Chapter 13 Answers - BISD Moodlemoodle.bisd303.org/file.php/18/Solutions/Chapter 13 Answers.pdf · concurrent can be found in two books by Howard Eves: Fundamentals of Modern Elementary

Chapter �� Lesson � ��

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

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

Exercises �� through �� illustrate a remarkablefact� Although it is easy to see why the linesthrough the pairs of intersections of three circlesare concurrent if the circles have equal radii� thelines are concurrent even if the circles havedifferent radii� If two circles intersect in twopoints� the line determined by those points iscalled the radical axis of the two circles� More onthe theorem that the radical axes of three circleswith noncollinear centers� taken in pairs� areconcurrent can be found in two books by HowardEves: Fundamentals of Modern ElementaryGeometry (Jones and Bartlett� ��) and CollegeGeometry (Jones and Bartlett� ��)�

Exercises �� through �� lead to anothersurprise� If four lines intersect to form fourtriangles� the circumcircles of the triangles alwaysintersect in a common point� named the Miquelpoint after an obscure nineteenthcenturymathematician� Auguste Miquel� In The PenguinDictionary of Curious and Interesting Geometry(Penguin� ��)� David Wells describes otherremarkable properties of this point� The centersof the four circumcircles also lie on a circle thatpasses through it� It is also the focus of theunique parabola tangent to the four lines�

The regions into which a map is divided as inexercise �� are called Dirichlet domains� Accordingto Jay Kappraff� “a Dirichlet domain of a pointfrom a set of points is defined to be the points ofspace nearer to that point than to any of theother points of the set�” A detailed discussion ofthe geometry of Dirichlet domains and theirconnection to soap bubbles� spider webs� andphyllotaxis can be found in Kappraff’s Connections—The Geometric Bridge Between Art and Science(McGrawHill� ��)�

Hot Tub.

1. Chords.

2. Inscribed angles.

3. Cyclic.

•4. Its circumcircle.

•5. ∆ABC is inscribed in circle O.

•6. They are equidistant from it.

Circumcircles.

•7. Minor arcs.

8. A semicircle.

9. A major arc.

10. An acute triangle.

11. A right triangle.

12. An obtuse triangle.

•13. A diameter (or a chord).

14. The midpoint of its hypotenuse.

RGB Color.

15.

16. They are concurrent.

•17. The lines are the perpendicular bisectors ofthe sides of ∆ABC, and the perpendicularbisectors of the sides of a triangle areconcurrent.

18. Example figure:

•19. No.

20. Yes.

21. By the Intersecting Chords Theorem. In thecircle at the top, AP ⋅ PD = BP ⋅ PE and, inthe circle on the left, BP ⋅ PE = CP ⋅ PF; soAP ⋅ PD = BP ⋅ PE = CP ⋅ PF.

Four Lines and Four Circles.

22. ∆ABF, ∆ACD, ∆BCE, ∆DEF.

�� Chapter �� Lesson �

23. They are their circumcircles.

•24. Three.

25. They intersect in a common point.

Equilateral Triangle.

26. Because AE, BF, and CD are altitudes, theyare perpendicular to the sides of ∆ABC.If a line through the center of a circle isperpendicular to a chord, it also bisects thechord.

27. 30°-60° right triangles.

•28. OA = 2OD because, in a 30°-60° righttriangle, the hypotenuse is twice the shorterleg.

29. The radius is the length of one of thealtitudes.

Nearest School.

30. The school at B.

31.

Folding Experiment.

33.

•32. The circumcenter of the triangle with A, B,and C as its vertices.

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

The trianglefolding exercises �� through �� arebased on the fact that� if one point is reflectedonto another� the line of reflecti�n is theperpendicular bisector of the line segmentconnecting the two points�

Exercises �� through �� provide anotherexample of a problem whose solution is based onPythagorean triples and the Pythagorean Theoremyet was created more than a thousand yearsbefore Pythagoras was born� The tablet containingit was found in Susa� about �� miles east ofancient Babylon�

34. The folds are on the perpendicular bisectorsof the sides of the triangle and they areconcurrent.

35. It is equidistant from them.

•36. It is approximately 6.4 cm.

37.

38. A right triangle because 92 + 122 = 152; if thesum of the squares of two sides of a triangleis equal to the square of the third side, it is aright triangle.

•39. They are concurrent on the hypotenuse ofthe triangle.

•40. 7.5 cm.

Inscribed Triangles.

41.

•42. Both ∠E and ∠ABD intercept AD. Inscribedangles that intercept the same arc are equal.

43. ∠ADB is a right angle because it is inscribedin a semicircle.

44. All right angles are equal.

45. If two angles of one triangle are equal totwo angles of another triangle, the third pairof angles are equal.

(


Babylonian Problem.

46.

Set IIISet IIISet IIISet IIISet III (page ��)

The Slipping Ladder.

1.

2. Along an arc of a circle.

3. From the fact that the midpoint of thehypotenuse of a right triangle is itscircumcenter, it follows that O is equidistantfrom A, B, and C; that is, OA = OB = OC.But OA and OB do not change in length asthe ladder slides down the wall, and so OCdoes not change either. Because for everyposition of the ladder Ollie’s feet remain thesame distance from C, their path is along anarc of a circle.


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

Theorem � of this lesson appears in the Elementsas Proposition �� of Book III: “The opposite anglesof quadrilaterals in circles are equal to two rightangles�” Strangely� although the converse is trueand very useful� Euclid does not state or prove it�

A proof that the circumcircles of theequilateral triangles constructed on the sides of atriangle intersect in a common point (assumed inexercises �� through ��) can be found on pages�� and �� of Geometry Revisited� by H� S� M�Coxeter and S� L� Greitzer (Random House� �� )�The point in which the circles intersect is calledthe Fermat point of the triangle� According toDavid Wells in The Penguin Dictionary of Curiousand Interesting Geometry (Penguin� ��)� Fermatchallenged Torricelli (of barometer fame) to findthe point the sum of whose distances from thevertices of a triangle is a minimum� (In otherwords� to find the general solution to the Spotter’sPuzzle�) That point turns out to be the Fermatpoint� According to Wells: “If all the angles of thetriangle are less than ��° the desired point � � � issuch that the lines joining it to the vertices meetat ��°� If the angle at one vertex is at least ��°�then the Fermat point coincides with that vertex�”Wells also remarks that “the Fermat point can befound by experiment� Let three equal weightshang on strings passing through holes at the

•47. Approximately 31 mm.

