DG5 Solution Manual CH04 - flourishkh.com · 68 CHAPTER 4 Discovering Geometry Solutions Manual ©2015 Kendall Hunt Publishing Therefore, d 47° 40° 180° (Triangle Sum Conjecture),

68 CHAPTER 4 Discovering Geometry Solutions Manual

©2015 Kendall Hunt Publishing

Therefore, d� 47° � 40° � 180° (Triangle SumConjecture), so d� 93°. Next look at the trianglecontaining the 40° angle. One of its angles is avertical angle of the angle with measure d, sothe measure of this angle is 93° (Vertical AnglesConjecture). The measure of the third angle of thistriangle is 180° � 40° � 93° � 47° (Triangle SumConjecture). Finally, look at the small triangle atthe top of the figure that includes the angle withmeasure e. The angle in the lower left measures 47°(vertical angle of angle of measure b or supplementof 133° angle), and the angle in the lower rightmeasures 47° (vertical angle of angle found inthe triangle containing the 40° angle). Therefore,e� 47° � 47° � 180° (Triangle Sum Conjecture),so e� 86°.

9. m� 30°, n � 50°, p� 82°, q� 28°, r� 32°,s � 78°, t � 118°, and u� 50°. First look at theright triangle that contains a 60° angle. Here,m� 60° � 90° � 180° (Triangle Sum Conjecture),so m� 30°. Now look at the right triangle thatcontains a 40° angle. Here, by the Triangle SumConjecture, the unmarked angle measures 180° �90° � 40° � 50°. Because this angle and the anglewith measure n are vertical angles, they have thesame measure (Vertical Angles Conjecture), so n �50°. Now look at the small triangle with threeunmarked angles, which is below the triangle thatincludes the angle with measure m. In this triangle,one angle measures 180° � 112° � 68° (Linear PairConjecture), and another measures 30° because it isa vertical angle with the angle of measure m (VerticalAngles Conjecture). Therefore, the third angle in thesmall triangle measures 180° � 68° � 30° � 82°.Because this angle is a vertical angle with the angleof measure p, p� 82° (Vertical Angles Conjecture).Next look at the triangle that includes angles withmeasures p and q. The unmarked angle in thistriangle measures 70° (Vertical Angles Conjecture),so 82° � q� 70° � 180° (Triangle Sum Conjecture),and q� 28°. Now look at the triangle that containsthe angle with measure n, the angle that forms alinear pair with the angle of measure p, and thevertical angle of the angle with measure r. The anglemeasures in this triangle are 50°, 180° � p� 98°,and r. 50° � 98° � r� 180°, so r� 32°. Next lookat the large triangle that includes the angles withmeasures n and q. In this triangle, the measure ofthe unmarked angle is 180° � n � q� 180° �50° � 28° � 102° (Triangle Sum Conjecture), so s �78° (Linear Pair Conjecture). Now look at thetriangle that includes the angle with measure n andthe angles that form linear pairs with the 112° angleand the angle with measure t. The supplement ofthe 112° angle measures 68°. Thus, the angle in this

CHAPTER 4

LESSON 4.1

EXERCISES1. The angle measures change, but the sum remains

180°.

2. 73°. x� 52° � 55° � 180°, so x� 107° � 180°, andx� 73°.

3. 60°. All three angles of the triangle are congruent,so 3v� 180°. Therefore, v� 60°.

4. x � 52°, y � 85°. x � 90° + 38° � 180°, so x �128° � 180°, and x � 52°. y � 42° � 53° � 180°,so y � 95° � 180°, and y � 85°.

5. 110°. Ignore the 100° angle and the line thatintersects the large triangle. Find the angle measuresfor the large triangle: The supplement of the 120°angle measures 60°, and the supplement of the 130°angle measures 50° (Linear Pair Conjecture). Then,by the Triangle Sum Conjecture, the measure of thethird angle is 180° � (60° � 50°) � 70°. Because z�70° � 180°, z� 110°.

6. 900°. At each vertex of the triangle, four angles areformed, the sum of whose measures is 360°. Thethree (interior) angles of the triangle are unmarked,so the sum of their measures, which is 180°, mustbe subtracted from the total. Therefore, the sum ofthe measures of the marked angles is 3 ? 360° �180° � 900°.

7. 360°. At each vertex of the triangle, a linear pair isformed by a marked angle and an interior angle ofthe triangle. Because the sum of the measures of theangles in a linear pair is 180°, the sum of the meas-ures of the marked angles is 3 ? 180° � 180° �540° � 180° � 360°.

8. a� 69°, b� 47°, c � 116°, d� 93°, and e� 86°.First look at the large triangle that contains angleswith measures a, 40°, and 71°. By the Triangle SumConjecture, a� 40° � 71° � 180°, so a� 69°. Theangle with measure b forms a linear pair with the133° angle, so b� 133° � 180°, and b� 47°. Nextlook at the triangle that includes angles with meas-ures a and b. By the Triangle Sum Conjecture, themeasure of the unmarked angle in this triangle is180° � a� b� 180° � 69° � 47° � 64°. The anglewith measure c forms a linear pair with the 64°angle, so c � 180° � 64° � 116° (Linear PairConjecture). Now look at the triangle that includesangles with measures d and 47°. The unmarkedangle in this triangle forms a linear pair with the140° angle, so its measure is 180° � 140° � 40°.

DG5_Solution_Manual_CH04_Printer.qxd 3/16/16 3:18 PM Page 68

Discovering Geometry Solutions Manual CHAPTER 4 69©2015 Kendall Hunt Publishing

For Exercise 11: Use the method described for Exer-cise 11, but duplicate �L by tracing it onto pattypaper, and then bisect the angle that forms a linearpair with �L by folding the patty paper.

For Exercise 12: Use the method described forExercise 12, but duplicate the side and two angles bytracing them onto patty paper. No folding is necessary.

14. By the Triangle Sum Conjecture, m�A� m�S�m�M� 180°. Because �M is a right angle,m�M� 90°. By substitution, m�A� m�S�90° � 180°. By subtraction, m�A� m�S� 90°.So, two wrongs make a right!

15. It is easier to draw �PDQ if the Triangle SumConjecture is used to find that the measure of �Dis 85°. Then PD� can be drawn to be 7 cm, andangles P and D can be drawn at each endpointusing the protractor.

16. Sample answer: Draw �ABC with m�A� 110°,m�B � 40°, and AB � 3 cm.

Now trace angles A and B and make the segmentconnecting the new angles 5 cm long. Call the newtriangle �DEF, where m�D � 110°, m�E � 40°,and DE � 5 cm.

ED

F

110°5 cm

40°

A B

C

110°3 cm

40°

85°

55°

40°P D7 cm

Q

/A/E

/R

A

E

R

/L/G

Fold

triangle that forms a linear pair with the angle ofmeasure t has measure 180° � n � 68° � 180° �50° � 68° � 62°, so t � 180° � 62° � 118°. Nowlook at the triangle that includes the angle withmeasure s and an angle that forms a linear pair withthe angle of measure t. Because s � 78° and 180° �t � 62°, the third angle of this triangle measures180° � 78° � 62° � 40°. Finally, look at the righttriangle that includes the angle with measure u. Theother acute angle in this triangle measures 40°(Vertical Angles Conjecture), so 90° � 40° � u�180°, and u� 50°.

10. Draw a line with a straightedge and duplicate �Awith one side along this line. Duplicate �R so thatangles R and A have a common vertex and acommon side that is not along the original line. Theside of �R that is not common with �A and the rayopposite the one that forms the first side of �A willform �M because m�M �m�R�m�A� 180°.

This construction could also be done byconstructing the given angles at the ends of a linesegment; then �M is the third angle formed wherethe sides of the angles meet.

11. Draw a line with a straightedge and duplicate �Lwith one side along this line. Bisect the angle thatforms a linear pair with �L. Either of the twocongruent angles formed can be used as �G.

12. First construct �E, using the method used in Exer-cise 10. Then duplicate AE� and �A. Extend thesides of �A and �E that are not along AE� untilthey intersect. The intersection point of these tworays will be R, the third vertex of the triangle.

13. For Exercise 10: Use the method described for Exer-cise 10, but duplicate �A and �R by tracing themonto patty paper rather than using compass andstraightedge. No folding is necessary.

/M /A

/R

E

RA

/L

/G

/M /A

/R




Complete the table, stopping with the largestnumber of cards that is less than 2 ? 52 � 104.

Thus, the tallest house you can build with two52-card decks is eight stories. It will take 100 cardsto build this house.

DEVELOPING MATHEMATICAL REASONINGFirst notice which digits are missing, and find whatnumber they have to equal when combined.

1. 1 � 2 � 3 � 4 � 5 � 6 � 78 � 9 � 100

2. 1 � 2 � 3 � 4 � 5 � 6 1 7 1 8 ? 9 � 100

3. 1 � 2 � 3 ? 4 ? 5 � 6 � 78 1 9 � 100

4. (�1 � 2 2 3 2 4) � 5 � 6 � 7 � 89 � 100

5. 1 � 23 � 4 � 56 1 7 1 8 � 9 � 100

Many other identities using all nine digits in order arepossible, including 12 � 34 � 5 ? 6 � 7 � 8 � 9 � 100,123 � 4 � 5 � 6 � 7 � 8 � 9 � 100, and �1 � 2 �3 � 4 ? 5 ? 6 � 7 � 8 � 9 � 100.

EXTENSIONSee the solution to Take Another Look activity 8 onpage 22.

LESSON 4.2

EXERCISES1. 79°. Because m�H� m�O� m�T � 180° andm�T � 22°, m�H� m�O� 180° � 22° � 158°.By the Isosceles Triangle Conjecture, m�H� m�O.Therefore, 2m�H� 158°, and m�H� 79°.

2. 54°. By the Isosceles Triangle Conjecture,m�O� m�D, so m�O� 63°. Thenm�G � 180° � (2 ? 63°) � 54°.

