Chapter Congruence and Proof 6 pages 192–226 000200010271723961_CH06_p192-226” 2013/2/22 13:6 page 192 Chapter 6.....Congruence and Proof pages 192–226

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Chapter

6Congruence and Proof pages 192–226.....................................................................

6.0 Triangle Congruence

Check Your Understanding

1. The triangles may not be congruent. You can constructmany different triangles with these angle measures. Thetriangles can be different sizes. One example of twonon-congruent triangles is the following:

C

B

A

8 cm

8 cm

C

B

A

5 cm

5 cm

If you knew one side length, and where it is in relation tothe angles, then there would be only one triangle youcould construct, because it would satisfy the ASATriangle Congruence Postulate.

2. There are exactly two different triangles that can be madewith this information. Triangles ABC1, and ABC2 shownbelow, fit the information:

A

B

60°

C1

C2

8 cm

7 cm

7 cm

To choose between these two possibilites, we wouldneed any other piece of information about the triangle.If we knew m � B or m � C, then the information wouldsatisfy ASA. If we knew AC, it would satisfy SSSand SAS.

3. The two triangles are congruent by SSS. They have twopairs of congruent sides (by the tick marks): AB ∼= AD,and BC ∼= CD. The triangles also share AC.

4. These triangles can be proven congruent, but you need toknow that AB ∼= AD. This additional fact makes�ABC ∼= �ADC by SSS or by SAS.

5. Given any square, you can cut along the diagonal,producing two right triangles. One of the triangles can bereflected about the diagonal and will fit exactly on top ofthe other. Here are two ways to prove they are congruent,using the triangle congruence postulates:

Using SSS: Suppose ABCD is a square, with diagonalDB. By the definition of a square, you know that there arefour congruent sides. Specifically AB ∼= CB andAD ∼= CD. Notice that �ABD and �CBD share side BD.Since they have three pairs of congruent sides,�ABD ∼= �CBD.

Using SAS: As before, by definition of a square,AB ∼= CB and AD ∼= CD. Also, by definition of a square� A and � C are right angles. This makes � A ∼= � C. Sincethe triangles have two pairs of congruent sides, and a pairof congruent, included angles, �ABD ∼= �CBD.

6. It is unclear whether the 14-inch and the 8-inch sides arethe two sides of the triangle adjacent to the 30◦ angle atthe tip of the pennant, or if one of those two sides isopposite the 30◦ angle.

7. Sides AO and BO are congruent, as they are both radii ofthe circle. Triangles COA and COB also share side OCand angle � OCB. The two triangles are definitely notcongruent, however, as one is contained inside the other(at the very least, � BOC is contained inside � AOC,making it smaller). Thus, knowing SSA information isnot enough to uniquely determine a triangle.

On Your Own

8. There is not enough information, as only two sides aregiven.

9. There is enough information. By the SSS postulate,�ABC ∼= �DEF.

10. There is enough information. By the SAS postulate�ABC ∼= �DBC. (Notice that BC is a common side.)

11. There is not enough information. Notice that the twotriangles have two pairs of congruent sides, and a pair ofcongruent angles. In �ABC, the marked angle is anincluded angle (between the two sides), while in�BCD itis a non-included angle (not between the two sides). Thismeans they are not corresponding angles - there is no wayyou could rotate, flip, or slide these two triangles to makethe congruent angles and the congruent sides match up.

Mathematics II Solutions Manual • Chapter 6, page 192

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It is possible the triangles are congruent, but you wouldneed to know that either � ABC ∼= � DBC or DC ∼= AC.

12. There is enough information. By the SAS postulate,�ABC ∼= �YXC.

13. The only congruence statements you can make about theparts of �ACD and � BCD are � ACD ∼= � BCD andCD ∼= CD. You would need at least one more pair ofcongruent sides, or one more pair of congruent angles toprove �ACD ∼= � BCD. The correct answer is A.

14. Triangles �ABD and �CBD are congruent. The fact thatBD is the perpendicular bisector of AC yields two piecesof information:

AD = DC and m� BDA = m� BDC = 90◦.

This, combined with the fact that both triangles share sideBD lets you apply the SAS postulate.

15. (a) You cannot prove any new triangles congruent.AB = AC is not actually new information. FromExercise 14, we know �ABD ∼= �ACD, and fromthis we can conclude that AB = AC. No newinformation means no new triangles are congruent.

(b) You can prove that �ADE ∼= �ADF by SAS.This is the same argument as the one used in

Exercise 14. Because AD is the perpendicularbisector of EF , we know that ED = FD. We alreadyknow AD is a common side and � ADF ∼= � ADE.

You can also prove that �AEB ∼= �AFC and�AEC ∼= �AFB. The two known congruences,�ADB ∼= �ADC and �ADE ∼= �ADF, yieldDB = DC and DE = DF by CPCTC. Thesestatements imply that BF = EC and BE = FC.CPCTC also gives that AB = AC and AE = AF.Now apply SSS to obtain the two congruences.

(c) You can prove �ADE ∼= �ADF by ASA.(� EAD ∼= � FAD, AD is a common side, and

� ADB ∼= � ADC.)As in the solution to part (b), you can also prove�AEB ∼= �AFC and �AEC ∼= �AFB.

16. (a) Yes, �ABC ∼= �DEF.m� C = 180− 72− 42 = 66◦ = m� F . So, the

triangles are congruent by the ASA Postulate.(b) Suppose you have two triangles with two pairs of

corresponding angles congruent and a pair ofnon-included sides congruent. Because all the anglesin the triangles must add up to 180◦, it follows thatthe third pair of angles is congruent. Now, the givencongruent sides are included between a pair ofcongruent angles. So, by the ASA Postulate, thetriangles are congruent. This shows that AAS provestriangles congruent.

Maintain Your Skills

17. A diagonal of a rectangle divides it into two congruenttriangles. Both of the arguments used for the square inExercise 5 apply to the rectangle as well.

A diagonal of a parallelogram divides it into twocongruent triangles as well. The SSS argument used forthe square in Exercise 5 applies here.

A diagonal of a trapezoid does not divide it intocongruent triangles. The bases of the trapezoid are notequal. This makes a pair of corresponding sides of thetriangles not congruent. Since all six pairs ofcorresponding parts must be congruent for the trianglesto be congruent, these triangles cannot be.

One of the diagonals of a kite will divide it intocongruent triangles. Consider kite KITE.

K

I

T

E

�KIT ∼= �KET, by the same SSS argument used for thesquare in Exercise 5. However, �KEI is not congruent to�TEI, as KE �= ET, by definition of a kite.

INVESTIGATION 6A INVARIANTS—PROPERTIESAND VALUES THATDON’T CHANGE

6.01 Getting Started

For You to Explore

1. (a) All of these squares end in 25. In fact, even thehundreds digit of the squares is partially determined.It must be one of 0, 2, or 6.

(b) Here are two conjectures you might have discovered.(They really both say the same thing! And there areother ways of wording this same discovery, too.)

Square numbers (1): If the units digits of a pair ofwhole numbers add up to 10, then the units digits oftheir squares will be equal.

Square numbers (2): If the sum of two wholenumbers is a multiple of 10, then the difference oftheir squares is also a multiple of 10.

One way to prove this conjecture is to let thenumbers be x and y. Then we remember thatx2 − y2 = (x+ y)(x− y). Since x+ y is a multipleof 10, so is this product.

(c) Because no rule is given for the set {1, 4, 7, 10, . . . },it could be a set of almost anything. But, if weassume these elements are representative, then the setappears to be every third whole number, startingwith 1. It would be tedious to check 301 that way, butevery third number, starting with 0, is simply themultiples of 3. So another way to describe this set is:“numbers of the form 3n+ 1, where n is a wholenumber.” Then it is easy to see that 301 is a numberof that form. The product of any two of the numbersbelongs to the set, as does the sum of any four ofthem. Their difference does not. These statements are


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easy to prove algebraically. Here’s one of the proofs.

(3a+ 1)(3b+ 1) = 9ab+ 3a+ 3b+ 1

= 3(3ab+ a+ b)+ 1

2. The triangle invariably contains a hexagon, but otherwisethe strict doubling pattern is not reliable.

On Your Own

3. Answers will vary. Possible answer:

zw= 8

w z

2 163 244 325 40

g+ h = 8

g h

0 82 64 46 2

m− n = 8

m n

10 220 1230 2240 32

4. (a) If the outer figure is any quadrilateral, the inner figurewill have four distinct vertices, so it must be aquadrilateral.

(b) The inner figure appears to be a parallelogram.


5. (a) Check students’ work.(b) Some invariants:

• Lines � and m will always be parallel since theywere constructed that way.• If angles are numbered as in the picture below,

then the following angle pairs will be equal inmeasure: 1 and 6, 1 and 3, 1 and 8, 5 and 2, 5 and7, 5 and 4, 2 and 7, 6 and 3.

�

m

1 5

2 6

3 7

4 8

• Some pairs of angles are supplementary. The sumsof the angles 1 and 4, 2 and 3, 5 and 8, and 6 and 7are each 180◦.

6. (a) Check students’ work.(b) Some invariants:

• All four sides of the square will stay equal to eachother.• All four angles of the square will remain 90◦.• The diagonals are equal.• The diagonals meet at a 90◦ angle.• The diagonals interesect at their midpoints.• The area of one of the small triangles formed is 1

4the area of the whole square.

7. (a) Check students’ work.(b) Perhaps the most surprising result is that all these

circles seem to pass through the same point on line �.

6.02 Numerical Invariants in Geometry


1. In the sketch below, AB is a diameter of the circle.

A

B

D

(a) • m � ADB is constant, (90◦)• AD < AB• BD < AB• m � BDA > m� BAD• AD+ DB > AB• AD2 + DB2 = AB2

• m � BAD+m � ABD is constant (the sum isalways 90◦)• AB ≥ AD, AB ≥ BD

Here are some things which are NOT invariant:

• AD < BD in our diagram, but not always.• m� A < m� B in our diagram, but not always.• chord BD cuts off more of the circumference of the

circle than chord AD, but not always.

(b) • m� ADB is constant (90◦)• ratio of any two segment lengths (e.g., AB

AD ) isconstant

• Circumference(⊙

)

AB is constant (π)

• Area(�AB)

Area(⊙

)is constant

2. (a) No(b) Yes, ratio = 1(c) No(d) Yes, ratio = 1(e) No

3. AC + BC = AB, m� ACD+m � DCB = 180◦.4. The area is invariant; the perimeter is not.5. The two lengths vary in opposite directions. Their sum

cannot be constant because they add up to the full lengthof the chord, which changes. Checking the product to seewhere that might lead produces the remarkable result thatCE × CD is constant.

6. The constant product associated with a given point C istraditionally called the power of the point, and it dependson the distance of that point from (the center of) thecircle. When C is the center of the circle, the power of thepoint C achieves a maximum.

7. There are many invariants to find. Here are a few.

• The ratio AEAD is invariant, as are DE

AD and AEDE . The fact

that these are invariant also makes their sums,differences, products, squares, and so on, invariant.• AD+ DC is invariant. (It equals AC.)


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• BAEA = CA

DA = CBDE

• ADDC = AE

EB

• Area(�ABC)

Area(�AED)= (

BCED

)2

• Never overlook the obvious: The lengths of the sidesof the large triangle, �ABC, do not change as youmove D.• The problem didn’t ask about angles, but they, too, do

not change. And the shape of �ADE does not change,even though its size does.

On Your Own

8. Since the ratio of circumferencediameter is slightly more than 3,

the circumference (about 3.1d) will be slightly greaterthan the height (3d) of the can.

9. To make a right triangle with the same area, construct aperpendicular to EF through E. Mark the point where itintersects the other line P . Triangle �PEF will be a righttriangle since EP is constructed perpendicular to EF . Itsarea will be the same as the area of �DEF .

10. (a) The two ratios are the same.(b) To find the numerical value of this ratio, we set up the

equality described in the problem:long sideshort side =

long sideshort side . Using the picture, we get

x

1= 1

x− 1x2 − x = 1

x2 − x− 1 = 0

By the quadratic formula, we have

x = 1±√5

2

One of these numbers is positive, and the othernegative. Since we take the ratio of lengths to be apositive number, take the positive value.

11. (a) Yes, the area of the triangle will always be half thearea of the square.

(b) Yes, the area of the circle will be about 3.1 times thearea of the square and that ratio will be invariant.

(c) No, there will be no invariant relationship betweenthe perimeter and area of the triangle. In particular,we can make the perimeter very large while the areagets very small, or we can make the perimeter getvery small while the are gets very small.

12. 6 · 180◦ − 6 · 60◦ = 1080◦ − 360◦ = 720◦. The correctanswer is C.


13. If you measure carefully, the ratio should be about 3 : 1for each of the cylindrical objects.

6.03 Spatial Invariants: Shape,

Concurrence, and Collinearity


1. (a) It is possible that three or more of the perpendicularbisectors intersect at a point, but in general it is notthe case for an arbitrary convex pentagon.

(b) Yes.(c) It is possible. In fact, this happens when all the

vertices of the pentagon lie on a circle.2. It may take patience to adjust your figure, but there are

many pentagons whose angle bisectors all concur at asingle point. Again, a circle can help, but this time thecircle must be inside the pentagon, just touching (tangentto) each side.

3. Experimentation suggests that concurrence ofperpendicular bisectors is an invariant of regularpolygons.

4. Experimentation suggests that concurrence of anglebisectors is an invariant of regular polygons.

5. The perpendicular bisectors of these sides are concurrentat the center of the circle. Here’s why: Perpendicularbisectors of a segment are lines containing all the pointsthat are equidistant from the endpoints of the segment. Ifa segment’s endpoints are on a circle, then the center ofthe circle (which is certainly equidistant from thoseendpoints) must lie on the segment’s perpendicularbisector.

The angle bisectors of the inscribed pentagon show noparticular relationship.

6. In this situation, all the angle bisectors are concurrent atthe center of the circle, but the perpendicular bisectorsshow no special relationship.

Here’s an easy way to construct a figure with whichyou can experiment. Draw a circle, construct fiveradii, and then construct perpendiculars at thecircumference. Those tangent lines can define the sides ofthe polygon.

On Your Own

7. The medians of triangles seem to be concurrent, nomatter what the triangle looks like. The proof of thisstatement is much harder than the proofs of concurrenceof perpendicular bisectors or angle bisectors.

8. AD is a median since D is the midpoint of BC, sooption C is always true. AD is an altitude, a median, andan angle bisector, so the concurrences of altitudes,medians, and angle bisectors must necessarily be on AD.So option A is always true. Because CH is an anglebisector, we have � BCH ∼= � HCA, so option D is alwaystrue. Option B is true only if �ABC is equilateral. Thecorrect answer is B.

9. There are many nonsurprising collinearities in the figurebelow. For example (A, E, B), (E, G, C), and (B, J, D).But the collinearity of H, J , and G is most surprising.


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A

B

CD

E

F

GHJ

10. Here is a claim, and part of a strategy for proving ordisproving it. This solution leaves a lot of room to dobetter.

Claim: In any hexagon with all diagonals drawn in,there can be at most one concurrence of three diagonals.

