This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Step 1: Complete the proof for the case where one side of the inscribed angle passes through the center O. Let m( OBA) = . Since OAB is an isosceles triangle with OA = OB (radii), the base angles are equal: m( OAB) = m( OBA) = . The central angle AOC is an exterior angle to OAB. m( AOC) = m( OAB) + m( OBA) m( AOC) = + m( AOC) = 2 The measure of the central angle AOC is equal to the measure of the arc it subtends, arc AC. m(arc AC) = m( AOC) = 2 Since m( ABC) = m( OBA) = , we can substitute = (1)/(2) m(arc AC). Therefore, for this case: m( ABC) = (1)/(2) m(arc AC) Step 2: Prove the theorem for the case where the center O is inside the inscribed angle (as shown in Figure 5.66). Draw a diameter BD from vertex B through the center O. This divides ABC into two angles: ABD and DBC. m( ABC) = m( ABD) + m( DBC) Both ABD and DBC have one side as a diameter (BD). Applying the result from Step 1: m( ABD) = (1)/(2) m(arc AD) m( DBC) = (1)/(2) m(arc DC) Substitute these into the equation for m( ABC): m( ABC) = (1)/(2) m(arc AD) + (1)/(2) m(arc DC) m( ABC) = (1)/(2) (m(arc AD) + m(arc DC)) By the Arc Addition Postulate, $m(arc AD)