•48. 31.25 mm.

[Because ∆ACD is a right triangle withAC = 50 and AD = 30, CD = 40. In right∆BDO, (40 – r)2 + 302 = r2; so1,600 – 80r + r2 + 900 = r2, 2,500 = 80r,r = 31.25.]

Diameters and Sines.

49. ∆BCD is a right triangle because ∠BCD isinscribed in a semicircle.

50. Inscribed angles that intercept the same arcare equal.

•51. sin D = .

52. = 2r. (Because sin A = sin D,

sin A = ; so 2r sin A = a and = 2r.)

53. It reveals that it is equal to the diameter ofthe triangle’s circumcircle.


vertices of the triangle� the strings being knottedat one point� The knot will move to the Fermatpoint�”

Quilt Quadrilaterals.

•1. By seeing if a pair of opposite angles aresupplementary.

2. The diamond.

Cyclic and Noncyclic.

3. The lines appear to be their perpendicularbisectors.

•4. They must be concurrent.

5. It is equidistant from them.

Euclid’s Proof.

6. Two points determine a line.

7. The sum of the angles of a triangle is 180°.

•8. Inscribed angles that intercept the same arcare equal.

9. Betweenness of Rays Theorem.

•10. Substitution.

11. Addition.

12. Substitution.

A Different Proof.

13.

Isosceles Trapezoid.

•18. AB || DC because they are the bases of atrapezoid.

19. ∠A and ∠D are supplementary. Parallellines form supplementary interior angleson the same side of a transversal.

20. ∠A = ∠B. The base angles of an isoscelestrapezoid are equal.

•21. ∠B and ∠D are supplementary.

22. ABCD is cyclic. A quadrilateral is cyclicif a pair of its opposite angles are supple-mentary.

23. They prove that isosceles trapezoids arecyclic.

Equilateral Triangles on the Sides.

•24. They intersect in a common point, P.

25. They are cyclic.

26. They are each equal to 120°. They aresupplementary to the angles opposite themin the quadrilaterals. The angles oppositethem are 60° because they are also angles ofthe equilateral triangles.

27. Example figure:

•14. They are isosceles because all radii of acircle are equal.

15. 360°. The sum of the angles of aquadrilateral is 360°.

16. 180°.

17. The sum of each pair of opposite angles ofABCD is equal to w + x + y + z.

28. (Student answer.) (Yes.)


Brahmagupta’s formula for the area of a cyclicquadrilateral� discovered in the seventh century�is an extension of the formula for the area of atriangle in terms of its sides discovered by Heronof Alexandria six centuries earlier� (Heron’sTheorem was introduced in the Set III exercises ofChapter � Lesson ��) H� S� M� Coxeter and S� L�Greitzer include a derivation of Brahmagupta’s


formula on page �� of Geometry Revisited�Although they refer to it as “one of the simplestmethods for obtaining Brahmagupta’s formula�”the method is based on several trigonometricidentities including the Law of Cosines and is noteasy�

J� L� Heilbron in Geometry Civilized—History�Culture and Technique (Clarendon Press� ��)�refers to Claudius Ptolemy as the astronomer“who was to astronomy what Euclid was togeometry” and refers to the proof of his theoremas “one of the most elegant in all plane geometry�”From a special case of Ptolemy’s Theorem comesthe trigonometric identity

sin(x – y) � sin x cos y – cos x sin y�

Cyclic or Not.

29. ABCD is cyclic, because its vertices areequidistant from point E. A circle is the setof all points in a plane equidistant from agiven point; so a circle can be drawn with itscenter at E that contains all of the vertices.

•30. ABCD is not cyclic, because its oppositeangles are not supplementary. (The sum of aright angle and an obtuse angle is more than180°, and the sum of a right angle and anacute angle is less than 180°.)

31. ABCD is cyclic. Because ∠BCE is an exteriorangle of ABCD, it forms a linear pair with∠DCB; so ∠BCE and ∠DCB are supplemen-tary. Because ∠BCE = ∠A, it follows that ∠Aand ∠DCB are also supplementary; soABCD is cyclic.

Brahmagupta’s Theorem.

32. Example figure:


36. Substitution.

37. Point Y is the midpoint of AC. Because∠2 = ∠1, AY = YE and, because ∠3 = ∠4,YC = YE; so AY = YC.

38. If a cyclic quadrilateral has perpendiculardiagonals, then any line through their pointof intersection that is perpendicular to a sideof the quadrilateral bisects the opposite side.

Area Formula.

39. The perimeter of a rectangle with sides aand b is 2a + 2b; so, for a rectangle, s = a + b.

A = =

= = ab.

•40. The perimeter of this quadrilateral is 176,so for it, s = 88.

A = =

= 1,764.

41. No. It is not cyclic and its area is evidentlymuch smaller.

Ptolemy’s Theorem.

•42. The Protractor Postulate.


44. AA.

45. Corresponding sides of similar trianglesare proportional.

•46. Multiplication.

47. Addition.

48. Betweenness of Rays Theorem andsubstitution.


50. AA.


52. Multiplication.

53. Addition.

•33. Vertical angles are equal.

•34. Supplements of the same angle are equal.(∠7 and ∠5 are supplements of ∠6[∠8] and∠6 and ∠8 are supplements of ∠7[∠5].)



55. Multiplication.

Two Applications.

56. Because a rectangle is a cyclic quadrilateraland its opposite sides are equal, Ptolemy’sTheorem becomes a ⋅ a + b ⋅ b = c ⋅ c, ora2 + b2 = c2, the Pythagorean Theorem!

57. Applying Ptolemy’s Theorem to cyclicquadrilateral APBC,PA ⋅ BC + PB ⋅ AC = PC ⋅ AB. Because∆ABC is equilateral, AB = BC = AC. Bysubstitution, PA ⋅ AB + PB ⋅ AB = PC ⋅ AB;dividing by AB gives PA + PB = PC.


