If a tangent segment and a secant segment are drawn to a

In the preceding construction, not shown is a radius of the smaller circle Q. In the larger circle M, is an inscribed angle that intercepts a semicircle. Thus, is a right angle and . Because the line drawn perpendicular to the radius of a circle at its endpoint on the circle is a tangent to the circle, is a tangent to circle Q. INEQUALITIES IN THE CIRCLE The remaining theorems in this section involve inequalities in the circle. ET ET ⬜ TQ ∠ETQ ∠ETQ QT M E Q b M E V Q T c Figure 6.56 E Q a Construction 9 To construct a tangent to a circle from an external point. GIVEN: and external point E [See Figure 6.56a.] CONSTRUCT: A tangent to , with T as the point of tangency CONSTRUCTION: Figure 6.56a: Consider and external point E. Figure 6.56b: Draw . Construct the perpendicular bisector of , to intersect at its midpoint M. Figure 6.56c: With M as center and MQ or ME as the length of radius, construct a circle. The points of intersection of circle M with circle Q are designated by T and V. Now draw , the desired tangent. NOTE: If drawn, would also be a tangent to . }Q EV ET EQ EQ EQ }Q }Q ET }Q Exs. 1–3 In a circle or in congruent circles containing two unequal central angles, the larger angle corresponds to the larger intercepted arc. THEOREM 6.4.2 GIVEN: with central angles and in Figure 6.57; PROVE: PROOF: In , . By the Central Angle Postulate, and By substitution, The converse of Theorem 6.4.2 follows, and it is also easily proved. m AB ¬ 7 mCD ¬ . m ∠2 = mCD ¬ . m ∠1 = mAB ¬ m ∠1 7 m∠2 }O m AB ¬ 7 mCD ¬ m ∠1 7 m∠2 ∠2 ∠1 }O D B A C O 1 2 Figure 6.57 In a circle or in congruent circles containing two unequal arcs, the larger arc corre- sponds to the larger central angle. THEOREM 6.4.3 GIVEN: In Figure 6.57, with and PROVE: The proof is left as Exercise 35 for the student. m ∠1 7 m∠2 m AB ¬ 7 mCD ¬ CD ¬ AB ¬ }O In the proof of Theorem 6.4.5, the positive numbers a and b represent the lengths of line segments. If , then ; the converse is also true. a 2 6 b 2 a 6 b EXAMPLE 2 GIVEN: In Figure 6.58, with a Using Theorem 6.4.3, what conclusion can you draw regarding the measures of and ? b What does intuition suggest regarding RS and TV? Solution a b RS 7 TV m ∠RQS 7 m∠TQV ∠TQV ∠RQS mRS ¬ 7 mTV ¬ . }Q 쮿 Before we apply Theorem 6.4.4 and prove Theorem 6.4.5, consider the Discover activity at the left. The proof of Theorem 6.4.4 is not provided; however, the proof is similar to that of Theorem 6.4.5. T S R V Q Figure 6.58 Discover In Figure 6.59, measures the distance from center P to chord . Likewise, measures the distance from P to chord . Using a ruler, show that . How do the lengths of chords and compare? ANSWER EF AB PR 7 PT AB PR EF PT AB6 EF In a circle or in congruent circles containing two unequal chords, the shorter chord is at the greater distance from the center of the circle. THEOREM 6.4.4 EXAMPLE 3 In circle P of Figure 6.59, any radius has a length of 6 cm, and the chords have lengths , , and . Let , , and name perpendicular segments to these chords from center P. a Of , , and , which is longest? b Of , , and , which is shortest? Solution a is longest, according to Theorem 6.4.4. b is shortest. PT PR PT PS PR PT PS PR PT PS PR EF = 10 cm DC = 6 cm AB = 4 cm 쮿 T B E D P F C A R S Figure 6.59 In a circle or in congruent circles containing two unequal chords, the chord nearer the center of the circle has the greater length. THEOREM 6.4.5 GIVEN: In Figure 6.60a, with chords and and PROVE: PROOF: In Figure 6.60b, we represent the lengths of and by a and c, respectively. Draw radii , , , and , and denote all lengths by r. is the perpendicular bisector of , and is the perpendicular bisector of , because a radius perpendicular to a chord bisects the chord and its arc. Let and . NC = d MB = b CD QN AB QM QD QC QB QA QN QM AB 7 CD QM 6 QN QN ⬜ CD QM ⬜ AB CD AB }Q With right angles at M and N, we see that and are right triangles. According to the Pythagorean Theorem, and , so and . If , then and . Multiplication by reverses the order of this inequality; therefore, . Adding , we have or , which implies that . If , then . But and . Therefore, . It is important that the phrase minor arc be used in our final theorems. The proof of Theorem 6.4.6 is left to the student. For Theorem 6.4.7, the proof is provided because it is more involved. In each theorem, the chord and related minor arc share common endpoints. AB 7 CD CD = 2d AB = 2b 2b 7 2d b 7 d b 7 d b 2 7 d 2 r 2 - a 2 7 r 2 - c 2 r 2 -a 2 7 - c 2 - 1 a 2 6 c 2 a 6 c QM 6 QN d 2 = r 2 - c 2 b 2 = r 2 - a 2 r 2 = c 2 + d 2 r 2 = a 2 + b 2 䉭QNC 䉭QMB B N Q C D A M a B N Q C D A M b r r r c d r a b Figure 6.60 Exs. 4–9 In a circle or in congruent circles containing two unequal chords, the longer chord corresponds to the greater minor arc. THEOREM 6.4.6 If in Figure 6.61, then m AB ¬ 7 mCD ¬ . AB 7 CD In a circle or in congruent circles containing two unequal minor arcs, the greater minor arc corresponds to the longer of the chords related to these arcs. THEOREM 6.4.7 GIVEN: In Figure 6.61a, with and chords and CD AB m AB ¬ 7 mCD ¬ }O B O D C A a b B O D C A c B O D A C d B O D A C E Figure 6.61 PROVE: PROOF: In circle O of Figure 6.61b, draw radii , , , and . Because , it follows that because the larger arc in a circle corresponds to the larger central angle. In Figure 6.61c, we rotate to the position on the circle for which D coincides with B. Because radii and are congruent, is isosceles; also, . In , . Because is positive, we have and by substitution. Therefore, . m ∠C 6 90° 2 m ∠C 6 180° m ∠C + m∠CDO 6 180° m ∠COD m ∠COD + m∠C + m∠CDO = 180° 䉭COD m ∠C = m∠ODC 䉭COD OB OC 䉭COD m ∠AOB 7 m∠COD m AB ¬ 7 mCD ¬ OD OC OB OA AB 7 CD 쮿 Now construct the perpendicular segment to at point C, as shown in Figure 6.61d. Denote the intersection of the perpendicular segment and by point E. Because is a right with hypotenuse , . Because and , we have . By the Transitive Property, the starred statements reveal that . NOTE: In the preceding proof, must intersect at some point between A and B . If it were to intersect at A, the measure of inscribed would have to be more than 90°; this follows from the facts that is a minor arc and that the intercepted arc for would have to be a major arc. ∠BCA AB ¬ ∠BCA AB CE AB 7 CD AB 7 EB AE 7 AB = AE + EB EB 7 CD EB 䉭 䉭DCE AB CD Exs.10–16 Exercises 6.4 In Exercises 1 to 8, use the figure provided.

