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10-4 Inscribed Angles PROOF Write the specified type of proof. 22. two-column proof Given: Prove: SOLUTION: Statements (Reasons): 1. (Given) 2. (Inscribed intercepting same arc are .) 3. (Vertical are .) 4. (AA Similarity) ALGEBRA Find each value. 23. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle is 180. So, 24. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle is 180. So, Therefore, 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, eSolutions Manual - Powered by Cognero The sum of the measures of the angles of a triangle is 180. So, Page 1 The sum of the measures of the angles of a triangle is 180. So, 10-4 Inscribed Angles Therefore, 25. x SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle is 180. So, 26. SOLUTION: An inscribed angle of a triangle intercepts a diameter or semicircle if and only if the angle is a right angle. So, The sum of the measures of the angles of a triangle is 180. So, Therefore, CCSS STRUCTURE Find each measure. 27. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore, 28. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore, CCSS STRUCTURE Find each measure. eSolutions Manual - Powered by Cognero 29. SOLUTION: Page 2 10-4 Inscribed Angles Therefore, CCSS STRUCTURE Find each measure. 29. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore, 30. SOLUTION: If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary. Therefore, In , FL = 24, HJ = 48, and . Find each measure. 55. FG SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, Therefore, bisects 56. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 57. NJ SOLUTION: eSolutions Manual - Powered by Cognero If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, Therefore, Page 3 bisects 56. SOLUTION: a diameterAngles (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, 10-4IfInscribed Therefore, bisects 57. NJ SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, 58. SOLUTION: If a diameter (or radius) of a circle is perpendicular to a chord, then it bisects the chord and its arc. So, bisects Therefore, Find x. 59. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 60. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, eSolutions Manual - Powered by Cognero Page 4 SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, 10-4 Inscribed Angles 61. SOLUTION: The sum of the measures of the central angles of a circle with no interior points in common is 360. So, eSolutions Manual - Powered by Cognero Page 5