Consider the first diagram and draw the auxiliary chord $\overline{BC}$ and point $P.$

By the Inscribed Angle Theorem, the measure of $\angle ABC$ is half the measure of $\overgroup{AC}.$ Also, the Tangent and Intersected Chord Theorem tells that the measure of $\angle PCB$ is half the measure of $\overgroup{BC}.$ Next, the Triangle Exterior Angle Theorem says that $m\angle PCB$ equals the sum of the measures of $\angle ABC$ and $\angle 1.$ Express the measure of $\angle 1$ using this information. Next, substitute the two angle measures found before into the right-hand side. Finally, the proof can be completed by factoring $\frac{1}{2}$ out.