1. If write

an inequality that compares and .

2. If write

an inequality that compares CD and AB.

3. If

write an inequality that compares QM and QN.

4. If

write an inequality that compares and .

5. If

, write an inequality that compares CD to AB.

6. If

, write an inequality that compares QM to QN.

7. If

write an inequality that compares QM to QN.

8. If

, write an inequality that compares to

9. Construct a circle O and choose some point D on the

circle. Now construct the tangent to circle O at point D.

10. Construct a circle P and choose three points R, S, and T on

the circle. Construct the triangle that has its sides tangent to the circle at R, S, and T.

11. X, Y, and Z are on circle O such

that , , and . Suppose that triangle XYZ is drawn and that the triangle ABC is constructed with its sides tangent to circle O at X, Y, and Z. Are and similar triangles? 䉭ABC 䉭XYZ m XZ ¬ = 110° m YZ ¬ = 130° m XY ¬ = 120° mCD ¬ . m AB ¬ QN :QM = 5:6 mCD ¬ :m AB ¬ = 3:2, m ∠CQD 6 m∠AQB m ∠CQD 6 m∠AQB m ∠C m ∠A mCD ¬ 6 mAB ¬ , mCD ¬ 6 mAB ¬ , mCD ¬ 6 mAB ¬ , m ∠AQB m ∠CQD mCD ¬ 6 mAB ¬ ,

12. Construct the two tangent segments to circle P not

shown from external point E.

