# The Proof of the Third Case of the Inscribed Angle Theorem

This is the third and the last case of the Proofs of the Inscribed Angle Theorem. In the first case, we used an auxiliary line between the sides of the angles to come up with the proof. In the second case, we drew a line segment from the center to the intersection of the circle and one of the sides to form an isosceles triangle. In this case, we cannot do both. To prove the third case, we will draw an auxiliary line outside the circle.  We draw segment $latex \overline{BD}$ through the center of the circle. The detail of the proof is as follows.

Given

$latex \angle ABC$ is inscribed in circle O.

Prove

$latex m \angle ABC = \frac{1}{2} m \widehat {AC}$

Proof

Draw diameter $latex \overline{BD}$. Now, $latex m \angle ABC = m \angle ABD – m \angle CBD$. by the consequence of the Angle Addition Postulate.

$latex = (\frac{1}{2} m \widehat{AD} – \frac{1}{2} m \widehat{CD})$ by Case 2 of the Inscribed Angle Theorem. (Hint: Draw $latex \overline{OA}$ and $latex \overline{OC}$).

Factoring out $latex \frac{1}{2}$, we have $latex = \frac{1}{2} (m \widehat{AD} – m \widehat{CD})$.

By the consequence of the Angle Addition Postulate and the equality of the central angle and intercepted arc, we have

$latex m \angle ABC = \frac{1}{2} m \widehat{AC}$

which is what we want to prove.

Note: The notation $latex m \widehat {AC}$ means measure of arc $latex AC$.