Explanation: In any pentagon, it is impossible to get aconcurrence of three diagonals. At most, two diagonalswill intersect in any point. Draw a pentagon and itsdiagonals. Next try to convert the pentagon into ahexagon by adding a new point somewhere and drawingin the new sides and diagonals, as in the picture below. Ifit is possible to find a place for that new point that allowsmore than one of the new diagonals to pass through thealready existing intersections of other diagonals, then thestatement is false. But it is not possible. Only one of thenew diagonals can pass through already existingintersections. So the statement is true.

A

B C

D

E

newpoint


11. (a)

(b) The midpoints of the parallel segments areconstructed to be collinear with the endpoints of thesegments, but they turn out to be collinear with theintersection of the diagonals as well.

6A MATHEMATICAL REFLECTIONS

1. By drawing segments from one vertex of the pentagon toeach other nonadjacent vertex, you can subdivide thepentagon into three triangles. The sum of the angles ofthese three triangles is the same as the sum of the anglesof the pentagon, so the angles of a pentagon must sum to540◦. The hexagon can be subdivided in the same wayinto four triangles, so the sum of its angles must be 720◦.

2. CDCB = 1

2 and CFCH = 1

2 ; Since D is the midpoint of CB,CB = 6. Since ED is a midline of �ABC, F is themidpoint of CH . Therefore CF = 2 and CH = 4.

3. The points A and B divide the circle into two arcs. Theangle at point C is invariant wherever you place C withineach of these two arcs, though the angle for each arc willbe different, unless A and B are two ends of a diameter ofa circle.

4. Yes, the medians of an equilateral triangle areconcurrent. Because all three sides are congruent, anymedian cuts the triangle into two congruent triangles.Two congruent angles, back-to-back along a line, mustmeasure 90◦. So the median is also an altitude (and anangle bisector). Since altitudes of any triangle areconcurrent, the medians of an equilateral triangle alsohave to be concurrent.

5. Possible responses include

• Sum of angles is 720◦.• Internal angles each measure 120◦.

6. An invariant is a feature of a problem or diagram thatdoesn’t change as other things around it change. Ingeometry you can look for spatial invariants, numericalinvariants, measure invariants, as well as others.

7. A line drawn parallel to the base of a triangle willdivide the other two sides proportionally. If the line isdrawn through the midpoints of the other two sides, thesegment will be half the length of the base. Otherwise, itwill be in the same proportion as the other part of thetriangle.

8. When you connect midpoints of a quadrilateral in orderyou end up with a parallelogram.

INVESTIGATION 6B PROOF AND PARALLELLINES


For You to Explore

1.1

2

34

56

78


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(a) Number the eight angles formed as shown in thediagram.

� 1 ∼= � 3 ∼= � 5 ∼= � 7� 2 ∼= � 4 ∼= � 6 ∼= � 8

(b) There are many sums that are invariant. Here are afew:

• The sum of the four adjacent angles around avertex is 360◦.

m � 1+m � 2+m � 3+m � 4 = 360◦

m � 5+m � 6+m � 7+m � 8 = 360◦

• The sum of any pair of adjacent angles is 180◦.(e.g., m� 1+m � 2 = 180◦; m � 7+m � 8 = 180◦.)• m � 3+m � 6 = 180◦. Likewise, m � 4+m � 5 =

180◦.• m� 2+m � 7 = 180◦. Likewise, m � 1+m � 8 =

180◦.• m� 3+m � 4+m � 5+m � 6 = 360◦• m � 1+m � 2+m � 7+m � 8 = 360◦

You could create more invariant angle sums bysubstituting the angles in these sums with congruentangles.

2. There are many invariant relationships that exist nomatter which line you move.

12

34

56

7 8

Here are some of them:

• The sum of the four adjacent angles around a vertex is360◦.• The sum of any pair of adjacent angles is 180◦.• The sum of all eight angles is 720◦.• Vertical angle pairs are congruent (angles 1 and 3;

angles 2 and 4; angles 5 and 7; angles 6 and 8).• m� 3+m � 4+m � 5+m � 6 = 360◦• m � 1+m � 2+m � 7+m � 8 = 360◦

Notice that all of these relationships existed in Problem 1as well.

When you move the transversal (but not the parallellines), four other sums remain invariant. They arem � 4+m � 5; m � 3+m � 6; m � 2+m� 7; and m� 1+m� 8.Notice that these sums were invariant in Problem 1,but in that exercise, the sum was always 180◦. Here, theconstant value of the sum depends on the orientation ofthe intersecting lines. When you move one of theintersecting lines, this relationship does not hold.

This last invariant relationship shows theinterconnectedness of the angles along the transversal. Itis because � 4 and � 5 share the transversal as a side thatwhen you move the transversal, they change in a way that

keeps their sum the same. To see this more clearly,consider the triangle formed by the two intersecting linesand the transversal.

xa

180°–x–a

180°–a

x+a

Let x and a denote the measures of the angles indicated.The invariant sums listed above tell us what all the otherangles must be, in terms of a and x (see diagram). Sincethe intersecting lines are not moving, x does not change.a may change as the transversal moves, but when youadd up the angle pairs, the sums do not depend on a:

a+ (180◦ − x− a) = 180◦ − x

(180◦ − a)+ (x+ a) = 180◦ + x

Since x does not change, the sums are invariant. Now, ifyou move one of intersecting lines, x may change and sothe sums will not be invariant.

Notice that as you move one of the intersecting lines sothat it becomes closer to being parallel to the other line, x

will become closer to 0, making the sums closer to 180◦,as you found in Exercise 1.

On Your Own

3. There are many observations to be made, such as:

• Each midline is parallel to the side it does not intersectand half the length of that side.• The four smaller triangles formed by the midlines are

congruent. (The sides are all half the lengths of thesides of the original triangle. Their angles are the sameas those of the original triangle.) Their areas are thesame: they all are 1

4 the area of the originaltriangle.• The four smaller triangles are similar to the original

triangle.

Explanations for why these observations hold true willfollow later.

(a) From the length of one midline, you can determinethe length of the side that the midline is parallel to. Itis twice the length of the midline.

(b) The areas of the four smaller triangles are 14 of the

area of the original triangle.

4. Since M is the midpoint of BC, BM = MC. Also,m� AMC = m� RMB because they are vertical angles.So, �AMC ∼= �RMB by SAS. Since � C and � CBR arecorresponding parts of these two congruent triangles, theyare congruent (have equal measures).


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5. If you know that the sum of the measures of the angles ina triangle is 180◦, then

m � A+m � C +m � CBA = 180 and

m� CBD+m � CBA = 180

Now, subtract m� CBA from both sides to get:

m � A+m � C = 180−m � CBA and

m � CBD = 180−m � CBA

This shows that m� A+m � C = m� CBD. So, m � CBDis greater than the measures of � A and � C because it isequal to their sum.

What if you didn’t know that the sum of the measuresof the angles in a triangle is 180? Then use Exercise 4.Notice that the diagram in Exercise 4 is the same as thediagram for this exercise, but AR is added throughmidpoint M. You know m� C = m� CBR. Since � CBDcontains � CBR, m � CBD must be larger than m � CBR.And m� CBR = m� C, so m � CBD > m� C.

A similar argument can be made to show that m � CBDis also larger than m � A. To do this, you need a differentdiagram.

A

C

B DN

PQ

6. For these problems, you can use the following facts(in many different ways) to find the missingangles:

• m � ADB+m � BDC = 180◦• m � CDQ+m � BDC = 180◦• m � CDQ+m � ADQ = 180◦• m � ADB+m � ADQ = 180◦• All four angles add up to 360◦.

(a) m� BDA = 118◦; m � ADQ = 62◦; m � CDQ = 118◦(b) m � BDA = 108◦; m � ADQ = 72◦; m � CDQ = 108◦(c) m � BDC = 125◦; m � ADQ = 55◦; m � CDQ = 125◦(d) m � BDA = (180◦ − x); m � ADQ = x;

m � CDQ = (180◦ − x)


7. The software will only let you construct one line parallelto line � through point P . There is only one such parallelline.

8. The software will only let you construct one lineperpendicular to line � through point P . There is onlyone such perpendicular line.

9. It is impossible to find the intersection of lines � and m,because these two lines are parallel. One way to envisionthis is to imagine sliding P and line m along line n, untilP is on line �.

Pm

n

�

At this point, line m will match perfectly with line �,because of the right angles made by the perpendicularlines.

6.05 Deduction and Proof


1. (a) m� COB = 90◦ − 25◦ = 65◦m � COD = 90◦ − 65◦ = 25◦

(b) m � BOA = 90◦ − 63◦ = 27◦m � COD = 90◦ − 63◦ = 27◦

(c) m � COB = 90◦ − 31◦ = 59◦m � BOA = 90◦ − 59◦ = 31◦

(d) m � DOA = 90◦ + 31◦ = 121◦(e) m � COB = 90◦ − x

m� COD = 90◦ − (90◦ − x) = x

2. Here is one possible proof:

Statement Reason1. m � COB+m � BOA = 90◦ This is evident from the

diagram, given that� COA is a right angle.

2. m � DOC+m � COB = 90◦ This is evident from thediagram, given that� DOB is a right angle.

3. m � COB+m � BOA = Basic moves of equations.m � DOC+m � COB

4. m � BOA = m� DOC Basic moves of equations.

On Your Own

3. � BCA and � ECD are vertical angles. The correct answeris D.

4. (a) Reason is already given.(b) Reason: Given.(c) Reason: Any segment is congruent to itself.


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(d) Reason: SAS(e) Reason: CPCTC

5. Here is one proof:

Statement Reason

1. PL ∼= RQ Given in the problem statement.PQ ∼= RL

2. QL = QL A segment is congruent to itself.3. �QPL ∼= �LRQ SSS4. � P ∼= � R CPCTC

6. (a) You can prove m� FCA = m� FCB.(b) Proof:

Statement Reason

1. FC ⊥ ED Given2. m � FCD = 90◦ Definition of

m� FCE = 90◦ perpendicular lines3. m � FCA+m � ACE =

m � FCE = 90◦ Evident from diagramm� FCB+m � BCD =m � FCD = 90◦

4. m � FCA+m � ACE =m � FCB+m � BCD Basic moves of equations.

5. m � ACE = m� BCD Given6. m � FCA = m� FCB Basic moves of equations.


7. (a) m� 1 = 50◦; m � 2 = 130◦; m � 3 = 50◦; m � 4 = 130◦;m� 5 = 140◦; m � 6 = 40◦; m � 7 = 140◦; m � 8 = 40◦.

One way to find these angle measures is to startwith m� 7 = 140◦. From this, you can determine thatm� 6 = 180◦ − 140◦ = 40◦, as does m � 8. Andm � 5 = m� 7, because they are vertical angles.

Now that you know m � 6 = 40◦, you can use thefact that m� 1+m � 6 = 90◦ to find m � 1 =90◦ − 40◦ = 50◦. Then, m � 2 = 180◦ − 50◦ = 130◦,as does m� 4. And, m � 3 = m� 1, because they arevertical angles.

(b) All of the angles at E are right angles.We know m � 1+m � 6 = 90◦. Since all the angles

in the triangle must add up to 180◦, the remainingangle in the triangle, at E, must measure180◦ − 90◦ = 90◦. This makes all the other angles atE measure 90◦ as well.

6.06 Parallel Lines


1. Definitions may vary from this. When two lines are cutby a transversal, 8 angles are formed at the intersections.

These angles can be identified in pairs:

m

j

k

1 2

34

5 6

8 7

(a) Alternate Interior Angles are a pair of angles, one oneach of the lines. They are on opposite sides of thetransversal, between the two lines. In the diagramabove, they are angles � 3 and � 5, or � 4 and � 6.

(b) Alternate Exterior Angles are a pair of angles, one oneach of the lines. They are on opposite sides of thetransversal, and they are not between the two lines. Inthe diagram, they are angle pairs � 1 and � 7, or � 2and � 8.

(c) Corresponding Angles are a pair of angles, one oneach of the lines. These angles are in the same relativeposition on each line (for instance, if one is above theline and to the right of the transversal, the other oneis as well.) In the diagram, they are angle pairs � 1and � 5, or � 4 and � 8, or � 2 and � 6, or � 3 and � 7.

(d) Consecutive Angles are a pair of angles, one on eachof the lines. They are on the same side of thetransversal, between the two lines. In the diagram,they are angle pairs � 4 and � 5, or � 3 and � 6.

2. (a) �PXU ∼= �UMP by SSS (UP is congruent to itself).(b) Number the angles as follows:

Because the triangles are congruent, � 1 ∼= � 4,� 2 ∼= � 5, and � 3 ∼= � 6, by CPCTC.

There are vertical angles that are congruent, like� 8 ∼= � 1.

A congruent pair that is not so easy to see is� 1 ∼= � 7. To prove this, you need to use the fact thatthe sum of the angles in a triangle is 180◦. So, m � 1+m � 2+m � 3 = 180◦. Notice that m � 7+m � 3+m� 5 = 180◦ as well. By the basic moves ofequations,


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m � 1+m � 2+m � 3 = m� 7+m � 3+m � 5

You can subtract m � 3 from both sides to get

m� 1+m � 2 = m� 7+m � 5

And, m � 2 = m� 5 because the triangles arecongruent. This means m� 1 = m� 7. A similarargument can be used to show many other pairs arecongruent, such as m � 4 = m� 9 or m � 1 = m� 10.

(c)←→XU ‖ ←→PM and

←→XP ‖ ←→UM, by the AIP Theorem. To see

why←→XU ‖ ←→PM, remove line UM and line XP from

the diagram:

Line UP is a transversal, cutting line XU and linePM. � 2 and � 5 are alternate interior angles. Since� 2 ∼= � 5, the lines must be parallel.

To see why←→XP ‖ ←→UM, remove line XU and line

PM from the diagram:

Line UP is still a transversal; this time it cuts lineXP and line UM. � 6 and � 3 are alternate interiorangles. Since they are congruent, the lines must beparallel.

3. (a) �XUP ∼= �MUP by SSS (UP is congruent toitself).

(b) Number the angles in the diagram:

� 1 ∼= � 4, � 2 ∼= � 5, and � 3 ∼= � 6 by CPCTC. Thereare also many vertical angle pairs that are congruent,like � 1 ∼= � 7.

(c) There are not necessarily any parallel lines in thisdiagram. As it is drawn, line XU looks like it couldbe parallel to line PM, and line XP looks like it could

be parallel to line UM. But, the diagram could beredrawn to look like this:

Notice that XP ∼= PM and XU ∼= UM . But, it is nowclear the lines are not parallel. Just having these pairsof congruent segments is not enough information toguarantee the lines will be parallel.

4. (a) Yes.

m

j

k

1 2

34

5 6

8 7

Suppose � 1 ∼= � 5. � 3 ∼= � 1 because they are verticalangles. This makes � 3 ∼= � 5, by transitivity. So, � 3and � 5 are a pair of congruent alternate interiorangles. This makes the lines parallel, by the AIPTheorem.

