chapter 6 answers - bisd moodlemoodle.bisd303.org/file.php/18/solutions/chapter 06 answers.pdf ·...

Chapter �� Lesson � ��

Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �Chapter �� Lesson �

Set I Set I Set I Set I Set I (pages ��–��)

The exercises on folding a paper chain suggestsome possible exploration on the student’s part�Cutting and manipulating a paper chain of thetype illustrated reveals examples of how the stripcould have been folded any number of times fromthree (a “binary” folding approach) to seven (an“accordion” folding approach)� In general� �n

penguins require at least n folds but could beproduced by as many as �n – � folds� Even exploring the physical limit on the possible value of nfor the binaryfolding approach yields an answerthat surprises most students�

Rope Trick.

1.

7.

8.

9.

10.

Baseball Diamond.

11.2. Two points each equidistant from the

endpoints of a line segment determine theperpendicular bisector of the line segment.

3. Yes.

•4. Yes, because W and E are equidistant fromthe endpoints of NS.

Lines of Symmetry.

5.

6.

•12. 6 cm.

•13. 90 ft. (6 ⋅ 15 = 90.)

14. 127.5 ft. (8.5 ⋅ 15 = 127.5.)

•15. In the west.

16. So that the sun doesn’t get in the batter’seyes.

17. Because, as he faces the batter, his left arm ison the south.

Paper Chain.

18. (Student answer.) (Any number from three toseven is correct.)

•19. Seven.

� Chapter �� Lesson �

20. Four times. (24 = 16. Unlike exercise 18, thisexercise asks for the minimum rather thanfor simply the possible number of folds.)

Stealth Bomber.

•21. Line l is the perpendicular bisector of ST.

•22. If a line segment is bisected, it is dividedinto two equal segments.

23. Perpendicular lines form right angles.

24. All right angles are equal.

25. Reflexive.

26. SAS.

27. Corresponding parts of congruent trianglesare equal.

Set IISet IISet IISet IISet II (pages ��–��)

The “clairvoyance” test problem is anotherversion of the surfer’s puzzle but with a regularpentagon instead of an equilateral triangle� Anexplanation of why the sum of the five distancesis independent of the choice of the point is basedon area� For this figure� the “exact” sum is

� or approximately �� inches� We will

return to the surfer’s puzzle later�

In his book titled Inversions (Byte Books��)� Scott Kim wrote of his amazing PROBLEM/SOLUTION figure: “Reflecting on this PROBLEMyields a surprising SOLUTION� Notice that reflection about a diagonal axis preserves a pen angleof �� degrees� so that calligraphic conventions aremaintained�”

The figure drawn by Kepler is the horoscopethat he cast for Albrecht von Wallenstein� one ofthe most famous military commanders of histime� He was Emperor Ferdinand II’s majorgeneral during Europe’s Thirty Year War� CharlesBlitzer in his book titled Age of Kings (TimeLifeBooks� ��) wrote that� although hundreds ofbooks have been written about Wallenstein� “themost illuminating analysis of this complex mancomes from a horoscope cast for him [on his ��thbirthday] by the great ��th century astronomer�Johannes Kepler�”

Race to the Fence.

28.

29. It would have been the point midwaybetween them.

Clairvoyance Test.

30. Example answer:

•31. 9.5 in.

32. Because the answer is predictable; it doesn’tdepend on the choice of the point. The“clairvoyant” can predict the subject’sanswer.

Linkage Problem.

33. Because points A and C are equidistant fromthe endpoints of BD (two points eachequidistant from the endpoints of a linesegment determine the perpendicularbisector of the line segment.)


•34. C and E; D and F.

35.

points are the vertices of a square� and the fourouter points are the vertices of four equilateraltriangles erected on the sides of the square�)Although there are � possible pairings of theeight points� the symmetries of the figure aresuch that only six of them are distinct� These sixpossibilities are illustrated in the figures below�

•36. Because they are corresponding parts ofcongruent triangles (∆ABC ≅ ∆ADC by SSS).

37. Yes. ∠B = ∠D (exercise 36) and ∠F = ∠Dbecause ∆EFC ≅ ∆EDC; so ∠B = ∠F bysubstitution.

One Problem, Two Solutions.

38. At a 45° angle from the lower left to theupper right.

•39. N.

40. T.

41. The word SOLUTION is readable on bothsides of the symmetry line.

Kepler’s Diagram.

42. (There are various ways to construct this figure.)

