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12.2 HW Solutions. a) PL b) PM c) All radii of a circle are congruent d) Triangle LPN e) SAS f) CPCTC a) All radii of a circle are congruent b) AB ≅ CD c) Given d) SSS e) angle AEB ≅ CED f) Thm 12.1. 14 2 7 50 8 10 a) CE b) DE c) angle CEB d) angle DEA
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12.2 HW Solutions a) PL b) PM c) All radii of a circle are congruent d) Triangle LPN e) SAS f) CPCTC a) All radii of a circle are congruent b) AB ≅ CD c) Given d) SSS e) angle AEB ≅ CED f) Thm 12.1 • 14 • 2 • 7 • 50 • 8 • 10 • a) CE b) DE c) angle CEB d) angle DEA • The Center of the circle • 6 • 5.4 • 8.9 • 12.5 • 9.9 • 20.8 • 108 • 90 • 123.9 or 124
Inscribed Angles Section 12.3
Three high-school soccer players practice kicking goals from the points shown in the diagram. All three points are along an arc of a circle. Player A says she is in the best position because the angle of her kicks towards the goal is wider than the angle of the other players’ kicks. Do you agree? Player B Player A Player C
Vocabulary • Inscribed Angle • An angle whose vertex is on the circle and whose sides are chords of the circle • Intercepted Arc • The portion of the circle intercepted by an inscribed angle
Theorem 12-9: Inscribed Angle Theorem • The measure of an inscribed angle is half the measure of its intercepted arc A B ? ° 45 ° 60° ?° C
Three Cases for 12-9 Q R F 1) The center is on a side of the angle 2) The center is inside of the angle 3) The center is outside of the angle
Circle O, with inscribed B and diameter BC mB = ½mAC Given: Prove: C A O B Statement Reason Given Definition of AC Tri. Ext. Ang. Thm. Isosceles Tri. Thm Substitution Division Prop. Of Equal. 1. BC is a diameter 2. mAC = mAOC 3. mAC = mA + mB 4. mA = mB 5. mAC =2mB 6. ½mAC = mB
Find a and b P a˚ = 120° T 60˚ 30˚ Q S b˚ = 75° R
Corollaries: Inscribed Angle Theorem • Two inscribed angles that intercept the same arc are congruent x 70˚
Corollaries: Inscribed Angle Theorem 2) An angle inscribed in a semicircle is a right angle ( A semicircle is 180, ½ of that is 90) x° = 90°
Corollaries: Inscribed Angle Theorem 3) The opposite angles of a quadrilateral inscribed in a circle are supplementary (opposite angles intercept the entire circle, ½ of 360 is 180) 70˚ x
Corollaries: Inscribed Angle Theorem • Two inscribed angles that intercept the same arc are congruent • An angle inscribed in a semicircle is a right angle ( A semicircle is 180, ½ of that is 90) • The opposite angles of a quadrilateral inscribed in a circle are supplementary (opposite angles intercept the entire circle, ½ of 360 is 180)
Find the measure of angles 1 and 2 40˚ 2 = 38° 70˚ 38˚ 70˚ 1 = 90°
Theorem 12-10 • The measure of an angle formed by a tangent and a chord is half the measure of the intercepted arc. B B D D O O C C GEOGEBRA
Find x and y 110° J Q 90° 35˚ x˚ = 35° 70° y˚ = 55° L K
Homework • Section: 12-3 • Pages: 681-683 • Questions: 4-22, 36, 38
C P Circle O, with inscribed ∠ABC Given: Prove: O mABC = ½mAC A B 1. Circle O, inscribed ∠ABC 2. Draw diameter BP 3. mABP = ½AP 4. mCBP = ½PC 5. mABC = mABP + mABP 6. mABC = ½AP + ½PC 7. mABC = ½(AP+PC) 8. mABC = ½AC Given Def. of Diameter Thm. 12-9 Thm. 12-9 Ang. Add. Post. Substitution Distributing (factoring) Arc Addition Post.
Post-Homework Review • CHAPTER REVIEW • Page: 707 • Questions: 6-14 • Section: 12-3 • Pages: 681-683 • Questions: 25