(b) Yes. Suppose � 1 ∼= � 7. � 3 ∼= � 1 and � 5 ∼= � 7,because they are vertical angles. By transitivity,� 3 ∼= � 5. This makes the lines parallel, by AIPTheorem.

(c) Let �PAB be an isosceles triangle with PA ∼= PB. IfM is the midpoint of AB, then �PMA ∼= �PMB bySSS. Hence � A ∼= � B. But AB is a transversal ofintersecting lines PA and PB, and there are congruentconsecutive angles at A and B. So congruentconsecutive angles do not guarantee parallel lines.

(d) Consider the situation described in the answer forpart (c). For the transversal AB of PA and PB, eachpair of alternate exterior angels are supplementary,but PA and PB are not parallel.

(e) Yes. Suppose � 4 and � 5 are supplementary: m � 4+m� 5 = 180◦. Notice that m � 4+m � 3 = 180◦ aswell. By basic moves of equations, we get:

m � 4+m � 5 = m� 4+m � 3

Subtracting m � 4 from both sides gives:

m� 5 = m� 3


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� 3 and � 5 are congruent alternate interior angles. Sothe lines are parallel.

On Your Own

5. Maybe. If the lines are coplanar, then they are parallel.

In the diagram above, two lines are perpendicular to thesame line. All eight angles in the diagram are rightangles, so they are all congruent. In particular, thealternate interior angle pairs are congruent. By the AIPTheorem, the lines are parallel.

If the lines are not coplanar, then they are called skewlines. Skew lines do not intersect, but as they lie indifferent planes they are not considered parallel.

plane m

plane n

Skew lines do not intersect.

6. In a plane, when two lines are not parallel, they mustintersect (eventually). In space, this is not the casebecause lines can lie on different planes. Imagine a linegoing North-South on the ceiling of your room. Now,imagine another line going East-West on your floor. Evenif you could extend your ceiling and floor forever, theselines would never cross (because your ceiling and yourfloor should be parallel planes). But, these lines are notparallel — one travels North-South, another travelsEast-West. Parallel lines need to travel in the samedirection, to keep the distance between them the same.These lines are skew. Notice that skew lines wouldintersect, if they were in the same plane (e.g., if the lineon your ceiling fell onto the floor); but because they lie indifferent planes, they do not intersect.

7. Line MP is parallel to line NQ.Because O is the midpoint of NP, NO ∼= OP.

Likewise, MO ∼= OQ. Also, � MOP ∼= � NOQ, becausethey are vertical angles. So, �MOP ∼= �QON by SAS.

By CPCTC, � MPO ∼= � ONQ. These are alternateinterior angles. (You can redraw the diagram without MQto visualize this.) Thus, the lines are parallel by AIP.

8. NQ ‖ MP. To prove this, it suffices to prove that a pair ofalternate interior angles are congruent. We will show that� QNP ∼= � NPM.

Start by numbering the angles as shown in the diagram.

N Q

O

M P

N

M

O

Q

P1 2

3

654

Since the sum of the angles in a triangle is 180◦, we havethe following equation:

m� 1+m � 2+m � 3 = m� 4+m � 5+m � 6

m � 3 = m� 6, because they are vertical angles. By basicmoves of equations, we get

m� 1+m � 2 = m� 4+m � 5

By the given information, MO = OP. This makes �MOPan isosceles triangle and m� 1 = m� 2. We will laterprove that the base angles in an isosceles triangle arecongruent. Likewise, �NOQ is an isosceles triangle andm � 4 = m� 5. Substituting this information, we get:

m� 2+m � 2 = m� 4+m � 4

2(m� 2) = 2(m� 4)

m� 2 = m� 4

With this pair of congruent alternate interior angles,NQ ‖ MP by AIP.

9. � ABG and � DEB are consecutive interior angles. Thecorrect answer is B.

10. FI ‖ EJ and DE ‖ FH .Proof that FI ‖ EJ : By the given information, � HFI

and � FGE are congruent alternate interior angles. So thelines are parallel by AIP.

Proof that DE ‖ FH : Putting the two given factstogether, we get

m � DEG+m � FGE = 180◦

These two angles are supplementary same-side interiorangles. By the proof in Exercise 4, this shows the linesare parallel.


11. (a) m � 2 = 180◦ − 58◦ = 122◦; m � 3 = m� 2 = 122◦;m� 4 = m� 1 = 58◦; m � 6 = 180◦ − 58◦ = 122◦;m� 7 = m� 6 = 122◦; m � 8 = m� 5 = 58◦.

(b) m� 2 = 180◦ − 58◦ = 122◦; m � 3 = m� 2 = 122◦;m� 4 = m� 1 = 58◦; m � 5 = 180◦ − 125◦ = 55◦;m� 7 = m� 6 = 125◦; m � 8 = m� 5 = 55◦.

(c) m� 1 = m� 4 = 55◦; m � 2 = 180◦ − 55◦ = 125◦;m� 3 = m� 2 = 125◦; m � 5 = 180◦ − 125◦ = 55◦;m� 7 = m� 6 = 125◦; m � 8 = m� 5 = 55◦.

(d) m� 1 = 180◦ − 55◦ = 125◦; m � 2 = m� 3 = 55◦;m� 4 = m� 1 = 125◦; m � 5 = 180◦ − 55◦ = 125◦;m� 6 = m� 7 = 55◦; m � 8 = m� 5 = 125◦.

(e) m� 2 = 180◦ − 125◦ = 55◦; m � 3 = m� 2 = 55◦;m� 4 = m� 1 = 125◦; m � 5 = 180◦ − 55◦ = 125◦;m� 6 = m� 7 = 55◦; m � 8 = m� 5 = 125◦.


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12. Statements (a), (c), (d), (e), (g), and (i) will guaranteem ‖ n.

(a) � 1 and � 5 are congruent corresponding angles. ByExercise 4, this guarantees m ‖ n.

(b) Since m � 1 = 58◦, m � 3 = 180◦ − 58◦ = 122◦.� 3 and � 6 are alternate interior angles, but they arenot congruent. So by Exercise 4, the lines are notparallel.

(c) � 4 and � 6 are supplementary same-side interiorangles. By Exercise 4, this guarantees m ‖ n.

(d) � 1 and � 5 are congruent corresponding angles. ByExercise 4, this guarantees m ‖ n.

(e) Since m � 1 = 55◦, m � 3 = 180◦ − 55◦ = 125◦.� 3 and � 7 are congruent corresponding angles. ByExercise 4, this guarantees m ‖ n.

(f) � 7 and � 6 are vertical angles, so of course they arecongruent. But this does not give any informationabout the measures of the angles on line m. So thereis not enough information to determine if the linesare parallel.

(g) � 7 and � 2 are congruent alternate exterior angles.This means their vertical angles, � 6 and � 3respectively, are also congruent. Since � 6 and � 3 arecongruent alternate interior angles, we know m ‖ n

by AIP.(h) m� 7 = m� 6 because they are vertical angles. So,

m� 6+m � 3 = 180◦. However, � 3 and � 6 arealternate interior angles. By Exercise 4, in order forthe lines to be parallel, these angles need to becongruent. There is not enough information todetermine this. (They could be congruent, 90◦ each;or they could have different measures, like 50◦ and130◦.)

(i) � 4 and � 6 are supplementary congruent consecutive(or same-side interior) angles. By Exercise 4, thisguarantees m ‖ n.

6.07 The Parallel Postulate


1.m � 1 m � 2 m � 3 m� 4 m � 5 m� 6 m � 7 m � 8

(a) 72◦ 108◦ 108◦ 72◦ 72◦ 108◦ 108◦ 72◦(b) 46◦ 134◦ 134◦ 46◦ 46◦ 134◦ 134◦ 46◦(c) x 180◦ 180◦ x x 180◦ 180◦ x

−x −x −x −x

(d) 60◦ 120◦ 120◦ 60◦ 60◦ 120◦ 120◦ 60◦(e) 60◦ 120◦ 120◦ 60◦ 60◦ 120◦ 120◦ 60◦

In (d), we need to find the measure of � 4 before we canproceed. From (c), we know that m � 7 = 180◦ −m � 4.Substituting this into the equation, we get:

2m� 4 = 180◦ −m � 4

3m � 4 = 180◦

m � 4 = 60◦

(e) The angles are supplementary, so x+ 2x = 180◦,which leads to x = 60◦.

2. (a) Yes, they must be parallel for the angles to beequal.

(b) If both pairs of lines are parallel, m � 2 = m� 3 at alltimes.

(c) If lines a and b are not parallel, the aboverelationships do not hold. It would be possible form � 2 = m� 3 without c ‖ d, and it would be possiblefor c ‖ d without m� 2 = m� 3.

3. Perhaps the most interesting invariant in this situation isthat the measure of the angle at P equals the sum of theacute angles the two segments from P make with theparallel lines. One way to explain this is to draw a thirdparallel line through P :

123

4

A

P

B

This new line at P divides the angle at P into two smallerangles, � 2 and � 3. By PAI, � 1 ∼= � 2 and � 4 ∼= � 3.Adding these together, we get:

m � 1+m � 4 = m� 2+m � 3

4. Start with a line � and a point P not on �. Assume thatthere are two lines m and n through P that are bothperpendicular to �. Lines m and n intersect � at points M

and N respectively.

Look at the triangle �PMN. Angles M and N bothmeasure 90◦. Since the sum of the angles in a triangleis always 180◦, that means that angle P measures 0◦.Since angle P measures 0◦, we know that points M

and N must coincide and there is only one perpendicularto � through P .

5. The sum of the measures of the angles in a quadrilateralis 360◦. This is because a quadrilateral can be dividedinto two triangles, whose angles match up perfectly withthose of the quadrilateral. Given quadrilateral ABCD,draw diagonal AC and number the angles as showbelow.

A

BC

D

1 2

34

5

6


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By the Triangle Angle Sum Theorem, m � 1+m � 2+m� 3 = 180◦ and m � 4+m � 5+m � 6 = 180◦. So,

(m � 1+m � 2+m � 3)+ (m � 4+m � 5+m � 6) =180◦ + 180◦ = 360◦

Regroup the angles:

m � 1+ (m � 2+m � 5)+m � 6+ (m � 3+m � 4) = 360◦

m � 2+m � 5 = m� BCD and m � 3+m � 4 = m� BAD.This shows the sum of the angles of the quadrilateral is360◦.

6. The difference between the two theorems is in what youuse them for. The AIP theorem is used to prove that linesare parallel, whereas the PAI theorem is used to provethat angles are congruent.

7. Add points G and H to the diagram as shown:

Then, the following congruences hold:

(a) • � CBD ∼= � CDB• � DCE ∼= � DEC• � ACD ∼= � DEG• � ABC ∼= � CDH

(b) It is not possible to make BC parallel to DE. To provethis, suppose BC is parallel to DE. � ACB and � AEDare corresponding angles for these two lines, so byPAI, � ACB ∼= � AED. � A ∼= � A, because an angle iscongruent to itself. And, BC ∼= DE by the giveninformation. This makes �ACB ∼= �AED by AAS.Now, by CPCTC, AC ∼= AE and AB ∼= AD. But thiscannot be, since AE > AC and AD > AB by the waywe constructed the diagram. As we have reached acontradiction, it cannot be true that BC is parallel toDE.

On Your Own

8. Because � ABD and � DBC are supplementary, andbecause m � ABD = 125◦, the measure of � DBC is 55◦.By the Triangle Angle Sum Theorem, m � DBC+m � BCD+m � BDC = 180◦. Therefore, 55◦ + 25◦+m� BDC = 180◦, or m � BDC = 180◦ − 55◦ − 25◦ =100◦. The correct answer is C.

9. (a) By PAI, � 1 and � 2 are alternate interior angles ofparallel lines.

(b) They are vertical angles.(c) By part (b) � 2 ∼= � 3, and by (a), � 1 ∼= � 2. So,

� 1 ∼= � 3.

10. Suppose � ‖ m and n ‖ m. We will prove that � ‖ n.Begin by constructing a transversal that cuts through �,m, and n, as shown in the diagram:

�

m

n

12

34

Since � ‖ m, � 1 ∼= � 2 by PAI. Since n ‖ m, � 3 ∼= � 4 byPAI. And, � 2 ∼= � 3 because they are vertical angles. So,by transitivity, � 1 ∼= � 4. � 1 and � 4 are alternate interiorangles for lines � and n. So, � ‖ n by AIP.

This proof assumes that m is between � and n. Asimilar proof follows for other placements of �, m, and n.

11. Let’s start with �ABC and �XYZ, with AB ∼= XY ,� B ∼= � Y , and � C ∼= � Z, as shown in the diagram:

We will prove �ABC ∼= �XYZBy the Triangle Angle Sum Theorem, m � A+m � B+

m � C = 180◦ and m � X+m � Y +m � Z = 180◦. Puttingthis together, we get

m � A+m � B +m � C = m� X+m � Y +m � Z

By the given information, we know m � B = m� Y andm � C = m� Z. By basic moves of equations, we getm� A = m� X. So, �ABC ∼= �XYZ by ASA.

12. Given �ABC with exterior angle at C, we will prove thatm� 4 = m� 1+m � 2.

A

B

C1

2

3 4

Since the sum of the angles in a triangle is 180◦, m � 1+m � 2+m � 3 = 180◦. Also, m � 3+m � 4 = 180◦. Usingbasic moves of equations, we get

m � 3+m � 4 = m� 1+m � 2+m � 3

m� 4 = m� 1+m � 2

13. (a) The sum of the angles in a pentagon is 540◦.Consider pentagon PENTA. Draw diagonal ET ,

dividing the pentagon into quadrilateral PETA andtriangle ENT. The sum of the angles in quadrilateralPETA is 360 (from Exercise 5) and the sum of theangles in triangle ENT is 180 by the Triangle Angle


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Sum Theorem. Since the angles of the quadrilateraland the triangle perfectly match up to the angles ofthe pentagon, the sum of the pentagon’s angles is360◦ + 180◦ = 540◦.

(b) The sum of the angles in a hexagon is 720◦.Consider hexagon HEXAGO. Draw diagonal XG,

dividing the hexagon into pentagon HEXGO andtriangle XAG. Using an argument similar to the onefrom part (a), the sum of the angles in the hexagon is540◦ + 180◦ = 720◦.

14. (a) Impossible. From the diagram, m � 3+m � 4 = 180◦.By the given information m � 3+m � 6 = 180◦.Putting this together, we get:

m � 3+m � 4 = m� 3+m � 6

m� 4 = m� 6� 4 and � 6 are congruent alternate interior angles. ByAIP, n and p must be parallel.

(b) Possible. This follows directly from PAI.(c) Possible, but only if line q is perpendicular to both

lines n and p.(d) Impossible. � 4 and � 2 are vertical angles, and

therefore have equal measure. The same is true for� 5 and � 7. The sums of equals are equal.