3. 107.5°. By the Triangle Sum Conjecture, m�S�m�SLO� 35° � 180°. By the Isosceles TriangleConjecture, m�S� m�SLO. Therefore, 2m�SLO�35° � 180°, so 2m�SLO� 145° and m�SLO�72.5°. �SLO and �OLE form a linear pair, so, bythe Linear Pair Conjecture, m�OLE � 180° �72.5° � 107.5°.

4. m�R � 44°, RM� 35 cm. Use the Triangle SumConjecture to find m�R � 180° � (2 ? 68°) � 44°.By the Converse of the Isosceles TriangleConjecture, �ARM is isosceles with RA� � RM��.Therefore, RM � RA� 35 cm.

5. m�Y� 76°, RD � 3.5 cm. m�Y� m�R � 180° �28° � 152° and m�R � m�Y, so m�Y� 76°.RD � YD � 3.5 cm.

Number ofstories 1 2 3 4 5 6 7 8

Number ofcards 2 7 15 26 40 57 77 100

Measure the third angle of each triangle with aprotractor to find that m�C � 30° and m�F� 30°,so the third angles are also equal in measure.Conjecture: If two angles of one triangle have thesame measures as two angles of another triangle,then the third angles are have the same measure(Third Angle Conjecture).

17. You know from the Triangle Sum Conjecture thatm�A� m�B � m�C � 180°, and m�D �m�E � m�F� 180°. Thus, m�A� m�B � x�180° and m�D � m�E � y � 180°, so by the tran-sitive property, m�A� m�B � x� m�D �m�E � y. Because the figures show that m�A�m�D and that m�B � m�E, subtracting equalterms from both sides results in the equation x� y,that is, m�C � m�F.

18. For any triangle, the sum of the angle measures is180° by the Triangle Sum Conjecture. Because thetriangle is equiangular, each angle has the samemeasure, say x. So x� x� x� 180°, and x� 60°.

19. False. The angles between thetwo given sides may differ,which will result in trianglesthat differ in both size andshape.

20. False. The triangles will be the same shape, but maynot be the same size.

21. False. The other sides and angles of the trianglemay differ.

22. False. The triangles will be the same shape, but maynot be the same size.

23. True

24. Draw the next house in the series. A four-storyhouse will need 15 � 11 � 26 cards. Use inductivereasoning to find a pattern.

The number of cards added when a new story isadded to the tower increases by 3 each time, so afive-story house will need 26 � 14 � 40 cards.

15 18

2, 7, 15,

111

26



c � 56° by the AIA Conjecture. Now look at thelarge triangle with angle measures 66°, d, and 2e.Here, 66° � d� 2e� 180°, but d� e, so 66° �3d� 180°, and d� e� 38°. Now look at the talltriangle on the left: 66° � 38° � f � 180°, so f �76°. Next look at the isosceles triangle containingthe angle with measure d. Because d� 38°, theother base angle (not labeled) also measures 38°.Then 76° � 38° � g� 180°, so g� 66°. Also, fromthe isosceles triangle, h � 180° � (2 ? 38°) � 104°.The angles with measures k and h form a linearpair, so k� 180° � 104° � 76°. d� c � n � 180°,so n � 180° � 38° � 56° � 86°. d and p are themeasures of corresponding angles, so p� d by theCA Conjecture; thus, p� 38°.

11. a� 36°, b� 36°, c � 72°, d� 108°, e� 36°; none.There are 10 congruent central angles at the centerof the star decagon, and a is the measure of one ofthese angles, so a�

31600° � 36°. Because b� a, b�

36°, and the triangle containing the angles withmeasures a and b is isosceles by the Converse of theIsosceles Triangle Conjecture. Now look at thesmaller isosceles triangle in which the angle ofmeasure a is the vertex. (You also know that thistriangle is isosceles by the Converse of the IsoscelesTriangle Conjecture.) Here, the measure of eachbase angle is

12(180° � 36°) �

12(144°) � 72°. This

tells you that the measure of the third angle in thetriangle containing angles with measures b and c isalso 72°. Thus, b� c � 72° � 180° or 36° � c �72° � 180°, so c � 72°. Therefore, this triangle hastwo congruent angles and is thus also isosceles. Theangle with measure d forms a linear pair with a 72°angle, so d� 108°. The measure of the third anglein the triangle containing the angles of measures dand e is 180° � 72° � c � 180° � 72° � 72° � 36°.Therefore, e� 180° � 36° � d� 180° � 36° �108° � 36°. Thus, this triangle has two 36° anglesand is also isosceles.

Every triangle in the design is isosceles. (In otherwords, none of the triangles is not isosceles.)

12. A triangle is equilateral if and only if it isequiangular.

13. A triangle is isosceles if and only if the base anglesare congruent.

14. �GEA � �NCA. From the information given inthe figure, the correct correspondence of vertices isG to N, E to C, and A to A, and all pairs of corre-sponding parts are congruent, so �GEA � �NCA.

15. �JAN � �IEC. Notice that m�N� 40° andm�I� 50°. From the information given in thefigure, the correct correspondence of vertices is Jto I, A to E, and N to C, and all pairs of correspon-ding parts are congruent, so �JAN � �IEC.

6. m�D � 72°, m�U � 36°, MD � 8.6 cm. The mark-ings on the figure show that UD�� UM��, so �MUDis isosceles with �M and �D as the base angles.Therefore, by the Isosceles Triangle Conjecture,�D � �M, so m�D � 72°. By the Triangle SumConjecture, m�U � m�M� m�D � 180°, som�U � 180° � 2 72° � 36°. BecauseUD�� UM��, UD � 14 cm. The perimeter of�MUD is 36.6 cm, so MD � 36.6 cm� (2 14 cm) � 8.6 cm.

7. m�T � 78°; 93 cm. By the Triangle SumConjecture, m�T � m�B � m�S� 180°, som�T � 180° � 24° � 78° � 78°. This shows that�T � �S, so the triangle is isosceles by theConverse of the Isosceles Triangle Conjecture.Because �T and �S are the base angles, the twocongruent sides (or bases) of �TBS are SB� and TB�.Therefore, y � 22.5 cm � 38.5 cm, so y � 16 cm.Thus, the perimeter of �TBS is 16 cm � 2 38.5 cm � 93 cm.

8. NB � 75 m; m�N� 81°. From the figure,CN�� CB�, so CN� 2x� 90 m. Because theperimeter of �NBC is 555 m, 2(2x� 90 m) �x� 555 m. Solve this equation.

2(2x� 90) � x� 555

4x� 180 � x� 555

5x� 180 � 555

5x� 375

x� 75 m

Therefore, NB � 75 m.

By the Triangle Sum Conjecture, m�N� m�B �m�C � 180°. Because �NBC is isosceles withCN�� CB�, by the Isosceles Triangle Conjecture,m�N� m�B. Therefore, 2(m�N) � 18° � 180°,so m�N� 81°.

9. MV� 160 in., m�M� 126°. Using the fact that theperimeter is 605 in., MV� VT � TM� 605 in.Therefore, MV� 605 � 285 � 160 � 160 in.Because MV� TM� 160 in., �MTV is isosceleswith MV�� and TM�� as the legs and angles V and T asthe base angles. Therefore, by the Isosceles TriangleConjecture, m�T � m�V� x. Apply the TriangleSum Conjecture to get x� x� (x� 99°) � 180°,so x� 27°. Then x� 99° � 126°, so m�M� 126°.

10. a� 124°, b� 56°, c � 56°, d� 38°, e� 38°,f � 76°, g� 66°, h � 104°, k� 76°, n � 86°, andp� 38°. a� 56° � 180° by the Linear Pair Conjec-ture, so a� 124°. b� 56° by the Vertical AnglesConjecture. The angle with measure c and the anglemarked 56° are alternate interior angles formedwhen the parallel lines are cut by a transversal, so




Slope GH�� 4��4(��

61) �

�510 � �2

The slopes are neither equal nor oppositereciprocals, so DE�� and GH�� are neither parallelnor perpendicular.

23. Parallelogram. In quadrilateral FGCD, FG� and CD��are opposite sides, and GC� and FD� are oppositesides. From Exercise 20, FG� � CD��. Find the slopesof the other two sides:

Slope GC�� 4 �3 �

(�61) �

�53 � �

35

Slope FD�� 1��2(��

14) �

�53 � �

35

Because their slopes are equal, GC� � FD�.

Both pairs of opposite sides of FGCD are parallel, soFGCD is a parallelogram.

24. x � 45°, y � 55°, z � 45°. x � 90° � 45° � 180°,so x � 135° � 180°, and x = 45°. y � 70° � 55° �180°, so y � 125° � 180°, and y � 55°. z � 90° +45° � 180°, so z � 135° � 180°, and z � 45°.

25. Move each point of the original triangle to the right5 units and down 3 units to obtain the new triangle.Original triangle: (1, 0), (�3, �2), (�2, 3). Newtriangle: (6, �3), (2, �5), (3, 0). The originaltriangle was translated (moved to the right anddownward), but its size and shape were notchanged, so the two triangles are congruent.

26. Reflect each point of the original triangle across thex-axis to obtain the new triangle. Original triangle:(3, 3), (�3, 1), (�1, 5). New triangle: (3, �3),(�3, �1), (�1, �5). The original triangle has beenflipped over the x-axis, but its size and shape werenot changed, so the two triangles are congruent.

x

y

(–1, 5)

(–1, –5)

(–3, 1)

(–3, –1)

(3, 3)

(3, –3)

x

y

(3, 0)

(6, –3)

(2, –5)

(–3, –2)

(–2, 3)

(1, 0)

16. Possible answer: The three anglesof �ABC will be congruent to thecorresponding angles of �DEF,but the corresponding sides shouldnot be congruent. To construct�ABC, duplicate �D to make �A,and duplicate �E to make �B, butmake AB DE.

17. Possible answer: Duplicate MN��to make GH�� and duplicate �M tomake �G. Draw an arc with centerat H and length NP, intersecting theray that forms the side of �G thatdoes not lie along GH��. This arc will intersect the rayin two points, giving two possible positions for K,which are labeled as K1 and K2.

�GHK2 � �MNP, so choose K1for K, and complete the triangle.

18. Possible answer: Construct a60° angle (one angle of anequilateral triangle) anda 45° angle (bisect a rightangle). 60° � 45° � 105°, soyou obtain a 105° angle.

