from sides of the circle, andalso intersect the circle at one point.

d.34. The equation of the perpendicular bisector of

is y � �2x � 4 and the equation of theperpendicular bisector of is y � �1.

�1 � �2x � 4

2x � �3

x �

The circumcenter of �ABC is (�1.5, �1), whichis also the center of the circle. The radius equalsthe distance between (�1.5, �1) and

A(�1, 3) � .The equation of the circle is

.

35. a. ft b. c.d. 6 ft e. 12 ft

36. a. (x � 13)2 � (y � 13)2 � 169b. (8, 1) and (8, 25); (18, 1) and (18, 25)c. (8, 13) or (18, 13)d. 1,800 ft and 800 ft

13-8 Tangents and Secants in theCoordinate Plane (pages 592–593)

Writing About Mathematics1. Yes. The equation of the circle is

(x � r)2 � (y � k)2 � r2. The equation of the y-axis is x � 0. Solving for y gives:

Thus, the y-axis intersects the circle in the point(0, k), so it is tangent to the circle.

2. No. Since the slope of is not the negative reciprocal of the slope of , it is not perpen-dicular to at the point of intersection.

Therefore, is not tangent at A.Developing Skills

3. a. (0, 6) b. Tangent4. a. (6, 8) and (8, 6) b. Secant5. a. (3, 4) and (4, 3) b. Secant6. a. (1, 3) and (�1, �3) b. Secant7. a. (0, �3) and (3, 0) b. Secant8. a. (2, 2) and (�2, �2) b. Secant9. a. (4, 3) and (�3, �4) b. Secant

10. a. (2, 4) and (4, 2) b. Secant11. a. (�3, 3) b. Tangent

ABg

CACA

ABg

y 5 k

(y 2 k)2 5 0

r2 1 (y 2 k)2 5 r2

(0 2 r)2 1 (y 2 k)2 5 r2

A s, s!33 B(s, s!3)12!3

(x 1 1.5)2 1 (y 1 1)2 5 16.25

"(21 1 1.5)2 1 (3 1 1)2 5 !16.25

232

BCAB

(x 2 3)2 1 (y 1 1)2 5 9QR

PQ

340

12. a. (5, 5) b. Tangent13. a. (2, 2) b. Tangent14. a. (0, �4) and (2, �2) b. Secant15. x � 3 16. y � �417. y � x � 4 18. y � 2x � 10Applying Skills19. a. y � 3x � 5 b. x � 0 c. P(0, �5)

d. PB � 0 and PE � 0, so(PA)(PB) � (PD)(PE) � 0.

20. a. y � x � 2 b. y � 10 c. P(8, 10)d. , , PD � 8

✔

21. a. y � �x � 6 b. y � x � 6c. (6, 0) d. PA � PB � 18

22. The line and the circle intersect at a single point:(18, �4).

23. a. Both points are solutions to the equation.

b. 424. a. The equation of is . The

equation of is x � 2. The equation of is

y � 10. intersects the circle at .

intersects the circle at (2, 7). intersectsthe circle at (�1, 10).

b. The sides of the triangle each intersect thecircle at exactly one point.

c. Yes. Each side of the triangle is tangent to thecircle.

Review Exercises (pages 598–599)1. a. 100 2. a. 100 3. a. 40

b. 50 b. 110 b. 130c. 80 c. 75 c. 65d. 90 d. 30 d. 50e. 90 e. 100 e. 95

4. PB � 6, BC � 18 5. 106. AE � 3, EC � 8 7. (1)8. (3) 9. 60

10. 15 cm 11. AB � 4, AC � 12

12. ,13. Let ABCD be inscribed in circle O with .

Draw . If parallel lines are cut by a transversal, then alternate interior angles are congruent, so �ACD � �BAC or m�ACD � m�BAC. The measure of aninscribed angle of a circle is equal to one-half the measure of its intercepted arc, so

m�ACD � and m�BDC � . By 12mBCC1

2mADC

ACAB y CD

mBCC 5 160mABC 5 mACC 5 100

ACBC

(2325, 51

5)AB

ACBCy 5 24

3x 1 23AB

(x 2 2)2 1 (y 2 3)2 5 25

64 5 64(2!2)(16!2) 5 82

(PA)(PB) 5? (PD)2

PB 5 16!2PA 5 2!2

the substitution postulate, or

. Therefore, . In a circle, congruent arcs have congruent chords, so

and ABCD is an isosceles trapezoid.