The Set III puzzle is adapted from one by StephenBarr� who has created many clever puzzles� Barr’spuzzle was originally included by Martin Gardnerin one of his “Mathematical Games” columns forScientific American and is reprinted on page ��of Mathematical Carnival (Knopf� � �)� Readersof Gardner’s column discovered an extra set ofpoints through which a circle could be drawn thatneither Barr nor Gardner had noticed!

Overlapping Cards Puzzle.

1. These points are the vertices of the tworectangles. Because the opposite angles of arectangle are supplementary, all rectanglesare cyclic.

2.

Another set of points consists of W, Y, F,and A. If WY is drawn and a circle is drawnwith WY as its diameter, then it can beproved that points A and F lie on this circleowing to the fact that ∠A and ∠F are rightangles.



Exercises � through �� are related to a resultknown as Poncelet’s porism� Jean Victor Poncelet�a nineteenthcentury French mathematician� iscredited for establishing the development ofprojective geometry as an independent subject�As Howard Eves explains in An Introduction tothe History of Mathematics (Saunders� ��)� aporism is “a proposition stating a condition thatrenders a certain problem solvable� and then theproblem has infinitely many solutions� Forexample� if r and R are the radii of two circles andd is the distance between their centers� theproblem of inscribing a triangle in the circle ofradius R� which will be circumscribed about thecircle of radius r� is solvable if and only ifR

� – d � � �Rr� and then there are infinitely many

triangles of the desired sort�” The general versionof Poncelet’s porism states that� if given twoconics� an ngon can be inscribed in one conic andcircumscribed about the other� there are aninfinite number of such ngons�

Postage Stamp.

•1. The angle bisectors of a triangle areconcurrent.

2. The incenter.

One set of points consists of W, F, Z, and D.Drawing the hidden parts of the edges WXand YZ produces quadrilateral WFZD.Because ∠F and ∠D are right angles, theyare supplementary; so WFZD is cyclic.


Circumcircle and Incircle.

•3. Two.

•4. They are perpendicular bisectors of thesides.

•5. The vertices of the triangle.

6. Three.

•7. Two angle bisectors.

8. A perpendicular from the triangle’s incenterto one of its sides.

9. The sides of the triangle.

Circumscribed Quadrilateral.

•10. Its incircle.

•11. Its incenter.

12.

23. Isosceles (obtuse) triangles and 30°-60°right triangles.

24. 2.

•25. 2π and 4π units.

26. π and 4π square units.

Construction Problem.

27. Example figure:

•13. They are perpendicular to the sides. If a lineis tangent to a circle, it is perpendicular tothe radius drawn to the point of contact.

14. They bisect the angles. They form pairs oftriangles that are congruent by HL.

•15. They must be concurrent.

16. They must be equal.

•17. Only if the rectangle is a square.

18. Only if the parallelogram is a rhombus.

19. Only if the sum of the lengths of the bases isequal to the sum of the lengths of the legs.


20. That circles have the same center.

•21. They are the perpendicular bisectors of thesides.

22. They are the bisectors of the angles.

•28. EF seems to be tangent to the incircle.

29. (Student answer.) (Probably not.)

30. Yes.

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

In Dutton’s Navigation and Piloting (NavalInstitute Press� ��)� Elbert S� Maloney remarksthat the LOP (line of position) is probably themost important concept in navigation� Whenthree LOPs are taken� they are usually notconcurrent� but intersect to form a triangle�The navigator ordinarily takes the point that isequidistant from the sides of this triangle (itsincenter) as the “fix” or position of the ship� Ifthere is a “constant error” in the LOPs� themethod of “LOP bisectors” outlined in exercises ��through � is used� As exercises �� through ��suggest and exercises �� through � prove� thebisectors of an angle and the two remote exteriorangles of a triangle are concurrent� In the methodof LOP bisectors� the fix of the ship is taken at thepoint of concurrency�

In exercises �� through �� two special casesof a remarkable theorem are considered and� inexercises � through �� the general case isdeveloped� The theorem states that� for anypolygon and its inscribed circle� the ratio of the


areas is equal to the ratio of the perimeters� Avery nice treatment of this theorem is included inGeometrical Investigations—Illustrating the Artof Discovery in the Mathematical Field� by JohnPottage (AddisonWesley� ��)� Pottage presentshis material in the form of a dialogue betweenGalileo’s characters Salviata� Sagredo� andSimplicio� His book� like the work of GeorgePolya� is a wonderful example of teachingmathematics by the Socratic method�

Ship Location.

31. l5.

•32. Exterior angles.

33. They appear to be concurrent.

34.

45. 4r2 square units.

•46. . ( = .)

47. . ( = .)

Incircle Problem 2.

48. ECDO is a square (because it is equilateraland equiangular).

•49. AE = 3 – r.

50. DB = 4 – r.

•51. AB = 7 – 2r. (3 – r + 4 – r.)

52. 7 – 2r = 5, 2r = 2, r = 1.

53. 2π units.

54. 12 units.

•55. π square units.

56. 6 square units. ( 3 ⋅ 4.)

57. . ( = .)

58. .

Incircle Problem 3.

•59. The area of a triangle is half the product ofits base and altitude.

60. αABCD =

AB ⋅ r + BC ⋅ r + CD ⋅ r + DA ⋅ r =

r(AB + BC + CD + DA) = rp.

•61. .

62. . ( = .)

Trisector Challenge.

63. ∠1 = ∠2. Because BE and CE bisect twoangles of ∆BDC, E is its incenter. Becausethe angle bisectors of a triangle areconcurrent, DE must bisect ∠BDC.

35. ∆PFB ≅ ∆PGB (AAS) and ∆PGC ≅ ∆PHC(AAS).

36. They are equal because PF = PG andPG = PH.

37. ∆FAP ≅ ∆HAP (HL). (These triangles areright triangles whose hypotenuse is APand whose legs PF and PH are equal.)

•38. It proves that AP bisects ∠DAE (because∠DAP = ∠PAE).

39. They prove that the three bisector lines areconcurrent.

Incircle Problem 1.

•40. FDEO is a square (because it is equilateraland equiangular).