•43. Three.

44. Isosceles right triangles.

45. 16.

Set IIISet IIISet IIISet IIISet III (page ��)

In the reference cited in the footnote� MartinGardner reports that Hallard Croft� a mathematician at Cambridge University� “asked if thereexisted a finite set of points on the plane suchthat the perpendicular bisector of the line segment joining any two points would always passthrough at least two other points of the set�” Thefigure in the Set III exercise is the solution usingonly eight points discovered by Leroy Kelly ofMichigan State University� (The four central

An Unusual Set of Points.

Every perpendicular bisector passes through twoother points of the figure.



The figure for exercises �� through �� illustrates atheorem of projective geometry� The theoremstates that� if the vertices of a [selfintersecting]hexagon lie alternately on two lines and if twopairs of opposite sides are respectively parallel�then the third pair of opposite sides also areparallel� If the two lines are parallel� this theorem


is easily proved� In contrast� the fact that it is trueeven if the lines are not parallel seems surprisingfrom the perspective of Euclidean geometry� andhow it might be proved is not obvious�

•1. (1) ∠1 = ∠2.(2) ∠2 = ∠3.(3) ∠1 = ∠3.(4) a || b.

2. (1) ∠1 and ∠2 are supplementary.(2) ∠2 and ∠3 are supplementary.(3) ∠1 = ∠3.(4) a || b.

3. (1) a ⊥ c and b ⊥ c.(2) ∠1 and ∠2 are right angles.(3) ∠1 = ∠2.(4) a || b.

Folded Paper.

•4. Alternate interior angles.

5. Equal alternate interior angles mean thatlines are parallel.

•6. In a plane, two lines perpendicular to athird line are parallel.

Snake Track.

•7. Corresponding angles.

8. They are parallel.

•9. Equal corresponding angles mean that linesare parallel.

10. Interior angles on the same sides of thetransversal.

11. They are supplementary.


13. Supplementary interior angles on the sameside of a transversal mean that lines areparallel.

I-Beam.

•14. No.

15. Yes.

16. Yes.

Parallel Lines.

17.

•18. AE || BF.

19. Equal corresponding angles mean that linesare parallel.

20. DB || EC(PC).


22. AD || CF.


Set II Set II Set II Set II Set II (pages ��–��)

Drafting Triangles.

24. ∆ABC ≅ ∆FED because the drafting trianglesare identical; so ∠A = ∠F. AB || EF becauseequal alternate interior angles mean thatlines are parallel.

25. ∠BCA = ∠EDF; so BC || DE because equalalternate interior angles mean that lines areparallel.

Optical Illusion.

26.

•27. No. These angles are not formed by atransversal. They could be equal and yetlines x and y could intersect either to the leftor to the right.

28. Yes. If BD = BC, ∠2 = ∠BDC (if two sides ofa triangle are equal, the angles opposite themare equal). Because ∠1 = ∠2, ∠1 = ∠BDC(substitution). So x || y (equal alternateinterior angles mean that lines are parallel).


Different Proofs.

29. (2) Two lines are perpendicular if they forma right angle.

(3) In a plane, two lines perpendicular to athird line are parallel.

30. (2) All right angles are equal.(3) Equal alternate interior angles mean

that lines are parallel.

Construction Exercise.

31.

Set III Set III Set III Set III Set III (page ��)

The crossbar figures are included in a sectiontitled “The Importance of Rules” in the chapter onhuman perception in Lindsay and Norman’sHuman Information Processing� They write:

“When a rectilinear object is viewed straight on�the intersecting lines formed by its contours formright angles� As the object is tilted and rotated inspace� the angles in the retinal image divergefrom right angles� The degree of divergencedepends on the orientation of the object in space�By interpreting that divergence from right anglesas a result of depth� distance information can beextracted�”

What Do You See?

1. (Student answer.) (The top bar.)

2. (Student answer.) (The bottom bar.)

3. The first figure is viewed “in two dimensions,”seen as lying in the plane of the paper. Thesecond figure is viewed “in three dimensions,”seen as in perspective on the side of a box.

4. Yes. A third possibility is that the two crossbars do not lie in a common plane, in whichcase both could be perpendicular to thevertical bar.

5. It doesn’t follow that they are parallel. Ourtheorem says that in a plane, two linesperpendicular to a third line are parallel.



