Prove the theorem for the case where one side of the inscribed angle passes through the center O.

The theorem states that the measure of an angle inscribed in a circle is half the measure of the arc it subtends. Let $\angle ABC$ be an inscribed angle subtending arc $AC$ in a circle with center $O$.

Step 1: Prove the theorem for the case where one side of the inscribed angle passes through the center $O$.
Assume side $BC$ is a diameter. Draw radius $OA$.
In $\triangle OAB$, $OA = OB$ because they are both radii of the same circle, making $\triangle OAB$ an isosceles triangle.
Therefore, the base angles are equal: $m(\angle OAB) = m(\angle OBA)$.
The central angle $\angle AOC$ is an exterior angle to $\triangle OAB$.

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