Chapter 10.3 Using Chords

10.3 Using Chords Exercise 1 To bisect a chord means to divide the chord into two segments that have same length. Exercise 2 For example, look at this figure! 𝐴𝐴𝐴𝐴 is a chord that is not a diameter. Then, rotate 𝐴𝐴𝐴𝐴 to get 𝐶𝐶𝐶𝐶 . 𝐴𝐴𝐴𝐴 is perpendicular to 𝐶𝐶𝐶𝐶 and they are congruent, but they are not a diameter. So, if two chords of a circle are perpendicular and congruent, one of them does not have to be a diameter. Exercise 3 Form the figure, 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐷𝐷 . By Congruent Corresponding Chords Theorem, 𝐴𝐴𝐴𝐴 � ≅ 𝐶𝐶𝐷𝐷 � . So, 𝑚𝑚𝐴𝐴𝐴𝐴 � = 𝑚𝑚𝐶𝐶𝐷𝐷 � = 75 ∘ . Exercise 4 Form the figure, 𝑚𝑚𝑇𝑇𝑇𝑇 � = 𝑚𝑚𝑇𝑇𝑈𝑈 � = 34 ∘ . So, 𝑇𝑇𝑇𝑇 � ≅ 𝑇𝑇𝑈𝑈 � . By Congruent Corresponding Chords Theorem, 𝑇𝑇𝑇𝑇 ≅ 𝑇𝑇𝑈𝑈 . So, UV=TU=5. Exercise 5 Form the figure, 𝑋𝑋𝑋𝑋 ≅ 𝑊𝑊𝑊𝑊 . By Congruent Corresponding Chords Theorem, 𝑋𝑋𝑋𝑋 � ≅ 𝑊𝑊𝑊𝑊 � . So, 𝑚𝑚𝑋𝑋𝑋𝑋 � = 𝑚𝑚𝑊𝑊𝑊𝑊 � = 𝑚𝑚𝑊𝑊𝑋𝑋 � + 𝑚𝑚𝑋𝑋𝑊𝑊 � = 110 ∘ + 60 ∘ = 170 ∘ . Exercise 6 The radius of ⊙ 𝐶𝐶 and ⊙ 𝑃𝑃 have same length. So, ⊙ 𝐶𝐶 ≅⊙ 𝑃𝑃 . Then, from the figure, 𝑚𝑚𝑄𝑄𝑄𝑄 � = 𝑚𝑚𝐿𝐿𝐿𝐿 � = 120 ∘ . Because 𝑄𝑄𝑄𝑄 � and 𝐿𝐿𝐿𝐿 � have same measure and lie on the congruent circles, so 𝑄𝑄𝑄𝑄 � ≅ 𝐿𝐿𝐿𝐿 � . By Congruent Corresponding Chords Theorem, 𝑄𝑄𝑄𝑄 ≅ 𝐿𝐿𝐿𝐿 . So, QR=LM=11. Exercise 7 By Perpendicular Chord Bisector Theorem, 𝐷𝐷𝐸𝐸 ≅ 𝐷𝐷𝐸𝐸 . So, EG=EJ or x=8. Exercise 8 By Perpendicular Chord Bisector Theorem, 𝑄𝑄𝑅𝑅 � ≅ 𝑅𝑅𝑇𝑇 � . So, 𝑚𝑚𝑄𝑄𝑅𝑅 � = 𝑚𝑚𝑅𝑅𝑇𝑇 � or 𝑥𝑥 = 40 ∘ .

