The circular geometry is really vast. A circle consists of many parts and angles. These parts and angles are mutually supported by certain Theorems, e.g., t**he Inscribed Angle Theorem**, Thales’ Theorem, and Alternate Segment Theorem.

**We will go through the inscribed angle theorem**, but before that, let’s have a brief overview of circles and their parts.

Circles are all around us in our world. There exists an interesting relationship among the angles of a circle. To recall, a chord of a circle is the straight line that joins two points on a circle’s circumference. Three types of angles are formed inside a circle when two chords meet at a common point known as a vertex. These angles are the central angle, intercepted arc, and the inscribed angle.

For more definitions related to circles, you need to go through the previous articles.

*In this article, you will learn:*

- The inscribed angle and inscribed angle theorem,
- we will also learn how to prove the inscribed angle theorem.

## What is the Inscribed Angle?

**An inscribed angle is an angle whose vertex lies on a circle, and its two sides are chords of the same circle.**

On the other hand, a central angle is an angle whose vertex lies at the center of a circle, and its two radii are the sides of the angle.

The intercepted arc is an angle formed by the ends of two chords on a circle’s circumference.

Let’s take a look.

In the above illustration,

**α** = The central angle

**θ** = The inscribed angle

**β** = the intercepted arc.

## What is the Inscribed Angle Theorem?

*The inscribed angle theorem, which is also known as the arrow theorem or the central angle theorem, states that:*

**The size of the central angle is equal to twice the size of the inscribed angle. The inscribed angle theorem can also be stated as:**

**α = 2****θ**

The size of an inscribed angle is equal to half the size of the central angle.

**θ = ½****α**

Where α and θ are the central angle and inscribed angle, respectively.

## How do you Prove the Inscribed Angle Theorem?

*The inscribed angle theorem can be proved by considering three cases, namely:*

- When the inscribed angle is between a chord and the diameter of a circle.
- The diameter is between the rays of the inscribed angle.
- The diameter is outside the rays of the inscribed angle.

**Case 1: When the inscribed angle is between a chord and the diameter of a circle:**

**To prove α = 2θ:**

- △
*CBD*is an isosceles triangle whereby*CD = CB*= the radius of the circle. - Therefore, ∠ CDB = ∠ DBC = inscribed angle = θ
- The diameter AD is a straight line, so ∠
*BCD*= (180**–**α) ° - By triangle sum theorem, ∠
*CDB*+ ∠DBC + ∠BCD = 180°

θ + θ + (180 **–** α) = 180°

Simplify.

⟹ θ + θ + 180 **–** α = 180°

⟹ 2θ + 180 – α = 180°

Subtract 180 on both sides.

⟹ 2θ + 180 – α = 180°

⟹ 2θ – α = 0

**⟹**** 2θ = α. Hence proved.**

**Case 2: when the diameter is between the rays of the inscribed angle.**

To prove 2θ = α:

- First, draw the diameter (in dotted line) of the circle.

- Let the diameter bisects θ into θ
_{1}and θ Similarly, the diameter bisects α into α_{1 }and α_{2}.

⟹ θ_{1} + θ_{2} = θ

⟹ α_{1 }+ α_{2} = α

- From the first case above, we already know that,

⟹ 2θ_{1 }= α_{1}

⟹ 2θ_{2} = α_{2}

- Add the angles.

⟹ α_{1 }+ α_{2} = 2θ_{1 }+ 2θ_{2}

⟹ α_{1 }+ α_{2} = 2 (θ_{1 }+ 2θ_{2})

**Hence, ****2θ = α:**

**Case 3: When the diameter is outside the rays of the inscribed angle.**

To prove 2θ = α:

- Draw the diameter (in dotted line) of the circle.

- Since 2θ
_{1}= α_{1}

⟹ 2 (θ_{1 }+ θ) = α + α_{1}

⟹ But, 2θ_{1 }= α_{1 }and 2θ_{2} = α_{2}

⟹ By substitution, we get,

2θ = α:

**Solved examples about inscribed angle theorem**

*Example 1*

Find the missing angle x in the diagram below.

__Solution__

By inscribed angle theorem,

The size of the central angle = 2 x the size of the inscribed angle.

Given, 60° = inscribed angle.

Substitute.

The size of the central angle = 2 x 60°

= 120°

*Example 2*

Given that ∠*QRP* = (2x + 20) ° and ∠*PSQ *= 30°, find the value of x.

__Solution__

By inscribed angle theorem,

Central angle = 2 x inscribed angle.

∠*QRP =2*∠*PSQ*

∠*QRP *= 2 x 30°.

= 60°.

Now, solve for x.

⟹ (2x + 20) ° = 60°.

Simplify.

⟹ 2x + 20° = 60°

Subtract 20° on both sides.

⟹ 2x = 40°

Divide both sides by 2.

⟹ x = 20°

So, the value of x is 20°.

*Example 3*

Solve for angle x in the diagram below.

__Solution__

Given the central angle = 56°

2∠*ADB =*∠*ACB*

2x = 56°

Divide both sides by 2.

x = 28°

*Example 4*

If ∠ *YMZ* = 150°, find the measure of ∠*MZY* and ∠ *XMY.*

__Solution__

Triangle MZY is an isosceles triangle, Therefore,

∠*MZY =*∠*ZYM*

Sum of interior angles of a triangle = 180°

∠*MZY = *∠*ZYM = *(180° – 150°)/2

= 30° /2 = 15°

Hence, ∠*MZY = *15°

And by inscribed angle theorem,

2∠*MZY = *∠ *XMY*

∠ *XMY *= 2 x 15°

= 30°