Euclid’s assumption that “if two lines forminterior angles on the same side of a transversalwhose sum is less than � °� then the two linesmeet on that side of the transversal” is his fifthpostulate (his version of the Parallel Postulate)�

In his commentary on Euclid’s Elements� SirThomas Heath wrote: “When we consider thecountless successive attempts made through morethan twenty centuries to prove the Postulate�many of them by geometers of ability� we cannotbut admire the genius of the man who concludedthat such a hypothesis� which he found necessaryto the validity of his whole system of geometry�was really indemonstrable�

From the very beginning� as we know fromProclus� the Postulate was attacked as such� andattempts were made to prove it as a theorem or

•32. Four.

33. ASA (∠ECF = ∠GCF, CF = CF,∠CFE = ∠CFG.)


•35. SAS. [CF = FD and CF ⊥ FD (construction),EF = FG (exercise 34).]



•38. No.

39. Proof.(1) AE = AD and ∠E = ∠BCE. (Given.)(2) ∠E = ∠ADE. (If two sides of a triangle

are equal, the angles opposite them areequal.)

(3) ∠ADE = ∠BCE. (Substitution.)(4) AD || BC. (Equal corresponding angles

mean that lines are parallel.)

40. Proof.(1) AB = CD and AD = BC. (Given.)(2) BD = BD. (Reflexive.)(3) ∆ABD ≅ ∆CDB. (SSS).(4) ∠ABD = ∠CDB. (Corresponding parts

of congruent triangles are equal.)(5) AB || DC. (Equal alternate interior

angles mean that lines are parallel.)

�� Chapter �� Lesson �

to get rid of it by adopting some other definitionof parallels; while in modern times the literatureof the subject is enormous�”

An entire chapter of David Henderson’svaluable book titled Experiencing Geometry(Prentice Hall� � �) is on the subject of parallelpostulates� In his historical notes� Hendersonremarks:

“Playfair’s Parallel Postulate got its currentname from the Scottish mathematician� JohnPlayfair (��–��)� who brought out successfuleditions of Euclid’s Elements in the years following�� After � many commentators referred toPlayfair’s [parallel] postulate (PPP) as the beststatement of Euclid’s [fifth] postulate (EFP)� so itbecame a tradition in many geometry books touse PPP instead of EFP�”

Henderson goes on to explain that in absolutegeometry� the part of geometry that can bedeveloped without using a parallel postulate�Playfair’s postulate is equivalent to Euclid’s� Hethen points out:

“Because absolute geometry was the focusof many investigations the statement ‘PPP isequivalent to EFP’ was made and is still beingrepeated in many textbooks and expositorywritings about geometry even when the contextis not absolute geometry � � � � EFP is true onspheres and so cannot be equivalent to PPP�which is clearly false�”

Optical Illusion.

•1. 12.

•2. If they form a right angle (or if they formfour equal angles).

3. If they lie in the same plane and do notintersect.

4. (Student answer.) (Possibly white circles orsquares at the “breaks” in the grid. Alsobroad white diagonal strips.)

Windshield Wipers.

•5. ∠3.


7. Interior angles on the same side of thetransversal.


9. ∠2 and ∠4.

10. Yes. ∠1 and ∠2 (or ∠3 and ∠4) are a linearpair; so they are supplementary. If ∠1 and∠4 are supplementary, then ∠2 = ∠4 (or∠1 = ∠3) because supplements of the sameangle are equal. It follows that the wipersare parallel because they form equalcorresponding angles with the transversal.

Exactly One.

11. That a cat has at least one flea.

•12. At least, no more than, exactly.

13. At least, no more than, exactly.

•14. No more than.

15. No more than.

16. At least.



Parallelogram.

19.

•20. They form equal corresponding angles witha transversal.

21. They seem to be equal.

•22. They seem to be supplementary.

23. They seem to be equal.

24. No. It can’t be folded so that the two halvescoincide.

Euclid’s Assumption.

•25. All four of them.


26. 41. No. (It will be possible to prove thesetriangle congruent by methods learned infuture lessons, however.)

Theorem 18.

•42. a and b are not parallel.

•43. Given.

44. Through a point not on a line, there isexactly one line parallel to the line.

45. a || b.

Proclus’s Claim.

46. Yes.

47. If c doesn’t intersect b, then c is parallel to b.This would mean that, through P, lines c anda are both parallel to b, which contradictsthe Parallel Postulate.