Exercise 9 By Perpendicular Chord Bisector Theorem, 5x-6=2x+9 5x-2x=9+6 3x=15 x=5 Exercise 10 By Perpendicular Chord Bisector Theorem, (5 𝑥𝑥 + 2) ∘ = (7 𝑥𝑥 − 12) ∘ 5 𝑥𝑥 + 2 = 7 𝑥𝑥 − 12 5 𝑥𝑥 − 7 𝑥𝑥 = − 12 − 2 − 2 𝑥𝑥 = − 14 𝑥𝑥 = 7 Exercise 11 It will be true if 𝐴𝐴𝐶𝐶 is a diameter. So, by Perpendicular Chord Bisector Theorem, 𝐴𝐴𝐶𝐶 bisects 𝐶𝐶𝐴𝐴 and 𝐴𝐴𝐶𝐶 � ≅ 𝐶𝐶𝐶𝐶 � . Exercise 12 Let the control panels as chord 𝐴𝐴𝐴𝐴 and 𝐶𝐶𝐶𝐶 . They have the perpendicular bisector that shown on the figure below. By Perpendicular Chord Bisector Converse, 𝑃𝑃𝑄𝑄 lies on the diameter of the circle. Let point O is the center of the circle. Because the control panels have the same length, so 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . By Equidistant Chords Theorem, OP=OQ. So, the center of the cross section should lie on the middle of 𝑃𝑃𝑄𝑄 . Exercise 13 Because 𝐴𝐴𝐶𝐶 ≅ 𝐴𝐴𝐶𝐶 , triangle ACD is an isosceles triangle. So, ∠ 𝐴𝐴𝐶𝐶𝐶𝐶 ≅ ∠ 𝐴𝐴𝐶𝐶𝐶𝐶 . Then, ∠ 𝐶𝐶𝐴𝐴𝐴𝐴 ≅ ∠ 𝐶𝐶𝐴𝐴𝐴𝐴 . So, by the ASA Congruence Theorem, 𝐴𝐴𝐴𝐴 will divide triangle ACD to be two congruent triangles. So, 𝐴𝐴𝐴𝐴 is perpendicular to and bisects 𝐶𝐶𝐶𝐶 . By Perpendicular Chord Bisector Converse, 𝐴𝐴𝐴𝐴 is a diameter of the circle. Exercise 14

By Pythagorean Theorem, 𝐶𝐶𝐷𝐷 = �𝐴𝐴𝐶𝐶 2 − 𝐴𝐴𝐷𝐷 2 𝐶𝐶𝐷𝐷 = � 5 2 − 3 2 𝐶𝐶𝐷𝐷 = √ 25 − 9 𝐶𝐶𝐷𝐷 = √ 16 𝐶𝐶𝐷𝐷 = 4 Because 𝐴𝐴𝐴𝐴 is perpendicular to 𝐶𝐶𝐶𝐶 , but does not bisect 𝐶𝐶𝐶𝐶 , by the contraposition of Perpendicular Chord Bisector Theorem, 𝐴𝐴𝐴𝐴 is not a diameter of the circle. Exercise 15 From the figure, EF=GH=16. So, 𝐷𝐷𝐸𝐸 ≅ 𝐸𝐸𝐺𝐺 . By Equidistant Chords Theorem, QA=QB. So, QA=QB 4x+3=7x-6 4x-7x=-6-3 -3x=-9 x=3 So, 𝑄𝑄𝐴𝐴 = 4 𝑥𝑥 + 3 = 4 ⋅ 3 + 3 = 12 + 3 = 15. Then, 𝑄𝑄𝐴𝐴 is perpendicular to 𝐷𝐷𝐸𝐸 and 𝑄𝑄𝐴𝐴 passes through center Q. So, 𝑄𝑄𝐴𝐴 bisects 𝐷𝐷𝐸𝐸 . Then, 𝐴𝐴𝐷𝐷 = 𝐴𝐴𝐸𝐸 = 12 ⋅ 16 = 8 . 