15. The following facts are useful in finding the missingangles:

• m� 1 = m� 2, because PR bisects � WPQ. Likewise,m� 6 = m� 8.• m� 1+m � 2+m � 3 = 180◦.• m � 3+m � 4 = 180◦, because they are same-side

interior angles and n ‖ m.• m � 4+m � 7 = 180◦.• m � 7 = m� XQP because they are vertical angles.• m � 2+m � 5+m � 8 = 180◦ by the Triangle Angle

Sum Theorem.

m � 1 m� 2 m� 3 m � 4 m� 5 m � 6 m � 7 m � 8(a) 64◦ 64◦ 52◦ 128◦ 90◦ 26◦ 52◦ 26◦(b) 68◦ 68◦ 44◦ 136◦ 90◦ 22◦ 44◦ 22◦(c) 25◦ 25◦ 130◦ 50◦ 90◦ 65◦ 130◦ 65◦(d) 30◦ 30◦ 120◦ 60◦ 90◦ 60◦ 120◦ 60◦(e) 20◦ 20◦ 140◦ 40◦ 90◦ 70◦ 140◦ 70◦(f) x x 180◦ 2x 90◦ 90◦ 180 90◦

−2x −x −2x −x

In part (f), m � 1 = x because m� 1 = m� 2. Since m � 1+m� 2+m � 3 = 180◦, m � 3 = 180◦ − 2x. Because m � 3+m � 4 = 180◦,

m � 4 = 180◦ −m � 3

= 180◦ − (180◦ − 2x)

= 2x

m � 4+m � 7 = 180◦, so m � 7 = 180◦ − 2x.Now, � XQP ∼= � 7, and because QR bisects � XQP,

m� 6 = m� 8 = 12m � XQP. So,

m � 6 = m� 8 = 1

2(180◦ − 2x) = 90◦ − x

The only angle left is � 5. By the Triangle Angle SumTheorem, m � 2+m � 5+m � 8 = 180◦.

m � 5 = 180◦ − (m � 2+m � 8)

= 180◦ − (x+ 90◦ − x)

= 180◦ − 90◦

= 90◦

So, � 5 will always be a right angle in this situation.

6B MATHEMATICAL REFLECTIONS

1.angle measure

1 88◦2 92◦3 126◦4 54◦5 126◦6 88◦7 92◦8 88◦9 38◦

10 54◦11 88◦12 54◦

2. If the lines were parallel, the sum of the measures of apair of consecutive angles would be 180◦. 103◦ + 75◦ =178◦, which is close but not close enough. The lines arenot parallel.

3. Here is the figure described in the exercise, withcongruences marked:

You also know that � AOC ∼= � BOD because they arevertical angles. This allows you to conclude that�AOC ∼= �BOD by SAS. This means that AC ∼= BD byCPCTC.

4. Refer to the following figure:


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Draw any line m through P intersecting �. Call theintersection A, and choose any other point on � and callit B. Construct an angle � APC so that it is congruent to� PAB, with

−→AB and

−→PC on opposite sides of m. The line n

through P and C is parallel to � because the twocongruent angles are a pair of alternate interiorangles.

5. Mathematics seeks to establish truths beyond thosewhich can just be measured or observed. Proof allowsmathematicians to justify general facts, beyond onespecific example or many examples. Proof allowsmathematics to build upon itself.

6. Answers may include: alternate interior angles arecongruent, corresponding angles are congruent,consecutive angles are supplementary, alternate exteriorangles are congruent.

7. 180◦

INVESTIGATION 6C WRITING PROOFS


For You To Explore

1. If the square you start with has sides of length a, you cancreate four smaller squares by connecting the midpointsof opposite sides. Each new square will have sides oflength a

2 . So all four squares will have the same area. Youcould also match up the four squares, one on top of theother, to see they have the same area.

Another possibility is to draw both diagonals, dividingthe square into four triangles. By the PythagoreanTheorem, you can show that the lengths of the diagonalsof the square have length

√2a. So, the four small

triangles have side lengths√

22 a,

√2

2 a, and a. By SSS, thetriangles are congruent and have the same area.

2. One way to do this is to divide the square into fivecongruent rectangles. If the sides of the square havelength a, divide one side into five congruent pieces, eachof length a

5 . Then draw segments from the divisionpoints, perpendicular to the opposite side. This will letyou form five rectangles, each having dimensions a

5 by a.

3. (a) Below is a sample argument to convince a fourthgrade student:

You can think of even numbers as beingmade out of pairs. Think of pairs of shoes. If

every shoe has one partner to make a pair,you have an even number of shoes. Oddnumbers are made out of pairs, too, but theyalso have one leftover shoe. When you addan even number to an odd number, you’llend up with some pairs and one leftovershoe from the odd number, so the sum has tobe odd.

Below is a picture argument.

All even numbers look like this, becausethey can be divided by 2:

All odd numbers look like this, because theyare one more than an even number:

Adding together an even number and an oddnumber, we get:

This is one more than an even number, so itmust be odd.

(b) Below is an argument to convince someone whoknows algebra:

Any even number can be written in the form2a for some integer a, while any odd numbercan be written as 2b+ 1 for some integer b.So, the sum of an odd number and an evennumber has the form (2b+ 1)+ (2a). Thissimplifies to 2(a+ b)+ 1 which has theform of an odd number. Therefore, whenyou add any odd number to any evennumber, the sum is odd.

4. Take any quadrilateral and construct one of the twodiagonals, forming two triangles. Suppose the angles in


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one triangle are labeled � 1, � 2, and � 3, while the anglesin the other are labeled � 4, � 5, and � 6.

1

2

3

4

5

6

A

B

C

D

Since you know that the sum of the measures of theangles in a triangle is 180◦, you know that

m � 1+m � 2+m � 3 = 180◦

and

m � 4+m � 5+m � 6 = 180◦.

The four angles in the original quadrilateral are � A,� ABC, � C, and � CDA. The sum of the measures of thesefour angles is simply the same as the sum of the measuresof all six angles; by the equations above, this is equal to180◦ + 180◦ = 360◦.

For concave quadrilaterals, only one choice ofdiagonal will form two triangles from the quadrilateral,but the argument still works if you choose the appropriatediagonal.

On Your Own

5. Yes, Ruth’s argument is correct. She could apply the SASpostulate to conclude that �ABC ∼= �DBC.

6. Ruth correctly finds that two sides of these triangles arecongruent, but the congruent angle pair given( � ABC ∼= � ACB) is not the included angle pair. In orderto use SAS, she would need to know that � DCB ∼=� EBC, which we cannot conclude from the giveninformation. So she cannot use SAS to prove the trianglescongruent.

7. You know that

m � DAB+m � BAC = 180◦

because they form a straight line (they aresupplementary). Also,

m� B +m � C +m � BAC = 180◦

because the sum of the angle measures in a triangleis 180◦. Combining these two equations, we seethat

m � DAB = m� B +m � C = 180◦ −m � BAC.


8. To do proofs like these, you need algebraic ways to writean even number, an odd number, and any number.Variables like a or b will certainly do for any number, butyou cannot restrict them to mean only an even number.Even numbers must have a factor of 2; you can express aneven number as 2× (some other quantity). It could be assimple as 2a or more complicated, like 2(a+ b+ 1).Whatever the other quantity is, the end result must beeven because 2 is one of its factors. Adding 1 to any evennumber produces an odd number. So, any odd number canbe expressed as 2× (some other quantity)+ 1. Again, itcould be as simple as 2a+ 1, or more complicated, like2(b+ 1)+ 1.

(a) Two even numbers can be written as 2a and 2b.Their sum is 2a+ 2b = 2(a+ b), which is an evennumber.

(b) Two odd numbers can be written as 2a+ 1and 2b+ 1. Their sum is 2a+ 1+ 2b+ 1 = 2a+ 2b+2 = 2(a+ b+ 1), which is an even number.

(c) An even number can be written as 2a. The othernumber can be written as b (we don’t know if it’s evenor odd). The product is (2a)b = 2(ab), which is aneven number.

(d) Two odd numbers can be written as 2a+ 1 and2b+ 1. Their product is (2a+ 1)(2b+ 1) =4ab+ 2a+ 2b+ 1 = 2(2ab+ a+ b)+ 1, which is anodd number.

6.09 What Does a Proof Look Like?


1. (a) You could build polygons with different numbers ofsides using dynamic geometry software. Thesoftware will measure each of the interior angles andcalculate the sum. You can then move the vertices ofeach polygon to see if the sum remains constant asthe polygon changes its size and shape. Thisexperiment could give you a lot of evidence for a lotof different polygons, but you could never test everypossible polygon.


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(b) A deductive proof of this has two steps: First, noticethat every n-gon can be divided into (n− 2) triangles.

3 triangles from a pentagon 6 triangles from an octagon2 triangles from a quadrilateral

Second, since the sum of the angle measures of the(n− 2) triangles is the same as the sum of the anglemeasures of the polygon, the total is (n− 2)180◦.

2. � ABD ∼= � CBE, as they are vertical angles. Since it isgiven that AB ∼= BC and BD ∼= BE, it follows that�ABD ∼= �CBE by SAS.

3. �STV and �UVT share side VT . This, combined with thegiven information SV ∼= UT and ST ∼= UV , shows thatthe two triangles are congruent by SSS.

4. Because the angles in a square are all right angles, itfollows that � SWB ∼= � EBW. Since the sides of a squareall have the same length, SW ∼= EB. And, WB iscongruent to itself. Therefore, by the SAS postulate,�SWB ∼= �EBW.

On Your Own

5. B6. Suppose that �ABC is isosceles with vertex angle A, and

let AD be the angle bisector of � CAB. By the definition ofisosceles triangle, we know that AC ∼= AB. Because ADis an angle bisector, � CAD ∼= � BAD. Also, the twotriangles share side AD. The SAS postulate shows that�ACD ∼= �ABD.

A

B

C

D

7. Because XE is a median, it follows that ME ∼= YE. XE iscongruent to itself. It is given that XY ∼= XM . Therefore,by the SSS postulate, �XEM ∼= �XEY.

8. It is true that � ELM and � HML are alternate interiorangles, but this does not mean they are congruent.

9. (a) The only mistake in this proof is the use of the ASApostulate. The congruent sides (AB ∼= ED) are notincluded sides.

(b) Two pairs of corresponding angles and a pair ofcorresponding sides have been shown to becongruent, so the triangles are congruent by AAS.

10. To prove a statement is not true, you just need onecounter example. So in this case you just need two horsesthat are not the same color.

11. The picture below has some labels added to help with thesolution.

A

F

E

D

C

B

P

Each triangle has two sides that are radii of the circle, sothe following sides are all congruent:

AP ∼= BP ∼= CP ∼= DP ∼= EP ∼= FP

Also, all of the included angles are congruent by themarkings, so we can conclude that �APB ∼= �CPD ∼=�EPF, all by SAS.


12. (a) A prime number is a whole number that has only twofactors, 1 and itself.

(b) 1+ 12 + 41 = 43; prime(c) 2+ 22 + 41 = 47; prime(d) 3+ 32 + 41 = 53; prime(e) −1+ (−1)2 + 41 = 41; prime(f) −2+ (−2)2 + 41 = 43; prime(g) −10+ (−10)2 + 41 = 131; prime(h) 15+ 152 + 41 = 281; prime(i) 100+ 1002 + 41 = 10, 141; prime(j) This conjecture is not true. If we make n = 41, then

n2 + n+ 41 = (41)2 + 41+ 41 = 41(41+ 1+ 1) =41 · 43. So, since n2 + n+ 41 can be factored whenn = 41, its result is not always prime.

6.10 Analyzing the Statement to Prove


1. The statement, “Two triangles with the same area arecongruent.” is false; a right triangle with legs of lengths4 and 8 has the same area as a right triangle with legs oflengths 2 and 16, but the triangles are not congruent.

The statement, “People with large hands have largefeet.” is probably false. There are likely to be somepeople in the world with large hands but small feet.

The last statement, “Out of sight, out of mind.” isprobably false as well. There have likely been timeswhen you couldn’t see a thing, but were able to thinkabout it anyway.

Notice that the Hypothesis statements and theConclusion statements have no truth value, outside of a


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particular situation. To say, “Two lines are parallel.” hasno validity to it until you refer to a particular pair oflines.

2. (a) Hypothesis: Two lines form congruent alternateinterior angles with a transversal. Conclusion: Thelines are parallel.

(b) Hypothesis: n is any whole number. Conclusion:n2 + n+ 41 is prime.

(c) Hypothesis: Three sides of one triangle are congruentto three sides of another triangle. Conclusion: Thetwo triangles are congruent.

(d) Hypothesis: Two lines are parallel to a third line.Conclusion: The two lines are parallel to each other.

On Your Own

3. The hypothesis follows the “if”. The correct answer is A.4. The statement is true. Suppose two lines are both

perpendicular to a third line. Then angles � 1 and � 2 arecongruent, as they are both right angles. These angles arealternate interior angles formed by the third line as atransversal. Therefore, the first two lines are parallel byAIP.

12

5. This statement is false. In the picture below, the anglebisector of the right angle is drawn, but this segmentclearly does not bisect the opposite side of the triangle. Inother words, an angle bisector is not necessarily a median.

6. The statement is false. An equilateral quadrilateral is arhombus, and there are many rhombuses which are notequiangular. For example, connect two equilateraltriangles. The result is an equilateral quadrilateral (orrhombus) with angles measuring 60◦ and 120◦.

7. The statement is true. A proof is shown below.Given: �ABC with � B ∼= � C.Prove: AB ∼= AC.Construct the bisector of � BAC. Call the point of

intersection with BC point P .

We were given that � CBA ∼= � BCA, so we also knowthat � PBA ∼= � PCA. We constructed AP so that � BAP ∼=� CAP. AP is congruent to itself. Therefore �BAP ∼=�CAP by AAS. Now we can conclude that AB ∼= ACby CPCTC.

8. The statement is true. Let �ABC be an equiangulartriangle. Construct the altitude AD, forming two rightangles, � ADB and � ADC. Then triangles �ADB and�ADC are congruent by AAS, and AB ∼= AC by CPCTC.Repeat this same argument, this time starting with thealtitude drawn from � B. This will show that AB ∼= CB,and you will be able to conclude that all three sides of�ABC are congruent.

A

B CD

9. The statement is false. Consider any rectangle that is not asquare. All angles measure 90◦, but the rectangle is notequilateral.


10. If one of the legs of a right triangle is half as long as theother, then the square of the length of the hypotenuse isfive times the square of the length of the shorter leg.Proof: There is a right triangle �ABC with legs AB andAC such that AB = 1

2 AC. If we call the length of AB x,then the length of AC is 2x. Then we can use thePythagorean Theorem to find the length of thehypotenuse. CB2 = x2 + (2x)2. When we simplify thisequation, we get CB2 = 5x2 where x is the length of theshorter leg of the right triangle.

6.11 Analysis of a Proof


1. (a) The fact that AC and BH bisect each other at M doesnot mean that all four segments AM , CM , BM , andHM are congruent. What it means is that AC dividesBH in half, so that BM ∼= HM , and also that BHdivides AC in half, giving that AM ∼= CM . It is


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still possible for BH and AC to have differentlengths.

(b) The error in the markings could lead someone toconclude that ABCH is a rectangle. This is notnecessarily the case. However, there is enoughinformation given to conclude that ABCH is aparallelogram. Look at the picture below. Becausethe diagonals bisect each other, and it is given thatAB ∼= CH , you can conclude that �ABM ∼= �CHMby SSS. This implies that � CHM ∼= � ABM, so CH isparallel to AB. Since these two opposite sides arecongruent and parallel, we can conclude that ABCHis a parallelogram.