The “dragon curve” was discovered by NASAphysicist John Heighway� Martin Gardnerexplains in the chapter titled “The Dragon Curveand Other Problems” in his book titled MathematicalMagic Show (Knopf� ��) that the curve gets itsname from the fact that� when it is drawn witheach right angle rounded� it “vaguely resembles asea dragon paddling to the right with clawed feet�his curved snout and coiled tail just above animaginary waterline�”

The curve can be generated in various waysincluding paper folding� by geometric construction�and from a sequence of binary digits� The versionin the text is called an order� dragon because itis produced by four folds� The order� dragonshown below with its right angles rounded makesthe origin of the name more evident�

•27. Down, below the figure.

28. To the left.


The restriction to the plane in Theorem � is� ofcourse� unnecessary� because the theorem is alsotrue in space� (The proof in solid geometry is basedon theorems on perpendicular lines and planes�)

Gateway.

29.

•30. One.

31. One.



34. They seem to be parallel.

35. One.

•36. Through a point not on a line, there isexactly one line parallel to the line.

•37. Yes. (SAS; AC = BD, ∠CAB = ∠DBA,AB = AB.)

38. Yes. (Corresponding parts of congruenttriangles are equal.)

39. Yes. (SSS; AC = BD, AD = BC, CD = CD.)

40. Yes. (Corresponding parts of congruenttriangles are equal.)


The length of the strip of paper suggested forfolding the dragon was chosen so that� if a studentuses graph paper of � units per inch� the unfoldeddragon will fit nicely on the graph� Each time thepaper is unfolded� the half produced is a � °rotation� The original fold point has been placedat the origin so that someone studying the graphmay discover the connection between the � °clockwise turn and the coordinate transformation:(a� b) → (b� –a)�

A Dragon Curve.

1. (The folded paper should look like thedragon curve shown in the text.)


Set ISet ISet ISet ISet I (pages ��–��)

2.

3. It looks like the edge of the folded strip.

4. Point I.

5. Points E and M.

6. It has the same shape but is rotated 90°clockwise.

7. For each point with coordinates (a, b) in thefirst half, there is a corresponding pointwith coordinates (b, –a) in the second half.

The fishpole puzzle is based on an illusion createdby J� C� Poggendorff in �� The original illusionis illustrated above� Although the oblique segmentsare aligned� the upper one appears raised inrelation to the lower one� The Poggendorffillusion and the theories developed to explain itare discussed in depth in Perception� by Irvin Rock(Scientific American Library� ��)�

Which Fishpole?

•1. They are equal.

•2. Parallel lines form equal correspondingangles.

3. Fishpole A.

4. It doesn’t look like fishpole A.

Corollary Proofs.

•5. (1) Given.(2) Parallel lines form equal corresponding

angles.(3) Vertical angles are equal.(4) Substitution.

6. (1) Given.(2) Parallel lines form equal corresponding

angles.(3) The angles in a linear pair are

supplementary.(4) The sum of two supplementary angles

is 180°.(5) Substitution.(6) Two angles are supplementary if their

sum is 180°.


7. (1) Given.(2) Perpendicular lines form right angles.(3) A right angle is 90°.(4) Given.(5) Parallel lines form equal corresponding

angles.(6) Substitution.(7) A 90° angle is a right angle.(8) Lines that form a right angle are

perpendicular.

Bent Pencil.

8. Parallel lines form equal correspondingangles.

•9. Betweenness of Rays Theorem.

10. Substitution.

11. Subtraction.


Find the Angles.

•13. ∠1 = 115°, ∠2 = 65°, ∠3 = 115°.

14. ∠1 = 90°, ∠2 = 90°, ∠3 = 90°.

15. ∠1 = 70°, ∠2 = 55°.

16. ∠1 = 110°, ∠2 = 70°, ∠3 = 110°.

City Streets.

•17. They must be perpendicular because, in aplane, a line perpendicular to one of twoparallel lines is also perpendicular to theother.

18. They must be parallel because, in a plane,two lines perpendicular to a third line areparallel.

19. 113°, 67°, and 113°.

•20. 67°, 113°, 67°, and 113°.


T Puzzle.

•22. ∠1 and ∠2 are supplementary becauseparallel lines form supplementary interiorangles on the same side of a transversal.

23. ∠2 = ∠3 because parallel lines form equalcorresponding angles.

•24. ∠3 = ∠4 because parallel lines form equalalternate interior angles.

25. ∠2 = ∠4 by substitution.