19. Perpendicular

Slope AB�� 06

��

31 �

�53 � �

35

Slope CD�� 12��43

� ��

53 �

53

The slopes, �35 and

53, are opposite reciprocals of

each other, so by the perpendicular slope property,AB�� and CD�� are perpendicular.

20. Parallel

Slope FG�� 1

6��

(1�4) �

53

From Exercise 19, slope CD�� 53.

The slopes are equal, so by the parallel slopeproperty, FG�� and CD�� are parallel.

21. Parallel. AD�� is a vertical line because thex-coordinates of points A and D are equal. CH�� isalso a vertical line because the x-coordinates ofpoints C and H are equal. Any two vertical linesare parallel, so AD�� and CH�� are parallel. (Note thatslopes cannot be used here because the slope of avertical line is undefined.)

22. Neither

Slope DE�� 8��3(��2

1)

� �104 � �

52

ED

BA

F

C

Fold 2

Fold 1

Fold 4

Fold 3

105° 60°45°

M N

G

K

H

P

G

K1

H

K2



B. See the solutions to Take Another Look activities 1and 9 on page 22.

COORDINATE GEOMETRY 4

EXERCISES1. y � �

16x. The median from R to ES� is the segment

whose endpoints are R and the midpoint of ES�.

The midpoint of ES� is �4 �2

8, �6

2� 4� � (6, �1).

To find the equation of the line through R(0, 0)

and (6, �1), first find its slope: �61��00

� �16. The

line containing the median from R to ES� goes

through the origin, so its y-intercept is 0. Therefore,

the equation of this line is y � �16x.

2. y � �2x � 2. First find the slope of RS�, and then

use this slope to find the slope of the altitude from

E to RS�. The slope of RS� is 48

��

00 �

48 �

12, so the

slope of the altitude to RS� is �2, the opposite

reciprocal of 12. Find the equation in the form

y � mx� b of the line through E(4, �6) with

slope �2.

y �x�

(�46)

� �2

y � 6 � �2(x � 4)

y � 6 � �2x � 8

y � �2x � 2

3. Centroid is �2, 23�. There are two methods for

finding the centroid.

Method 1: Find the equations of the lines containingtwo of the medians, and solve a system of equationsto find their point of intersection, which will be thecentroid.

Midpoint of MO�� 4 �2

10,

0 �2(�3)�

� �3, �32�

Find the equation of the line through N(0, 5) and

�3, �32�. The slope of this line is

� � �163

Any line through N(0, 5) has y-intercept 5, so anequation of the line containing the median from Rto RO� is y � �

163x� 5.

5 � ��32�

0 � 3

123

�3

Q RU

PT

S

PERFORMANCE TASK: ROSS’ CONSTRUCTIONx � 30

According to the Isosceles Triangle Conjecture, �CAB ��ABD; therefore, they both equal x. By the Triangle SumConjecture, m�CAB � m�ABD � m�ADB � 180°. Bysubstitution, we can write x � x � m�ADB � 180° orm�ADB � 180° − 2x. Since they form a linear pair,m�ADB � m�CDB � 180°. By substitution, 180° −2x � m�CDB � 180° and as a result m�CDB � 2x. Inisosceles triangle CBD, the base angles are equal by theIsosceles Triangle Conjecture, therefore, m�CDB �m�BCD � 2x. By the Triangle Sum Conjecture,m�CAB � m�ACB � m�ABC � 180°. With substitu-tion, x � 2x � 90° � 180°. Solving for x, we get x �30°. Construction: Draw AB � 8 cm. Draw a 30° angleat point A. Construct a perpendicular at point B. Markthe intersection of the 30° angle and the perpendicularas point C. You now have acute right triangle ABC.

EXTENSIONSA. The conjecture is false; any scalene triangle is a

counterexample. It is possible to divide any triangleinto a kite and two triangles, but the triangles willbe isosceles only if the original triangle is isosceles.(Recall that equilateral triangles are isosceles.)

This extension can be explored with geometry soft-ware or with compass-and-straightedge or patty-paper constructions. Here is one possible approachusing compass and straightedge: Starting with�PQR, to construct the kite, make an arc to findtwo points, S and T, one on each side of �P, thatare equidistant from point P. Connect these twopoints with ST� and then construct the perpendi-cular bisector of ST�. U is the point where theperpendicular bisector intersects the third side ofthe triangle, QR��. Draw SU� and TU�. By construc-tion, PS� PT and SU � TU, so PSUT is a kite.

The figures below show this construction for anisosceles triangle (using the vertex angle of thetriangle as one vertex of the kite) and a scalenetriangle. Observe that the two smaller trianglesthat are formed are each isosceles when theoriginal triangle is isosceles, but are scalenewhen the original triangle is scalene.

Q

S T

P

RU




Mean of y-coordinates � 6 � 0 �

3(�6)

� 0

The centroid is (4, 0).

5. �1, 43�. Sketch the triangle and find the coordinates

of its vertices. Graph the line 12x � 9y � 36by finding its intercepts: If x � 0, y � 4, so they-intercept is 4. If y � 0, x � 3, so the x-interceptis 3.

From the graph, you can see that the vertices

of the triangle are (0, 0), (0, 4), and (3, 0). The

mean of the x-coordinates is 0 � 3

3� 0 � 1,

and the mean of the y-coordinates is 4 � 0

3� 0 �

43.

Therefore, the centroid of the triangle is �1, 43�.

6. (�1, �1). Find the vertices of the triangle bysolving three systems of equations, each madeup of two of the three equations. First solvethe system formed by the first and secondequations:

�To solve this system by elimination, rewrite thesecond equation as �7x � 6y � 24, multiply thefirst equation by 2, and then subtract one resultingequation from the other.

16x � 6y � 24

�7x � 6y � 24

23x � 0

x � 0

Substitute 0 for x in the first original equation, andsolve for y.

8(0) � 3y � 12

y � 4

One vertex is (0, 4).

Now solve the system formed by the first and thirdequations.

�8x� 3y � 12x� 9y � 33 � 0

8x� 3y � 126y � 7x� 24

x

y

12x 1 9y 5 36

(0, 0) (3, 0)

(0, 4)

Midpoint of NO�� 0 �210

, 5 �

2(�3)� � (5, 1)

Find the equation of the line through M(�4, 0) and(5, 1). The slope of this line is 5 �

1 �(�

04) �

19. Find

the equation of the line passing through (�4, 0)with slope

19.

x�y �

(�04) �

19

x�y

4 � 19

9y � x � 4

y � 19x�

49

Solve the system formed by the equations of the linecontaining the two medians.

�To eliminate fractions, multiply both sides of thefirst equation by 9 and both sides of the secondequation by 6.

�To solve this system by substitution, solve the firstequation for x; then substitute the resulting expres-sion in the second equation and solve for y.

x� 9y � 4

6y � �13(9y � 4) � 30

6y � �117y � 52 � 30

123y � 82

y � 18223 �

23

Now substitute 23 for y in the equation x � 9y �4

and solve for x.

x� 9�23� � 4 � 2

Therefore, the coordinates of the centroid are �2, 23�.

Method 2: Find the mean of the x-coordinates andthe mean of the y-coordinates of the three verticesof the triangle.

The mean of the x-coordinates is �4 �30 � 10 � 2.

The mean of the y-coordinates is 0 � 5 �

3(�3)

� 23.

Therefore, the coordinates of the centroid are �2, 23�.

4. Centroid is (4, 0). First use the means of the coordi-nates of the vertices of �CDE to find the centroid.

Mean of x-coordinates � 0 � 1

32 � 0 � 4

9y � x� 46y � �13x� 30

y � 19x�

49

y � �163 x� 5



The equations of the medians are:

OF�: y � x

To find the intersection of OF� and AD�, create asystem of equations from the equations for thesegments:

y � x

y � x �

Substitute for y in the second equation and solve for x.

x � x � Multiply bothsides by(a + b)(b – 2a)

c(b – 2a)x � c(a � b)x – 2ac(a � b) Subtractc(a + b)x fromboth sides

c(b – 2a)x − c(a � b)x � – 2ac(a � b) Apply the Distri-butive Propertyto the left side

cbx � 2acx − cax � bcx � � 2ac(a � b) Collect liketerms

– 3acx � – 2ac(a � b) Divide bothsides by – 3ac

x � Simplify

x �

Substitute for x in the first equation and solve for y.

y � x

y � � � � �

y �

So the ordered pair � , � is the intersection of

OF� and AD�.

2(a�b)

32c3

cb � 2a

2c3

2(a�b)3

ca�b

ca�b

2(a�b)

3

�2ac(a�b)

�3ac

2acb�2a

cb�2a

ca�b

2acb � 2a

ca � b

AD: y �c

b � 2ax �

2ac

b � 2a

ca�b

BE: y ��2c

a � 2bx �

2ac

a � 2b

Multiply the first equation by 3, and rewrite thesecond equation by subtracting 33 from both sides.Then one resulting equation from the other.

24x � 9y � 36

x � 9y � �33

23x � 69

x � 3

Substitute 3 for x in the equation x � 9y � 33 � 0,and solve for y.

(3) � 9y � 33 � 0

9y � �36

y � �4

The second vertex is (3, �4).

Now solve the system formed by the second andthird equations, with both of them rewritten in thesame ways as before.

�Multiply the second equation by 7, and add theresult to the first equation.

�7x � 6y � 24

7x � 63y � �231

69y � �207

y � �3

Substitute �3 for y in the equation x � 9y � �33,and solve for x.

x � 9(�3) � �33

x� 27 � �33

x � �6

The third vertex is (�6, �3).

Finally, use the coordinates of the three vertices,(0, 4), (3, �4), and (�6, �3), to find thecoordinates of the centroid.

Mean of x-coordinates � 0 � 3 �

3(�6)

� �33 ��1

Mean of y-coordinates � 4 � (�4

3) � (�3)

� �33 � �1

The centroid is (�1, �1).