A

B

C

H

M

2. Here is one possible flow chart:

∠1 ≅ ∠ 2

∠3 ≅ ∠ 4

XY bisects ∠ MXT

∠5 ≅ ∠ 6 XY ≅ XY

MY ≅ YT

MXY ≅ ΔΔ TXY

3. (a)

(b) Below is a proof written in paragraph style:Because ABCD is a square, all its sides are

congruent. So AB ∼= BC and AB ∼= DA. Also, all its

angles are congruent. So � MAD ∼= � NBA. SinceAB ∼= BC and M and N are the midpoints of AB andBC respectively, then AM ∼= BN .

From all this you can conclude that �ABN ∼=�DAM by SAS. By CPCTC, you can conclude thatAN ∼= DM .

On Your Own

4. The correct answer is B. The fact that D, E, and F , aremidpoints allows you to make the three congruencystatements AD ∼= DB, AF ∼= FC, and BE ∼= EC.

5. • HJ ∼= HL and JK ∼= LK (given)• HK is congruent to itself• �JHK ∼= �LHK (SSS)• � JHM ∼= � LHM (CPCTC)• HM is congruent to itself• �HJM ∼= �HLM (SAS)

H

J

K

LM

6. Consider the picture below.

E

K

H

J

F

G

12 3

45

Because GF ⊥ GH it follows that

m � 3+m � 4 = 90◦

and because GJ ⊥ GK , it also follows that

m� 2+m � 3 = 90◦.

Putting the above together, you can see that

m � 3+m � 4 = m� 2+m � 3

So it must be that m � 4 = m� 2, implying that� JGF ∼= � KGH.

7. Since FACG and DABE are squares, FA ∼= CA andAD ∼= AB. Also, � FAC ∼= � DAB since both are rightangles. Since these two angles have the same


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measurement, we can add the measure of � BAC to both,obtaining equal quantities. In other words,

m � FAC+m � BAC = m� DAB+m � BAC

This means that m � FAB = m� CAD. Therefore, by SAS,�FAB ∼= �CAD.

8. To prove that MK ∼= NJ , we will first prove that�MLK ∼= �JLN. Because triangles LJM and LNK areequilateral, ML ∼= JL and LK ∼= LN . Furthermore,m� MLJ = m� NLK, since all angles in all equilateraltriangles measure 60◦.

Now, we use a strategy similiar the one used inExercise 7 to show m � MLK ∼= m� NLJ. Add m � JLK toboth of the angle measures above to get:

m� MLJ+m � JLK = m� NLK+m � JLK

This means that m� MLK = m� NLJ.Therefore, �MLK ∼= �NLJ by SAS. Then, by

CPCTC, MK ∼= NJ .9. We will first prove that �URQ ∼= �VRQ.

Because �QRT is isosceles, its base angles � TRQ and� TQR are congruent. Since VR is a median, it divides QTinto two congruent parts. So, VQ = 1

2 (QT). Likewise,UR = 1

2 (TR). It is given that QT = TR. So, VQ = UR aswell.

V

R

Q

R

Q

U

The triangles also share side QR. Therefore, triangles�URQ and �VQR are congruent by SAS. By CPCTC,UQ = VR.

10. Construct segments PL and PM forming triangles �PLNand �PMN.

P

MNL

We will prove PL = PM, by proving the triangles arecongruent.

The fact that P is on the perpendicular bisector of LMgives two pieces of information: m� PNL = m� PNM =90◦, and NL ∼= NM . The triangles share side PN , so�PLN ∼= �PMN by SAS. By CPCTC, PL = PM.


11. Here is a picture of �AOB with A = (6, 0), O = (0, 0),and B on x = 3. Assume that B is not on the x-axis, orthere is no triangle, so call the point (3, 0) where the linex = 3 intersects the x-axis Q.

�OQB ∼= �AQB by SAS. OQ = AQ = 3,m� OQB = m� AQB = 90◦, and the triangles share QB.This means that OB ∼= AB by CPCTC, so �AOB isisosceles.

6.12 The Reverse List


1. (a) Let the intersection of the perpendicular bisector andAB be C.Below is a reverse list:NEED �APB is isoscelesUSE isosceles triangles have two congruent sides

• NEED AP ∼= PB• USE corresponding parts of congruent triangles

– NEED �APC ∼= �BPC– USE SAS

* NEED PC ∼= PC* USE the triangles share this side* NEED � PCB ∼= � PCA* USE both right anlges because

−→CP is a

perpendicular bisector* NEED AC ∼= BC* USE C is a midpoint because

−→CP is a

perpendicular bisector

(b) Here is the proof:

Because←→CP is a perpendicular bisector of AB, C is

the midpoint of AB, so we know that AC ∼= BC.

Also because←→CP is a perpendicular bisector of AB,

we know that m� PCB = m� PCA = 90◦.Since �ACP and �BCP share side PC, we can

conclude that they are congruent by SAS.Since �ACP ∼= �BCP, AP ∼= BP because they are

corresponding parts of congruent triangles. Thismeans that �APB is isosceles because it has twocongruent sides.


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2. (a) Triangles AFC and AFB can be proven congruentusing SAS.

(b) The proof is exactly the same as the one used inExercise 16 to prove that �ACP ∼= �BCP. Thecongruent parts are AF congruent to itself, BF ∼= FC,and � AFB ∼= � AFC.

On Your Own

3. In the picture below, �ABC is isosceles with AB = AC.Points D and E are midpoints.

A

B

CD

EA

B

CD

E

Look at �ADC and �AEB. We know that AB = AC.Also, these triangles share � A. Can we complete one ofthe congruence postulates?

Since D is a midpoint, AD = 12AB. Since E is a

midpoint, AE = 12 AC. But AB = AC, so AD = 1

2 AB =12 AC = AE. By SAS, the triangles are congruent.

The medians of the isosceles triangle, CD and BE mustbe congruent because they are corresponding parts ofcongruent triangles.

4. In the picture below, �ABC is isosceles with AB = AC.CD and BE are angle bisectors. We need to show they arecongruent.

A

B

C

D

E

A

B

C

D

E

Look at �ABE and �ACD. We know that AB = AC.Also, these triangles share � A. Can we complete one ofthe congruence postulates? We need either AD ∼= AE or� ACD ∼= � ABE.

Since BE is an angle bisector, m � ABE = 12m � ABC.

Since CD is an angle bisector, m � ACD = 12m � ACB.

Now, � ABC ∼= � ACB since the triangle is isosceles. So,

m � ABE = 1

2m � ABC = 1

2m � ACB = m� ACD

By ASA, the triangles are congruent.

The angle bisectors of our isosceles triangle, CD andBE, must be congruent because they are correspondingparts of congruent triangles.

5. First assume that the triangle is acute. In the picturebelow, CD ∼= BE, and both segments are altitudes. Weneed to show AB ∼= AC.

A

B

D

E

A

B

CC

D

E

Look at �ACD and �ABE. We know that CD = BE.Also, these triangles share � A. Can we complete one ofthe congruence postulates? We need another pair ofcongruent angles.

Since BE and CD are altitudes, m � AEB = m� ADC =90◦. By AAS, �ACD ∼= �ABE. By CPCTC, AB ∼= AC.This makes �ABC an isosceles triangle.

If the triangle is a right triangle, the altitudes coincidewith two of the side of the triangle. And since the altitudesare congruent, the sides are congruent as well. Therefore,the triangle is isosceles.

If the triangle is obtuse, the altitudes CD and BE lieoutside the triangle. Refer to the picture below.

A

CB

E D

� AEB ∼= � ADC (both are right angles) and � EAB ∼=� DAC (vertical angles). But BE ∼= CD (given). So�AEB ∼= �ADC (AAS). Hence AB ∼= AC (CPCTC),which means �ABC is isoceles.

6. First assume that the triangle is acute. In the picturebelow, �ABC is isosceles with AB = AC, and CD and BEare altitudes. We need to show CD ∼= BE.

A

B

D

E

A

B

CC

D

E

Look at �ACD and �ABE. We know that AC = AB.Also, these triangles share � A. Can we complete one of


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the congruence postulates? We need to show AD ∼= AE orwe need to find another pair of congruent angles.

Well, since BE and CD are altitudes,

m� AEB = m� ADC = 90◦

By AAS, �ACD ∼= �ABE. By CPCTC, CD ∼= BE.So the altitudes to the legs of an isosceles triangle arecongruent.

If the triangle is a right triangle, the altitudes coincidewith two of the sides of the triangle. And since the sidesare congruent, the altitudes are congruent as well.

If the triangle is obtuse, � AEB ∼= � ADC (right angles)and � EAB ∼= � DAC (vertical angles). It is given thatAB ∼= AC. So �AEB ∼= �ADC (AAS). Hence BE ∼= CD(CPCTC).

7. Since M is the midpoint of AC, we know that AM ∼= MC.Similarly, CN ∼= NB. Also, since AC ∼= CB, all four of thesmaller segments are congruent to each other. AM ∼=MC ∼= CN ∼= NB From this, we can see that CM ∼= CNso that �CMN is isosceles.

8. For these two triangles:

We know that AC ∼= DF . We also know that medians AMand DN are congruent. Because CB ∼= FE and M and N

are the midpoints of those segments, respectively, we canconclude CM ∼= MB ∼= FN ∼= NE. From this we knowthat CM ∼= FN , so �AMC ∼= �DNF by SSS.

Since � ACM ∼= � DFN by CPCTC, then �ABC ∼=�DEF by SAS.

9. The correct answer is A. Options B and C use the wrongtriangles for I, which do not include the segments EFand DF . Option D would be correct if part III read“CE = BD”.


10. (a) �ABC is half of an equilateral triangle ifm� ABC = 2m� ACB or m � ABC = 1

2m � ACB.�ABC is half of an equilateral triangle if AB = 1

2 BC

or AC = √3AB.(b) �ABC is half of a square if m � ABC = m� ACB.�ABC is half of a square if AC = AB orCB = √2AB.

(c) �ABC is half of an isosceles triangle no matter whatits other angles are because every right triangle is halfof an isosceles triangle. �ABC is half of an isoscelestriangle no matter what its sides measure becauseevery right triangle is half of an isosceles triangle.

6.13 Practicing Your Proof-Writing Skills


1. To prove these angles are congruent, we will first provethat �SRP ∼= �SQP and �TPR ∼= �TPQ. Since ST is aperpendicular bisector of RQ, it makes four right anglesat P , and RP ∼= PQ. Since SP is shared by �SPR and�SPQ, these triangles are congruent by SAS. Likewise,as �TPR and �TPQ share side TP they are congruent bySAS.

Now, by CPCTC, � SRP ∼= � SQP and � TRP ∼= � TQP.It follows that

m � SRP+m � TRP = m� SQP+m � TQP

This implies that m � SRT = m� SQT; hence these twoangles are congruent.

2. The side opposite the larger angle will be longer than theside opposite the smaller angle.

To understand this, consider �ABC with fixed lengthsAB and and AC. Imagine a hinge is placed at vertex A, sothat you could adjust the angle to any size you wantedwithout changing the lengths of the sides. The distancefrom B to C will be larger when the angle is wider, andsmaller when the angle is narrower.

3. Look at the picture below.

A B

CD

E

F

G

H

1

23

45

67

8

We need to prove that quadrilateral EFGH is equilateraland equiangular in order to prove it is a square.


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Because ABCD is a square, AB ∼= BC ∼= CD ∼= DA.By construction, AH ∼= BE ∼= CF ∼= DG.Moreover, since each angle in the original square is a

right angle, it follows that � HAE ∼= � EBF ∼= � FCG ∼=� GDH, since they are also all right angles. Therefore, wecan apply the SAS postulate to conclude that

�AHE ∼= �BEF ∼= �CFG ∼= �DGH.

By CPCTC,

HE ∼= EF ∼= FG ∼= GH

showing that quadrilateral EFGH is equilateral.Next, look at the labeled angles in the figure. By

CPCTC, you know that � 1 ∼= � 3 ∼= � 5 ∼= � 7 and� 2 ∼= � 4 ∼= � 6 ∼= � 8. Putting the above two statementstogether, you see that

m� 1+m � 8 = m� 2+m � 3 = m� 4+m � 5 = m� 6+m � 7

which implies that EFGH is also equiangular, so it mustbe a square.

4. Suppose that in the quadrilateral shown here, sides ABand DC are congruent, as well as diagonals AC and DB.With this information, �DCB ∼= �ABC by SSS, sincethey share side CB. Now, by CPCTC, � ACB ∼= � DBC.So, �MBC is isosceles since it has two congruentangles.

C

D

B

A

M

5. In isosceles triangle �ABC, let P be any point on thebase BC, and construct segments PR and PS parallel tothe congruent two sides of the triangle, as in thefigure.

A

B CP

R

S

We will show that you can express the perimeter of PRASas a quantity which does not depend on P , R, or S. This

way, the perimeter will not change as point P movesalong the base BC (notice that as P moves, R and S alsomove).

Because RP ‖ SC, it follows that � SCP ∼= � RPB, sincethey are corresponding angles for the parallel lines. And,since �ABC is isosceles, it also follows that � SCP ∼=� RBP. Putting these two statements together, we find� RPB ∼= � RBP. This means that �RPB is isosceles andRB = RP. Now write the perimeter of the parallelogramPRAS as follows:

Perimeter = 2(RA)+ 2(RP)

= 2(RA)+ 2(RB)

= 2(RA+ RB)

= 2(BA)

Thus, the perimeter of PRAS is equal to 2BA, which willnot change as the location of P changes.

6. There will be six proofs; basically each proof is centeredaround proving that �ACD ∼= �BCD and usingCPCTC.

• Suppose statements 1 and 2 are given. Since �ABC isisosceles with base AB, you know that CB ∼= CA.Further, since CD is a median, you know that DB ∼=DA. By SSS, �ACD ∼= �BCD, implying that� ACD ∼= � BCD; thus CD bisects � ACB, which isstatement 4.