14. a. Since , alternate interior angles �B and �C are congruent so m�B � m�C.

m�B � and m�C � . By the

substitution postulate, . If two inscribed angles of a circle intercept the samearc, then they are congruent. Therefore,�A � �C and �D � �B. By the transitiveproperty of congruence, �A � �B and �C � �D. If two angles of a triangle arecongruent, then the sides opposite these angles are congruent, so and

. Therefore, �ABE and �CDE areisosceles.

b. From part a, . Therefore,

.

c. From part a, �A � �C and �D � �B.Therefore, �ABE � �CDE by AA�.

15. We are given that .

Therefore, � 90.The measure of an inscribed angle of a circle isequal to one-half the measure of its intercepted

arc, so m�A � .Therefore, m�A is a right angle. In a circle, if twoarcs are congruent, then their chords are congruent. Therefore,and ABCD is a rhombus. A rhombus with a rightangle is a square, so ABCD is a square.

16. Let ABC be a triangle with right angle �B and M the midpoint of . Then, . Let �ABC be inscribed in a circle. Since a right angle

is inscribed in a semicircle, is a semicircle and is a diameter. Since , M is thecenter of the circle. All radii of the same circleare congruent, so and MA �

MC � MB, and the midpoint of the hypotenuseis equidistant from the vertices of the triangle.

17. Since , PAB � PCD. If two secant segments are drawn to a circle from an externalpoint, then the product of the lengths of onesecant segment and its external segment is equalto the product of the lengths of the other secantsegment and its external segment. Therefore,

PAB > PCD

MA > MC > MB

MA > MCACABCC

MA > MCAC

AB > BC > CD > DA

12(mBCC 1 mCDC) 5 1

2(180) 5 90

mABC 5 mBCC 5 mCDC 5 mDACABC > BCC > CDC > DAC

ACC > BDCmACC 5 mBDC

CE > DEAE > BE

mACC 5 mBDC12mBDC1

2mACC

AB y CD

AD > BC

ADC > BCCmADC 5 mBCC12mADC 5 1

2mBCC

341

(PAB)(PA) � (PCD)(PD). By the divisionpostulate, PA � PD. Then, by the subtractionpostulate, PAB � PA � PCD � PD, so AB � CD and . Two chords areequidistant from the center of a circle if and onlyif the chords are congruent. Therefore, and

are equidistant from the center of the circle.

18. 12 cm

Exploration (pages 599–600)a. Place the point of your compass on a point P on a

line . With any convenient radius, draw arcs

that intersect at C and at D. With C and D as centers and the compass open to a radiusequal to CD, draw arcs that intersect at a point E.Connect C, D, and E to form �CDE. Congruent radii were used to draw , , and , so

and �CDE is equilateral.

b. The interior angles of a regular hexagon eachmeasure 120°. The angle bisectors of the anglesof the polygon intersect in the center of thepolygon and form congruent equilateraltriangles. Use the construction from part a toconstruct equilateral �ABP. Then, constructequilateral triangle �BCP. Next, constructequilateral triangle �CDP. Continue this processuntil you have constructed equilateral �FAP.Since the equilateral triangles all share two sides,they are all congruent. Thus, the sides of theresulting hexagon, ABCDEF, are all congruent.Each interior angle of the hexagon is the unionof two angles of two adjacent equilateraltriangles. Since the angles of an equilateral allmeasure 60°, each interior angle of the hexagonmeasures 60 � 60 � 120°. Therefore, the hexagonformed is a regular hexagon.

c. The diagonals of a square bisect each other, areperpendicular, and are congruent. Let O be thecenter of the circle. Draw a diameter of the circle. Construct the perpendicular bisector ofthe diameter. Let C and D be the points wherethe perpendicular bisector intersects the circle.Since and are radii, , and theperpendicular bisector passes through O. But

and are also radii. Therefore, and bisect each other, are congruent, and areperpendicular, so ACBD is a square.

CDABOCODDC

AO > OBOBAO

AOB

CD > DE > CECEDECD

PBh

PAh

ABg

CDAB

AB > CD

d. From the construction in part c, , , ,

and are all congruent since their central angles are all right angles. Thus, the arcs formedby bisecting these arcs are all congruent (sincehalves of congruent arcs are congruent). Twochords are congruent if and only if their arcs arecongruent. Therefore, the chords formed by thearcs are all congruent, so the sides of the polygonformed are all congruent. Draw segments fromthe vertices of the polygon to the center of thecircle. These segments are all congruent becausethey are radii of the same circle. Thus, congruentisosceles triangles are formed by SSS. The baseangles of these triangles are congruent, so itfollows that the interior angles of the octagon are all congruent. Therefore, the octagon formed is a regular octagon.

f. Construct six congruent and adjacent equilateraltriangles with the center of the circle as a vertexand using two radii as sides.