7. By the Coordinate Midpoint Property, the coordi-nates of the midpoints are: D(b, c), E(a, 0), andF(a + b, c).

�7x� 6y � 24x� 9y � �33




are eight different placements of two identical cards ina 3-by-3 array.)

Solution: An ace is in the center.

LESSON 4.3

EXERCISES1. Yes. The sum of the lengths of any two sides is

greater than the length of the third side.

2. No. 4 � 5 � 9, so the sum of the two shorter sidesis not greater than the length of the longest side.

3. No. 5 � 6 � 11 � 12

4. Yes. The sum of the lengths of any two sides isgreater than the length of the third side.

5. a, c, b. The lengths of the sides, from longest toshortest, are 28 in., 17 in., 15 in., so the order of theangles opposite those sides is the same: a, c, b.

6. a, b, c. There are two isosceles triangles in the figure.In the larger triangle, the vertex angle measures 30°,so each base angle measures

12(180° � 30°) � 75°.

You can see by referring to the isosceles triangle onthe right that the length of the base of the triangleon the left is b. By the Side-Angle Inequality Conjec-ture, you know that a� b because 75° > 30°. In thesmaller triangle, one base angle measures 72°, so theother base angle must also measure 72°, and there-fore the vertex angle measures 180° � 72° � 72° �36°. Apply the same conjecture to the triangle on theright: Because 72° � 36°, you know that b� c. Thus,a� b� c, so the correct order is a, b, c.

7. v, z, y, w, x. Apply the Side-Angle Inequality Conjec-ture first to the larger triangle and then to thesmaller one, noticing that the two triangles share aside. In the larger triangle, the third angle measures122°. Therefore, the angle measures in this triangle,from largest to smallest, are 122°, 30°, 28°, and theorder of the side lengths, from largest to smallest, isv, z, y. Now look at the smaller triangle. Here, thethird angle measures 104°, so the angle measuresfrom largest to smallest are 104°, 42°, 34°, and theorder of the side lengths from largest to smallest isy, w, x. Putting together the results from the two

AQ Q

KJ J

AK K

4 5

9

5 6

12

To find the intersection of OF� and BE�, create a systemof equations from the equations for the segments:

y � x

y � x �

Substitute for y in the second equation and solve for x.

x � x � Multiply bothsides by (a + b)(a – 2b)

c(a – 2b)x � –2c(a � b)x � 2ac(a � b) Add 2c(a + b)xfrom both sides

c(a – 2b)x � 2c(a � b)x � 2ac(a � b) Apply the Distri-butive Propertyto the left side

cax � 2bcx � 2cax � 2bcx � 2ac(a � b) Collect liketerms

3cax � 2ac(a�b) Divide bothsides by 3ac

x � Simplify

x �

Substitute for x in the first equation and solve for y.

y � x

y ��

y �

So the ordered pair � , � is the intersection of

OF� and BE�.

Since the intersection points are the same, the threemedians are concurrent.

DEVELOPING MATHEMATICAL REASONINGIn effect, the problem states that there is a uniquesolution, so finding any solution is sufficient. If youare stuck, systematically try all possible locations of,for example, two jacks. You can use symmetry todiminish the number of possibilities; that is, for everyplacement of the jacks, you can rotate or reflect theentire square to account for other possibilities. (There

−2ca�2b

2aca�2b

2c3

2(a�b)

3

2c3

2(a�b)

3c

a�b

ca�b

2(a�b)

3

2ac(a�b)

3ac

ca�b

2aca�2b

�2ca�2b

ca�b



shown to be congruent using the Triangle SumConjecture.

18. �HEJ � �GOH. �HJE � �GHO by SAS Congru-ency Conjecture. Since all parts of congruent trian-gles are congruent, GO� � HE�, �HJE � �GOH,Other congruences can be established. For example,since JH� � HE� �HJE is an isosceles triangle, so�HJE � �HEJ. By congruent parts, �GHO ��HEJ and �HEJ � �GHO.

19. a� 52°, b� 38°, c� 110°, and d� 35°. Firsta� 38° � 90° � 180° (Triangle Sum Conjecture), soa� 52°. The angle with measure b and the 38° angleare alternate interior angles, so b� 38° (AIA Conjec-ture). Then, in the triangle with angles of measure cand 32°, the third angle also measures 38°, so c�32° � 38° � 180°, and c� 110° (Triangle SumConjecture). In the small isosceles triangle at the topof the diagram, the vertex angle measures 110°(Vertical Angles Conjecture) and the third anglemeasures d (Isosceles Triangle Conjecture), so 2d�110° � 180° (Triangle Sum Conjecture), and thusd� 35°.

20. a� 90°, b� 68°, c� 112°, d� 112°, e� 68°, f � 56°,g� 124°, and h � 124°. First, the angle with measure aand the angle marked as a right angle are corre-sponding angles formed when two parallel lines are cutby a transversal, so a� 90° (CA Conjecture). a� b�22° � 180° (Triangle Sum Conjecture), so b� 68°. b�c� 180° (Linear Pair Conjecture), so c� 112°. Theangles with measures c and d are alternate interiorangles, so d� c� 112° by the AIA Conjecture. Theangles with measures b and e are alternate exteriorangles, so e� b� 68°. In the isosceles trianglecontaining the angles with measures e and f, the anglewith measure e is the vertex angle, so 2f� e� 180°(Triangle Sum Conjecture); therefore, f� 56°. Then56° � g� 180° (Isosceles Triangle Conjecture andLinear Pair Conjecture), so g� 124°. Finally, the angleswith measures g and h are alternate interior angles, soh� g� 124° (AIA Conjecture).

21. By the Triangle Sum Conjecture, the third anglemust measure 36° in the small triangle, butmeasure 32° in the large triangle. These are thesame angle, so they can’t have different measures.

22. x� 23 � 38 � 61°

y � 90 � 29 � 61°

z� 622 � 31°

Possible method: Add auxiliary transversals that alsoform triangles. Use linear angles are supplementary,alternate interior angles are congruent, and sum ofthe angles of a triangle are 180° to find the meas-ures of x, y, and z.

triangles, the order of all the side lengths in thefigure is v, z, y, w, x.

8. By the Triangle Inequality Conjecture, the sum of21 cm and 25 cm should be greater than 48 cm,but 21 � 25 � 46 cm.

9. b� 55°, but 55° � 130° � 180°, which is impossibleby the Triangle Sum Conjecture. Another way to seethat this situation is impossible is to apply theTriangle Exterior Angle Conjecture. The exteriorangle shown in the figure measures 125°, which mustbe the sum of the two remote interior angles. Butone of the remote interior angles measures 130° anda must be a positive angle measure, so this isimpossible.

10. 135°. By the Triangle Exterior Angle Conjecture,t � p� 135°.

11. 72°. The triangle is isosceles with base angles ofmeasure x. By the Triangle Exterior Angle Conjec-ture, x� x� 144°, or 2x� 144°, so x� 72°.

12. 6 � length � 102. The third side must be greaterthan 54 � 48 � 6 and less than 54 � 48 � 102, sothat the sum of the lengths of any two sides will begreater than the length of the third side.

13. Probability is 0. Since the two pieces of spaghetti arethe same length, breaking one at any point would meanthe lengths of the two sides formed would be equal tothe length of the third side. By the Triangle InequalityConjecture the sum of the of any two sides of atriangle must be greater than the length of the thirdside, so the lengths given would not form a triangle.

14. Area is 0. 24 � 14 � 38. By the Triangle InequalityConjecture the sum of the of any two sides of atriangle must be greater than the length of the thirdside, so the lengths given would not form a triangleand there would be no area.

15. 45°. When the perpendicular distance is measured,the bearing will be 90°. Because this is double thebow angle when you begin recording, that anglemust measure 45°.

16. �BAR � �ABE. All pairs of corresponding partsare congruent. Two pairs of congruent sides aremarked and the shared side is congruent to itself.Two pairs of angles are congruent by the AIAConjecture, and the third pair of angles must becongruent by the Triangle Sum Conjecture.

17. �FAR � �FNK. All pairs of corresponding partsare congruent. Two pairs of congruent sides aremarked. Applying the Converse of the IsoscelesTriangle Conjecture to �ANF shows that AF� �NF�, which gives the third pair of congruent sides.One pair of angles is congruent because they areboth right angles and the other two pairs can be




congruent to two sides of another triangle, and theangles between those sides are also congruent, thenthe two triangles are congruent.

3. Answers will vary. Possible answer:

4. SAS. Notice that if one of the triangles is rotated180° in either direction, it will coincide with theother, with L coinciding with I, U with D, and Zwith A, which confirms the congruence.

5. SSS. Use the shared side, FD�, as the third pair ofcongruent sides. Every side is congruent to itself.

6. Cannot be determined. Matching congruent sidesand angles give �COT � �NAP, rather than �COT� �NPA.

7. SSS. Use the shared side, CV�, as the third pair ofcongruent sides. Every side is congruent to itself.

8. SAS. Use the shared side, KA�, for the second pair ofcongruent sides.

9. SSS. Because Y is a midpoint, AY� � RY�, and by theConverse of the Isosceles Triangle Conjecture,YB� � YN�.

10. Yes; SAS. Because the perimeter of �ABC is 180 m,x� (x� 11) � (x� 11) � 180. Solve this equationto get x� 60 m, so x� 11 � 71 m. Therefore,AC � 60 m and AB � 71 m, so AC�� AE� andAB� � AD��. Also, �BAC � �DAE by the VerticalAngles Conjecture. Hence, �ABC � �ADE by SAS.

11. Possible answer: Boards nailed diagonally in thecorners of the shelves form triangles in thosecorners. Triangles are rigid, so the triangles in theshelves's corners will increase the stability of thosecorners and keep them from changing shape.

12. �ANT � �FLE by SSS. Match congruent sides. Yes�N � �L. If triangles are congruent, then corre-sponding angles are congruent.

13. Cannot be determined. SSA is not a congruenceconjecture.

14. �GIT � �AIN by SSS. Or use the vertical angles asthe pair of included congruent angles for SAS. Yes�G � �A. If triangles are congruent, then corre-sponding angles are congruent.

15. Cannot be determined. Parts do not correspond.Notice that the marked side is included between thetwo marked angles in �BOY, but not in �MAN.