Now, � CDA ∼= � CDB by CPCTC as well. Since themeasures of these angles add up to 180 and are equal,they must each measure 90◦. This shows that CD is analtitude, proving statement 3.• Suppose statements 1 and 3 are given. The fact that�ABC is isosceles with base AB tells you thatCB ∼= CA and � A ∼= � B, while the fact that CD is analtitude tells you that � CDA and � CDB are rightangles. By AAS, triangles �ACD and �BCD arecongruent, implying that DB ∼= DA and � ACD ∼=� BCD; these statements show that CD is a median(statement 2) and CD bisects � ACB (statement 4).• Now suppose statements 1 and 4 are given. Together

they tell you that CB ∼= CA, � A ∼= � B, and� ACD ∼= � BCD. Then ASA lets you conclude that�ACD ∼= �BCD. This shows that CD is a median, asDB ∼= DA (statement 2). Repeating the argument in thefirst proof lets you conclude that CD is an altitude,proving statement 3.• This time assume statements 2 and 3. Therefore,

DB ∼= DA and � CDA and � CDB are right angles. Nowyou know that �ACD ∼= �BCD by SAS, which letsyou conclude statements 1 and 4.• Assume statements 2 and 4. This case may be the most

difficult. First we use an indirect proof that AB ⊥ CD.Suppose that AB is not perpendicular to CD. Draw theline � that is perpendicular to CD at D. Let M be thepoint where � intersects CB, and let N be the pointwhere � intersects CA. Since � ⊥ CD, � MDC and� NDC are right angles and hence are congruent. But


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� BCD ∼= � ACD (given) and CD ∼= CD. Hence�CMD ∼= �CND by AAS. By CPCTC, DM ∼= DN .But DB ∼= DA (given) and � BDM ∼= � ADN (verticalangles). Hence �BMD ∼= �AND by SAS. By CPCTC,� BMD ∼= � AND. However these are alternate interiorangels for lines CB and CA and the transversal �. TheAIP Theorem implies that lines CB and CA are parallel.This contradicts the fact that lines CB and CA intersectat C. Since the assumption that AB is not perpendicularto CD has led to a contradiction, these segments areperpendicular. Hence CD is an altitude and theperpendicular bisector of AB. By the PerpendicularBisector Theorem, CA = CB.• Finally, assume statements 3 and 4. You thus know that� CDA and � CDB are right angles, and � ACD ∼=� BCD. This time it is ASA which lets you concludethat �ACD ∼= �BCD, which is enough to provestatements 1 and 2.

On Your Own

7. The statement “triangles ACD and BCD are congruenttriangles” guarantees that all four given statements arecorrect. Because the two right triangles are congruent,you have that AD ∼= BD. So in �ABC, CD is a median.You also have that AC ∼= BC, so �ABC is isosceles, and� ACD ∼= � BCD, so CD is an angle bisector. Finally,since � ADC and � CDB are right angles because they arecongruent and must add up to 180. So, it follows that CDis an altitude of �ABC.

8. (a) Only isosceles triangles have a perpendicular bisectorthat passes through a vertex. Only isosceles trianglesthat are not equilateral will have exactly one suchperpendicular bisector.

(b) No triangles have exactly two perpendicular bisectorsthat pass through vertices.

(c) Only equilateral triangles have three perpendicularbisectors that pass through all three vertices.

The important idea here can be summarized in thefollowing statement:

The perpendicular bisector of a side of a trianglepasses through the opposite vertex if and only ifthe remaining two sides of the triangle arecongruent.

This means that if exactly one side of a triangle is to havethis property, then the other two sides are congruent, sothe triangle is isosceles. The triangle cannot beequilateral, however, for the above statement would thenimply that all sides of the triangle would have thisproperty.

It is not possible to have a triangle with exactly twosides having this property that their perpendicularbisectors pass through the opposite vertex. To see this,suppose you have a triangle with vertices A, B, and C,and sides a, b, c (see picture).

A

B

C

a

b

c

Suppose that a and c are the two sides whoseperpendicular bisectors pass through the opposite vertex.Now, use the above statement. If the perpendicularbisector of side a passes through vertex A, it follows thatsides b and c have the same length. However, if theperpendicular bisector of side c passes through C, thesame reasoning says that sides a and b have the samelength. This means that sides c and a are congruent,which implies that the perpendicular bisector of b passesthrough vertex B, by the converse reasoning above. Butwe wanted only two sides of the triangle to have thisproperty, not all three.

Finally, the set of triangles with all three sides havingthe property that their perpendicular bisectors passthrough the opposite vertex is the same as the set ofequilateral triangles. We saw above that if even two sideshave this property, the triangle must be equilateral.

9. To find the center of a circle, first construct any twononparallel chords. Then construct the perpendicularbisectors of these chords. These bisectors will intersect inone point, and this point will be the center of the circle.

Here’s why: Suppose the two chords are AB and CD.Any point on the perpendicular bisector of AB will beequidistant from A and B, while any point on theperpendicular bisector of CD will be equidistant from C

and D. The center of the circle is equidistant from allpoints on the circle, so it is certainly equidistant from A,B, C, and D. But any point with that property must lie onboth bisectors, so it must be the intersection point of thetwo perpendicular bisectors.

With a circle drawn on paper, you can also fold thecircle in half in two ways. Each fold that divides thecircle exactly in half marks a diameter of the circle. Thepoint where the two fold lines intersect is the center ofthe circle, because the only point that is on more than onediameter of a circle is the center.

10. The correct answer is C.11. The solution to this exercise uses the same reasoning as

the solution to Exercise 9. Think of each side of thetriangle as a chord of the circle you are looking for. Thecenter, O, of this circle will need to be equidistant fromall three vertices. O must lie on the perpendicularbisector of AB, in order for AO = OB. Similiarly, O mustlie on the perpendicular bisector of AC, since we needOA = OC.

Therefore, construct these two perpendicular bisectorsand find their intersection point; this point will be thecenter of the circle. Next, using a compass, construct thecircle with center O and radius OA (or radius OBor OC).


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A

B

C

O

12. The tips of each point of the blade all lie on a circle.Using four of these points, two chords can beconstructed. These chords can be used to find the centerof the circle, as described in the solution for Exercise 9.

Or, three points can be chosen to determine a triangle.The center can be found by the intersection of theperpendicular bisectors of two sides of thetriangle.

13. Suppose you have two right triangles with hypotenuseand one leg congruent. This means that the triangles canbe positioned next to each other to form a largertriangle—the two right angles together form a straightangle, and the two congruent legs line up to form thealtitude of the larger triangle.

Because the triangles have congruent hypotenuses, thelarge triangle is isosceles, and hence has congruent baseangles. Now, the AAS postulate shows that the two righttriangles are congruent.

14. It is true that the perpendicular bisectors of all three sidesof a triangle will meet at a point. This point is called thecircumcenter of the triangle. (The circumcenter is thecenter of a circumscribed circle.) Suppose that in �ABC,point O is the point of intersection of the perpendicularbisectors of sides AB and AC (the two lines will alwaysmeet in a point). Since O lies on both perpendicularbisectors, it follows that OA = OC and OA = OB,implying that OC = OB. This means that O is equidistantfrom C and B, and so it must lie on the perpendicularbisector of BC. Thus, all three perpendicular bisectorsintersect at the same point O. Recall that in the Exercise11, this point O was used as the center of the circlepassing through all three vertices.

15. To show that any point on CD is the same distance fromA as from B, it suffices to show that CD is theperpendicular bisector of AB. Let P be the point whereAB and CD intersect. P will be the midpoint of AB,because the fold represented by CD was obtained byfolding AB in half, that is, by connecting A to B.

It remains to show that CD intersects ABperpendicularly. Consider the angle � APB (where P isstill the intersection point). Since this is a straight angle,its measure is 180◦. The fold along CD bisects the angle.We know this because the two angles which make up thestraight angle fit exactly on top of each other. Hence, theyare each right angles, and so CD ⊥ AD.

16. Suppose you are given two sidelengths, say a and c, andthe measure of � A. You want to determine if you can findmore than one triangle with two sides of length a and c,with the additional property that � A is the non-includedangle, and � A is also the largest angle in the triangle.

In the picture below, A and B are two vertices of sucha triangle. The third vertex must lie somewhere on theray, and its distance from B must be equal to a. What youcan do is consider all the points in the plane which lie adistance of a away from B. This produces a circle ofradius a centered at B. A circle can intersect a ray at atmost two locations, so this gives at most two possibilitiesfor the third vertex; call them C and C′.

B

A C C'

a acx

If � A is the largest angle, then a is the longest side.Specifically, a > c, so the picture will look like this:

ca

B

AC

Because A is inside the circle, the ray will only intersectit in one place, producing one unique triangle. Therefore,there can be at most one triangle satisfying all of thenecessary properties, so SSA is a congruence test.

17. There is no SSa congruence test. Consider the picturebelow.


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Ax

B

C C'

ca a

Both �ABC and �ABC′ have sides of length c and a

with a non-included angle A, where A is the smallestangle in the triangle.


18. Draw lines through C and D perpendicular to AB at P

and Q, respectively. Since CP and DQ are perpendicularto the same segment, they are parallel. If you draw DP,you can use ASA to prove �DQP ∼= �PCD. It followsthat CP ∼= QD (CPCTC). The HL Congruence test showsthat �ADQ ∼= �BCP, and hence that � A ∼= � B. Itfollows by the converse of the Isosceles TriangleTheorem that �ABE is isosceles.

19. If lC is not the largest angle, then either AC or CB aloneis bigger than AB, so certainly their sum is. If lC is thelargest angle, draw an altitude from C to a point P on AB.Since a hypotenuse is greater than a leg in a right triangle(it is opposite the largest angle), it follows that AC > APand CB > PB. Hence, AC+ CB > AP+ PB = AB.

6C MATHEMATICAL REFLECTIONS

1. Consider �APC and �BPC. You know that AC ∼= BC,because �ABC is isosceles with base AB. You also know

that � ACP ∼= � BCP because−→CP is the bisector of

� ACB. Certainly, CP ∼= CP. So, �APC ∼= �BPC bySAS. Then, AP ∼= BP by CPCTC and �APB isisosceles.

2. The two triangles must be congruent. Because you knowtwo of the angles of either triangle (the right angle andone other) you also know the measure of the third,because the three measures must sum to 180◦. Thismeans that you have ASA congruence between the twotriangles.

3. Draw in the median of �ABC from C to M on AB.AC ∼= BC because the triangle is isosceles with base AB.AM ∼= BM because M is the midpoint of AB.CM ∼= CM because every segment is congruent to itself.So, �ACM ∼= �BCM by SSS. Then you know that� CBA ∼= � CAB by CPCTC.� CBA ∼= � FBT and � CAB ∼= � EAS because they are

pairs of vertical angles. From this, you can conclude that� FBT ∼= � EAS because they are each congruent tocongruent angles. Then, using the given information thatEA ∼= BF and AS ∼= BT , you can conclude that �AES ∼=�BFT by SAS.

4. For any isosceles triangle, the base angles must becongruent to each other. You can see this if you draw inthe median to the base of the isosceles triangle. Then the

two smaller triangles that are formed are congruent bySSS. (One pair of sides are the sides that make thetriangle isosceles. Another are the sides formed bycutting the base of the triangle by its midpoint. The final“pair” of congruent “sides” is the median congruent toitself.) The base angles are corresponding parts of thesetwo congruent triangles.

Since the base angles must be congruent in anyisosceles triangle, if you know the angle at the apex ofsome isosceles triangle, you also know the two baseangles. The measures of these three angles must sum to180◦, so each base angle measures half of the differencebetween 180◦ and the apex angle.

For this exercise, you know that the bases and apexangles of two isosceles triangles are congruent. From theabove, you know that their base angles also must becongruent to each other and to the base angles in theother isosceles triangle. Together with the basecongruence, this means that the two isosceles triangleshave ASA congruence.

5. Refer to this figure:

A B

C

T

Because �ATC is isosceles with base AC, � ACT ∼=� TAC because they are the base angles of an isoscelestriangle. Their measures are marked in the figure as α.

Because �BTC is isosceles with base BC, � BCT ∼=� TBC because they are the base angles of an isoscelestriangle. Their measures are marked in the figure as β.

The sum of the angles of �ABC must equal 180◦, so

α+ β + α+ β = 180◦

and

2α+ 2β = 180◦,

so

α+ β = 180◦

This means that m � ACB = 90◦ and �ABC is a righttriangle.

6. You can use an outline, a flow chart, a paragraph-styleproof, or a two-column proof.

7. The Perpendicular Bisector Theorem states that any pointon a perpendicular bisector of a line segment isequidistant from the endpoints of the segment.

8. Rewrite the statement in “if, then” form, and then youcan see the hypothesis and conclusion more clearly. If anobject is a tree, then it is green. The hypothesis is anobject is a tree and the conclusion is it is green.


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INVESTIGATION 6D QUADRILATERALS ANDTHEIR PROPERTIES


For You to Explore

1. Check students’ work.2. Properties of the sides of a square include:

• The opposite sides of the square are parallel.• All four sides of the square are congruent.• Adjacent sides of the square are perpendicular.

Properties of the angles of a square include:

• All four angles of the square are congruent.• All four angles of the square are right angles.

Properties of the diagonals of a square include:

• The diagonals of the square are congruent.• The diagonals of the square are perpendicular.• The diagonals of the square bisect its vertex angles.• The diagonals of the square bisect each other.

Some properties of the sides of a rectangle are:

• The opposite sides of the rectangle are parallel.• The opposite sides of the rectangle are congruent.• Adjacent sides of the rectangle are perpendicular.

Some properties of the angles of a rectangle are:

• All four angles of the rectangle are congruent.• All four angles of the rectangle are right angles.

One property of the diagonals of a rectangle is that thediagonals of the rectangle are congruent.

3. Answers will vary. Some examples are:

• Not all of the sides of the rectangle are congruent.• The diagonals of the rectangle do not bisect the vertex

angles.• The diagonals of the rectangle are not perpendicular.

On Your Own

4. The shape described is a rhombus, or “diamond”.5. One way to think about this problem is to draw the

diagonals first, then build the quadrilateral around them.For example, you can think of two congruent diagonalsas “sticks”, and put them down on the plane so that theyintersect (but do not bisect each other). Then, connectingthe endpoints of the sticks in order, you will have drawn ascalene quadrilateral—one with no pairs of parallelsides—that has congruent diagonals.

6. A quadrilateral with four congruent angles must be arectangle.

7. Your construction should be congruent to this figure:

8. This figure is a square.9. One way to do this is to fold two adjacent sides together,

forming a diagonal crease. The point where this diagonalcrease meets the longer side of the rectangular paper willmark off a segment on the longer side congruent to theshorter side. Now you have two congruent adjacent sideswhich include a right angle, and you can complete thesquare as shown.


10. The area of the rectangle is 12 in.× 8 in. or 4 square

inches.11. The area of the square is 4 cm × 4 cm or 16 square

centimeters.12. The area of the rectangle is 2 ft × 10 ft or 20 square feet.

6.15 General Quadrilaterals


1. Because each vertex must be the endpoint of exactly twosides, any figure satisfying this definition must be closed.

2. In fact, you can draw a figure which satisfies thisdefinition, but is not planar. Imagine connecting fourtoothpicks with marshmallows to form a quadrilateral.Now, lift one of the marshmallows up off of the table.Such a quadrilateral is called a skew quadrilateral.

3. (a) The line segments in this figure do not intersect attheir endpoints.

(b) Some of the endpoints of the segments in this figureare not shared by exactly two of the segments.

(c) There are three, not four, line segments in this figure.(d) Some of the endpoints of the segments in this figure

are not shared by exactly two of the segments.

On Your Own

4. D; Counterexamples include a concave quadrilateral anda kite.


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5. The usual proof given that the sum of the angles of aquadrilateral is 360◦ involves drawing a diagonal whichdivides the figure into two triangles. If the diagonal youchoose does this, you can use the usual proof. However,in a concave quadrilateral, the diagonal you choose maylie outside the quadrilateral, as in this figure:

You can still prove that the angles A, B, C, and D of thequadrilateral sum to 360◦. First, see that the measures ofangle D of the quadrilateral and � ADC must sum to 360◦because they completely surround point D. This meansthat we can write m � D = 360−m � ADC. Looking at�ABC, you can see that m � A+m � DAC+m � B+m� C +m � DCA = 180◦. You also know that m � DAC+m� ACD+m � ADC = 180◦.