1. 3 2. 1 3. 44. 3 5. 1 6. 37. 3 8. 2 9. 1

10. 3Part II11. a. (2, 2) b. (�2, 0)

c. Yes. The slope of is and the slope of

is .d. Yes. By the reflexive property of congruence,

�B � �B. If two parallel lines are cut by atransversal, then the corresponding angles arecongruent. Since , �CAD � �EDBand �ABC � �DBE by AA�.

12. Statements Reasons1. ABCD with 1. Given.

, and and bi-

sect each other.2. ABCD is a 2. If the diagonals of a

parallelogram. quadrilateral bisect eachother, then the quadri-lateral is a parallelogram.

3. ABCD is a 3. If the diagonals of arectangle. parallelogram are con-

gruent, then the parallel-ogram is a rectangle.

BDACAC > BD

DE y AC

12

AC12DE

DACBDCCBCACC

342


1. and 1. Given.

2. �AEF and �CEF 2. Definition of are right angles. perpendicular lines.

3. �AEF � �CEF 3. Right angles arecongruent.

4. 4. Reflexive propertyof congruence.

5. 5. Given.6. �AEF � �CEF 6. SAS.7. 7. Corresponding


14.

x � 15 � x � 3

x cannot be 3 since 3 � 7 � �4.The lengths of the sides of the triangle are 15, 8,and 17.

Part IV15. a. (�5, 2); ry=x (5, 2) � (2, 5), R90 (2, 5) � (�5, 2)

b. ry-axisc. No. Rotation is a direct isometry and line

reflection is an opposite isometry.

16. a. We are given that AD : DC � 1 : 3 and BE :

EC � 1 : 3. Let AD � x, so DC � 3x. Let BE

� y, so EC � 3y. AC � AD � DC � 4x and

BC � BE � EC � 4y. and

. Therefore, AC : DC � BC : EC.

b. From part a, AC : DC � BC : EC. By thereflexive property of congruence, �C � �C.Therefore, �ABC � �DEC by SAS�.

BCEC 5

4y3y 5 4

3

ACDC 5 4x

3x 5 43

(x 2 15)(x 2 3) 5 0

x2 2 18x 1 45 5 0

x2 1 x2 2 14x 1 49 5 x2 1 4x 1 4

x2 1 (x 2 7)2 5 (x 1 2)2

FA > FC

EA > EC

EF > EF

EFg

' CDg

EFg

' ABg

14-1 Constructing Parallel Lines (pages 607–609)

Writing About Mathematics1. The distance between two parallel lines is

defined as the length of the perpendicular fromany point on one line to the other line. Therefore,since CE is the perpendicular with length 2PQ,

every point on is at a distance 2PQ from .2. If two coplanar lines are each perpendicular to

the same line, then they are parallel. Thus,

and . The distance between two parallel lines is defined as the length of theperpendicular from any point on one line to theother line. Thus, the distance from any point on

to and is .Developing Skills

3. a. Use Construction 7.b. Mark any point Q on l. Use Construction 6 to

construct the perpendicular to m through Q.Let R be the intersection of this perpendicularand line m. Then use Construction 3 to construct the perpendicular bisector of .

c. Construct a line parallel to l (and above l) that

is the same distance away as m. Extend to a point S such that QR � RS. Then useConstruction 7 to construct a line parallel to lthrough S.

d. Yes. If two of three lines in the same plane areeach parallel to the third line, then they areparallel to each other.

4. a. Use Construction 1 to construct a segment congruent to the segment with length b. Use

Construction 5 to construct and perpendicular to with C and D on thesame side of . Set the compass radius to a.

With A as the center, mark off point X on .With B as the center and using the same

radius, mark off point Y on . Draw .ABYX is a rectangle.

b. Use Construction 1 to construct a segmentcongruent to the segment with length a.

Use Construction 5 to construct and perpendicular to with C and D on thesame side of . Set the compass radius to a.