𝑄𝑄𝐸𝐸 is one of the radius of the circle. To get the length of radius, we can use Pythagorean Theorem on triangle AFQ. 𝑄𝑄𝐸𝐸 = �𝑄𝑄𝐴𝐴 2 + 𝐴𝐴𝐸𝐸 2 𝑄𝑄𝐸𝐸 = � 15 2 + 8 2 𝑄𝑄𝐸𝐸 = √ 225 + 64 𝑄𝑄𝐸𝐸 = √ 289 𝑄𝑄𝐸𝐸 = 17 So, the radius of ⊙ 𝑄𝑄 is 17. Exercise 16 From the figure, by Equidistant Chords Theorem, AD=BC 4x+4=6x-6 4x-6x=-6-4 -2x=-10 x=5 So, 𝐴𝐴𝐶𝐶 = 4 𝑥𝑥 + 4 = 4 ⋅ 5 + 4 = 20 + 4 = 24 . To get the length of radius, we can use Pythagorean Theorem, 𝑄𝑄𝐴𝐴 = � 5 2 + � 12 ⋅ 24 � 2 𝑄𝑄𝐴𝐴 = � 5 2 + 12 2 𝑄𝑄𝐴𝐴 = √ 25 + 144 𝑄𝑄𝐴𝐴 = √ 169 𝑄𝑄𝐴𝐴 = 13 So, the radius of ⊙ 𝑄𝑄 is 13. Exercise 17

The radius of the circular part is 𝑟𝑟 = � 6 2 + � 12 ⋅ 7 � 2 𝑟𝑟 = � 6 2 + � 72 � 2 𝑟𝑟 = � 36 + 49 4 𝑟𝑟 = � 144 4 + 49 4 𝑟𝑟 = � 193 4 𝑟𝑟 = √ 193 2 So, the diameter is 𝑑𝑑 = 2 𝑟𝑟 = 2 ⋅ √193 2 = √ 193 ≈ 13.9 inches. Exercise 18 a. By Perpendicular Chord Bisector Converse, 𝐴𝐴𝐴𝐴 is a diameter. b. By Congruent Corresponding Chords Theorem, 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . c. By Perpendicular Chord Bisector Theorem, 𝐺𝐺𝐸𝐸 bisects 𝐸𝐸𝐸𝐸 . d. By Equidistant Chords Theorem, 𝐿𝐿𝐿𝐿 ≅ 𝑁𝑁𝑃𝑃 . Exercise 19 a. Given that 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . 𝑃𝑃𝐴𝐴 , 𝑃𝑃𝐴𝐴 , 𝑃𝑃𝐶𝐶 , and 𝑃𝑃𝐶𝐶 are radii of the ⊙ 𝑃𝑃 , so they are congruent to each other. Let's see triangle PAB and PCD. From the triangles, we know that 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 , 𝑃𝑃𝐴𝐴 ≅ 𝑃𝑃𝐶𝐶 , and 𝑃𝑃𝐴𝐴 ≅ 𝑃𝑃𝐶𝐶 . By the SSS Congruence Theorem, triangle PAB and PCD are two congruent triangles. So, ∠ 𝐴𝐴𝑃𝑃𝐴𝐴 ≅ ∠ 𝐶𝐶𝑃𝑃𝐶𝐶 . By Congruent Central Angles Theorem, 𝐴𝐴𝐴𝐴 � ≅ 𝐶𝐶𝐶𝐶 � . b. Given that 𝐴𝐴𝐴𝐴 � ≅ 𝐶𝐶𝐶𝐶 � . By Congruent Central Angles Theorem, ∠𝐴𝐴𝑃𝑃𝐴𝐴 ≅ ∠𝐶𝐶𝑃𝑃𝐶𝐶 . 𝑃𝑃𝐴𝐴 , 𝑃𝑃𝐴𝐴 , 𝑃𝑃𝐶𝐶 , and 𝑃𝑃𝐶𝐶 are radii of the ⊙ 𝑃𝑃 , so they are congruent to each other. Let's see triangle PAB and PCD. From the triangles, we know that ∠𝐴𝐴𝑃𝑃𝐴𝐴 ≅ ∠𝐶𝐶𝑃𝑃𝐶𝐶 , 𝑃𝑃𝐴𝐴 ≅ 𝑃𝑃𝐶𝐶 , and 𝑃𝑃𝐴𝐴 ≅ 𝑃𝑃𝐶𝐶 . By the SAS Congruence Theorem, triangle PAB and PCD are two congruent triangles. So, 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . Exercise 20 From the figure, 𝐴𝐴𝐶𝐶 ≅ 𝐴𝐴𝐶𝐶 . By Congruent Corresponding Chords Theorem, 𝐴𝐴𝐶𝐶 � ≅ 𝐴𝐴𝐶𝐶 � . Let 𝑚𝑚𝐴𝐴𝐶𝐶 � = 𝑚𝑚𝐴𝐴𝐶𝐶 � = 𝑦𝑦 ∘ . Then, 𝑚𝑚𝐴𝐴𝐶𝐶𝐴𝐴 � = 𝑚𝑚𝐴𝐴𝐶𝐶 � + 𝑚𝑚𝐴𝐴𝐶𝐶 � = 𝑦𝑦 ∘ + 𝑦𝑦 ∘ = 2 𝑦𝑦 ∘ . From the figure, 𝑚𝑚𝐴𝐴𝐴𝐴 � = 360 ∘ − 𝑚𝑚𝐴𝐴𝐶𝐶𝐴𝐴 � 𝑥𝑥 ∘ = 360 ∘ − 2 𝑦𝑦 ∘ Because all the arcs have integer measures, so 360 ∘ − 2 𝑦𝑦 ∘ must be even. So, 𝑥𝑥 ∘ must be even. Exercise 21 Let's see at this figure!

Let 𝑃𝑃𝐶𝐶 is perpendicular to 𝐴𝐴𝐷𝐷 and 𝐴𝐴𝐶𝐶 . Because 𝑃𝑃𝐶𝐶 passes through the center of the circle, by Perpendicular Chord Bisector Theorem, 𝐴𝐴𝐸𝐸 ≅ 𝐷𝐷𝐸𝐸 and 𝐴𝐴𝐸𝐸 ≅ 𝐶𝐶𝐸𝐸 . Because QR=20, AE=16, and BD=12, so PA=PB=PC=PQ=PR=10, AF=EF=8, and BG=DG=6. From triangle APF, we have 𝑠𝑠𝑠𝑠𝑠𝑠∠𝐴𝐴𝑃𝑃𝐸𝐸 = 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 = 8 10 = 45 So, 𝑚𝑚∠𝐴𝐴𝑃𝑃𝐸𝐸 = arcsin 45 ≈ 53.13 ∘ . Then, from triangle BPG, we have 𝑠𝑠𝑠𝑠𝑠𝑠∠𝐴𝐴𝑃𝑃𝐸𝐸 = 𝐵𝐵𝐵𝐵𝐵𝐵𝐴𝐴 = 6 10 = 35 . So, 𝑚𝑚∠𝐴𝐴𝑃𝑃𝐸𝐸 = arcsin 35 ≈ 36.87 ∘ . From the figure, we have 𝑚𝑚𝐴𝐴𝐴𝐴 � = 𝑚𝑚𝐴𝐴𝐶𝐶 � − 𝑚𝑚𝐴𝐴𝐶𝐶 � 𝑚𝑚𝐴𝐴𝐴𝐴 � = 𝑚𝑚∠𝐴𝐴𝑃𝑃𝐸𝐸 − 𝑚𝑚∠𝐴𝐴𝑃𝑃𝐸𝐸 𝑚𝑚𝐴𝐴𝐴𝐴 � ≈ 53.13 ∘ − 36.87 ∘ 𝑚𝑚𝐴𝐴𝐴𝐴 � ≈ 16.26 ∘ Exercise 22 Given 𝐷𝐷𝐸𝐸 is a diameter of ⊙ 𝐿𝐿 and 𝐷𝐷𝐸𝐸 ⊥ 𝐶𝐶𝐸𝐸 . Let's see triangle DFL. 𝐿𝐿𝐶𝐶 and 𝐿𝐿𝐸𝐸 are radii of ⊙ 𝐿𝐿 , so 𝐿𝐿𝐶𝐶 ≅ 𝐿𝐿𝐸𝐸 . So, triangle DFL is an isosceles triangle and ∠ 𝐶𝐶𝐶𝐶𝐿𝐿 ≅ ∠ 𝐶𝐶𝐸𝐸𝐿𝐿 . Then, let's see triangle CDL and CFL. 𝐶𝐶𝐿𝐿 from triangle CDL is congruent to 𝐶𝐶𝐿𝐿 from triangle CFL and ∠ 𝐶𝐶𝐶𝐶𝐿𝐿 ≅ ∠ 𝐸𝐸𝐶𝐶𝐿𝐿 . Because ∠ 𝐶𝐶𝐶𝐶𝐿𝐿 ≅ ∠ 𝐶𝐶𝐸𝐸𝐿𝐿 , ∠ 𝐶𝐶𝐶𝐶𝐿𝐿 ≅ ∠ 𝐸𝐸𝐶𝐶𝐿𝐿 , and ( 𝐶𝐶𝐿𝐿 ≅ 𝐶𝐶𝐿𝐿 , by AAS Congruence Theorem, triangle CDL and CFL are two congruent triangles. So, 𝐶𝐶𝐶𝐶 ≅ 𝐸𝐸𝐶𝐶 and ∠ 𝐶𝐶𝐿𝐿𝐸𝐸 ≅ ∠ 𝐸𝐸𝐿𝐿𝐸𝐸 or 𝐶𝐶𝐸𝐸 � ≅ 𝐸𝐸𝐸𝐸 � . Exercise 23 Plot the center L and draw triangle LPT and LPR. Because point L is the center of the triangle, 𝐿𝐿𝑇𝑇 and 𝐿𝐿𝑄𝑄 are the radii of the circle, so 𝐿𝐿𝑇𝑇 ≅ 𝐿𝐿𝑄𝑄 . Because 𝑄𝑄𝑅𝑅 is a perpendicular bisector of 𝑄𝑄𝑇𝑇 , then 𝑇𝑇𝑃𝑃 ≅ 𝑄𝑄𝑃𝑃 . Because 𝐿𝐿𝑇𝑇 ≅ 𝐿𝐿𝑄𝑄 , 𝑇𝑇𝑃𝑃 ≅ 𝑄𝑄𝑃𝑃 , and 𝐿𝐿𝑃𝑃 ≅ 𝐿𝐿𝑃𝑃 , by SSS Congruence Theorem, triangle LPT and LPR are two congruent triangles. Then, ∠ 𝐿𝐿𝑃𝑃𝑇𝑇 ≅ ∠ 𝐿𝐿𝑄𝑄𝑇𝑇 and 𝐿𝐿𝑃𝑃 ⊥ 𝑇𝑇𝑄𝑄 . So, point L is on the perpendicular bisector of 𝑄𝑄𝑇𝑇 and L lies on 𝑄𝑄𝑅𝑅 . Because 𝑄𝑄𝑅𝑅 passes through center L, so 𝑄𝑄𝑅𝑅 is a diameter of the circle. Exercise 24 The ∠ APB and ∠ CPD are vertical angles, so ∠ 𝐴𝐴𝑃𝑃𝐴𝐴 ≅ ∠ 𝐶𝐶𝑃𝑃𝐶𝐶 . Then, the ∠ BAD and ∠ BCD are two inscribed angles with the same arc. So, they are congruent and ∠ 𝐴𝐴𝐴𝐴𝐶𝐶 ≅ ∠ 𝐴𝐴𝐶𝐶𝐶𝐶 . By AA Similarity Theorem, triangle ABP and triangle CDP are similar. So, 𝐴𝐴𝐴𝐴𝐵𝐵𝐴𝐴 = 𝐶𝐶𝐴𝐴 𝐷𝐷𝐴𝐴 .