41. AB = 2r.

•42. 2πr units.

43. 8r units.

44. πr2 square units.

π π

π π

πr

πr

π π π

π

πr

πr πr πr


Set IIISet IIISet IIISet IIISet III (page �� )

Every triangle has not only an incircle� but alsothree excircles� Each excircle is tangent to oneside of the triangle and the lines of the other twosides� The center of each excircle is the point ofconcurrence of the bisectors of an angle and thetwo remote exterior angles of the triangle� Thecenters of the excircles are the vertices of a largertriangle whose altitudes lie on the lines thatbisect the angles of the original triangle�

Excircles.

1. They appear to be its altitudes.

2. AY bisects the exterior angles of ∆ABC at A;so ∠1 = ∠2. Because O is the incenter of∆ABC, AO bisects ∠CAB and so ∠3 = ∠4.∠1 + ∠2 + ∠3 + ∠4 = 180°; so∠2 + ∠2 + ∠3 + ∠3 = 180°,2(∠2 + ∠3) = 180°, ∠2 + ∠3 = 90°, and soAO ⊥ YA. A line segment from a vertex ofa triangle that is perpendicular to the lineof the opposite side is an altitude of thetriangle.



Martin Gardner points out a rather surprising fact(Mathematical Circus� Knopf� � ): that thealtitudes of a triangle are concurrent does notappear in Euclid’s Elements� Gardner observesthat� “although Archimedes implies it� Proclus� afifthcentury philosopher and geometer� seems tohave been the first to state it explicitly�” The factthat the medians of a triangle are concurrentdoes not appear in the Elements either� butArchimedes knew that the point in which theyare concurrent is the triangle’s center of gravity�More specifically� it is the center of gravity of atriangular plate of uniform thickness and density;it is usually not the center of gravity of a wire inthe shape of the triangle�

“Ortho” Words.

1. Orthodontist.

2. Orthodox.

3. Orthopedic.

•4. The fact that the altitudes of a triangle formright angles with the lines of its sides.

Theorem 71.

•5. Two points determine a line.

•6. The Ruler Postulate.

7. Two points determine a line.

•8. A midsegment of a triangle is parallel to thethird side.

9. A quadrilateral is a parallelogram if itsopposite sides are parallel.

10. The diagonals of a parallelogram bisect eachother.

•11. A line segment that connects a vertex of atriangle to the midpoint of the opposite sideis a median.

12. Lines that contain the same point areconcurrent.

Theorem 72.

13. An altitude of a triangle is perpendicular tothe line of the opposite side.

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

15. A quadrilateral is a parallelogram if itsopposite sides are parallel.

16. The opposite sides of a parallelogram areequal.

17. Substitution.

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

19. The perpendicular bisectors of the sides of atriangle are concurrent.


Median Construction.

20. (The construction lines and arcs have beenomitted from the figure below.)

30. Yes. We have proved that the lines thatcontain the altitudes of a triangle areconcurrent.

31. Its orthocenter and its circumcenter.


Exercises �� through � reveal Carnot’s observationconcerning the special relation between fourpoints that are the three vertices of a triangle andits orthocenter: each point is the orthocenter ofthe triangle whose vertices are the other three�According to Howard Eves in An Introduction tothe History of Mathematics (Saunders� ��)�Lazare Carnot was a general in the French Armyand supported the French Revolution� In � ��however� he opposed Napoleon’s becomingemperor and had to flee to Switzerland� Whileliving in exile� he wrote two important books ongeometry� In one of them� he introduced severalnotations still used today� including ∆ABC torepresent the triangle having the vertices A� B�and C and AB to represent arc AB� One of Carnot’ssons became a celebrated physicist for whom theCarnot cycle of thermodynamics is named� andone of his grandsons became President of theFrench Republic�

Other Triangles, Other Orthocenters.

•32. GE, AF, and CD.

•33. At point B.

34. GF, AE, and BD.

35. At point C.

36. GD, BF, and CE.

37. At point A.

Medians as Bisectors.

38. The green line segments appear to beparallel to BC.

39. It appears to bisect them.

40. They are similar by AA. (Each pair oftriangles has a common angle, and parallellines form equal corresponding angles.)

•41. Corresponding sides of similar triangles areproportional.


43. BM = MC.

21. ED || AB and ED = AB. GH || AB and

GH = AB. A midsegment of a triangle is

parallel to the third side and half as long.

22. EDHG is a parallelogram. A quadrilateralis a parallelogram if two opposite sides areboth parallel and equal.

•23. GD and EH bisect each other. The diagonalsof a parallelogram bisect each other.

24. AG = GF and BH = HF because G and H arethe midpoints of AF and FB, respectively.GF = FD and HF = FE because GD and EHbisect each other.

25. = = 2 and = = 2.

26. Points of trisection.

Altitude Construction.

27.

•28. ∆ABC is obtuse.

•29. No. The altitude segments do not intersect.

(


44. Substitution.

45. Multiplication.

46. They suggest that the triangle wouldbalance on the edge of the ruler.

Doing Without a Compass.

47. Example figure:

56. Constructing perpendiculars from A to linen and from B to line m produces two altitudesof the triangle formed by AB, m, and n. Thepoint in which these altitudes intersect isthe orthocenter of the triangle. A line drawnthrough this point perpendicular to ABtherefore contains the third altitude of thetriangle and passes through the point inwhich lines m and n intersect.


The centroid of a uniform sheet in the shape of aquadrilateral was discovered by a Germanmathematician� Ferdinand Wittenbauer (�� –��)� The parallelogram whose sides lie in thelines through the points that trisect the sides ofa quadrilateral is known as its “Wittenbauerparallelogram�”

Balancing Point.48. ∠ACB and ∠ADB are right angles. An angle

inscribed in a semicircle is a right angle.

•49. PD and EC are altitudes of ∆APE.

50. Its orthocenter.

51. AF is the third altitude of ∆APE. The linesthat contain the altitudes of a triangle areconcurrent.

52. PE ⊥ l.

53. Example figure:

54. All of them.

Engineering Challenge.

55.

