Inscribed Angle Theorem Explained

Grades 9-10

Question: .

In circle geometry, one of the most fundamental relationships is the Inscribed Angle Theorem. This theorem describes the relationship between an angle subtended at the center of a circle and an angle subtended at the circumference by the same arc.

The Theorem

For any circle with center

O

Central Angle: The angle $\angle AOC$ is formed by two radii $OA$ and $OC$ .
Inscribed Angle: The angle $\angle ABC$ is formed by two chords $BA$ and $BC$ , where $B$ is any point on the circle's circumference.

The theorem states that the central angle subtended by an arc at the center is twice the inscribed angle subtended by the same arc at any point on the circumference.

\begin{align*} \angle AOC = 2 \times \angle ABC \end{align*}

Visual Properties

In the accompanying diagram, you can observe several key properties:

Constant Inscribed Angle: If you move point $B$ along the major arc $AC$ (the larger part of the circle), the measure of $\angle ABC$ remains constant. This is known as the "angles in the same segment" theorem.
The 2:1 Ratio: The central angle $\angle AOC$ (shown in orange) is always exactly double the inscribed angle $\angle ABC$ (shown in blue), provided $B$ remains on the same side of the chord $AC$ .
Semicircle Case: If the chord $AC$ happens to be a diameter (passing through $O$ ), then $\angle AOC = 180^\circ$ , which implies that $\angle ABC = 90^\circ$ . This special case is known as Thales' Theorem.

Geometric Proof Sketch

The theorem can be proven by dividing the inscribed angle into two parts using a line from

B

through the center

O

. This creates two isosceles triangles (

\triangle OAB

and

\triangle OBC

) because the segments

OA, OB,

and

OC

are all radii. Using the property that the exterior angle of a triangle is equal to the sum of the two opposite interior angles:

\begin{align*} \text{Let } \angle OBA = x \text{ and } \angle OBC = y \\ \angle OAB = x \text{ (base angles of isosceles } \triangle OAB) \\ \angle OCB = y \text{ (base angles of isosceles } \triangle OBC) \\ \text{Exterior angle at } O \text{ for } \triangle OBA = 2x \\ \text{Exterior angle at } O \text{ for } \triangle OBC = 2y \\ \angle AOC = 2x + 2y = 2(x + y) = 2 \angle ABC \end{align*}

This simple yet powerful relationship is a cornerstone of circle geometry and is used extensively in solving complex geometric proofs and trigonometric problems.