Using standard moves of equations, you can show thatm � A+m � B +m � C = 180◦ −m � DAC−m � ACD andthat m � D = 360◦ −m � ADC. Now, you can see that:

m � A+m � B +m � C +m � D =180◦ + 360◦ − (m � DAC+m � ACD+m � ADC)

Since the angle sum in the parentheses above is 180◦,you’ve shown that m� A+m � B +m � C +m � D =360◦.

6. For a self-intersecting quadrilateral, the diagonals bothlie outside the figure. And, in fact, the sum of its anglesneed not be 360◦.

7. For a skew quadrilateral, the definition of “inside” and“outside” is difficult to make clearly And, in fact, the sumof its angles need not be 360◦

8. Label the vertices and sides of your quadrilateral in thisway:

By the Triangle Inequality, you know that c + d > BDand that b+ BD > a. Therefore, by standard moves ofinequalities, b+ c + d > a.

• This proof still works if the quadrilateral is concave,because points B, C, and D will still form a triangle aswill points A, B, and D, so both instances of theTriangle Inequality are valid.• Even if the quadrilateral is self-intersecting, any set of

three points will form a triangle for which the TriangleInequality must hold.

• Although students have had less experience withgeometry outside of the plane, some may know that forany three points in space there is a plane that containsall three. The Triangle Inequality will still work for thetriangles formed by any three vertices of a skewquadrilateral, so the proof still works.


9. (a) The four angles sum to 360◦, so 6x = 360◦, andx = 60◦.

(b) The four angles are 60◦, 60◦, 120◦, and 120◦(c) The diagram shows the two different possibilities for

the shape of the figure, depending on whether a pairof congruent angles is adjacent or opposite. Onefigure is an isosceles trapezoid, and the other aparallelogram.

6.16 Properties of Quadrilaterals


1. A. This figure is a trapezoid.B. This figure is a trapezoid.C. This figure is a trapezoid.D. This figure is NOT a trapezoid.E. This figure is NOT a trapezoid.F. This figure is NOT a trapezoid.G. This figure is a trapezoid.H. This figure is a trapezoid.I. This figure is a trapezoid.J. This figure is NOT a trapezoid.

K. This figure is a trapezoid.L. This figure is a trapezoid.

2. always3. sometimes4. sometimes5. sometimes6. sometimes7. sometimes8. never9. The distance between two parallel lines is usually defined

as the length of a segment perpendicular to both lines withan endpoint on each line. This is the smallest possiblelength for a segment with an endpoint on each line.

10. Let ABCD be an isosceles trapezoid with parallel basesAB and CD with (AB > BD), and BC ∼= AD. Draw linesthrough each of the four vertices perpendicular to theparallel lines. Let the intersection of AB and theperpendicular line through D be X, and let theintersection of AB and the perpendicular line through C


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be Y . Extend CD so that CD intersects the perpendicularline through A at M, and the perpendicular line throughB at N. XAMD and BYCN are rectangles. Opposite sidesof rectangles are congruent. It follows by SSS that thefour right triangles in the figure are congruent. ByCPCTC, � ADX and � CBY. And since � BCY ∼= � ADX byCPCTC, we have m� BCY+m � XDC = m� ADX+m� YCD. Therefore, m � BCD ∼= � ADC.

11. We begin with trapezoid ABCD with parallel bases ABand CD, and BC ∼= DA. From the previous exercise, wealso know that � ADC ∼= � BCD. This means that�ACD ∼= �BDC by SAS. (The triangles share BC.)From this, we can conclude that AC ∼= BD by CPCTC.

On Your Own

12. All of the figures except E appear to be kites.13. always14. always15. always16. sometimes17. always18. All three statement are always true. The correct

answer is D.19. Given kite ABCD with AB ∼= BC and CD ∼= DA. The

symmetry diagonal for this kite is BD.We can show that �DAB ∼= �DCB by SSS. This

means that � BDA ∼= � BDC and � DBA ∼= � DBC byCPCTC so that BD bisects both � ABC and � CDA.

20. As in the previous exercise, we have kite ABCD withAB ∼= BC and CD ∼= DA. From that exercise we alsoknow that � BDA ∼= � BDC and � DBA ∼= � DBC.

We can show that �ABE ∼= �CBE by SAS. (They shareBE.) So � BEA ∼= � BEC by CPCTC. We also know thatm� BEA+m � BEC = 180◦. This means thatm � BEA = m� BEC = 90◦ and AC is perpendicular toBD.


21. False. You could live in Toronto.22. true23. False. The figure could be a 2× 8 rectangle.24. true25. true

6.17 Parallelograms


1. • If a quadrilateral has both pairs of opposite sidescongruent, then it is a parallelogram.• This converse is true.• Draw a diagonal, and prove the two triangles

congruent by SSS. Then pick out congruent alternateinterior angles for each pair of opposite sides, and youcan show that the opposite sides are parallel.

2. • If pair of consecutive angles a quadrilateral has eachsupplementary, then it is a parallelogram.• This converse is true.• Given quadrilateral ABCD with � A supplementary to� D and � A supplementary to � B.

We can conclude that AB and CD are parallel and ADand BC are parallel. So the figure is a parallelogram.

3. • If a quadrilateral has both pairs of opposite anglescongruent, then it is a parallelogram.• This converse is true.• If we call the measure of one pair of congruent angles

x, and the measure of the other pair y, then 2x+ 2y

represents the sum of the four angles of thequadrilateral, so 2x+ 2y = 360◦, and x+ y = 180◦.But this means that pairs of consecutive angles foropposite sides are supplementary, so the figure is aparallelogram by Exercise 2.

4. • If both diagonals of a quadrilateral divide it into twocongruent triangles, then the figure is a parallelogram.• This converse is true.• Match up any congruent side of a pair of congruent

triangles. One finds that the opposite angles of thefigure are congruent. Then, by Exercise 3, the figuremust be a parallelogram.

5. • If the diagonals of a quadrilateral bisect each other,then the figure is a parallelogram.• This converse is true.• The diagonals divide the figure into four small

triangles. We can prove pairs of these congruent (bySAS). Then we know that opposite sides of thequadrilateral are congruent, and by Exercise 1 thefigure is a parallelogram.

6. Answers may vary. Sample: If a pair of consecutiveangles of a parallelogram are congruent, then theparallelogram is also a rectangle. To prove this, note thatconsecutive angles of a parallelogram are supplementary.Then each congruent angle must have measure 90◦. Theother two angles must then also have measure 90◦ since


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they are also consecutive with the first two anglesconsidered.

7. (a) Theorem 6.7 states that each diagonal of aparallelogram divides it into two congruent triangles.

So, using diagonal AC, we know that �ABC ∼=�CDA, and can conclude that AB ∼= CD and BC ∼=DA by CPCTC.

(b) Again, Theorem 6.7 and diagonal AC give us�ABC ∼= �CDA, so � B ∼= � D by CPCTC.Similarly, diagonal BD gives us �DAB ∼= �BCDand � A ∼= � C by CPCTC.

8. Start with a parallelogram ABCD. Call the intersection ofthe diagonals E. We can assume that a line through E

intersects DA and BC. (If it intersected the other twosides, we could just change the way we labeled theparallelogram and keep the proof.)

We know that AE ∼= CE because the diagonals of aparallelogram bisect each other. We know that � EAF ∼=� ECG because AD is parallel to CB. We also know that� AEF ∼= � CEG because they are vertical angles. Thismeans that�AEF ∼= �CEG by AAS, so we can concludethat EF ∼= EG.

Two small questions remain. First, what if the linedoesn’t intersect two opposite sides of the parallelogram,but rather two adjacent sides? Look at the situation wherea line intersects DA and AB. The only line like that alsointersects point E is the diagonal BD. Second, what if theline intersected vertices of the parallelogram? The resultwould still be true, because then the line would be one ofthe diagonals, and we already know that E is themidpoint of both diagonals.

On Your Own

9. Always10. Sometimes. A parallelogram could have four congruent

sides, so it could have as many as three congruent sides.But it doesn’t always.

11. Never. If it has three congruent sides, the fourth must becongruent to the one opposite it, so is congruent to theother three.

12. Sometimes. Such a parallelogram is a rhombus.13. Sometimes. Such a parallelogram is a rectangle.14. Sometimes15. Sometimes. If the other diagonal also creates two

congruent triangles, the figure must be a parallelogram.In a kite, which is not a parallelogram, one diagonalcreates congruent triangles, but the other does not.

16. Sometimes. In any trapezoid, you can find a pair ofsupplementary consecutive angles. You need at least onemore pair (and not just any pair) to show that the figure isa parallelogram.

17. Sometimes. For example, any rectangle is a quadrilateralwith one right angle which is a parallelogram. But thereare many quadrilaterals with one right angle which arenot parallelograms.

18. Never. You can show that if a parallelogram has one rightangle, then it must have four right angles.

19. Sometimes. A rectangle is an example of a such aquadrilateral which is a parallelogram. However, thereare quadrilaterals with two right angles which are notparallelograms.

20. Never. See Exercise 18.21. Never. See Exercise 20.22. Always. If a quadrilateral has three right angles, then it

must have four right angles, which means that it is arectangle. All rectangles are also parallelograms.

23. Always. See Exercise 5.24. Sometimes. Any parallelogram has this property, but

some kites also have this property.25. Sometimes. It could be a parallelogram, but only if the

pair are opposite each other, and the other pair of sides isalso congruent.

26. Sometimes. A rhombus is a parallelogram with three (infact, four) congruent sides. One can also construct anisosceles trapezoid with one base congruent to the twolegs, which is not a parallelogram.

27. Never. If three sides of a parallelogram were congruent,the fourth side would also have to be congruent to them.

28. Always. In fact, it’s a rhombus.29. A; a parallelogram that is not a rectangle or a rhombus

will not have any lines of symmetric.30. (a) One can prove, for example, that triangles �XMS

and �TPW are congruent, as are triangles �XWQand �STN. Many more statements can be deducedfrom these facts.

(b) Quadrilateral STWX is a parallelogram. This can beproven using the congruent triangles above to showthat opposite sides of the figure are congruent.

(c) Segments SW and TX are the diagonals of aparallelogram, and so they bisect each other. Notethat this would be very difficult to prove withoutlooking at quadrilateral STWX and seeing that it mustbe a parallelogram.

31. Suppose, in quadrilateral ABCD, AD and BC are bothcongruent and parallel.


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Drawing diagonal AC, we can prove �ABC ∼= �CDA(by SAS, using alternate interior angles of the parallellines). Then � BAC ∼= � DCA, which shows that segmentsAB and CD are parallel.

We have proven that both pairs of opposite sides ofquadrilateral ABCD are parallel, so ABCD is aparallelogram.

32. (a) Triangles �AMN and �CPN are congruent bySAS.

(b) From these congruent triangles, we see that � AMN ∼=� NPC. Then PC is parallel to AB because a pair ofalternate interior angles are congruent.

Also, AM ∼= CP. Since M is the midpoint of ABwe know that AM ∼= MB. From this, we canconclude that MB ∼= CP.

This means that quadrilateral MPCB is aparallelogram, because one pair of sides is bothcongruent and parallel. See Theorem 6.10.

(c) From parallelogram MPCB we conclude thatMP = BC. Since MN ∼= NP, this means thatMN = 1

2 BC.(d) We also know that MN is parallel to BC, since MPCB

is a parallelogram.


33. (a) If a person lives in New York City, then that personlives in New York State.

(b) True; New York City is contained in New York State.(c) If a person lives in New York State, then that person

lives in New York City.(d) False; Many parts of New York State are not in New

York City.34. (a) If an animal is a lizard, then it is a reptile.

(b) True; All lizards are reptiles.(c) If an animal is a reptile, then it is a lizard.(d) False; It might be a snake.

35. (a) If an animal is a bird, then it has feathers.(b) True; Unless a particular bird has had a terrible

accident, all birds have feathers.(c) If an animal has feathers, then it is a bird.(d) True; All feathered animals are birds.

36. (a) If a person is a cowboy, then that person comes fromTexas.

(b) False; There are many cowboys who come fromplaces other than Texas.

(c) If a person comes from Texas, then that person is acowboy.

(d) False; There are many people who come from Texaswho have made career choices other than that ofcowboy.

37. (a) If something is a piece of fruit, then it is an apple.(b) False; It might be a banana.(c) If something is an apple, then it is a piece of fruit.(d) True; All apples are fruit.

6.18 Classifying Quadrilaterals


1. The figure below shows a rectangle with diagonalsAC, BD drawn in. Triangles �ABD,�ABC arecongruent (by SAS). So AC ∼= BD.

2. If two adjacent sides of a parallelogram are congruent,the sides opposite them must also be congruent. Thatmeans all four sides are congruent and the parallelogramin question is also a rhombus.

3. A rhombus is sometimes a rectangle. A rhombus is arectangle only when it is a square.

4. A square is always a parallelogram.5. A rectangle is sometimes a rhombus. A rectangle is a

rhombus only when it is a square.6. A square is always a rectangle.7. The opposite sides of a rectangle are always congruent,

since a rectangle is a type of parallelogram.8. The diagonals of a rectangle are always congruent (see

Exercise 1).9. If the diagonals of a quadrilateral are congruent, then the

figure is sometimes a rectangle. Diagonals of a rectangleare congruent, but the diagonals of an isosceles trapezoidare also congruent.

10. Two adjacent sides of a rectangle are sometimescongruent. They will only be congruent if the rectangle isalso a square.

11. The diagonals of a rectangle sometimes bisect its vertexangles. They will only bisect the vertex angles if therectangle is also a square.

12. A rhombus sometimes has a right angle. It can only have aright angle if it has four right angles and is also a square.

13. Opposite sides of a rhombus are always congruent. Infact, the definition of a rhombus requires all four sides tobe congruent.

On Your Own

14. A square is a rhombus with four congruent angles. Thecorrect answer is A.

15. The opposite angles of a parallelogram are alwayscongruent, so the angle opposite the right angle must also


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be right. Consecutive angles in a parallelogram arealways supplementary, so the other two angles must alsobe right angles. This means that all four angles in thefigure are right angles, so it is a rectangle.

16. The figure below shows a parallelogram with diagonalsAC ∼= BD.

�ABD ∼= �ABC by SSS. This means that � DAB ∼=� CBA. Since these angles are also supplementary(consecutive angles of any parallelogram aresupplementary), they must both be right angles. SoABCD is a rectangle.

17. In the figure, M, N, P, Q are midpoints of the sides ofrectangle ABCD.

Q

P

N

MA

D C

B

�AMQ ∼= �MBN ∼= �PCN ∼= �PDQ by SAS, soQM ∼= MN ∼= NP ∼= PQ by CPCTC. This means thatMNPQ is a rhombus.