16. �SAT � �SAO by SAS. Use the shared side for onepair of congruent sides. Yes, AO�� AT�. If trianglesare congruent, then corresponding parts arecongruent.

23. The sum of the interior angles of a quadrilateral is360°. �TAU is isosceles so m�TUA is 20°.60° � 80° � 130 � x� y � 360°. The sum of theangles of �TQU is 180° so y is 50°, which meansx� 40°. x� 40˚, y � 50˚

24. x� 22.5°, y � 67.5°. To get the relationship betweenx and y set up and solve a system of equations.

x � y � z� 180°

4x � z � 180°

�3x � y � 0

y � 3x

DEVELOPING MATHEMATICAL REASONING

EXTENSIONSA. Research results will vary.

B. See the solution to Take Another Look activity 2on page 22.

LESSON 4.4

EXERCISES1. Answers will vary. Possible answer: If three sides of

one triangle are congruent to three sides of anothertriangle, then the triangles are congruent. (All corre-sponding angles are also congruent.)

2.

Answers will vary. Possible answer: The picturestatement means that if two sides of one triangle are

If you know this: then you also know this:

8 9 6

7 5

4

18 16

17

13

12

14 15

1911 36

30

31

10

20

2434

2325262728 22

21233

32 35 1

29

3

A

D

B

C

x

y

z

z

23°29°

62°38°

38°

29°z

C

BA8 cm

67.5°

22.5°



25.

The number of struts needed to make a polygonwith n sides rigid is n � 3. This is the number ofdiagonals needed to divide the polygon intotriangles by connecting one of the n vertices to allof the other vertices that are not adjacent to it.

DEVELOPING MATHEMATICAL REASONINGMove g and j to the second row, and move a to make anew bottom row:

EXTENSIONSee the solution to Take Another Look activity 7 onpage 22.

LESSON 4.5

EXERCISES1. If two angles and the included side of one triangle

are congruent to two angles and the included sideof another triangle, then the triangles are congruent.

2.

If two angles and a non-included side of onetriangle are congruent to two angles and a non-included side of another triangle, then the trianglesare congruent.

3. Answers will vary. Possible answer:

Number of sides 3 4 5 6 7 12 n 20

Number of strutsneeded to makepolygon rigid 0 1 2 3 4 9 n 2 3 17

If you know this:

then you also know this:

17. Cannot be determined. Parts do not correspond.Notice that the shared side is included between thetwo marked angles in �WOM, but not in �WTO.

18. �SUN � �RAY by SAS. UN� AY� 4,SU � RA� 3, and m�U � m�A� 90°.Yes �RYA � �SNU. If triangles are congruent, thencorresponding angles are congruent.

19. �DRO � �SPO by SAS. The midpoint of both SD�and PR� is (0, 0), giving two pairs of congruentsides. Use the vertical angles as the pair of includedcongruent angles.

20. Yes. Students could draw right triangles withcongruent sides 2 units and 7 units and usecongruent triangles by SAS. The segments wouldthen be corresponding parts. Or some studentscould recall the distance formula from algebra.

21. Because the LEV is marking out two triangles thatare congruent by SAS, measuring the length of thesegment leading to the finish will also approximatethe diameter of the crater.

22. Draw a baseline with your straightedge and copythe length of one side along this baseline to formone side of the new triangle. Place the point of yourcompass at one endpoint of this segment anddraw an arc with radius equal to the length of asecond side of the original triangle. Then place yourcompass point at the other endpoint of the samesegment and draw an arc using a radius equal to thelength of the third side of the original triangle. Theintersection point of the two arcs will be the thirdvertex of the new triangle.

23. Duplicate one side of the triangle.Using the resulting segmentas one side of the angle, duplicatean angle adjacent to this side.Then duplicate a second side ofthe original triangle so that theduplicate angle will be includedbetween the two duplicate sides.

24. a. Yes the two triangles are congruent by SSS.

b. EFD

c. (x, y) �� (� x� 4, y)

d. The two triangles are congruent by SSS.




equation to get x� 46. Therefore, AC � 46 m, soAC� � AE�. Because BC� � DE� and EC�� is a transversalthrough BC�� and DE��, �E � �C by the AIA Conjec-ture. Similarly, BD�� is also a transversal through BC��and DE��, so �B � �D by the AIA Conjecture.Therefore, �ABC � �ADE by SAA.

Alternate solution: Notice that �BAC � �DAE bythe Vertical Angles Conjecture. Using this pair ofangles, AC� � AE�, and �E � �C, �ABC � �ADEby ASA.

19. �ABC � �CDA by SAA. Find the slopes of thesides of quadrilateral ABCD: slope AB�� slopeCD�� 3 and slope BC�� slope DA��

13. Therefore,

AB� � BC� and CD�� DA�, which tells us that �Band �D are both right angles, so they arecongruent. Also, from the slopes, BC� � DA�, so�BCA � �CDA by the AIA Conjecture. Usingthese two pairs of congruent angles and the sharedside, AC�, �ABC � �CDA by SAA.

20. Duplicate one side of the original triangle. Thenduplicate the two angles that include that side. Useone endpoint of the new segment as the vertex ofone of these angles and the other endpoint of thesegment as the vertex of the second angle. Find theintersection of the rays of the two angles that youduplicated.

Because a side and the angles that include thatside in the new triangle are congruent to the corre-sponding parts of the original triangle, the secondtriangle is congruent to the first by ASA.

21. The construction is thesame as the constructionin Exercise 20 once youfind the third angle, whichis used here. (Finding thethird angle is not shown.)

22. Draw any triangle. Duplicate any two angles of theoriginal triangle, but make the included side adifferent length from the one in the original triangle.By the Third Angle Conjecture, the two triangles willhave three pairs of congruent angles. The newtriangle will have the same shape, but not the samesize, as the original. Because the sides of the newtriangle are not congruent to those of the original,the triangles are not congruent. Possible answer:

A C

B

D F

E

4. �AMD � �RMC by ASA. Use the vertical angles atM as one pair of congruent angles.

5. Cannot be determined. Using the shared side as apair of congruent sides, you have SSA, which is nota congruence shortcut.

6. �GAS � �IOL by SAA. To confirm the congruenceand correspondence of vertices, flip over one of thetriangles and slide it to coincide with the other.

7. Cannot be determined. There is only one pair ofcongruent sides given, and the only pair of angles thatyou know are congruent are the vertical angles. Thisis not enough information to determine congruence.

8. �BOX � �CAR by ASA.

9. Cannot be determined. The parallel lines and twotransversals give two pairs of congruent angles, andthe vertical angles at T are congruent, but you don’thave any information about the sides. AAA is not acongruence shortcut.

10. �FAD � �FED by SSS. Use the shared side as thethird pair of congruent sides.

11. �WHO � �WTA by ASA or SAA. HT�� and OA�� aretransversals of the parallel sides, so �H � �T and�O � �A by the AIA Conjecture. Also, �HWO ��TWA by the Vertical Angles Conjecture. If you use�H and �T and the pair of vertical angles, thecongruence is true by ASA, or if you use both pairsof alternate interior angles, or �O and �A andthe pair of vertical angles, the congruence is trueby SAA.

12. �LAT � �SAT by SAS. Because AT� is an anglebisector, �LAT � �SAT, and the shared side, AT�,is congruent to itself.

13. �POE � �PRN by ASA. It is given that �O ��R. Since PO � PR, PO� � PR� All right angles arecongruent so �OPE � �NPR.

14. Cannot be determined. Parts do not correspond.Note that �A is adjacent to AM� but �C isopposite RM�.

15. �RMF � �MRA by SAS. �FMR � �ARMbecause both are right angles, and the shared side,MR��, is congruent to itself.

16. Cannot be determined. Using alternate interiorangles formed by the two transversals to the parallellines and the vertical angles, you have three pairs ofcongruent angles, but no information about thesides. AAA does not guarantee congruence.

17. �LAW � �WKL by ASA. Use the shared side, theright angles, and the other marked angles.

18. Yes; SAA or ASA. Because the perimeter of �ABC is138 m, x� (x� 4) � (x� 4) � 138. Solve this


29. a.

b. A�(�4, 3), B�(�2, 7), C�(�7, 5)

c. (�y, x)

d.

e. A�(�3, �4), B �(�7, �2), C �(�5, �7)

f. (�y, x)

g. (x, y) → (�x, �y)

EXTENSIONSA. See the solutions to Take Another Look activity 4 on

page 22.

B. See the solution to Take Another Look activity 7 onpage 22.

DEVELOPING MATHEMATICAL REASONINGPuzzle A transversalPuzzle B isosceles triangle

LESSON 4.6

EXERCISES1. Yes. BD� � BD� (same segment), �A � �C (given),

and �ABD � �CBD (given), so �DBA � �DBCby SAA. Therefore, AB� � CB� by CPCTC.

2. Yes. CN�� WN�� and �C � �W (given), and�RNC � �ONW (Vertical Angles Conjecture),so �CRN � �WON by ASA. Therefore, RN�� ON��by CPCTC.

y

x

B

A

C

C'

A'

B'

y

x

B

A

C

C'

A'

B'

B''

A''

C''

23. Draw a line segment. Construct a perpendicular.Bisect the right angle. Construct a triangle withtwo congruent sides and with a vertex thatmeasures 135°.

24. 125. Look for a pattern. Two concurrent linesdivide the plane into 4 parts, 3 concurrent linesdivide it into 6 parts, and 4 concurrent lines divideit into 8 parts. Each time another line is drawn,two more parts are added. Thus, n concurrentlines divide the plane into 2n parts; if 2n � 250,then n � 125.

25. False. One possible counterexample is a kite.

26. None. Because the triangle is isosceles, both legswill have length KM. By SAS, the two pairs ofcongruent sides and the included right angleswill guarantee that all isosceles right triangles witha given length for one of the legs will becongruent.

27.

28. a. About 100 km southeast of San Francisco

b. Yes. No, two towns would narrow it down totwo locations. The third circle narrows it downto one.

