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

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

Strategy: prove it in three cases by where the center $O$ lies relative to the inscribed angle. Case 1 (center on a side) uses an isosceles triangle and the exterior-angle relation. Cases 2 and 3 reduce to Case 1 by drawing diameter $CD$: when $O$ is inside, the two parts add; when $O$ is outside, they subtract.

Step 1: The Inscribed Angle Theorem

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

We prove this in three cases based on where center $O$ lies relative to the inscribed angle.

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 $△OBC$, 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:

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

In all three cases:

All inscribed angles subtending the same arc are therefore equal. ∎