26. ∠4 and ∠5 are complementary because theytogether with the right angle add up to 180°.

27. ∠5 and ∠6 are supplementary becauseparallel lines form supplementary interiorangles on the same side of a transversal.


In his book titled The Flying Circus of Physics(Wiley� ��)� Jearl Walker reports: “Lookingdown on desert sand dunes from a high altitudeairplane� one sees curious long� narrow dune beltsrunning across the desert� roughly from north tosouth� in almost straight lines� as if one wereviewing welldesigned parallel streets� The dunebelts are characteristic of every major desert inthe world�” Walker explains that the belts arecaused by air currents and that the dominantwinds in every desert of the world are north orsouth�

Parallel Construction.

28.


•30. Through a point not on a line, there isexactly one line parallel to the line (theParallel Postulate).

•31. Parallel lines form equal alternate interiorangles.

32. 180°.

33. 180°.

Overlapping Angles.

34. Yes. ∠B = ∠DGC and ∠DGC = ∠E becauseparallel lines form equal correspondingangles. So ∠B = ∠E by substitution.


35. No. If one angle were turned with respect tothe other, their sides would obviously notbe parallel.

Sand Dune Lines.

36.

50. Because a || b, ∠PBA = ∠2 (parallel linesform equal alternate interior angles). Also,PA = PB, and so ∠1 = ∠PBA (angles oppositeequal sides in a triangle are equal). So∠1 = ∠2 by substitution.


The context of Kepler’s method for determiningthe shape of Earth’s orbit is set forth by DavidLayzer in Constructing the Universe (ScientificAmerican Library� ��)� The parallel lines arein the direction of an arbitrary star� The anglenumbered � is measured when Mars is “atopposition�” that is� when Earth is in alignmentbetween Mars and the sun (MES)� The point E’is the position of Earth � Martian year later�Layzer explains: “By means of observations ofMars and the sun at intervals separated by oneMartian year� Kepler calculated the shape of theearth’s orbit and its dimension relative to thereference distance SM�”

Plotting successive positions of Earth in thisway led to Kepler’s realization that the orbits ofthe planets are elliptical rather than circular andto his three laws of planetary motion to describethis motion�

1. ∠4 = ∠1 because parallel lines form equalcorresponding angles.

2. ∠5 = 180° – ∠3 because parallel lines formsupplementary interior angles on the sameside of a transversal.

3. ∠6 = ∠4 – ∠5.

4. ∠M is the third angle of ∆MSE’; so∠M = 180° – ∠2 – ∠6.

5. ∠4 = ∠1 = 84°.∠5 = 180° – ∠3 = 180° – 122° = 58°.∠6 = ∠4 – ∠5 = 84° – 58° = 26°.∠M = 180° – ∠2 – ∠6 = 180° – 124° – 26° = 30°.

6.

•37. Two points determine a line.

38. Parallel lines form equal alternate interiorangles.



41. Reflexive.

42. ASA. (∠CAD = ∠BDA, AD = AD,∠BAC = ∠CDA.)


•44. The length of a perpendicular segmentbetween them.

SAT Problem.

45. x + (2x – 30) = 180.

•46. 3x = 210; so x = 70. The acute angle is 70°.

47. 2x – 30 = 2(70) – 30 = 110°. The obtuse angleis 110°.

Construction Problem.

48.

49. ∠1 and ∠2 seem to be equal.




It is strange to think that a steel rod bent into theshape of a triangle would make a good musicalinstrument and to realize that geometry is usefulin describing the sounds produced� They dependin part on where the triangle is struck� Accordingto Thomas Rossing in The Science of Sound(AddisonWesley� �� )� single strokes playedon the base create vibrations in the plane of thetriangle; tremolos are played between the sides ofthe vertex angle and create vibrations perpendicularto the plane of the triangle�

The primary appearance of geometry ineconomics is in graphs� The consumption functionexample is about the “break even” point� Asdescribed in Economics by Paul Samuelson andWilliam Nordhaus (McGrawHill� ��)� “at anypoint on the ��° line� consumption exactly equalsincome � � � � The ��° line tells us immediately�therefore� whether consumption spending isequal to� greater than� or less than the level ofincome� The point on the consumption schedulewhere it intersects the ��° line shows us the levelof disposable income at which households justbreak even�”

Corollary Proofs.

•1. The sum of the angles of a triangle is 180°.

•2. Substitution.

•3. Substitution.

•4. Subtraction.