With A as the center, mark off point X on .With B as the center and using the same

ACh

ABAB

BDg

ACg

AB

XYBDh

ACh

ABAB

BDg

ACg

AB

QRh

QR

12GHCD

gABg

GHg

GHg

y CDg

GHg

y ABg

ABg

CDg

343

radius, mark off point Y on . Draw .ABYX is a square.

c. Use Construction 1 to construct a segment congruent to the segment with length a. UseConstruction 2 to construct an angle congruent

to �A on . On the side of this angle not containing point B, use Construction 1 toconstruct a segment congruent to thesegment with length b. Use Construction 7 toconstruct a line parallel to through pointC. On this line, use Construction 1 to construct

congruent to the segment with length asuch that is on the same side of . Draw

. ABDC is a parallelogram.

d. Repeat part c, but construct to be congruent to the segment with length a.ABDC is a rhombus.

5. a. Construct the perpendicular bisector of atC. Construct the perpendicular bisector of at D and of at E. Then D, C, and E divide

into four congruent parts.b. Set the compass radius to AB. Draw a circle.c. Set the compass radius to AC. Draw a circle.

6. a. Use Construction 7.b. Use Construction 3 to find the midpoint M of

. Draw .Then is the median to .c. Use Construction 3 to find the midpoint N of

. Draw . Then is the median to .d. Yes. Any two medians of a triangle intersect in

the same point, and two points determine aline. Thus, the line drawn from C to the median of is the same as that drawn fromC to P.

7. a. Use Construction 3.b. Use Construction 3.c. Use Construction 3 to find the midpoint M of

.d. P and M. Any two perpendicular bisectors of

the sides of a triangle intersect in the samepoint. Thus, the line containing P and M is the perpendicular bisector of .

8. a. Use Construction 6 to construct a line perpendicular to through B intersecting

at D. is the altitude to .b. Use Construction 6 to construct a line

perpendicular to through A intersectingat E. is the altitude to .BCAEBC

BC

ACBDACAC

AB

AB

AB

BCANANBC

ACBMBMAC

ABCB

ACAB

AC

BDACCD

CD

AB

AC

ABh

AB

XYBDh

Chapter 14. Locus and Construction

c. P and C. Any two altitudes of the sides of atriangle intersect in the same point, and two

points determine a line. Thus, contains the altitude to .

9. a. Use Construction 4 to construct the angle

bisector, , of �CBA. Let D be the

intersection of and . Draw . Then is the angle bisector from B in �ABC.

b. Use Construction 4 to construct the angle

bisector, , of �CAB. Let E be the

intersection of and . Draw . Then

is the angle bisector from A in �ABC.c. P and C. Any two angle bisectors of a triangle

intersect in the same point, and two pointsdetermine a line. Thus, the line determined byP and C contains the angle bisector from C in�ABC.

10. b. Draw any point D on . Set the compass radius to AD. With D as the center, mark off a

point X on . With X as the center and using the same compass radius, mark off point

E on . Then DE � 2AD.c. Use Construction 7 to construct a line parallel

to through D.d. If a line is parallel to one side of a triangle and

intersects the other two sides, then the pointsof intersection divide the sides proportionally.

Therefore, since is parallel to and intersects the sides of �ABE,

.

11. a. Answers will vary.

b. Draw any point N on . Set the compass radius to PN. With L as the center, mark off

point L2 on . With L2 as the center and using the same compass radius, mark off point

L3 on . Repeat this procedure until you have drawn L8. The point Q is L8, and thepoint S is L5.

c. Use Construction 7 to draw a line parallel tothrough S.

d. If a line is parallel to one side of a triangle andintersects the other two sides, then the pointsof intersection divide the sides proportionally.

Thus, since is parallel to and intersects the sides of �PQR, PS : SQ � PT : TR � 3 : 5.If two line segments are divided propor-tionally, then the ratio of the length of a partof one segment to the length of the whole is

QRSTg

QR

PLh

PLh

PLh

AF : FB 5 AD : DE 5 1 : 2

EBDFg

EB

ADh

ADh

ACh

AE

AEBCAYh

AYh

BDBDACBX

hBXh

ABPCg

344

equal to the ratio of the corresponding lengthsof the other segment. Thus, PS : PQ � PT : PR� 3 : 8. Since �P � �P, �PST ~ �PQR bySAS~ with a constant of proportionality of .

14-2 The Meaning of Locus (pages 612–613)Writing About Mathematics

1. No. The locus of all points equidistant from theendpoints of a segment is the line that is theperpendicular bisector of the segment. Differentpoints on the line will be different distances fromthe endpoints.

