The Inscribed Angle Theorem

A proof that an inscribed angle is half the central angle subtending the same arc, covering three cases based on the position of the center.

Keywords: inscribed angle, central angle, isosceles triangle, circle theorem

Prerequisites: circle-angles-intro, isosceles-triangle-proof · Difficulty: intermediate

The Inscribed Angle Theorem

Theorem: An inscribed angle equals half the central angle subtending the same arc.

Proof Strategy (Three Cases)

Case 1 (center on a side): Diameter from $C$ through $O$. Triangle $OCB$ is isosceles ($OC = OB$). The exterior angle $\angle AOB = 2 \cdot \angle ACB$.

Case 2 (center inside): Diameter $CD$ splits the angle into two Case 1 sub-problems. Add them.

Case 3 (center outside): Diameter $CD$ gives a difference of two Case 1 results. Subtract them.

The Proof

Step 1: The Inscribed Angle Theorem

We prove $\angle ACB = \tfrac{1}{2} \cdot \angle AOB$ in three cases depending on the position of center $O$.

Step 2: Case 1: Isosceles Triangle Argument

$CA$ passes through $O$, so it is a diameter. Triangle $OBC$ is isosceles ($OB = OC$ = radius), giving $\angle OBC = \angle OCB$.

The central angle $\angle AOB$ is an exterior angle of $\triangle OBC$, so $\angle AOB = 2 \cdot \angle OCB = 2 \cdot \angle ACB$.

Step 3: Case 2: Center Inside — Split by Diameter

Draw diameter $CD$ through $O$. It splits the inscribed angle into two parts. By Case 1 applied to each part:

$\angle AOD = 2 \cdot \angle ACD$ and $\angle DOB = 2 \cdot \angle DCB$.

Adding: $\angle AOB = 2 \cdot \angle ACB$.

Step 4: Case 3: Center Outside — Subtract

Draw diameter $CD$ through $O$. The inscribed angle is the difference of two Case 1 angles:

$\angle ACB = \angle ACD - \angle BCD$

By Case 1, central angles are double, so subtracting: $\angle AOB = 2 \cdot \angle ACB$.

Step 5: The Theorem Holds in All Cases

In every configuration:

\angle \text{inscribed} = \tfrac{1}{2} \cdot \angle \text{central}
    

Corollary: All inscribed angles subtending the same arc are equal.

Conclusion

In all three cases:

\angle ACB = \tfrac{1}{2} \cdot \angle AOB
    

All inscribed angles subtending the same arc are therefore equal.