5. A right angle equals 90°.

6. Substitution.

7. Subtraction.

8. Two angles are complementary if their sumis 90°.

•9. An equilateral triangle is equiangular.

•10. An equiangular triangle has three equalangles.

11. Substitution (and algebra).

12. Division.

13. Substitution.

Exterior Angle Theorem.

•14. Through a point not on a line, there isexactly one line parallel to the line.



17. Betweenness of Rays Theorem.

18. Substitution.

UFO Angles.

•19. ∠2 > ∠1 and ∠2 > ∠A.

20. ∠2 = ∠1 + ∠A.

21. 28°.

Right and Wrong.

22. Acute.

23. The acute angles of a right triangle arecomplementary; so their sum is a rightangle.

Another Kind of Triangle.

•24. An equilateral triangle.

25. As a musical instrument.

26. 60°.

Economics Graph.

•27. It is a right triangle.

•28. They are complementary.

29. 45°.

30. If two angles of a triangle are equal, thesides opposite them are equal.

Set II Set II Set II Set II Set II (pages �� –��)

The direction of the North Star is treated inexercises �� through �� as if it coincides exactlywith that of celestial north even though there is aslight discrepancy� Peculiar� a small town near thewestern border of Missouri� got its name byaccident� When the residents applied for a postoffice� several names that they proposed wererejected because other towns in the state hadalready taken them� Finally the appointed postmaster for the town wrote to Washington D� C�saying� if the names previously submitted would


not do� then please assign some “peculiar” nameto the town� The Post Office Department didexactly that�

The catch question about the height of thetriangle was created by Leo Moser of the University of Alberta�

North Star and Latitude.


32. ∠4 and ∠3 are complementary.

33. The acute angles of a right triangle arecomplementary.

•34. ∠2 and ∠3 are complementary.

35. ∠4 = ∠2.

•36. Complements of the same angle are equal.

37. Substitution. [∠1 = ∠4 (exercise 31) and∠4 = ∠2 (exercise 35).]

Angle Bisectors.

38.

46. Proof.(1) ∆ABC is a right triangle with right

∠ABC; ∠ABD and ∠C are complemen-tary. (Given.)

(2) ∠A and ∠C are complementary.(The acute angles of a right triangle arecomplementary.)

(3) ∠ABD = ∠A. (Complements of thesame angle are equal.)

(4) AD = BD. (If two angles of a triangleare equal, the sides opposite them areequal.)

(5) ∆ABD is isosceles. (A triangle is isoscelesif it has at least two equal sides.)

Catch Question.

47.

39. (Student answer.) (∠B seems to be twice aslarge as ∠E.)

•40. An exterior angle of a triangle is equal to thesum of the remote interior angles.

•41. Substitution.

42. Multiplication.

43. Substitution.

44. Subtraction.

45. Proof.(1) In ∆ABC and ∆ADE, ∠ADE = ∠B.

(Given.)(2) ∠A = ∠A. (Reflexive.)(3) ∠AED = ∠C. (If two angles of one

triangle are equal to two angles ofanother triangle, the third angles areequal.)

48. Because AX ⊥ BY, ∆ABP is a right triangle;so ∠PAB and ∠PBA are complementary.Therefore, ∠PAB + ∠PBA = 90°. BecauseAX and BY bisect two of the angles,∠YAP = ∠PAB and ∠XBP = ∠PBA. Ittherefore follows that ∠YAB + ∠XBA = 180°.If ∠YAB and ∠XBA are supplementary,however, then AY || BX (supplementaryinterior angles on the same side of atransversal mean that lines are parallel).

The problem is a “catch” question becauseAY and BX do not intersect to form atriangle.


As this proof without words establishes� the sumof the corner angles of any fivepointed star� nomatter how irregular� is � °� A possible area ofexploration would be to draw stars with othernumbers of points and try to discover if anythingcomparable is true of them� (We will return tothis problem in our study of circles� Using the factthat an inscribed angle is half the measure of itsintercepted arc makes it easy to show for thespecial case of symmetric stars with an oddnumber of sides that the sum of the corner anglesis � °�)


1. The sum of the angles that are the corners ofa five-pointed star is 180°.

2. A parallel line is drawn through one cornerof the star to form angles with one of thesides extended. Applying the exterior angletheorem to two of the large overlappingtriangles gives 1 + 3 and 2 + 4 in the smallertriangle on the right. Equal correspondingangles and equal alternate interior anglesformed by the parallels reveal that5 + (2 + 4) + (1 + 3) = 180°.