135°

K

L

M





13. Draw AC� and DF� to form �ABC and �DEF.AB� � CB� � DE� � FE� because all were drawnwith the same radius. AC� � DF� for the samereason. �ABC � �DEF by SSS. Therefore,�B � �E by CPCTC.

14. Cannot be determined. You have two pairs ofcongruent sides (CM�� PM�� from the median, andAM�� AM��), but that is not enough information toshow that the triangles are congruent.

15. �HEI � �KEI by ASA. The shared side isincluded between the two pairs of given congruentangles.

16. �ULF � �UTE by SAS. Because U is the midpointof FE�, UF� � UE�, and because U is also themidpoint of LT�, UL� � UT�. The included angles arevertical angles.

17. Copy one side of the original triangle, then one ofthe angles adjacent to that side using an endpointof the segment as a vertex. Then use a compass tomark off an arc whose length is equal to the otherside of the original triangle that is adjacent to thesame angle. Connect the endpoints of the twosegments you have constructed that are not thevertex of the angle between them to form the thirdside of the new triangle.

18. Make the included anglesdifferent.

19. a� 112°, b� 68°, c � 44°, d� 44°, e� 136°,f � 68°, g � 68°, h � 56°, k� 68°, l � 56°, and m�124°. a� 68° � 180° (Linear Pair Conjecture), soa� 112°. b� 68° (AIA Conjecture). 68° � 68° �c � 180° (Linear Pair Conjecture), so c � 44°. d�c � 44° (AIA Conjecture). d� e� 180° (Linear PairConjecture), so e� 136°. In the triangle with anglemeasures f and g, the unmarked angle measures 44°(Vertical Angles Conjecture). This is an isoscelestriangle with base angles of measures f and g, so2f � 44° � 180°. Therefore, f � 68° and g� 68°. Inthe isosceles triangle with base angle of measure h,the measure of the vertex angle is 68° (VerticalAngles Conjecture), so 2h � 68° � 180° and h �56°. k� 68° (AIA Conjecture). k� h � l � 180°,so l � 56°. Finally, m� k� h (AIA Conjecture),so m� 124°.

3. Cannot be determined. The given congruentparts and CH�� CH�� lead to SSA for trianglesCHS and HCR, but SSA does not prove trianglecongruence.

4. Yes. �S � �I, �G � �A (given), and TS� � IT�(definition of midpoint), so �ATI � �GTS by SAA.Therefore, SG� � IA� by CPCTC.

5. Yes. Draw VF� to form �FIV and �FEV. ThenFI� � FE� and VI�� VE� (given), and VF� � VF�(same segment), so �FIV � �FEV by SSS.Therefore, �I � �E by CPCTC.

6. Yes. MN�� MA�� (given), ME�� MR�� (given), and�M � �M (same angle), so �EMA � �RMN bySAS. Therefore, �E � �R by CPCTC.

7. Yes. Draw UT� to form triangles TUB and UTE.BT� � EU� (given), BU� � ET� (given), and UT� �UT� (same segment), so �TUB � �UTE by SSS.Therefore, �B � �E by CPCTC.

8. Cannot be determined. �HLF � �LHA by ASA (usingtwo pairs of alternate interior angles and the sharedside), but HA�� and HF� are not corresponding sides.

9. Cannot be determined. AAA does not guaranteecongruence. There is no information given aboutthe sides.

10. Yes. RO � GE � 2, FO� TE � 4, and the includedangles are congruent because they are both rightangles. Therefore, �FRO � �TGE by SAS andFR� � GT� by CPCTC.

11. Yes. �DNO � �RCO by SAS, and �OND ��OCR by CPCTC.

12. Yes. Using the sides of the triangles as thehypotenuse, construct right triangles on the sides ofeach triangle.

�PMR � �UXT, so by CPCTC PR� � UT�

�QNR � �SVT, so by CPCTC QR� � ST�

�PNQ � �UWS, so by CPCTC PQ� � UW��

Therefore, �PRQ � �UTS so by CPCTC �P � �U.

xR

T

U

y

S

Q

P7

74

22WV

3

4

42

2

O

NX3M4



whose midpoint is that point. Then construct theperpendicular to the segment.

Label the points of intersection. Draw AB� and AC�to form �ABD and �ACD.

By construction, we know BD�� DC� and AB�� AC�.AD�� AD� because it is the same segment, and anysegment is congruent to itself. So �ABD � �ACDby SSS. If �ABD � �ACD then �ADB � �ADC byCPCTC. �ADB and �ADC are linear angles, som�ADB � m�ADC � 180°. If they add to 180° andare congruent then they each measure 90°. SoAD�� BC� by definition of perpendicular lines.

7. The angle bisector does not go to the midpoint ofthe opposite side in every triangle, only in isoscelestriangles.

8. NE�. Apply the Side-Angle Inequality Conjecturetwice, noticing that EA� is a side of both triangles.First in �EAL, EA� is the shortest side becauseit is opposite the smallest angle. Now look at�NAE. The measure of the unmarked angle is180° � 61° � 58° � 60°. In this triangle, �EAN isthe smallest angle, so the side opposite this angle,which is NE�, is the shortest side. Because EA� is alsoa side of this triangle, you are able to compare theside lengths of the two triangles and conclude thatNE� is the shortest segment in the figure.

9. The triangles are congruent by SSS, so the twocentral angles cannot have different measures.

10. �POE � �PRN by ASA, �SON � �SRE by ASA.For the first pair of triangles, use the shared rightangle and ignore the marks for OS� � RS�. For thesecond pair, use the vertical angles and ignore theright-angle marks and the given statement, PO� � PR�.

11. Cannot be determined. The parts do notcorrespond.

12. �RCK � �ACK by SSS. KA� KR because CK� is amedian, and CK� is the shared side.

CB

A

D

20. ASA. The “long segment in the sand” is a sharedside of both triangles.

21. �� , � �. In Coordinate Geometry 4, you

discovered that the easiest way to find the coordinatesof the centroid of a triangle is to find the mean of thevertex coordinates.

x � � � and

y � � � .

22. In any size grid, there will be 4 elbows for the corners,4(n� 1) Ts, and (n� 1)2 crosses. So in a 15-by-15grid there are 4 elbows, 56 Ts and 196 crosses.

23. Value C always decreases. The distance from point Pto point C decreases until PB� � �; then the distancebegins to increase. The distance from C to AB�remains constant; it is always half the distancebetween the two parallel lines. Because AP isincreasing and AB is not changing, the ratio

AABP is

decreasing. Because C is the midpoint of BP�, theratio

BBCP is always equal to

12.

24. x� 3, y � 10. Add or subtract given lengths:x� 5 � 2 � 3 and y � 3 � 4 � 3 � 10.

DEVELOPING MATHEMATICAL REASONINGA possible solution:

E 1 B 2 G � 1 X 6

K � 1 C 2 K � 2

D 3 F � 1 J � 2

LESSON 4.7

EXERCISES1. 1. Given; 2. Given; 3. Vertical Angles Conjecture;

4. �ESM � �USO; 5. CPCTC; 6. Definition ofIsosceles Triangle

2. 2. Given; 4. Definition of midpoint; 5. VerticalAngles Conjecture; 6. �CIL � �MIB by SAS;7. CL� � MB��

3. 5. Given; 6. �WSN � �ESN by SSS; 7. �W � �Eby CPCTC

4. 2. �1 � �2, Definition of angle bisector; 3. Given;4. Same segment; 5. �WNS � �ENS by SAA;6. CPCTC; 7. Definition of isosceles triangle

5. 1. Given; 2. Given; 4. �1 � �2 by AIA Conjecture;6. �ESN � �ANS by ASA; 7. SA� � NE� by CPCTC.This proof shows that in a parallelogram, oppositesides are congruent.

6. Start with a line and a point on the line. From thepoint, swing arcs on the line to construct a segment

2�(�3)�0

3

133

13

13

�8�(�5)�0

3133

�

�

�

��




48, so AB � 12. By the Vertex Angle BisectorConjecture, altitude CD�� is also a median, soAD �

12(AB) � 6.

2. m�ADC � 90° ; m�ACD � 18° . By the VertexAngle Bisector Conjecture, the angle bisector ofthe vertex angle of isosceles triangle ABC is alsothe altitude to the base, so m�ADC� 90°.By the Triangle Sum Conjecture, �ACD mustmeasure 18°.

3. 45°. Because DB � DA, CD�� is the median from thevertex of the isosceles triangle, so it is also analtitude. Therefore, m�CDA� 90°. �CAD and�CAB are the same angle, so m�CAD � 45°. Thus,m�ACD � 180° � 90° � 45° � 45°.

4. Orthocenter. Point F is the point of concurrency ofthe triangles three altitudes.

5. Centroid. Point V is the point of concurrency of thetriangles three medians.

6. Incenter. Point N is the point of concurrency of thetriangles three angle bisectors.

7.

8. Conjecture A: The median to the base of anisosceles triangle is also an angle bisector.

Given: �ABC is isosceles AC� � BC�, CD�, is a median

Show: CD� is an angle bisector

Conjecture C: The altitude to the base of anisosceles triangle is also a median.

SSS

4 DACD DBCD

CPCTC

5 ÐACD ÐBCD 6 CD is angle bisector

Definition ofangle bisector

2AD DB

1 DABC is isosceles withAC BC and CD is a median.

GivenDefinitionof median

3CD CD

Samesegment

1 Isosceles nABCwith AC > BC

and CD bisects /C

Given

2 nADC > nBDC

Conjecture A(Exercise 4)

3 AD > BD

CPCTC

4 CD is a median

Def. of median

C

A BD1 2

13. a� 72°, b� 36°, c� 144°, d� 36°, e� 144°, f � 18°,g � 162°, h� 144°, j� 36°, k� 54°, and m� 126°.a� 72° (AIA Conjecture). 2a� b� 180°, so b� 36°.b� c� 180°, so c� 144°. 72° � 72° � d� 180°(Triangle Sum Conjecture), so d� 36°. d� e� 180°,so e� 144°. The triangle with the angles with meas-ures e and f is isosceles, so 2f� e� 180°, and f� 18°.g� 18° � 180°, so g� 162° (Linear Pair Conjecture).h� e� 144° (CA Conjecture). j� b� 36° (CAConjecture). The unmarked angle in the right trianglecontaining the angle with measure k is 180° � c� 36°,so k� 36° � 90° � 180°, and therefore k� 54°.Finally, in the isosceles triangle with vertex angle ofmeasure a, the measure of each base angle is

12(180° �

72°) � 54°, so m� 54° � 180° and m� 126°.