2. The perpendicular to the line or ray through thepoint

Developing Skills3. The circle with the point as the center and a

radius of 10 centimeters4. The perpendicular bisector of 5. The line that is parallel to both lines and midway

between them

6. A pair of lines parallel to each 4 inches away

from such that is midway between them7. The line that is parallel to both lines and midway

between them8. The line that is parallel to both sides and midway

between them9. The line containing the diagonal between the two

other vertices10. The point that is the intersection of both

diagonals11. The circle with the same center and a radius of 1

inch12. The circle with the same center and a radius of 5

inches13. A pair of circles, both with the same center as the

given circle and one with a radius of 1 inch andthe other with a radius of 5 inches

14. a. The circle with the same center and a radius of(r � m)

b. The circle with the same center and a radiusof (r � m)

c. A pair of circles, both with the same center asthe given circle and one with a radius of (r � m) and the other with a radius of (r � m)

15. The circle with the same center as the givencircles and a radius of 14 centimeters

16. The perpendicular bisector of . (Note: Thetriangles are drawn on both sides of .)

17. A pair of lines parallel to each 3 feet away

from such that is midway between themABg

ABg

ABg

ABAB

ABg

ABg

ABg

AB

38

Applying Skills18. The circle centered at the base of the hour hand

with a radius equal to the length of the hour hand19. The line parallel to the track at a distance from

the track equal to the radius of the train wheel20. The line parallel to the curbs that is equidistant

from the curbs21. The circle with the stake as the center and a

radius of 6 meters22. The line that bisects the angle formed by the two

roads23. The line that is the perpendicular bisector of the

two floats on the lake24. The line parallel to the horizontal line at a

distance from the horizontal line equal to theradius of the dime

25. The circle with the same center as the circulartrack and a radius of 32 feet

14-3 Five Fundamental Loci (pages 615–616)Writing About Mathematics

1. Yes. The locus of points equidistant to and

is the line that is parallel to both and midway between them. This line is the perpendicularbisector of , and so every point on this line isequidistant to P and S.

2. Two lines that are equidistant from twointersecting lines are the angle bisectors of theangles formed by the two intersecting lines. Letone of the angles formed by the intersecting lineshave measure 2x. Then both of the adjacentangles have measure (180 � 2x). Thus, themeasures of the angles formed by the anglebisectors are x and (90 � x), and so the measureof the angle formed by the two angle bisectors isx � (90 � x), or 90°.

Developing Skills3. The perpendicular bisector of the segment

formed by the two points4. The circle with a radius of 6 inches and the center

equal to the midpoint of the segment5. The perpendicular bisector of the base of the

isosceles triangle6. The line containing the angle bisector of the

vertex angle of the isosceles triangle7. The two distinct perpendicular bisectors of the

sides of the square8. The line parallel to both bases and midway

between them9. The circle with a radius of 4 inches and the center

equal to the midpoint of the base

SP

RSg

PQg

345

10. The locus is the union of the two segments thatare both congruent to and parallel to the altitudeand 6 centimeters away from the altitude, and thetwo semicircles of radius 6 centimeters centeredat the endpoints of the altitude.

11. a. The locus is the line parallel to both lines andmidway between them.

b. Draw point P on any of the given lines. Thelocus is the circle centered at P with a radiusof 3 centimeters.

c. 212. a. The locus is the perpendicular bisector of .

b. The locus is a circle centered at M with aradius of .

c. 2d. A square. The diagonals bisect each other

since they are both radii of the same circle.Therefore, the quadrilateral formed is a paral-lelogram. The diagonals are congruent sincethey are both diameters of the same circle.Therefore, the parallelogram is a rhombus. Thediagonals are perpendicular since one is con-tained in the perpendicular bisector of theother. Therefore, the rhombus is a square.

14-4 Points at a Fixed Distance inCoordinate Geometry (pages 618–619)

Writing About Mathematics1. Yes. It intersects the circle in two points.2. Yes. The locus of points of two intersecting lines

is a pair of lines that bisect the angles formed bythe intersecting lines. A point is on the anglebisector of an angle if it is equidistant from the

sides of the angle. Since and are each tangent to the circle, they each intersect the circlein exactly one point. At these two points, the radiidrawn are perpendicular to the lines and meet in

the point O. Thus, O is equidistant from and

, and so O is on the locus.PBg

PAg

PBg

PAg

12AB

AB

Developing Skills3. 4.5.6.7.8.9. x � 12 and x � 2 10. x � 0 and x � 2

11. y � 6 and y � �2 12. y � 13 and y � 113. (�4, �3) and (3, 4) 14. (12, 5) and (�5, �12)15. (�8, �5) and (6, 9) 16. (�2, 0) and (0, 2)17. a. y � 1 and y � 7

b. x � 1 and x � 3c. (1, 7), (3, 7), (1, 1), (3, 1)