13. Point V is in the exterior of circle Q not shown such that

is equal in length to the diameter of circle Q. Construct the two tangents to circle Q from point V. Then determine the measure of the angle that has vertex V and has the tangents as sides.

14. Given circle P and points R-P-T such that R and T are in

the exterior of circle P, suppose that tangents are constructed from R and T to form a quadrilateral as shown. Identify the type of quadrilateral formed a when . b when .

15. Given parallel chords

, , , and in circle O , which chord has the greatest length? Which has the least length? Why?

16. Given chords

, , and in such that , which chord has the greatest length? Which has the least length? Why?

17. Given circle O with

radius , tangent , and line segments , , , and : a Which line segment drawn from O has the smallest length? b If , , , and , which line segment from point O has the greatest length? m ∠4 = 30° m ∠3 = 45° m ∠2 = 50° m ∠1 = 40° OD OC OB OA Í AD OT QZ 7 QY 7 QX }Q TV RS MN GH EF CD AB RP = PT RP 7 PT VQ C Q D A B M N Exercises 1–8 Y X C A B Z O T R P O A B C D E F G H Z N T R Q S M X Y V A B C D T O 1 4 2 3 쮿

18. a If , write an inequality

to compare with . b If , write an inequality to compare with .

19. a If , write an inequality to

compare the measures of minor arcs and . b If , write an inequality to compare the measures of major arcs and .

20. a If , write an

inequality to compare the measures of inscribed angles 1 and 2. b If , write an inequality to compare the measures of and .

21. Quadrilateral ABCD is inscribed in circle P not shown.

If is an acute angle, what type of angle is ?

22. Quadrilateral RSTV is inscribed in circle Q not shown.

If arcs , , and are all congruent, what type of quadrilateral is RSTV?

23. In circle O, points A, B, and C are on

the circle such that and . a How are and related? b How are AB and BC related?

24. In , and

. a How are and related? b How are and related?

25. In ,

and . See the figure above. a How are and related? b How are AB and BC related?

26. Triangle ABC is inscribed in circle O;

, , and . a Which is the largest minor arc of : , , or ? b Which side of the triangle is nearest point O?

27. Given circle O with and :

a Which angle of triangle ABC is smallest? b Which side of triangle ABC is nearest point O?

28. Given that in circle O:

a Which arc is largest? b Which chord is longest? mAC ¬ :mBC ¬ :m AB ¬ = 4:3:2 mAC ¬ = 130° mBC ¬ = 120° AC ¬ BC ¬ AB ¬ }O AC = 7 BC = 6 AB = 5 mBC ¬ m AB ¬ m ∠BOC = 30° m ∠AOB = 70° }O mBC ¬ m AB ¬ m ∠BOC m ∠AOB BC = 4 cm AB = 6 cm }O m ∠BOC m ∠AOB mBC ¬ = 40° m AB ¬ = 60° TV ¬ ST ¬ RS ¬ ∠C ∠A YZ ¬ XY ¬ m ∠1 6 m∠2 m XY ¬ 7 m YZ ¬ PMQ ២ MPN ២ MN 7 PQ PQ ¬ MN ¬ MN 7 PQ m TV ¬ m RS ¬ m ∠1 7 m∠2 m ∠2 m ∠1 m RS ¬ 7 m TV ¬ Q 1 2 R S T V N M Q P 2 1 X Y Z C O B A C O B A Exercises 23–25 Exercises 26–29

29. Given that in circle O:

a Which angle is largest? b Which chord is longest? Note: See the figure for Exercise 26.

30. Circle O has a diameter of length 20 cm. Chord has

length 12 cm, and chord has length 10 cm. How much closer is than to point O?

31. Circle P has a radius of length 8 in. Points A, B, C, and

D lie on circle P in such a way that and . How much closer to point P is chord than ?

32. A tangent is constructed to circle Q from external

point E. Which angle and which side of triangle QTE are largest? Which angle and which side are smallest?

33. Two congruent circles, and

, do not intersect. Construct a common external tangent for and .