14. The circumcenter is equidistant from all threevertices because it is on the perpendicular bisectorof each side. Every point of the perpendicularbisector of a segment is equidistant from theendpoints. The incenter is equidistant from all threesides because it is on the angle bisector of eachangle, and every point of an angle bisector is equi-distant from the sides of the angle.

15. ASA. The fishing pole forms the side. “Perpendicularto the ground” forms one angle. “Same angle on herline of sight” forms the other angle.

16.

17. (1, 1) The vertices of the original rectangle areB(1, 3), O(4, 3), X(4, 5), and Y(1, 5). The vertices ofthe rotated rectangle are B�(3, 1), O�(3, -3),X�(5, �2), and Y�(5, 1).To find the center of rota-tion, determine the perpendicular bisectors of BB'and OO'. The point of intersection of these twobisectors is the center of the rotation, (1, 1).

DEVELOPING MATHEMATICAL REASONING

LESSON 4.8

EXERCISES1. 6. The perimeter of �ABC is 48, so AB � BC �AC � 48. Because AC � BC � 18, AB � (2 ? 18) �

N

T U V W X Y Z F L I P N

F

L

T

X

P

Y



12.

Start with �A. Select a point on one of the sides ofthe angle, label it point D. Construct a perpendi-cular line k through D. Label the intersection of theside of �A and line k with point C. Reflect �A overline k and label the image of point A as point B.AC� � BC�, and AD� � DB� because they are imagesand reflections preserve length. CD� � CD� because itis the same segment and all segments are congruentto themselves. �ADC � �BDC by SSS. �A � �Bby CPCTC. If AC� � BC�, then �ABC is isosceles bydefinition of isosceles triangles. Therefore, if atriangle has two congruent angles, it is isosceles.

13. a� 128°, b� 128°, c � 52°, d� 76°, e� 104°,f � 104°, g� 76°, h � 52°, j � 70°, k� 70°, l �40°, m� 110°, and n � 58°. a� 52° � 180°(Linear Pair Conjecture), so a� 128°. b� 128°(AIA Conjecture with �1 and �2). b� c � 180°(Linear Pair Conjecture), so c � 52°. d � 180° �2c (Triangle Sum Conjecture), so d� 76°. Theangle directly below the angle with measure d hasmeasure 180° � d (Linear Pair Conjecture), so itsmeasure is 104°. Then e� 104° (AIA Conjecturewith �1 and �2). Next, f � e� 104° (AIA Conjec-ture with �3 and �4). g� d� 76° (CA Conjecturewith �3 and �4). h �

12(180° � g) � 52°. j � 70°

(marked congruent to other 70° angle). Look atthe triangle containing the angle with measure n.The unmarked angle in this triangle measures 70°(CA Conjecture with �1 and �2). Then k� 70°(Vertical Angles Conjecture). l � k� 70° � 180°(Triangle Sum Conjecture), so l � 40°. m� k�180° (Linear Pair Conjecture), so m� 110°.Finally, n � 70° � 52° � 180° (Triangle SumConjecture), so n � 58°.

14. Between 16 and 17 minutes. First find a five-minuteinterval during which this happens, and thennarrow it down to a one-minute interval. At 3:15,the hands have not yet crossed each other becausethe minute hand is at the 3, but the hour hand hasmoved one-fourth of the way between the 3 andthe 4. At 3:20, the hands have already crossedbecause the minute hand is on the 4, but the hourhand is only one-third of the way from the 3 to the4. To get closer to the time where the handsoverlap, try some times between 3:15 and 3:20. At3:17, the hour hand has gone

1670 of the way from

the 3 to the 4, while the minute hand has gone 25 of

the way from the 3 to the 4. Compare these frac-tions by changing them to decimals:

1670 � 0.283� and

25 � 0.4, so

25 �

1670. This means that at 3:17, the

A D

C

A D B

C

k k

Given: �ABC is isosceles AC� � BC�, CD�, is a altitude

Show: CD�� is a median

9.

10. Yes. First show that the three exterior triangles arecongruent by SAS. Because �SLN is equilateral,SN� NL � LS. The figure shows that TN� EL �IS, so by subtraction, ST � NE � LI. Therefore, youhave two sets of congruent sides in the three trian-gles: TN� � EL� � IS� and NE� � LI� � ST�. Also,because �SLN is equilateral, each of its anglesmeasures 60°, so �N � �L � �S. Therefore,�TNE � �ELI � �IST by SAS. Because all corre-sponding parts of congruent triangles are congruent,TE� � EI� � IT�, which means that �TIE isequilateral.

11. Drawing the vertex angle bisector as an auxiliarysegment, you have two triangles. You can show themto be congruent by SAS, as you did in Exercise 4.Then, �A � �B by CPCTC. Therefore, baseangles of an isosceles triangle are congruent.

C

A D B

5 nADC > nBDC

1 AC > BC

Given

4 AD > BD

Def. of midpoint

Def. of angle bisector

9 CD ' AB

Def. of altitude

SSS

6 /ACD > /DCB

CPCTC

2 CD > CD

Same segment

3 D ismidpoint of AB

Given

7 CD is anglebisector of /ACB

8 CD is altitudeof nABC

Conjecture B(Exercise 5)

3

2 CD

1 ABC is isosceles withAC BC and CD isan altitude.

GivenDef of altitude

6 AD BD

SAA CPCTC

5 ACD BCD

Definition ofmedian

⊥ AB

7 CD is a median

∠1 ∠2 ∠A ∠B

Definition ofIsosceles Triangle

Definition ofperpendicular

4

C

A BD1 2




CHAPTER 4 REVIEW

EXERCISES1. a. true, b. true, c. false, d. false, e. both

2. The Triangle Sum Conjecture states that the sum ofthe measures of the angles in every triangle is 180°.Possible answers: It applies to all triangles; manyother conjectures rely on it.

3. Possible answer: The angle bisector of the vertexangle is also the median and the altitude.

4. The distance between A and B is along the segmentconnecting them. The distance from A to C to Bcan’t be shorter than the distance from A to B.Therefore, AC � CB � AB. Points A, B, and C form atriangle. Therefore, the sum of the lengths of any twosides is greater than the length of the third side.

5. SSS, SAS, ASA, or SAA

6. In some cases, two different triangles can beconstructed using the same two sides and non-included angle.

7. 1

8. Annie is correct. Given �NOT. �N� �O� �T �180° by the Triangle Sum Conjecture. So 2y � 2x�2z� 180°, or y � x� z� 90°. Subtracting z fromeach side gives y � x� 90° � z. If the bisectors metat right angles, then y � x� 90° in �TPN.

9. Cannot be determined. SSA does not guaranteecongruence.

10. �TOP � �ZAP by SAA.

11. �MSE � �OSU by SSS.


13. �TRP � �APR by SAS.

14. �CGH � �NGI by SAS. Use the Converse of theIsosceles Triangle Conjecture to get HG�� IG�, anduse the vertical angles.

15. Cannot be determined. You have two pairs ofcongruent sides, but there is no informationabout the third pair of sides or about any of thepairs of angles.

16. �ABE � �DCE by SAA or ASA.

17. �ACN � �RBO � �OBR by SAS. In a regularpolygon, all sides are congruent and all interiorangles are congruent.

18. �AMD � �UMT by SAS; AD�� UT� by CPCTC.

19. Cannot be determined. AAA does not guaranteecongruence.

minute hand has already crossed the hour hand, sotry 3:16. At 3:16, the hour hand has gone

1660 of the

way from the 3 to the 4, while the minute handhas gone

15 of the way from the 3 to the 4. Again,

compare the fractions by changing them to deci-mals:

1660 � 0.26� and

15 � 0.2.

1660 �

15, so at 3:16, the

minute hand has not yet crossed the hour hand.Therefore, the hands overlap sometime between3:16 and 3:17, or between 16 and 17 minutesafter 3:00.

15. 120. n concurrent lines divide the plane into 2nparts; if n � 60, 2n � 120. (See solution forLesson 4.5, Exercise 24.)

16. Any point at which the x-coordinate is either 1 or 7and the y-coordinate does not equal 3, or the points(4, 6) or (4, 0).

17. 230°

18. Hugo and Duane can locate the site of the fireworksby creating a diagram using SSS.

19.

Because “hept” means 7, cycloheptane has 7 C’s.There are 2 H’s branching off each C, so there are14 H’s. In general, a cycloparaffin has n C’s and 2nH’s, so the general rule for a cycloparaffin is CnH2n.

PERFORMANCE TASKIf Michaels remembers which rope he tied to whichinitial point, he would be able to locate the treasure.The triangle congruence is SSS because he knows thelengths of the two ropes and the distance between SkullRock and Hangman’s tree stays the same. Using acompass setting for 55 m and 105m, the only intersec-tion on the island would be Freebooter Cemetery for theburied treasure.

EXTENSIONIf any two segments coincide, all three will. (Thishappens when the three segments are drawn from thevertex angle of an isosceles triangle or from any vertexof an equilateral triangle.)

CC C

CC

C C

H

HH

H

HH

H

HH

HHHH

H

Fireworks

Duane

Hugo

1.5 km

340 m/sec ? 5 sec = 1.7 km

340 m/sec ? 3 sec = 1.02 km



look at the triangle with side lengths d, e, and f. Theangle opposite the side with length e measures 60°,because the measure of this angle � 30° � 90° �180°. So the angle opposite the side with length fmust also measure 60°. Thus, this triangle is equilat-eral, and f � d� e. Finally, look at the right triangle.The unmarked angle measures 45°, so this is anisosceles right triangle with b� d. In this triangle, c(the hypotenuse) is the longest side, so b� d� c.Putting the inequalities from the three trianglestogether, you have a� g� f � d� e� b� c, so c isthe longest segment and a and g are the shortest.