18. a.b. y � �4 and y � 2c. (5, 2) and (�1, 2)

19. a.b. y � 3 and y � �3c. (�4, 3), (4, 3), (�4, �3), (4, �3)

20. a.b. x � 8 and x � �8c. (8, 6), (8, �6), (�8, 6), (�8, �6)

21. a.

b. x � 2 and x � �2c. (2, 0), , , (�2, 0)

22. a.b. x � �1 and x � 4c. (�1, 10), (�1, 0), (4, 5)

14-5 Equidistant Lines in CoordinateGeometry (pages 622–624)

Writing About Mathematics1. Yes. The locus of points equidistant from two

intersecting lines is a pair of lines that are theangle bisectors of the angles formed by theintersecting lines. The angle bisectors of theangles formed by y � x and y � �x are the x- and y-axes.

2. Yes. The locus of points equidistant from twoparallel lines is the line parallel to the lines andmidway between them. The slope of this line is x,and the y-intercept is the average of the y-intercepts of the given lines, that is,b � � 6. Therefore, the locus is y � x � 6.

Developing Skills3. x � 5 4. y � �15. y � x 6.

7. 8.9. x � 3 10. y � �3

11. y � x � 6 12. y � �x � 2

y 5 213x 2 3y 5 21

2x 2 12

y 5 12x 1 1

2 1 102

(x 1 1)2 1 (y 2 5)2 5 25(22, 24!2)(22, 4!2)

(x 1 4)2 1 y2 5 36(x 1 4)2 1 y2 5 4

x2 1 y2 5 100

x2 1 y2 5 25

(x 2 2)2 1 (y 2 2)2 5 9

(x 1 3)2 1 (y 2 5)2 5 18(x 2 3)2 1 (y 1 1)2 5 10(x 2 1)2 1 (y 2 1)2 5 49x2 1 (y 1 2)2 5 9

(x 1 1)2 1 y2 5 1x2 1 y2 5 16

346

13. y � 2x � 3 14. y � �2x � 815. (1, 6) and (1, 0) 16. (4, �1) and (�4, �1)Applying Skills

17. a.

b.

✔

c. Distance from (3, 1) to (�2, 6) �Distance from (5, 5) to (�2, 6) �

18. a. y � x � 1b.

✔

c. y � �x � 5 d. A(1, 4)e. B(5, 0) f.

19. The locus of points equidistant from twointersecting lines is a pair of lines that bisect theangles formed by the intersecting lines. Thus, thelocus of points equidistant from y � x and y � �x are the x- and y-axes. Translationspreserve angle measure. Thus, if the x- and y-axesare the angle bisectors of y � x and y � �x, thentheir images will be the angle bisectors of theimages of y � x and y � �x. Under , the images of y � x and y � �x are y � x � 1 and y � �x � 1, respectively. The images of the anglebisectors are the y-axis and the line y � 1 underthe same translation. Therefore, the y-axis andthe line y � 1 are the angle bisectors of y � x � 1and y � �x � 1, and so they are the locus ofpoints equidistant from these two lines.

20. The locus of points equidistant from twointersecting lines is a pair of lines that bisect theangles formed by the intersecting lines. Thus, thelocus of points equidistant from y � 3x and y � �3x are the x- and y-axes. Translationspreserve angle measure. Thus, if the x- and y-axisare the angle bisectors of y � 3x and y � �3x,then their images will be the angle bisectors ofthe images of y � 3x and y � �3x. Under ,the images of y � 3x and y � �3x are y � 3x � 2and y � �3x – 2, respectively. The images of the angle bisectors are the y-axis and the liney � �2 under the same translation. Therefore,the y-axis and the line y � �2 are the anglebisectors of y � 3x � 2 and y � �3x – 2, and sothey are the locus of points equidistant fromthese two lines.

T0,22

T0,1

PA 5 PB 5 2!2

2 5 2

2 5? 3 2 1

y 5 x 2 1

!50!50

6 5 66 5? 1 1 5

6 5? 212(22) 1 5

y 5 212x 1 6

y 5 212x 1 5

21. a. (0, b) b. (0, c)c.d. Since the first line is parallel to the second

line, alternate interior angles are congruent.Thus, . Since verticalangles are congruent, �BMA � �B�MA�.Since M is the midpoint of , .Therefore, by ASA, �ABM � �A�B�M.

e. Since corresponding parts of congruent triangles are congruent, . The distance between two lines is the length of aperpendicular segment joining both lines.Since is perpendicular to the given lines,BB� is the distance between the given lines.Thus, since M is midway between B� and B, itlies on the line equidistant to the given lines.