The Rogallo glider is named after Francis Rogallo�a NASA engineer who invented its triangularwinged design in the late �� s� According toJames E� Mrazek in Hang Gliding and Soaring(St� Martin’s Press� ��)� Rogallos have beentested by the U� S� Army at altitudes of � milesand speeds of �� miles per hour and havesuccessfully carried loads as great as three tons!NASA has even considered using radiocontrolledRogallos to bring � ton spacecraft boosters backto Earth for reuse�

The Roman alphabet is used almost withoutexception throughout the world to write thesymbols of mathematics� even in such languagesas Russian� Hebrew� and Chinese� No matterwhere you study geometry� the triangles areprobably named ABC!

Hang Glider.

•1. ASA.

2. SSS.

3. SAS.

4. AAS.

•5. HL.

Congruent or Not?


7. Three.

8. No.

•9. Three.

10. No.

11. Three.

•12. Yes.

13. They are congruent by AAS.

Korean Theorem.

14. That ∠C and ∠C’ are right angles.

15. That ∆ABC and ∆A’B’C’ are congruent.

16. The HL Theorem. If the hypotenuse and aleg of one right triangle are equal to thecorresponding parts of another righttriangle, the triangles are congruent.

17. Right, hypotenuse, and side.

Triangle Problem.

18.

•19. HL.


•21. If two angles of a triangle are equal, thesides opposite them are equal.

22. Isosceles.


Richard G� Pawley explored the question� “Underwhat conditions can two triangles have five partsof one congruent to five parts of the other� butnot be congruent?” in an article in The Mathematics Teacher (May ��)� He called suchtriangles “�con” and proved� among other things�that the sides of such triangles must be ingeometric sequence� The angle measures for thetwo triangles in exercise �� have been rounded tothe nearest degree� Their exact and more precise

values are: ∠A � Arccos ≈ � ��°;

∠B � Arccos ≈ ��°; ∠C � Arccos ≈

�� °� Calculating these values might make anamusing problem for trigonometry students todo�

Chapter �� Lesson �

Mirror Distances.

23.

38. One set of parts: AD = AE, ∠ADC = ∠AEB,and DP = PE. (Another set of parts:AD = AE, DP = PE, AP = AP.)

39. SAS (or SSS).

•40. Because ∠DAP = ∠EAP (correspondingparts of congruent triangles are equal).

Origami Frog.

41.





28. Reflexive.

•29. AAS.


Angle Bisection.

31.

•32. AB = AC, AD = AE, and ∠A = ∠A.

33. SAS.


35. BD = CE (because AD – AB = AE – AC),∠ADC = ∠AEB, and ∠BPD = ∠CPE.

36. AAS.


•42. SSS.


•44. 45°.

45. 45°.


47. Right triangles.

48. HL.



More on Equal Parts.

51. Two triangles that have these equal partsmust be congruent by SSS.

52.

53. Yes. ∆ABC and ∆DEF are an example.

Chapter �� Review �

54. No. Two triangles that have six pairs ofequal parts must be congruent by definition(or SSS).


An entire “Mathematical Games” column byMartin Gardner in Scientific American is on thetopic of geometrical fallacies� It is included in hisWheels� Life and Other Mathematical Amusements(W� H� Freeman and Company� ��)� Gardnerreports that Euclid wrote a book on the subject�Called Pseudaria� the book is lost and there is norecord of its content�

The proof that every triangle is isoscelesappeared for the first time in W� W� Rouse Ball’sMathematical Recreations and Essays� LewisCarroll included it in The Lewis Carroll PictureBook (��)� As Gardner explains: “The error isone of construction� E is always outside thetriangle and at a point such that� when perpendiculars are drawn from E to sides CA and CB� oneperpendicular will intersect one side of thetriangle but the other will intersect an extensionof the other side�”

Is Every Triangle Isosceles?

1.

2. AAS.


4. SAS.


6. HL. [EA = EB (exercise 5), EF = EG(exercise 3).]


8. Addition. (FC = GC because ∆CEF ≅ ∆CEG.)

9. Betweenness of Points Theorem.

10. Substitution.

11. The false statements appear at the end.Although it is true for this triangle thatA-F-C and so AF + FC = AC, it is not truethat B-G-C; so BG + GC ≠ BC. So the lastequation, AC = BC, does not follow.