1. It seems to be a parallelogram.

2. The point in which the diagonals of theparallelogram intersect.


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

Although Ceva’s Theorem is usually stated interms of ratios� it appears in Nathan AltshillerCourt’s College Geometry (Barnes & Noble� ��)�in the following form: “The lines joining thevertices of a triangle to a given point determineon the sides of the triangle six segments such thatthe product of three nonconsecutive segments isequal to the product of the remaining threesegments�” This version of the theorem makes itimmediately obvious that Obtuse Ollie’s variationsof it in exercises �� and �� are correct�


Which is Which?

•1. CF.

2. GF.

3. CD.

4. CE.

•5. No. GF is not a cevian, because neither ofits endpoints is a vertex of the triangle.

Using Ceva’s Theorem.

•6. 3.2. ( = 1, = 1, 15x = 48, x = 3.2.)

7. 2.4. ( = 1, = 1, 5x = 12, x = 2.4.)

•8. Not concurrent.

= ≈ 1.002 ≠ 1.

9. Not concurrent.

= ≈ 0.999 ≠ 1.


10.

16. + + = + + = 1.

17. + + = 3 + 3 + 3 = 9.

•18. + + = + + = 2.

19. ⋅ ⋅ = 1 (by Ceva’s Theorem.)

20. + + = 2 + 2 + 2 = 6.

Right Triangle.

•21. = = 3.

22. = = .

23. ⋅ ⋅ = 1, ⋅ 3 ⋅ = 1,

= .

•24. Because ∆ABC is a right triangle with legs 6

and 8, its hypotenuse is 10. Because =

and AX + XB = 10, AX = 4 and XB = 6.

Ollie’s Equations.

25. Alice is wrong. Ollie’s equation is correct.One way to show this is to start with Alice’s

equation, = 1. Clearing it of fractions

gives ace = bdf. Dividing both sides by ace

gives 1 = = .

26. Ollie’s second equation also is correct. Oneway to show this is to start with the expres-

sion , getting = = .

27. They suggest that you can begin with anyone of the six segments and go in eitherdirection around the triangle.


In his A Course of Geometry for Colleges andUniversities (Cambridge University Press� � �)�

•11. 2.

12. 2.

13. Because the cevians are concurrent,

2 ⋅ 2 ⋅ = 1; so = .

14. Because = , ZA = 4CZ. Also,

CZ + ZA = 7.5; so CZ + 4CZ = 7.5,5CZ = 7.5, CZ = 1.5, and AZ = 6.

Ratio Relations.

•15. ⋅ ⋅ = 2 ⋅ 2 ⋅ 2 = 8.


Dan Pedoe wrote: “The theorems of Ceva andMenelaus naturally go together� since the onegives the condition for lines through vertices of atriangle to be concurrent� and the other gives thecondition for points on the sides of a triangle tobe collinear�” Menelaus is known for his bookSphaerica� in which the concept of a sphericaltriangle appears for the first time� The sphericalversion of the theorem of Menelaus considered inexercises �� through �� is� in fact� the basis for thedevelopment of much of spherical trigonometry�

Exercises �� through � demonstrate� for thecase of an acute triangle� how Ceva’s Theoremcan be used to prove that the altitudes of atriangle are concurrent�

The Theorem of Menelaus.

•28. ∆AXZ ~ ∆CPZ and ∆BXY ~ ∆CPY.


30. Multiplication.

31. Multiplication and division.

•32. Division (and substitution).

33. The equation in Ceva’s Theorem.

Concurrent or Not?

•34. If a line parallel to one side of a triangleintersects the other two sides in differentpoints, it divides the sides in the same ratio.

35. ⋅ ⋅ = ⋅ ⋅ = 1. (By

substituting for and BY for YC.)

36. It proves that AY, BZ, and CX areconcurrent.

What Kind of Triangle?

37.

39. AA. (The parallel lines form equal alternateinterior angles in the triangles. Also, thevertical angles are equal.)

40. Corresponding sides of similar triangles areproportional.

•41. ⋅ ⋅ = 1 (by Ceva’s Theorem).

42. ⋅ ⋅ = 1 by substitution.

( = and = .)

43. = 1 because ⋅ ⋅ = 1.

44. Because = 1, AD = AE by multiplication.

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

46. ∆DYA ≅ ∆EYA by SAS. (AD = AE,∠DAY = ∠EAY, and AY = AY.)

47. It proves that ∆DYE is isosceles. DY = EYbecause corresponding sides of congruenttriangles are equal.

The Gergonne Point.

48.

38. Isosceles.

49. AD, BE, and CF appear to be concurrent.


50. According to Ceva’s Theorem, AD, BE, and

CF are concurrent if ⋅ ⋅ = 1.

The tangent segments to a circle from anexternal point are equal; so AF = AE,BF = BD, and CD = CE. Substituting in the

product ⋅ ⋅ gives

⋅ ⋅ = 1; so the three cevians

are concurrent.

51. The point of concurrence of AD, BE, and CF(or, the point in which the cevians connect-ing the vertices of a triangle to the points oftangency of the incircle to the opposite sidesare concurrent).

Another Look at Altitudes.

•52. ∆AXC ~ ∆AZB.

53. ∆BYA ~ ∆BXC.

54. ∆CZB ~ ∆CYA.

55. = .

56. = .

57. = .

58. ⋅ ⋅ = =

= ⋅ ⋅ =

⋅ ⋅ = 1; so AY, BZ, and CX are

concurrent.

59. The lines containing the altitudes of atriangle are concurrent.


If we go either clockwise or counterclockwisearound a triangle and three cevians divide each

side in the ratio � then the smaller triangle

formed by them has an area that is onesevenththe area of the original triangle� A generalization

of this result is that� if the three cevians divide

each side in the ratio � then the smaller triangle

has an area that is times the area of the

original triangle� In fact� in the chapter on affinegeometry in his Introduction to Geometry (Wiley��)� H� S� M� Coxeter considers an even broadergeneralization� If the three cevians divide the

sides in the ratios � � � then the smaller

triangle has an area that is

times the area

of the original triangle! The “seven congruenttriangles” approach to the version consideredin the text comes from Hugo Steinhaus’sMathematical Snapshots (Oxford UniversityPress� ��)�

Area Puzzle.