14-6 Points Equidistant from a Point and aLine (pages 629–630)

Writing About Mathematics1. No. The solutions to the equation

�x2 � 2x � 8 � 0 are x � �2 and x � 4. Thegraph of y � �x2 � 2x � 8 intersects the x-axis at (�2, 0) and (4, 0). When the graph intersectsthe y-axis, the x-coordinate is 0.

2. Yes. The tangent of a parabola of the formis a horizontal line passing

through its turning point. Since (1, 0) is theturning point and is on the x-axis, the x-axis is thetangent of the parabola.

Developing Skills3. (3, �8); x � 3 4. (1, 2); x � 15. (�2, �5); x � �2 6. (1, 6); x � 1

7. (�4, 20); x � �4 8.

9. (3, 5), (0, 2) 10. (�2, 3), (1, 0)

x

O�2

2

y

xO�1 1

y

A 52, 217

4 B ; x 5 52

y 5 ax2 1 bx 1 c

BBr

BM > BrM

AM > ArMAAr

/BrArM 5 /BAM

M 5 A 0 1 02 , b 1 c

2 B 5 A0, b 1 c2 B

347

11. (4, 3), (1, 0) 12. (�4, 5), (1, 0)

13. (5, 7), (1, �1) 14. (2, 2), (�2, �6)

15. (2, 3), (6, �5) 16. (2, 0), (�2, �8)

17. 18.

19. 20.

Hands-On Activity

1.2.

Review Exercises (pages 631–632)1. a. Answers will vary. Example: Draw any

segment and then construct its perpendicularbisector. The angles formed are right angles.

b. Use Construction 4 to bisect the angle formedin part a.

c. Use Construction 1 to construct congruent to the segment with length a. Use Construc-tion 2 to construct �ABX congruent to the

AB

y 5 ax2 2 2hax 1 ah2 1 ky 5 ax2 2 6ax 1 9a 1 5

y 5 2 112x2 1 12x 2 34y 5 x2

8

y 5 2x2y 5 x2

xO1�1

y

xO1

�1

y

xO2

�2

y

xO1

�1

y

xO�1

�1

y

xO�1

1

y

45° angle from part b. Set the compass radiusto b. With B as the center, mark off a point C

along . Use Construction 7 to construct a line parallel to through A and a lineparallel to through C. Let D be the intersection of these two lines. Then ABCD isa parallelogram with m�B � 45.

2. Draw a ray and a segment . Use Construction 1 to construct a segment congruent

to on . Call the other endpoint Q1.

Repeat this construction on , , ,

and . Let the final endpoint be called Q.Rename Q2 as S. Then PS : SQ � 2 : 3.

3. The perpendicular bisector of the segmentformed by the two points

4. The two points that are on the perpendicularbisector of and centimeters from thesegment

5. The two points that are on the perpendicularbisector of the segment and 2 centimeters fromthe segment

6. The two points that are the intersection of theline parallel to the two lines and 2.5 inches fromeach of them and the circle centered at the givenpoint with radius 4 inches

7.8. (4, 4) and (0, 4)9. x � 2

10. y � 511. x-axis and y-axis12.13. (4, 3) and (3, 4)14. a–b.

c. (3, �4) and (�1, 4)

xO�1

y

�1

y 5 212x

(x 1 1)2 1 (y 2 2)2 5 9

!7AB

Q4X›

Q3X›

Q2X›

Q1X›

PXh

YZ

YZPXh

ABBC

BXh

348

15. (4, 3) and (�3, �4)

16. (2, �3) and (2, 1)

17. (�2, �1) and (1, 2)

18. (4, 2) and (1, 5)

19. 1 point 20. (3, 1)

Exploration (page 632)a. An ellipse is the locus of points such that the sum

of the distances from two fixed points is aconstant, k. Let k represent the length of thestring. Let P be any point on the curve that isdrawn. Then, , the length of thestring. Since P is an arbitrary point on the curve,all of the points on the curve are such that thesum of the distances from and is k.Therefore, the curve drawn is an ellipse.

F2F1

F1P 1 PF2 5 k

xO1

1

y

xO 1

1

y

xO1

y

�1

xO1

y

�1

b. The ellipse becomes more circular.c. The distance between each point and the curve

can increase infinitely while keeping thedifference between the two distances constant.