34. Explain why the following statement is incorrect:

“In a circle or in congruent circles containing two unequal chords, the longer chord corresponds to the greater major arc.”

35. Prove: In a circle containing two unequal arcs, the larger

arc corresponds to the larger central angle.

36. Prove: In a circle containing two unequal chords, the

longer chord corresponds to the larger central angle. HINT: You may use any theorems stated in this section.

37. In , chord

chord . Radius is perpendicular to and at points M and N, respectively. If , , and , then the distance from O to is greater than the distance from O to . Determine how much farther chord is from center O than chord is from center O; that is, find MN. AB CD AB CD CD = 10 AB = 24 OE = 13 CD AB OE CD AB 7 }O }P }O }P }O ET CD AB m ∠CPD = 60° m ∠APB = 90° CD AB CD AB m ∠A:m∠B:m∠C = 2:4:3 O A M N E B C D

38. In

, whose radius has length 8 in., . Because , chord is longer than either of the congruent chords and . Determine how much longer is than ; that is, find the exact value and the approximate value of . AC - AB AB AC BC AB AC mAC ¬ = 120° m AB ¬ = mBC ¬ = 60° }P B C A P PERSPECTIVE ON HISTORY Circumference of the Earth By traveling around the earth at the equator, one would traverse the circumference of the earth. Early mathematicians attempted to discover the numerical circumference of the earth. But the best approximation of the circumference was due to the work of the Greek mathematician Eratosthenes 276–194 B . C .. In his day, Eratosthenes held the highly regarded post as the head of the museum at the university in Alexandria. What Eratosthenes did to calculate the earth’s circumference was based upon several assumptions. With the sun at a great distance from the earth, its rays would be parallel as they struck the earth. Because of parallel lines, the alternate interior angles shown in the diagram would have the same measure indicated by the Greek letter ␣. In Eratosthenes’ plan, an angle measurement in Alexandria would be determined when the sun was directly over the city of Syene. While the angle suggested at the center of the earth could not be measured, the angle in Alexandria formed by the vertical and the related shadow could be measured; in fact, the measure was . Eratosthenes’ solution to the problem was based upon this fact: The ratio comparing angle measures is equivalent to the ratio comparing land distances. The distance between Syene and Alexandria was approximately 5,000 stadia ␣ L 7.2° 1 stadium 516.73 ft. Where C is the circumference of the earth in stadia, this leads to the proportion Solving the proportion and converting to miles, Eratosthenes approximation of the earth’s circumference was about 24,662 mi, which is about 245 mi less than the actual circumference. Eratosthenes, a tireless student and teacher, lost his sight late in life. Unable to bear his loss of sight and lack of productivity, Eratosthenes committed suicide by refusing to eat. ␣ 360° = 5000 C or 7.2 360 = 5000 C L PERSPECTIVE ON APPLICATIONS Sum of Interior Angles of a Polygon Suppose that we had studied the circle before studying polygons. Our methods of proof and justifications would be greatly affected. In particular, suppose that you do not know the sum of interior angles of a triangle but that you do know these facts: 1. The sum of the arc measures of a circle is 360°. 2. The measure of an inscribed angle of a circle is the measure of its intercepted arc. Using these facts, we prove “The sum of the interior angles of a triangle is 180°.” Proof: In , , , and . Then 1 2 mBC ¬ + mAC ¬ + mAB ¬ = 1 2 360° = 180°. m ∠A + m∠B + m∠C = m ∠C = 1 2 m AB ¬ m ∠B = 1 2 mAC ¬ m ∠A = 1 2 mBC ¬ 䉭ABC 1 2 Alexandria Syene Figure 6.62 Using known facts 1 and 2, we can also show that “The sum of the interior angles of a quadrilateral is 360°.” However, we would complete our proof by utilizing a cyclic quadrilateral. The strategic ordering and association of terms leads to the desired result. Proof: For quadrilateral HJKL in Figure 6.64, or 1 2 360° + 1 2 360° 1 2 m LKJ ២ + m LHJ ២ + 1 2 mHLK ២ + m HJK ២ = 1 2 m LKJ ២ + 1 2 mHLK ២ + 1 2 m LHJ ២ + 1 2 m HJK ២ m ∠H + m∠J + m∠K + m∠L = B A C Figure 6.63