1. α∆ABD = 7x.(α∆ABD = α∆AFG + αFGHB + α∆BDH =x + 5x + x = 7x.)

2. α∆ADC = 14x. (Because ∆ABD and ∆ADChave equal altitudes and the base of ∆ADCis twice as long as the base of ∆ABD,α∆ADC = 2α∆ABD = 14x.)

3. α∆GHI = 3x.(α∆GHI = α∆ADC – αAGIE – αDHIC – α∆EIC =14x – 5x – 5x – x = 3x.)

4. Because α∆GHI = 3x andα∆ABC = α∆ABD + α∆ADC = 7x + 14x = 21x,

α∆GHI = α∆ABC.

5. The pairs of regions with the same numberappear to be congruent and hence have thesame area. From this, it appears that ∆ABCis equal in area to the sum of the areas of theseven colored triangles. If these triangles arecongruent, then α∆GHI (the area of theyellow triangle) is one-seventh of the area of∆ABC.




Exercises � through �� lead to the discovery ofthe Euler line� Euler� the most prolific writer onmathematics in history� noted the remarkablerelation of the circumcenter� orthocenter� andcentroid of every nonequilateral triangle in � ��Not only are the three points always collinear�but the distance from the centroid to theorthocenter is always twice the distance fromthe centroid to the circumcenter� It is interestingto note that Euler’s proof of this relation wasanalytic� It was not until �� that Lazare Carnotpublished the first proof by Euclidean methods�The exercises are developed around a trianglechosen on a coordinate grid so that the coordinatesof every point in the figure (except W) areintegers� This was done for two reasons: (�) tohelp the student draw the figure accurately enoughthat the Euler line theorem can be discovered and(�) to review basic ideas such as distance andslope� Unfortunately� this approach obscures thefact that the results obtained are not limited tothis particular triangle but are generally true�

Tilted Square.

•1. AB and DC.

•2. – .

3. Because the sides are perpendicular.

•4. units. ( .)

5. B(10, 6), C(8, 9), D(5, 7).

Euler’s Discovery.The Centroid.

6.

•7. M(9, 0), L(3, 6).

•8. (8, 4).

9. (12, 6).

10. The coordinates of N are the averages of(or midway between) the correspondingcoordinates of B and C.

•11. CN = = =

or ;

NB = = =

or .

12. Its centroid.

13. Medians.

The Circumcenter.

14. (9, 9).

15. (6, 0).

•16. mNT = 1 and mCB = –1.

17. Yes. NT ⊥ CB; so their slopes are theopposites of the reciprocals of each other(or the product of their slopes is –1).

•18. (9, 3).

19. Its circumcenter.

20. mAC = 2 and mZL = – .

21. Their slopes indicate that AC ⊥ ZL.

22. The perpendicular bisectors of its sides.

The Orthocenter.

23. CT ⊥ AB.

•24. At V.

25. mAV = 1.

26. Yes, because mCB = –1 and AV ⊥ CB.(Also, AV || NT because AV ⊥ CB andNT ⊥ CB; so mAV = mNT.)

•27. (6, 6).


29. Its altitudes.


Euler’s Conclusions.

30. They appear to be collinear.

31. mXY = –1 and mYZ = –1 (so, X, Y, and Z arecollinear).

32. XY = 2YZ because XY = = andYZ = .

33. Z.

34. ZA = ZB = ZC = = .


Students who did the Set III exercise of Chapter ��Lesson � (page ��)� may recall that the figure forexercises �� through �� is remarkable in that itshows an arrangement of eight points in theplane such that the perpendicular bisector of theline segment connecting any two of the pointspasses through exactly two other points of thefigure� For most pairs of points in the figure� suchas A and B or G and C� this is obvious from thesymmetries of its parts�

Exercises �� through �� explore anotherproperty possessed by all triangles� If squares areconstructed on the sides of a triangle as in thefamiliar figure for the Pythagorean Theorem� theline segments connecting the centers of thesquares to the opposite vertices of the triangleare always concurrent� Furthermore� these linesegments have a special relation to the trianglewhose vertices are the centers of the squares�Each one not only is perpendicular to one of itssides but also has the same length� Again� the useof the coordinate grid makes it easier to draw thefigure accurately (especially the squares and theircenters) but obscures the fact that the results aretrue for all triangles�

Exercises �� through �� concern a similar resultfor parallelograms� If squares are constructedoutward on the sides of any parallelogram� thetwo line segments connecting the centers of theopposite squares are both equal and perpendicularto each other� This is connected to the fact thatthe quadrilateral whose vertices are the centersof the squares also is a square�

A partial proof of Napoleon’s Theorem isconsidered in exercises �� through �� On theassumption that the three line segments connectingthe vertices of the original triangle to the oppositevertices of the equilateral triangles are equal� areconcurrent� and form equal angles with each

other� it is fairly easy to prove that the triangledetermined by the centers of the three equilateraltriangles also is equilateral� In Geometry Civilized(Clarendon Press� ��)� J� L� Heilbron presents acomplete trigonometric proof of the theorem basedon the Law of Cosines� In Hidden Connections�Double Meanings (Cambridge University Press��)� David Wells wrote concerning Napoleon’sTheorem: “To discover pattern and symmetrywhere there appears to be neither is alwaysdelightful and intriguing� It also suggests that weare not seeing all there is to be seen in the originaldiagram� There must certainly be another way oflooking at it which will be more symmetrical fromthe start� and therefore make the conclusionmore natural�” On pages �–�� of his book� Wellspresents another way� using tessellations� tounderstand this remarkable theorem�

Triangles on Four Sides.

35.

•36. 48 units. (8 × 6.)

37. 150°. (∠CBE = 90° + 60° = 150°.)

•38. 150°. (∠HAE = 360° – 60° – 90° – 60° = 150°.)

39. They appear to be perpendicular. (Also,DE appears to be the perpendicular bisectorof HC.)

40. They are congruent by SAS. (They are alsoisosceles.)

41. It proves that ∆HCE is equilateral.

42. They prove that HC and DE are perpendicu-lar because, in a plane, two points eachequidistant from the endpoints of a linesegment determine the perpendicularbisector of the line segment.


Squares on Three Sides.