1. 2 2. 1 3. 3 4. 15. 1 6. 3 7. 4 8. 39. 2 10. 3

Part II11. In �ABC, �A � �A by the reflexive property of

congruence. If two parallel lines are cut by a trans-versal, then the corresponding angles are congru-ent. Since , �ADE � �ABC. Therefore,�ABC � �ADE and the sides are in proportion.

EC � AC � AE � 20 � 8 � 1212. The circumference of the base � 75. Then the

radius is:

Part III13. a–b.

c. (3, 8) and (�1, 4)

y � 3 � 5 � 8y � �1 � 5 � 4

x 5 3, 21(x 2 3)(x 1 1) 5 0

x2 2 2x 2 3 5 0x2 2 x 1 2 5 x 1 5

xO1

1

y

< 1,800 cubic feet

55,625

p

5 4p A 5,6254p2 B

5 13p A 75

2p B 212

Volume 5 13pr2h

r 5 752p

2pr 5 75

AE 5 8

615 5 AE

20

ADAB 5 AE

AC

DE y BC

349

14. a.

y � 4 � 2 � 2

The line intersects the circle only at (2, 2) so istangent to the circle.

b. The slope of the tangent line is �1. The centerof the circle is (0, 0), so the slope of the radius to the point of tangency is . Since these slopes are negative reciprocals, the tangentand the radius to the point of tangency areperpendicular.

Part IV15. a. Radii of congruent circles are congruent.

If two points are each equidistant from theendpoints of a line segments, then the pointsdetermine the perpendicular bisector of theline segment.

b. Radii of congruent circles are congruent.�ABC � �DEF by SSS.Corresponding parts of congruent trianglesare congruent.

/ABC > /DEF/ACB > /DFE/FDE > /CAB

BC > EFAB > DEAC > DF

/AME > /EMB > /AMD > /BMD/EAD > /EBD/AEB > /ADB

/AEM > /BEM > /ADM > /BDM/EAM > /DAM > /EBM > /DBM

AE > AD > BE > BDDM > MEAM > MB

AE > AD > BE > BD

22 5 1

x 5 2

(x 2 2)2 5 0

x2 2 4x 1 4 5 0

2x2 2 8x 1 8 5 0

x2 1 16 2 8x 1 x2 5 8

x2 1 (4 2 x)2 5 8

y 5 4 2 xx 1 y 5 4

16. a. Statements Reasons1. ABCD inscribed in 1. Given.

circle O with and a

diameter2. m�D � 90 and 2. An angle inscribed

m�B � 90 in a semicircle is aright angle.

3. �A and �B are 3. If two parallel linessupplements. are cut by a trans-�C are �D are versal, then the in-supplements. terior angles on the

same side of thetransversal aresupplementary.

4. m�A � 90 and 4. Definition of sup-m�C � 90 plementary angles.

5. ABCD is a 5. If a quadrilateral is rectangle. equiangular, then it

is a rectangle.

b. Statements Reasons1. Diagonals and 1. Given.

of ABCDintersect at E.�ABE � �CDE

2. �EAB � �ECD 2. Corresponding an-gles of similar trian-gles are congruent.

3. 3. If two lines are cut by a transversal sothat the alternate in-terior angles formedare congruent, thenthe lines areparallel.

4. �ABE is not con- 4. Given.gruent to �CDE.

5. is not con- 5. The ratios of cor-gruent to . responding sides of

similar triangles arein proportion.

CEAE

AB y CD

BDAC

ACAB y CD

350

6. E is not the mid- 6. Definition of point of . midpoint.

7. does not 7. Definition of bisect . bisector.

8. ABCD is not a 8. The diagonals of a parallelogram. parallelogram

bisect each other.9. is not parallel 9. Definition of

to . parallelogram.10. ABCD is a 10. Definition of

trapezoid. trapezoid.

c. Statements Reasons1. �ABC is equi- 1. Given.

lateral. D, E, and Fare the midpoints of , , and

, respectively.2. and 2. A line segment

joining the mid-points of two sidesof a triangle is parallel to the thirdside.

3. DECF is a 3. Definition of parallelogram. parallelogram.

4. 4. Definition of equilateral triangle.

5. BD � and 5. Definition of

BE � midpoint.

6. 6. Halves of congru-ent segments arecongruent.

7. ABCD is a 7. Definition of rhombus. rhombus.

BD > BE

12BC

12AB

AB > BC

DF y ECDE y FCAC

BCAB

ADBC

ACBD

AC

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