Chapter 13 — Answers

Selected answers for Chapter 13: The Concurrence Theorems, transcribed from the book’s appendix. They are meant for checking your work, not for copying — the book prints answers to selected exercises only.

Back to Chapter 13 · Teacher’s answers: Chapter 13 · Complete key: Answers to Exercises PDF · Printed answers: appendices PDF

Lesson 1 (pages 532–535)

4. Its circumcircle.

5. △ABC is inscribed in circle O.

6. They are equidistant from it.

7. Minor arcs.

13. A diameter (or a chord).

17. The lines are the perpendicular bisectors of the sides of △ABC, and the perpendicular bisectors of the sides of a triangle are concurrent.

19. No.

24. Three.

28. OA = 2OD because, in a 30°-60° right triangle, the hypotenuse is twice the shorter leg.

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

36. It is approximately 6.4 cm.

39. They are concurrent on the hypotenuse of the triangle.

40. 7.5 cm.

42. Both ∠E and ∠ABD intercept $\widehat{AD}$. Inscribed angles that intercept the same arc are equal.

47. Approximately 31 mm.

48. 31.25 mm.

51. $\sin D=\dfrac{a}{2r}$.

Lesson 2 (pages 538–541)

1. By seeing if a pair of opposite angles are supplementary.

4. They must be concurrent.

8. Inscribed angles that intercept the same arc are equal.

10. Substitution.

14. They are isosceles because all radii of a circle are equal.

18. $AB\parallel DC$ because they are bases of a trapezoid.

21. ∠B and ∠D are supplementary.

24. They intersect in a common point, P.

30. ABCD is not cyclic, because its opposite angles are not supplementary.

33. Vertical angles are equal.

34. Supplements of the same angle are equal.

40. 1,764.

42. The Protractor Postulate.

46. Multiplication.

54. Substitution.

p. 762

Lesson 3 (pages 544–547)

1. The angle bisectors of a triangle are concurrent.

3. Two.

4. They are perpendicular bisectors of the sides.

5. The vertices of the triangle.

7. Two angle bisectors.

10. Its incircle.

11. Its incenter.

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

15. They must be concurrent.

17. Only if the rectangle is a square.

21. They are the perpendicular bisectors of the sides.

25. $2\pi$ and $4\pi$ units.

28. EF seems to be tangent to the incircle.

32. Exterior angles.

38. It proves that AP bisects ∠DAE.

40. FDEO is a square.

42. $2\pi r$ units.

46. $\dfrac{\pi}{4}$.

49. $3-r$.

51. $7-2r$.

55. $\pi$ square units.

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

61. $\dfrac{2\pi r}{p}$.

Lesson 4 (pages 550–553)

4. The fact that the altitudes of a triangle form right angles with the lines of its sides.

5. Two points determine a line.

6. The Ruler Postulate.

8. A midsegment of a triangle is parallel to the third side.

11. A line segment that connects a vertex of a triangle to the midpoint of the opposite side is a median.

14. Through a point not on a line, there is exactly one line parallel to the line.

18. In a plane, a line perpendicular to one of two parallel lines is also perpendicular to the other.

23. GD and EH bisect each other. The diagonals of a parallelogram bisect each other.

28. △ABC is obtuse.

29. No. The altitude segments do not intersect.

32. GE, AF, and CD.

33. At point B.

41. Corresponding sides of similar triangles are proportional.

42. Substitution.

49. PD and EC are altitudes of △APE.

Lesson 5 (pages 557–560)

1. CF.

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

6. 3.2.

8. Not concurrent.

11. 2.

15. That this product is 8.

18. That this sum is 2.

21. 3.

24. AX = 4; XB = 6.

28. △AXZ ∼ △CPZ and △BXY ∼ △CPY.

32. Division (and substitution).

34. If a line parallel to one side of a triangle intersects the other two sides in different points, it divides the sides in the same ratio.

41. $\dfrac{AX}{BX}\cdot\dfrac{BY}{CY}\cdot\dfrac{CZ}{AZ}=1$ (by Ceva’s Theorem).

45. In a plane, a line perpendicular to one of two parallel lines is also perpendicular to the other.

52. △AXC ∼ △AZB.

Lesson 6 (pages 562–565)

1. AB and DC.

2. $-\dfrac{3}{2}$.

4. $\sqrt{13}$ units.

7. M(9, 0); L(3, 6).

8. (8, 4).

11. CN = NB = $\sqrt{72}$ or $6\sqrt{2}$.

16. $m_{NT}=1$ and $m_{CB}=-1$.

18. (9, 3).

24. At V.

27. (6, 6).

36. 48 units.

38. 150°.

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

46. G(3, 14); I(17, 14).

53. AB = DC = 6; AD = BC = $\sqrt{10}$.

57. $m_{XZ}=9$; $m_{WY}=-\dfrac{1}{9}$.

Review (pages 566–570)

2. △ABC is inscribed in the circle.

6. △ABC is a right triangle because ∠ACB is inscribed in a semicircle.

12. F.

14. IJKL.

18. Midsegments.

24. AX = 4, YB = 3, and ZC = 2.

28. Cyclic.

34. $\dfrac{3}{4}$ in.

36. ∠AED + ∠AGD = 180°. ∠AED and ∠AGD are supplementary because they are opposite angles of a cyclic quadrilateral.

38. ∠EGD = ∠EAD because they are inscribed angles that intercept the same arc.

39. ∠EGD = 60° because ∠EGD = ∠EAD.

42. ∠BHC = 120° because it is supplementary to ∠BFC.

48. 125°.

53. They appear to bisect its angles.

55. DB and FB.

61. ABCD is concave.