Inscribed Angle Theorem and Thales' Semicircle Corollary

Step 1 (draw the diameter from the vertex and use isosceles triangles). Draw the diameter $CD$ passing through the vertex and the center. Because two radii have equal length $OA = OC = R$, the triangle $\triangle OAC$ is isosceles at the center, so its base angles are equal: $\angle OCA = \angle OAC$.

Step 2 (apply the exterior angle theorem to both halves). The exterior angle at the center equals the sum of the two opposite interior angles, giving $\angle AOD = \angle OCA + \angle OAC = 2\angle OCA$. Applying the exact same isosceles-triangle argument to $\triangle OBC$ gives $\angle BOD = 2\angle OCB$.

Step 3 (combine the two halves and deduce Thales' theorem). Adding the two central angles (or subtracting them when the center lies outside the inscribed angle) yields $\angle AOB = \angle AOD + \angle BOD = 2(\angle OCA + \angle OCB) = 2\angle ACB$. In the special case where the chord is a diameter, the central angle is straight ($\angle AOB = 180^\circ$), which immediately gives Thales' semicircle theorem $\angle ACB = \dfrac{1}{2} \cdot 180^\circ = 90^\circ$.