18. A rhombus is defined as a parallelogram with fourcongruent sides. We need to show that if a quadrilateralhas four congruent sides it must be a parallelogram. Sincethe quadrilateral has all four sides congruent, certainlyits opposite sides are congruent. This means that it mustbe a parallelogram. Now we know that our figure is aparallelogram with four congruent sides, so it is arhombus.

19. The triangles are isosceles because adjacent sides of arhombus are congruent. Note that the isosceles trianglesare also congruent because the diagonals of aparallelogram always divide the figure into twocongruent triangles.

20. In the diagram, triangles �FED and �FCD arecongruent because the diagonals of a parallelogramalways divide the figure into two congruenttriangles.

43

21

F E

DC

Because CDEF is a rhombus, the triangles are alsoisosceles. This means that � 1 ∼= � 2 ∼= � 3 ∼= � 4, so thevertex angles are bisected by the diagonals.

21. In rhombus ABCD, the diagonals AC and BD intersect atpoint E. In exercise rhomb4, we showed that thediagonals of a rhombus bisect the vertex angles. We alsoknow that opposite angles are congruent in the rhombusbecause it is a parallelogram. So we can mark thecongruent angles as in the diagram:

This means that �ABE ∼= �CBE ∼= �CDE ∼=�ADE by ASA. So we know that � AEB ∼= � CEB ∼=� CED ∼= � AED. Since any pair of these angles that areconsecutive are supplementary, we know that all four ofthe angles are right angles, and the diagonals areperpendicular.

22. We begin with a parallelogram ABCD withperpendicular diagonals AC and BD intersecting at pointE. We know that in a parallelogram, the diagonals bisecteach other, so we can mark the congruent segments in thediagram:

This means that �ABE ∼= �CBE ∼= �CDE ∼= �ADE bySAS. So we know that AB ∼= CB ∼= CD ∼= AD byCPCTC. Since the four sides of the parallelogram arecongruent, it is a rhombus.

23. Suppose that for parallelogram CDEF we know that� 1 ∼= � 2.

43

21

F E

DC

Since CF and DE are parallel, � 2 ∼= � 3. Then � 1 ∼= � 3,so triangle �FED is isosceles, and parallelogram CDEFhas two congruent adjacent sides. So it must be arhombus.


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24. See the figure below in which square ABCD is given,with midpoints M, N, P and Q.

AB

CD

M

N

P

Q

In triangle BCD, PN is equal to half of BD, which is adiagonal of the square. Using other triangles, we find thatNM , MQ and QP are all half one of the diagonals of thesquare, so the sides of MNPQ are all congruent, and it isa rhombus.

To show that MNPQ is a square, we note that PN isparallel to BD, and PQ is parallel to AC. Since AC andBD are perpendicular, consideration of alternate interior(or consecutive angles) shows that PQ and PN are alsoperpendicular. That fact, together with the fact thatMNPQ is a rhombus, tells us that MNPQ is a square.


25.

rhombus rectanglesquare

(a) (c)

parallel-ogram

rhombus kite

(b)

rectangle kitesquare

6D MATHEMATICAL REFLECTIONS

1. No, here is a counter-example:

No pairs of adjacent sides are congruent.


P T Q

S R

60

60 60 60

At point T , three angles come together with a sum of180◦. m � STP+m � STR+m � RTQ = 180◦. Sincem � STP = m� RTQ = 60◦, you know that m � STR =60◦. Now, �RST is isosceles with base RS, so its baseangles must be equal, and together with m � STR theirsum is 180◦. This means that m � RST = m� SRT = 60◦.� SRT and � STP are congruent alternate interior angles,so SR is parallel to PQ and PQRS it has one pair ofparallel sides.

You can conclude that PS is parallel to TR because� SPT and � RTQ are congruent corresponding angles.Since there is exactly one parallel to PS through R, QR isnot parallel to PS and PQRS has exactly one pair ofparallel sides.

3. A rectangle is always a parallelogram. It has two pairs ofparallel sides. However, a parallelogram is not always arectangle. It must have two pairs of parallel sides, but itneedn’t have adjacent sides perpendicular.

4. Yes it can. To be a kite, a quadrilateral must have a pair ofadjacent sides congruent, and the other pair must also becongruent. If both pairs of adjacent sides are congruent toeach other, the kite is a rhombus, and a rhombus is aparallelogram.

5. A kite must have a pair of adjacent congruent sides,and the other pair of sides must also be congruent. If thediagonal you draw has one of these pairs onone side and the other pair on the other side, it hasformed two isosceles triangles. Here is a figure thatshows this:

This can only be done in two ways if the kite is also arhombus. Otherwise, any side of the kite is congruent toonly one other side and only one diagonal will form twoisosceles triangles.


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6. Parallelograms have opposite sides parallel andcongruent, opposite angles congruent, and diagonals thatbisect each other. Kites have two pairs of adjacent sidesthat are congruent, its diagaonals are perpendicular, andat least one of its diagonals is a line of symmetry for thefigure and bisects the vertex angles at its endpoints.A trapezoid has a pair of parallel bases. Isoscelestrapezoids have other properties that are interesting, butnot all trapezoids are isosceles.

7. All squares are also considered parallelograms becausethey all have opposite sides parallel. Not allparallelograms are squares because many have no rightangles. There are also many parallelograms which do nothave four congruent sides.

8. No. For example, the statement, “If a figure is a square,then it is also a parallelogram,” is true, but its converse,“If a figure is a parallelogram, then it is also a square,” isfalse.

CHAPTER REVIEW

1. A perpendicular bisector for a segment is a lineperpendicular to the segment and that passesthrough its midpoint. All the points on the perpendicularbisector are equidistant from the endpoints of thesegment.

To see why the perpendicular bisectors of a triangle areconcurrent, start with any �ABC. Construct theperpendicular bisectors of sides AB and AC. Theseperpendicular bisectors must intersect because thesegments they are perpendicular to are not parallel. Callthe intersection of the two perpendicular bisectors D.D is equidistant from A and B because it is on theperpendicular bisector of AB. It is also equidistant fromA and C because it is on the perpendicular bisector ofAC. This means that D is equidistant from B and C, so itis on the perpendicular bisector of BC, and the threeperpendicular bisectors are concurrent at D.

2. (a)areaAEFD

areaABCDis invariant and equal to 1

2 . Because ABCD

is a rectangle, AEFD is also a rectangle. Thedimensions of ABCD are AB by BC. The dimensionsof AEFD are AE = 1

2 AB and EF = BC. The area ofAEFD is always half the area of ABCD no matterhow the original rectangle is stretched.

(b) The ratioareahexagon

perimeterhexagonis not invariant

but depends on the side length of the hexagon. Asthat length changes, so does the ratio. The area of oneof the triangles shown in the picture is

√3

4 r2 with r

representing the side length of the hexagon or theradius of the circle the hexagon is inscribed in. Thatmakes the area of the hexagon 3

√3

2 r2. The perimeterof the hexagon is 6r. That means the ratio has a valueof√

34 r and will change as r changes.


By construction, AC ∼= BD. � CEA ∼= � DEB because they

are vertical angles. Because←→AC is parallel to

←→BD,

� ACE ∼= � DBE because they are alternate interiorangles. This means that �ACE ∼= �DBE by AAS. So,AE ∼= DE by CPCTC, and E is the midpoint of AB.

4. 360◦. You can see this by drawing a diagonal of thequadrilateral that cuts the quadrilateral into two triangles.The angle sum of the quadrilateral is the same as the sumof the angle sums of the two triangles.

5. Angles 1 and 2 are congruent because they’re verticalangles. Similarly, angles 5 and 8 are congruent. Thismeans that angles 2 and 8 are congruent, because they’recongruent to congruent angles. Since these are alternateinterior angles, you know that � is parallel to m.


Both �RAB and �SAB are isosceles with base AB. Thebase angles of an isosceles triangle are congruent, so� RAB ∼= � RBA and � SAB ∼= � SBA. You can alsoconclude that � RAS ∼= � RBS because these angles areequal to the difference of two congruent angles. Fromthis, you can conclude that �RAS ∼= �RBS by SAS.

By CPCTC, � RSA ∼= � RSB. This means that←→RS is an

angle bisector for � ASB. The angle bisector of an

isosceles triangle is also an altitude and a median. So←→RS

is perpendicular to AB.7. (a) This triangle must be isosceles but not

equilateral. The angle bisector of the apex angle(between the two congruent sides) will be theperpendicular bisector of the base. This samerelationship will not hold for the other angles,because they are not between two congruent sides ofthe triangle.


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(b) If a triangle has three altitudes that coincide with theperpendicular bisectors of its sides it must beequilateral. In every case, if an altitude is also amedian, that implies that it goes through a vertexwhere two congruent sides meet. At every vertex ofan equilateral triangle the two adjacent sides arecongruent.

8. Consider �ATC and �DTB. AC ∼= DB and � CAT ∼=� BAT because they are corresponding parts of congruenttriangles �ABC and �DCB. You also know that � ATC ∼=� DTB because they are vertical angles. So,�ATC ∼= �DTB. This means that CT ∼= TB by CPCTC,and �CTB is isosceles.

9. “If the diagonals of a quadrilateral are perpendicular, thenit is a kite.” This statement is false, and here is acounter-example:

No pairs of adjacent sides are congruent.10. (a) A kite sometimes has four congruent sides. (A kite is

only required to have two distinct pairs of congruentadjacent sides, but it is possible for a kite to be arhombus with four congruent sides.)

(b) The diagonals of a parallelogram always bisect eachother. (This result was proven in the text.)

(c) The diagonals of a rectangle are sometimesperpendicular. (The diagonals of a rectangle will onlybe perpendicular to each other if the rectangle is alsoa square.)

(d) A kite never has exactly one pair of parallel sides. (Ifa kite has one pair of parallel sides, it must be arhombus and have two pairs of parallel sides.)

CHAPTER TEST

1. � 3 ∼= � 2 by the Vertical Angle Theorem, and � 2 ∼= � 7 byPAI. Therefore, � 3 ∼= � 7 by the transitivity of thecongruence relation. The correct answer is C.

2. The diagonals of parallelograms always bisect each other,but they need not be perpendicular. The same is true forrectangles. The diagonals of rhombuses are alwaysperpendicular. The diagonals of a square are alsoperpendicular, but this alone does not uniquely determinea square. The answer is B.

3. Choices A and B divide a segment into only twocongruent segments. Option C constructs four segmentsequal to the original segment. Option D is the only one

that divides the original segment into four congruentsegments.

4. The correct answer is A. Except for the vertices of thetriangles, medians of a triangle always lie entirely withinthe triangle. Furthermore, the point of concurrencecannot be at a vertex since there is only one median foreach vertex.

5. Sample: mn = 24

m n

2 1224 16 4−3 −848 0.5

6. The ratioareaMNP

areaABCis invariant, and it’s equal to 1

4 . One

way to see this is to notice that the sides of �MNP areeach parallel to and half the length of a side of �ABC byrepeated applications of Conjecture 6.1. This means that�MNP is similar to �ABC with a ratio of 1

2 . Since thearea of a triangle is equal to half the product of its baseand height, and the base and height of �MNP are bothhalf of the corresponding lengths in �ABC, the areashave the ratio 1

4 .7. (a) The triangles are congruent.

Since BC ‖ AD, then � BCA ∼= � DAC, and sinceAB ‖ DC, then � BAC ∼= � DCA, both because of thePAI Theorem. AC is congruent to itself, so thetriangles are congruent because of ASA.

(b) They are not congruent. The two triangles have threepairs of congruent angles since angle A is congruentto itself, but AAA is not a valid congruence test.

(c) The triangles are congruent. Since the larger triangleis equilateral, the midpoints of each side divide thesides into equal lengths, so AE ∼= CF . AC iscongruent to itself. And since the triangle isequilateral, it is also equiangular, so � BAC ∼= � BCA.So, �EAC ∼= �FCA by SAS.

8. (a) A geometry software experiment or construction byhand and measure should provide evidence, but notproof, that the altitudes of an equilateral triangle areall congruent.

(b) To prove that the altitudes of an equilateral triangleare all congruent you must show that three differenttriangles are congruent. Start by drawing threepictures of an equilateral triangle, with a differentaltitude drawn in each time.

A

P

Q

R

B

C

A

B

C

A

B

C

It can be proved that triangle APC is congruent totriangle BQA using one of the tests in this chapter.


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Since triangle ABC is an equilateral (AB = BC =AC), you could use the HL test for right triangles.Since equilateral triangles are equiangular, anglesABC, BAC, and ACB are all congruent. So, you canuse AAS in the proof. Since the altitude of anequilateral triangle is also a median, you could provethem congruent by SSS.

Similarly, you can prove triangle BQA is congruentto triangle CRB using any of the same trianglecongruence tests.

Since the three triangles are congruent, theircorresponding sides must be congruent, in particular,CP ∼= QA ∼= RB. These are the altitudes of thetriangle.

9. Number the angles as shown below. � 1 ∼= � 3 and� 2 ∼= � 4 because they are vertical angles. � 3 ∼= � 4 due tothe PAI theorem. � 1 ∼= � 2 because of transitivity.

m

n

1

2

3

10. Counterexamples will vary.(a) False(b) False(c) True(d) False(e) True(f) True(g) False

11. (a) The diagonals of a kite sometimes bisect each other.(This will happen only if the kite is also a rhombus.)

(b) The diagonals of a parallelogram always bisect eachother. (You can prove this by showing that anopposite pair of the four triangles formed by thediagonals of a parallelogram are congruent. Usealternate interior angles for each set of parallel linesand the fact that opposite sides of a parallelogram arecongruent to get ASA congruence. Then by CPCTCthe diagonals have bisected each other.)

(c) A median of a triangle is sometimes be an altitude ofthe triangle. (This will only happen if the median isdrawn to the base of an isosceles triangle.)

(d) The diagonals of a rhombus are alwaysperpendicular. (You can prove this using the previousresult for parallelograms to show that the diagonalsbisect each other. This means that one diagonal is themedian for the two triangles formed by the otherdiagonal. Since these triangles are isosceles, this

median is also the perpendicular bisector for theother diagonal.)

12. “If a triangle is equilateral, then the altitude to a side ofthe triangle is the perpendicular bisector of that side.”

The original statement is not true, because in anisosceles triangle the height relative to the base is theperpendicular bisector of that side, but the triangle neednot be equilateral.

However, the converse is true, and here is a proof:Choose any vertex of an equilateral triangle and draw

the altitude from that vertex to the opposite side. Thiscuts the equilateral triangle into two congruent righttriangles. They’re congruent because they have two pairsof congruent angles (the right angles and the 60◦ angles)and a pair of congruent sides (the original sides of theequilateral triangle). Because the two right triangles arecongruent, side cut by the altitude has been bisected andthe median is a perpendicular bisector.

Because you chose any vertex of the triangle, thisstatement is true for all three vertices. Therefore, if atriangle is equilateral, then the altitude to a side of thetriangle is the perpendicular bisector of that side.


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Chapter Congruence and Proof 6 pages 192–226 000200010271723961_CH06_p192-226” 2013/2/22 13:6 page 192 Chapter 6.....Congruence and Proof pages 192–226