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:

$$m(\angle ABC) = \frac{1}{2} m(\text{arc } AC)$$

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(arc AD)