43.

Squares on Four Sides.

52.

AX = YZ = = ;

BY = XZ = = ;

CZ = YX = = .

44. mAB = 0, mAC = 2, mCB = – .

•45. No. AC and CB are not perpendicular,because the product of their slopes is not –1.

•46. G(3, 14), I(17, 14).

47. X(12, 13), Y(4, 11), Z(9, 4).

48. They appear to be concurrent.

49. They appear to be perpendicular to thesides of ∆XYZ.

50. It is true because (mAX)(mYZ) = ( )(– ) =

–1, (mBY)(mXC) = (– )(3) = –1, and

(mCZ)(mYX) = (–4)( ) = –1.

51. They are equal to the sides of ∆XYZ towhich they are perpendicular.

•53. AB = DC = 6; AD = BC = .

54. It is a parallelogram.

55. Both pairs of opposite sides are equal (or,two opposite sides are equal and parallel).

56. W(12, 7), X(8, 12), Y(3, 8), Z(7, 3).

•57. mXZ = 9, mWY = – .

58. That XZ and WY are perpendicular.

59. XZ = WY = = .

60. WXYZ is a square.

Napoleon Triangles.

61. They are cyclic because their oppositeangles are supplementary. (For example, inAPCE, ∠E = 60° and ∠APC = 120°.)

62. They are perpendicular to the triangle’ssides because, in a plane, two points eachequidistant from the endpoints of a linesegment determine the perpendicularbisector of the line segment. (For example,XA = XP and YA = YP because all radii of acircle are equal; so XY is the perpendicularbisector of AP.)

63. ∆XYZ is equilateral because it is equiangu-lar. It is equiangular because each of itsangles is 60°. (For example, ∠X = 60°because, in quadrilateral PGXI,∠XGP = ∠XIP = 90° and ∠GPI = 120°.)

�� Chapter �� Review


Here� a surprising variation of Napoleon’s Theoremis considered�

Alternating Triangles.

1. If equilateral triangles are drawn alternatelyoutward and inward on the sides of aquadrilateral, their vertices determine aparallelogram.

2. Yes. The theorem still seems to apply.Example figure:

collinear with a midpoint of one of these arcs anda vertex of ABCD� A proof of the theorem thatthe incenters of the four triangles determined bythe vertices of a cyclic quadrilateral determine arectangle can be found on page �� of NathanAltshiller Court’s College Geometry�

Cyclic Triangle.

1. That there exists a circle that contains all ofits vertices.

•2. ∆ABC is inscribed in the circle.

3. They appear to be perpendicular to them.

4. They are collinear.

Circumcircle and Incircle.

5. Example figure:

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


Exercises � through � illustrate the fact that thefeet of the three perpendiculars to the sides of atriangle from a point on its circumcircle arecollinear� The line that they determine is calledthe Simson line� named after Robert Simson� aprofessor of mathematics at the University ofGlasgow� Simson produced an important edition ofEuclid’s Elements in � �� Proofs of the existenceof the Simson line based on cyclic quadrilateralscan be found in J� L� Heilbron’s Geometry Civilizedand Nathan Altshiller Court’s College Geometry�The Simson line leads to so many other ideas thatit is the subject of an entire chapter of Court’sbook�

The configuration in exercises �� through ��has additional interesting properties� Theperpendicular bisectors of the sides of the rectanglewhose vertices are the incenters of the fouroverlapping triangles also bisect the four arcs ofthe circle corresponding to the sides of ABCD�Furthermore� each vertex of the rectangle is

•6. ∆ABC is a right triangle because ∠ACB isinscribed in a semicircle.

7. AE (or CO).

8. AC (or BC).

9. A and D (or B and D or C and D).

10. D.

11. O.

•12. F.

13. C.

Three Trapezoids.

•14. IJKL.

15. EFGH.

16. A quadrilateral is cyclic iff a pair of itsopposite angles are supplementary. (ForABCD, we don’t know whether a pair ofopposite angles is supplementary or not.)

Chapter �� Review ��

Double Identity.

17. Its circumcenter.

•18. Midsegments.

19. The sides of ∆DEF are parallel to the sidesof ∆ABC and half as long.

20. Its altitudes.


22. A, B, and C.

Ceva’s Theorem.

23.

Centroid.

32.

•24. AX = 4, YB = 3, and ZC = 2.

25. = , = , and = .

26. ⋅ ⋅ = 1.

27. It indicates that AZ, BX, and CY areconcurrent.

Five Circles.

•28. Cyclic.

29. It is circumscribed about it.

30. They are the incircles of the (overlapping)triangles whose sides are the sides anddiagonals of the quadrilateral.

31. They seem to be the vertices of a rectangle.(This is surprising because ABCD, thequadrilateral from which it comes, has nospecial property other than being cyclic.)

33. 1 in.

•34. in.

35. They are one-third of the lengths.

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

The nineteenthcentury Swiss mathematicianJacob Steiner has been called by Howard Eves“one of the greatest synthetic geometers theworld has ever known�” The “Steiner problem”named after him concerns the problem ofconnecting a set of coplanar points with a set ofline segments having the least possible totallength� (The Spotter’s problem is the special caseof this problem for a set of three points situatedat the vertices of an equilateral triangle�) In TheParsimonious Universe (Copernicus� ��)� StefanHildebrandt and Anthony Tromba describe solvingthe Steiner problem by means of soap films:“Suppose we make a frame consisting of twoparallel glass or clear plastic plates; these areconnected by n parallel pins of the same size thatmeet both plates perpendicularly� If this frameworkis immersed in a soap solution and withdrawn� asystem of planar soap films is formed� These filmsare attached to the pins� and they have two kindsof liquid edges� both of which are straight lines�One type adheres to the glass (or plastic)� whichit meets at �° angles� � � � The other type is wherethree films meet� forming three angles of ��°� � � �Since the soapfilm system minimizes area� thesubsystem of edges on one plate must minimizelength among all connections between the n givenpoints�” The exercises demonstrate that� for foursuitably located points� equilateral triangles drawnon opposite sides of the quadrilateral determinedby them can be used to construct a network oflines that meet at equal (��°) angles� Repeatingthe method with the other pair of opposite sidesproduces a different network with a possiblydifferent total length� One of the two� however�

