Step 1: Complete the proof for the case where one side of the inscribed angle passes through the center $O$. Let $m(\angle OBA) = \theta$. Since $\triangle OAB$ is an isosceles triangle with $OA = OB$ (radii), the base angles are equal: $m(\angle OAB) = m(\angle OBA) = \theta$. The central angle $\angle AOC$ is an exterior angle to $\triangle OAB$. $$m(\angle AOC) = m(\angle OAB) + m(\angle OBA)$$ $$m(\angle AOC) = \theta + \theta$$ $$m(\angle AOC) = 2\theta$$ The measure of the central angle $\angle AOC$ is equal to the measure of the arc it subtends, arc $AC$. $$m(\text{arc } AC) = m(\angle AOC) = 2\theta$$ Since $m(\angle ABC) = m(\angle OBA) = \theta$, we can substitute $\theta = \frac{1}{2} m(\text{arc } AC)$. Therefore, for this case: $$m(\angle ABC) = \frac{1}{2} m(\text{arc } AC)$$ 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 $\angle ABC$ into two angles: $\angle ABD$ and $\angle DBC$. $$m(\angle ABC) = m(\angle ABD) + m(\angle DBC)$$ Both $\angle ABD$ and $\angle DBC$ have one side as a diameter ($BD$). Applying the result from Step 1: $$m(\angle ABD) = \frac{1}{2} m(\text{arc } AD)$$ $$m(\angle DBC) = \frac{1}{2} m(\text{arc } DC)$$ Substitute these into the equation for $m(\angle ABC)$: $$m(\angle ABC) = \frac{1}{2} m(\text{arc } AD) + \frac{1}{2} m(\text{arc } DC)$$ $$m(\angle ABC) = \frac{1}{2} (m(\text{arc } AD) + m(\text{arc } DC))$$ By the Arc Addition Postulate, $m(\text{arc } AD)