30. x� 20°. Apply the Triangle Sum Conjecture to boththe large triangle and the triangle containing theangle marked as 100°. From the large triangle,2a� 2b� x� 180°, and from the triangle with the100° angle, a� b� 100° � 180, or a� b� 80°,which is equivalent to 2a� 2b� 160°. Substituting160° for 2a� 2b in the equation 2a� 2b� x�180° gives x� 20°.

31. Yes

Given: RE� � AE�, �S � �T, �ERL � �EAL

Show: SA� � TR�

Flowchart Proof

32. Yes

Given: �A � �M, AF� � FR�, MR�� FR�

Show: �FRD is isosceles

Flowchart Proof

6 FR > RF

Same segment

10 RD > FD

Converse of Isosceles Triangle Conjecture

1 AF ' FR

Given

2 MR ' FR

Given

8 nFRM > nRFA

SAA

9 /RFM > /FRA

CPCTC

3 /AFR is a right angle

Definition of�perpendicular

11 nFRD is isosceles

Definition of�isosceles triangle

5 /AFR > /MRF

All right angles are congruent

7 /A > /M

Given

4 /MRF is a right angle

Definition of�perpendicular

1 RE > AE

Given

5 SA > TR

CPCTC

2 /S > /T

Given

3 /ERL > /EAL

Given

4 nSAE > nTRE

SAA


21. �TRI � �ALS by SAA; TR� � AL� by CPCTC. Usealternate interior angles to show the triangles arecongruent.

22. �SVE � �NIK by SSS; EI� � KV� by the overlap-ping segments property. EV� VI� EI and KI�IV� KV. Because EV� KI and VI� IV, EI� KVand thus EI� � KV�.

23. Cannot be determined. Parts don’t correspond.Notice that the shared side is not oppositecongruent angles.

24. Cannot be determined. There is not sufficient infor-mation to determine that the triangles arecongruent. By the AIA Conjecture, �MNT ��CTN. With the shared side, this gives SSA, whichdoes not guarantee that the triangles are congruent.You are not able to determine whether MT�� NC��, soyou cannot tell whether quadrilateral NCTM hasone pair of opposite parallel sides or two pairs, andthus cannot determine whether it is a parallelogramor a trapezoid.

25. �LAZ � �IAR by ASA, �LRI � �IZL by ASA, and�LRD � �IZD by ASA. There are actually twoisosceles triangles in the figure, �LAI and �LDI, andthere are three pairs of congruent triangles. For �LAZ� �IAR, use the shared angle, �A. For �LRI ��IZL, use the common side, LI�. Also, m�RLI�m�ZIL (Isosceles Triangle Conjecture), and m�RLZ� m�ZIR (given), so by subtraction, m�ZLI�m�RIL, and thus, �ZLI � �RIL. Thus, the trianglesare congruent by ASA. Because �ZLI � �RIL, DI� �DL� by the Converse of the Isosceles Triangle Conjec-ture. Using this pair of sides, the marked angles, andthe vertical angles, �LRD � �IZD by ASA.

26. �PTS � �TPO by ASA or SAA; yes. Use the sharedside to show that the triangles are congruent. By theConverse of the Parallel Lines Conjecture, you canshow that PS� � TO� and PO� � ST�, so STOP is aparallelogram.

27. �ANG is isosceles, so �A � �G. Then, m�A �m�N� m�G � 188°. The sum of the measures ofthe three angles of a triangle must be 180°, so anisosceles triangle with the given angle measures isimpossible.

28. �ROW � �NOG by ASA, implying that OW�� OG��. However, the two segments shown are not thesame length.

29. c is the longest segment, and a and g are the shortest.Apply the Side-Angle Inequality Conjecture to allthree triangles in succession. First, in the trianglewith side lengths a and g, the third angle measure is30°, so this triangle is isosceles with a� g� f. Now




For a patty-paper construction, see the solution forLesson 4.2, Exercise 17, which shows the constructionof a 105° angle. If you use just the lower half of the60° angle shown there, you will have a 30° angleadjacent to a 45° angle, which makes a 75° angle.

39. d, a� b, c, e, f. The figure shows that AC� � BD�, soa� b by the AIA Conjecture. It is given thatm�CAD � m�CBD, which means that a� f �b� e. Because a� b, we can substitute a for b inthis inequality to obtain a� f � a� e. Thensubtract a from both sides of the last inequality toobtain f � e. Now apply the Side-Angle TriangleInequality to each of the triangles. In �ABC,AC � AB gives e� c, and AB � BC gives c � a. In�ABD, BD � AD gives f � b, and AD � AB givesb� d. Putting all of this information together, wehave f � e� c � a� b� d.

40. True. The diagonals divide the quadrilateral into twotriangles. In each triangle, the sum of the twolengths of any two sides of the triangle (sides of theparallelogram) are greater than the length of thethird side (the diagonal).

41. True. In the figure, BD�� is the angle bisector of�CBA. By definition of an angle bisector �2 � �5.DE�� is constructed so that DE�� CD��1 � �5 byAIA. If �1 � �5 and �2 � �5, then �1 � �2. If�1 � �2, then BE� � DE� and �BDE is an isoscelestriangle.

42. Cannot be determined.

43. a.

b. �ABC rotated 90° around the origin. Yes.

y

xB

A'

B'

C

A

C'

EA

B

C

D

245

31

Fold 2

Fold 1

Fold 4

Fold 3

75°30°45°

33. x� 48°. The sum of the measures of the interiorangles of any quadrilateral is 360° because it can besubdivided into two triangles. Therefore, x� 90° �132° � 90° � 360°, so x� 48°.

34. The legs form two triangles that are congruent bySAS, using the vertical angles as the included angles.Alternate interior angles are congruent by CPCTC,so by the Converse of the Parallel Lines Conjecture,the seat must be parallel to the floor.

35. Construct �P and �A to be adjacent. The anglethat forms a linear pair with the conjunction of�P and �A is �L. Construct �A. Mark off thelength AL on one ray. Construct �L. Extend theunconnected sides of the angles until they meet.Label the point of intersection P.

36. Construct �P. Mark offthe length PB on one ray.From point B, mark offthe two segments thatintersect the other rayof �P at distance x.

37. 1. M is midpoint of TE� and IR�, Given; 2. �TMI ��RME, Vertical angles; 3. TM�� ME��, Definition ofmidpoint; 4. IM� � MR��, Definition of midpoint;5. �TMI � �EMR, SAS; 6. �T � �E or�R � �I, CPCTC; 7. TI� � RE�, Converse ofAIA Conjecture.

38. Possible method: With a compass and straightedge,construct an equilateral triangle, and bisect oneangle to obtain two 30° angles. Adjacent to thatangle, construct a right angle, and bisect it to obtaintwo 45° angles. The 30° angle and 45° angles thatshare a common side form a 75° angle.

30°45°

75°

z

xx

P B

S2

S1

A Ly

P

/P/A

/L


Proof: AC� � BC� � AD� � BD� and CD�� CD��,so �ACD � �BCD by SSS. Therefore, �ACE ��BCE. CE� � CE�, so �ACE � �BCE by SAS. AE�� BE� by CPCTC, so AB� is bisected by CD��. �AEC� �BEC. �AEC and �BEC form a linear pair.Therefore, �AEC and �BEC are right angles.Therefore, CD�� AB�.

5. The measure of an exterior angle of a convexquadrilateral is 180° less than the sum of the meas-ures of the remote interior angles. (In a concavequadrilateral, with appropriate interpretation, theexterior angle at an angle with measure more than180° is the supplement of the sum of the measuresof the remote interior angles.)

6. Two possible answers: (1) The sum of the lengths ofany three sides of a quadrilateral is greater than thelength of the remaining side. (2) The sum of thelengths of any two consecutive sides of a quadrilat-eral is greater than the length of the diagonaljoining their endpoints.

7. Neither SSSS nor SSSD guarantee congruence,but SSSDD does.

Draw diagonals and consider triangle congruence.Four sides and an angle will guarantee congruenceof quadrilaterals, and there are other possibilities.

SASA does not guarantee congruence, but SASASdoes. SASA is similar to the SSA case for triangles.

8. On a sphere or a globe, angles are measured alongtangent lines. It is indeed possible to draw a trianglewith two or more obtuse angles or with three rightangles. The sum of the interior angle measures ofany triangle drawn on a sphere is always greaterthan 180°.

9. Both conjectures are true on a sphere, but theangles in an equilateral triangle no longermeasure 60°.

>

À

c. �A�B�C� � �DEF by SSSd. They are congruent.

44. a.

b. Quadrilateral ABCD reflected across x = −1. No.c. They are not congruent.d. Point A' does not coincide with point F.

45. A rotation is a rigid transformation, so a figure andits image are congruent. Geometry software allowsstudents to check all relationships.

a. trueb. truec. trued. truee. true

TAKE ANOTHER LOOK

1. A triangle with the measure of one angle twicethat of another can be divided into two isoscelestriangles for certain only if the smallest angle meas-ures less than 45°. When the angle with twice themeasure is acute, it works, but when this anglemeasures, for example, 100°, it doesn’t work.

2. The claim must always be true. Let a and b be themeasures of interior angles A and B, respectively,and let x be the measure of the exterior angle to�C, with x� 2a. Then 2a� a� b, so a� b.

3. 900°. 900°. Around each vertex of a triangle, thereflex angle plus the vertex angle is 360°, a total of1080°. The sum of the measures of the angles ofevery triangle is 180°. 1080° − 180° = 900°

4. Use this construction to bisect AB�: From points Aand B, draw arcs that have the same radius and havelength greater than

12AB, intersecting at points C

and D. CD�� is the perpendicular bisector, bisectingAB� at point E.

A B

C

D

E

80°

40° 50°

100°

y

x

B

A'

B'

C

A

C'

D'D



Documents

DG5 Solution Manual CH04 - flourishkh.com · 68 CHAPTER 4 Discovering Geometry Solutions Manual ©2015 Kendall Hunt Publishing Therefore, d 47° 40° 180° (Triangle Sum Conjecture),