�� Chapter �� Review

is the absolute minimum� the one obtained by thesoapfilm method�

Exercises � through � imply that two anglebisectors of a triangle meet at an angle determinedentirely by the measure of the third angle of thetriangle� In terms of the figure and with the useof a similar but more general argument� it can be

shown that ∠D � ∠A � �°�

The two quadrilaterals in exercises �� through�� have another interesting relation� If perpendiculars are drawn from the point of intersectionof the diagonals of a cyclic quadrilateral to itssides� their feet are the vertices of the inscribedquadrilateral of minimum perimeter� EFGH istherefore the quadrilateral of minimum perimeterthat can be inscribed in ABCD�

Exercises �� through �� explore a moregeneral case of the relations introduced in exercises�� through �� of Lesson �� Martin Gardnerincluded it in one of his “Mathematical Games”columns for Scientific American� reprinted inMathematical Circus (Knopf� � )� The columnwas on the subject of “simplicity” in science andmathematics� and Gardner wrote: “Mathematiciansusually search for theorems in a manner not muchdifferent from the way physicists search for laws�They make empirical tests� In pencil doodlingwith convex quadrilaterals—a way of experimenting with physical models—a geometer may findthat when he draws squares outwardly on aquadrilateral’s sides and joins the centers ofopposite squares� the two lines are equal andintersect at � degrees� He tries it with quadrilaterals of different shapes� always getting the sameresults� Now he sniffs a theorem� Like a physicist�he picks the simplest hypothesis� He does not� forexample� test first the conjecture that the twolines have a ratio of one to �� and intersectat angles of � and � degrees� even though thisconjecture may equally well fit his crudemeasurements� He tests first the simpler guessthat the lines are always perpendicular and equal�His ‘test�’ unlike the physicist’s� is a search for adeductive proof that will establish the hypothesiswith certainty�” Gardner goes on to remark thatthis result is known as Von Aubel’s theorem andmentions several remarkable generalizations of it�

Exercises �� through � establish the niceresult that the area of a triangle is equal to theproduct of the lengths of its three sides dividedby four times the radius of its circumcircle� (It iseasier to begin with this conclusion and showthat it is equivalent to the familiar formula for

the area of a triangle than it is to do it the otherway around�)

Soap-Film Geometry.

•36. ∠AED + ∠AGD = 180°. ∠AED and ∠AGDare supplementary because they areopposite angles of a cyclic quadrilateral.

37. ∠AGD = 120° because ∠AED = 60°.

•38. ∠EGD = ∠EAD because they are inscribedangles that intercept the same arc.

•39. ∠EGD = 60° because ∠EGD = ∠EAD.

40. ∠DGH = 120° because ∠DGH forms alinear pair with ∠EGD.

41. ∠AGH = 120° because∠AGH = 360° – ∠AGD – ∠DGH.

•42. ∠BHC = 120° because it is supplementaryto ∠BFC.

43. ∠FHC = 60° because ∠FHC = ∠FBC.

44. ∠GHC = 120° because ∠GHC forms alinear pair with ∠FHC.

45. ∠BHG = 120° because∠BHG = 360° – ∠GHC – ∠BHC.

46. It suggests that they form equal angles.

Irrelevant Information.

47. 125°. (In ∆ABC, ∠A = 70° and ∠ABC = 60°;so ∠ACB = 50°. Because D is the incenter of∆ABC, BD bisects ∠ABC and so ∠DBC = 30°,and CD bisects ∠ACB and so ∠DCB = 25°.In ∆DBC, ∠DBC = 30° and ∠DCB = 25°; so∠D = 125°.)

•48. 125°. (In ∆ABC, ∠A = 70° and ∠ABC = 40°;so ∠ACB = 70°. Reasoning as in exercise 47,we find that ∠DBC = 20° and ∠DCB = 35°;so ∠D = 125°.)

49. 125°. (In ∆ABC,∠ACB = 180° – 70° – ∠ABC = 110° – ∠ABC.

∠DBC = ∠ABC and

∠DCB = (110° – ∠ABC) = 55° – ∠ABC;

so ∠D = 180° – ∠ABC – (55° – ∠ABC) =

125°.)

Medians Theorem.

50. Because the medians bisect the sides of thetriangle, AX = XB, BY = YC, and CZ = ZA;

so = = = 1. Therefore,

⋅ ⋅ = 1 ⋅ 1 ⋅ 1 = 1; so the

medians are concurrent.

Irregular Billiard Table.

51. Cyclic.

52. They appear to be perpendicular to its sides.

•53. They appear to bisect its angles.

54. It would travel around the sides ofquadrilateral EFGH (because its angles ofincidence and reflection at the points inwhich it hits the sides of ABCD are equal).

Perimeter Problem.

•55. DB and FB.

56. AE and AF.

57. AB.

58. AB.

59. The perimeter of ∆ABC isAB + BC + CA = 2R + a + b.AB = (a – r) + (b – r) = 2R; so a + b = 2R + 2r.Therefore, the perimeter of ∆ABC is2R + (2R + 2r) = 2r + 4R.

Chapter �� Review ��

Squares on the Sides.

60.

•61. ABCD is concave.

62. W(13, 12), X(10, 13), Y(4, 12), Z(10, 4).

63. XZ ⊥ WY and XZ = WY. (XZ = WY = 9.)

Area Problem.

64. ∆ABD ~ ∆AEC.

65. ∠B = ∠E because they are inscribed anglesthat intercept the same arc. ∠ADB is aright angle because AD ⊥ BC, and ∠ACE isa right angle because it is inscribed in asemicircle; so ∠ADB = ∠ACE.

66. Because ∆ABD ~ ∆AEC, =

(corresponding sides of similar trianglesare proportional). So AB ⋅ AC = AD ⋅ AEby multiplication.

67. = by substitution.

AE = 2r; so = .

is the area of ∆ABC.

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Chapter 13 Answers - BISD Moodlemoodle.bisd303.org/file.php/18/Solutions/Chapter 13 Answers.pdf · concurrent can be found in two books by Howard Eves: Fundamentals of Modern Elementary