Exercise 25 Let's see at this figure! First, given 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . Prove that EF=EG. From the figure, by Perpendicular Chord Bisector Theorem, 𝐴𝐴𝐸𝐸 ≅ 𝐴𝐴𝐸𝐸 and 𝐶𝐶𝐸𝐸 ≅ 𝐶𝐶𝐸𝐸 . Because 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 , so 𝐴𝐴𝐸𝐸 ≅ 𝐶𝐶𝐸𝐸 . Let's see triangle AEF and CEG below. Because 𝐴𝐴𝐷𝐷 and 𝐶𝐶𝐷𝐷 are radii of the circle, so 𝐴𝐴𝐷𝐷 ≅ 𝐶𝐶𝐷𝐷 . The hypotenuse and a leg of triangle AEF are congruent to the hypotenuse and a leg of triangle CEG, so by Hypotenuse-Leg Congruence Theorem, triangle AEF is congruent to triangle CEG. So, 𝐷𝐷𝐸𝐸 ≅ 𝐷𝐷𝐸𝐸 and EF=EG. Then, given EF=EG. Prove that 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . 𝐷𝐷𝐴𝐴 , 𝐷𝐷𝐴𝐴 , 𝐷𝐷𝐶𝐶 , and 𝐷𝐷𝐶𝐶 are radii of the circle, so 𝐷𝐷𝐴𝐴 ≅ 𝐷𝐷𝐴𝐴 ≅ 𝐷𝐷𝐶𝐶 ≅ 𝐷𝐷𝐶𝐶 . By Hypotenuse-Leg Congruence Theorem, Δ𝐴𝐴𝐷𝐷𝐸𝐸 ≅ Δ𝐴𝐴𝐷𝐷𝐸𝐸 ≅ Δ𝐶𝐶𝐷𝐷𝐸𝐸 ≅ Δ𝐶𝐶𝐷𝐷𝐸𝐸 . Then, 𝐴𝐴𝐸𝐸 ≅ 𝐴𝐴𝐸𝐸 ≅ 𝐶𝐶𝐸𝐸 ≅ 𝐶𝐶𝐸𝐸 . So, 𝐴𝐴𝐴𝐴 ≅ 𝐶𝐶𝐶𝐶 . Exercise 26 Let the point where the tire touches the ground as the point C. Let the point D on the tire so 𝐶𝐶𝐶𝐶 is a diameter of the tire. Let's see this figure.

𝐶𝐶𝐶𝐶 is perpendicular to the ground. Because the bottom edge of the panel is parallel to the ground, so 𝐴𝐴𝐴𝐴 is parallel to the ground and 𝐶𝐶𝐶𝐶 ⊥ 𝐴𝐴𝐴𝐴 . By Perpendicular Chord Bisector Theorem, 𝐴𝐴𝐶𝐶 � ≅ 𝐴𝐴𝐶𝐶 � . So, the point C bisects 𝐴𝐴𝐴𝐴 � . Then, my friend is correct because the point where the tire touches the ground bisects 𝐴𝐴𝐴𝐴 � . Exercise 27 The sum of the measure of the interior angles of a quadrilateral is 360 ∘ . So, 𝑚𝑚∠𝐸𝐸 + 𝑚𝑚∠𝐾𝐾 + 𝑚𝑚∠𝐿𝐿 + 𝑚𝑚∠𝐿𝐿 = 360 ∘ 32 ∘ + 25 ∘ + 44 ∘ + 𝑚𝑚∠𝐿𝐿 = 360 ∘ 101 ∘ + 𝑚𝑚∠𝐿𝐿 = 360 ∘ 𝑚𝑚∠𝐿𝐿 = 259 ∘ Exercise 28 Pentagon PQRST has 5 sides. By Polygon Interior Angles Theorem, 𝑚𝑚∠𝑃𝑃 + 𝑚𝑚∠𝑄𝑄 + 𝑚𝑚∠𝑄𝑄 + 𝑚𝑚∠𝑅𝑅 + 𝑚𝑚∠𝑇𝑇 = (5 − 2) ⋅ 180 ∘ 85 ∘ + 134 ∘ + 97 ∘ + 102 ∘ + 𝑚𝑚∠𝑇𝑇 = 3 ⋅ 180 ∘ 418 ∘ + 𝑚𝑚∠𝑇𝑇 = 540 ∘ 𝑚𝑚∠𝑇𝑇 = 122 ∘