- Home
- >
- Quadrilaterals in a Circle – Explanation & Examples

Contents

# Quadrilaterals in a Circle – Explanation & Examples

We have studied that a quadrilateral is a 4 – sided polygon with 4 angles and 4 vertices. For more details, you can consult the article “Quadrilaterals” in the **“Polygon” section.**

In **geometry exams**, examiners make the questions complex by inscribing a figure inside another figure and ask you to find the missing angle, length, or area. One example from the previous article shows how an inscribed triangle inside a circle makes two chords and follows certain theorems.

This article will discuss what a quadrilateral inscribed in a circle is and the inscribed quadrilateral theorem.

## What is a Quadrilateral Inscribed in a Circle?

**In geometry, a quadrilateral inscribed in a circle, also known as a cyclic quadrilateral or chordal quadrilateral, is a quadrilateral with four vertices on the circumference of a circle. In a quadrilateral inscribed circle, the four sides of the quadrilateral are the chords of the circle.**

In the above illustration, the four vertices of the quadrilateral* ABCD* lie on the circle’s circumference. In this case, the diagram above is called a quadrilateral inscribed in a circle.

## Inscribed Quadrilateral Theorem

There are two theorems about a cyclic quadrilateral. Let’s take a look.

### Theorem 1

The first theorem about a cyclic quadrilateral state that:

**The opposite angles in a cyclic quadrilateral are supplementary. i.e., the sum of the opposite angles is equal to 180˚.**

Consider the diagram below.

If a, b, c, and d are the inscribed quadrilateral’s internal angles, then

**a + b = 180˚ **and** c + d = 180˚.**

Let’s prove that;

- a + b = 180˚.

Join the vertices of the quadrilateral to the center of the circle.

Recall the inscribed angle theorem (the central angle = 2 x inscribed angle).

∠*COD *= 2∠*CBD*

∠*COD *= 2b

Similarly, by intercepted arc theorem,

∠*COD = 2 *∠*CAD *

∠*COD *= 2a

∠*COD + *reflex ∠*COD =* 360^{o}

2a + 2b = 360^{o}

2(a + b) =360^{o}

By dividing both sides by 2, we get

a + b = 180^{o}.

Hence proved!

### Theorem 2

The second theorem about cyclic quadrilaterals states that:

**The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.**

Consider the following diagram, where a, b, c, and d are the sides of the cyclic quadrilateral and D_{1} and D_{2} are the quadrilateral diagonals.

In the above illustration,

**(a * c) + (b * d) = (D _{1} * D_{2})**

### Properties of a quadrilateral inscribed in a circle

There exist several interesting properties about a cyclic quadrilateral.

- All the four vertices of a quadrilateral inscribed in a circle lie on the circumference of the circle.
- The sum of two opposite angles in a cyclic quadrilateral is equal to 180 degrees (supplementary angles)
- The measure of an exterior angle is equal to the measure of the opposite interior angle.
- The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.
- The perpendicular bisectors of the four sides of the inscribed quadrilateral intersect at the center O.
- The area of a quadrilateral inscribed in a circle is given by Bret Schneider’s formula as:

**Area = √[s(s-a) (s-b) (s – c) (s – c)]**

where a, b, c, and d are the side lengths of the quadrilateral.

s = Semi perimeter of the quadrilateral = 0.5(a + b + c + d)

Let’s get an insight into the theorem by solving a few example problems.

*Example 1*

Find the measure of the missing angles x and y in the diagram below.

__Solution__

x = 80^{ o }(the exterior angle = the opposite interior angle).

y + 70^{ o }= 180^{ o }(opposite angles are supplementary).

Subtract 70^{ o }on both sides.

y = 110^{o}

Therefore, the measure of angles x and y are 80^{o }and 110^{o,} respectively.

*Example 2*

Find the measure of angle ∠Q*PS *in the cyclic quadrilateral shown below.

__Solution__

∠*QPS* is the opposite angle of ∠*SRQ*.

According to the inscribed quadrilateral theorem,

∠*QPS *+ ∠*SRQ* = 180^{o }(Supplementary angles)

∠*QPS *+ 60^{o} = 180^{o}

Subtract 60^{o }on both sides.

∠*QPS *= 120^{ o}

So, the measure of angle ∠Q*PS* is 120^{o}.

*Example 3*

Find the measure of all the angles of the following cyclic quadrilateral.

__Solution__

Sum of opposite angles = 180^{ o}

(y + 2)^{ o }+ (y – 2)^{ o} = 180^{ o}

Simplify.

y + 2 + y – 2 =180^{ o}

2y = 180^{ o}

Divide by 2 on both sides to get,

y = 90^{ o}

On substitution,

(y + 2)^{ o }⇒ 92^{ o}

(y – 2)^{ o }⇒ 88^{ o}

Similarly,

(3x – 2)^{ o} = (7x + 2)^{ o}

3x – 2 + 7x + 2 = 180^{ o}

10x =180^{ o}

Divide by 10 on both sides,

x = 18^{ o}

Substitute.

(3x – 2)^{ o} ⇒ 52^{ o}

(7x + 2)^{ o }⇒ 128^{o}