Chapter �� ReviewChapter �� ReviewChapter �� ReviewChapter �� ReviewChapter �� Review

Set I Set I Set I Set I Set I (pages �� –��)

Exercises � through �� are based on the ratherremarkable dissection of an equilateral triangleinto five isosceles triangles such that none� one�or two of the triangles are equilateral� Other suchdissections are possible�

Flag Symmetries.

•1. Fold the figure along the proposed line ofsymmetry and see if the two halves coin-cide.

2. The Cuban flag. The star prevents it fromhaving a horizontal line of symmetry.

3. The Algerian flag has a horizontal line ofsymmetry. The flag of Barbados has avertical line of symmetry. The Jamaican flaghas two lines of symmetry: one horizontaland the other vertical.

4.

•5. It is the perpendicular bisector of AB.

Lots of Lines.

6. An overhead view of a parking lot.


� Chapter �� Review

8. 21.

Alphabet.

•9. To the sum of the remote interior angles.

10. Form equal corresponding angles.

11. Perpendicular to a third line are parallel.

•12. They are perpendicular.

13. Interior angles mean that lines are parallel.

•14. Are equal, their sides are perpendicular.

15. Are equal.

Ollie’s Mistakes.

•16. In a plane, two points each equidistant fromthe endpoints of a line segment determinethe perpendicular bisector of the linesegment.

17. If two angles and the side opposite one of themin one triangle are equal to the correspondingparts of another triangle, the triangles arecongruent.

18. If two angles of one triangle are equal to twoangles of another triangle, the third anglesare equal.

19. In a plane, a line perpendicular to one of twoparallel lines is also perpendicular to theother.

Triangle Division.

20.

22.

23. They are all isosceles.

24. First figure: no pieces. Second figure: onepiece. Third figure: two pieces.

Angle Trisection.

25.

•26. Equilateral.

27. ∠DAB = ∠EAC = 60°. Each angle of anequilateral triangle is 60°.

28. ∠EAB = 30°.∠EAB = ∠CAB – ∠DAB = 90° – 60° = 30°.

Surveyor’s Triangle.

29. Exactly one.


31. It would bisect the angle.

Chapter �� Review �

32.


Magnifying Glass.

•33. In a plane, a line perpendicular to one oftwo parallel lines is also perpendicular tothe other.

34. All right angles are equal.

•35. Concurrent.

36. SAS. (BD = DF, ∠BDF = ∠FDP, DP = DP.)


38. They are equal (by addition).

Measuring a Tree.

•39. ∠B = ∠CAD = 45°.

40. BA = AC and AC = CD.


•42. BA = CD by substitution.

Quadrilateral Problem.

43. ∠A = ∠D. (Or ∠B = ∠C.)

44. Parallel lines form supplementary interiorangles on the same side of a transversal.

45. No. (This answer may surprise some stu-dents. An example of why the answer is“no” is an isosceles trapezoid.)

Isosceles Triangle.

46.

47. ∠ECD = x°.

•48. ∠CDB = 2x°.

49. ∠DBC = 2x°.

•50. ∠ACB = 3x°. (∠ACB is an exterior angle of∆BCE; so ∠ACB = ∠E + ∠EBC = x° + 2x°.)

51. ∠A = 3x°.

52. ∠ABE = 3x°.

53. ∠E + ∠A + ∠ABE = x + 3x + 3x = 180; so

7x = 180 and x = ≈ 25.7. ∠E ≈ 25.7°.

Stair Steps.

54.

∆ABC ≅ ∆CDE (HL); so ∠BCA = ∠E (corre-sponding parts of congruent triangles areequal). BC || DE because equal correspond-ing angles (with a transversal) mean thatlines are parallel.

SAT Problem.

55. q = 40 because parallel lines form equalalternate interior angles. p = 140 becausetwo angles in a linear pair are supplemen-tary. So p – q = 140 – 40 = 100.

� Chapter �� Algebra Review

Algebra ReviewAlgebra ReviewAlgebra ReviewAlgebra ReviewAlgebra Review (page ��)

1. + = .

2. – = .

3. .

4. – = .

5. = .

6. .

7. ⋅ = .

8. ⋅ = .

9. = .

10. = = .

11. .

12. ⋅ = = .

13. + – = .

14. = .

15. – = .

16. = or .

17. – =

= .

18. + = .

19. ⋅ = 1.

20. + = = .

πr πr

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