# Inscribed Angle Theorem - Statement, Proof, and Examples

## TL;DR

The inscribed angle theorem states that an angle inscribed in a circle is half the central angle that subtends the same arc — so ∠inscribed = \tfrac{1}{2} ∠central. This article gives the exact statement, proves it in three cases, derives its main corollaries (angle in a semicircle is 90°; angles on the same arc are equal), and works through examples.

## Why Every Angle Drawn to a Semicircle Comes Out to Exactly a Right Angle

The **inscribed angle theorem** states that the measure of an **inscribed angle** (its vertex on the circle, its two sides being chords) is **half the measure of the central angle** that subtends the same arc. In symbols, if ∠ABC is inscribed and ∠AOC is the central angle on the same arc AC (with O the centre), then:

∠ABC = \tfrac{1}{2} ∠AOC

Equivalently, the inscribed angle equals half its intercepted arc.

## The Exact Statement

Fix a circle with centre O and an arc AC. Two angles "look at" that arc:

- The **central angle** ∠AOC, with vertex at the centre O.

- An **inscribed angle** ∠ABC, with vertex B on the circle and sides that are the chords BA and BC.

The theorem says the inscribed angle is always half the central angle on the same arc:

∠ABC = \tfrac{1}{2} ∠AOC

As long as two angles subtend the identical arc, the one at the centre is exactly twice the one on the circle.

## Proving The Theorem In Three Cases

**Case 1 - the centre lies on one side of the angle.**  Suppose side BC passes through the centre O, so BC is a diameter. Triangle OAB is isosceles (OA = OB), so its base angles are equal: ∠OAB = ∠OBA = θ. The central angle ∠AOC is the exterior angle of triangle OAB at O, so:

∠AOC = ∠OAB + ∠OBA = θ + θ = 2θ

Thus, ∠ABC = θ and ∠ABC = \tfrac{1}{2} ∠AOC.

**Case 2 - the centre lies inside the angle.**  Draw the diameter BD through B, splitting ∠ABC into ∠ABD and ∠DBC.  Case 1 applies to each half:

∠ABD = \tfrac{1}{2} ∠AOD, ∠DBC = \tfrac{1}{2} ∠DOC

Adding the two:

∠ABC = \tfrac{1}{2}(∠AOD + ∠DOC) = \tfrac{1}{2} ∠AOC.

**Case 3 - the centre lies outside the angle.**  Now ∠ABC is the _difference_ of two Case-1 angles:

∠ABC = ∠DBC - ∠DBA = \tfrac{1}{2} ∠DOC - \tfrac{1}{2} ∠DOA = \tfrac{1}{2} ∠AOC.

In all three positions the result is identical, which is why the theorem holds.

## Two corollaries worth memorising

- **Angle in a semicircle is 90° (Thales' theorem).** If AC is a diameter, the inscribed angle on the same arc is 90°.

- **Angles on the same arc are equal.** Two inscribed angles that subtend the same arc both equal half of the one central angle, so they are equal.

## Examples of Inscribed Angle Theorem

### Example 1

**A central angle subtends an arc of 80°. Find the inscribed angle on the same arc.**

The inscribed angle is half the central angle:

∠inscribed = \tfrac{1}{2} × 80° = 40°.

### Example 2

**A student sees an inscribed angle of 35° and reports the central angle on the same arc as 35°. Spot the error.**

The theorem says the central angle is _twice_ the inscribed angle:

∠central = 2 × 35° = 70°.

### Example 3

**In a circle, AC is a diameter and B is any other point on the circle. Find ∠ABC.**

Since AC is a diameter, the central angle ∠AOC = 180°. The inscribed angle ∠ABC is half of that:

∠ABC = \tfrac{1}{2} × 180° = 90°.

### Example 4

**Two inscribed angles, ∠ADB and ∠ACB, both subtend arc AB. If ∠ADB = 52°, find ∠ACB.**

Both angles subtend the same arc, so:

∠ACB = ∠ADB = 52°.

### Example 5

**An inscribed angle intercepts an arc of 130°. Find the inscribed angle, then the central angle on that arc.**

∠inscribed = \tfrac{1}{2} × 130° = 65°.

### Example 6

**A cyclic quadrilateral ABCD has ∠A = 95°. Find ∠C.**

In a cyclic quadrilateral, opposite angles sum to 180°:

∠C = 180° - 95° = 85°.

## Key Takeaways

- The **inscribed angle theorem** says an inscribed angle is half the central angle.

- Corollary: inscribed angles on the same arc are equal.