Max comes across a problem of geometry dealing with cyclic quadrilateral. He is struggling to figure out how to solve this geometric figure.
Here, Max supposed to find the value of $$\angle e$$, and the value of  $$\angle a=50^{\circ}$$ is given. His friend, Sam, drops in and helps him understand the cyclic quadrilateral, cyclic quadrilateral's properties, and important theorems related to the cyclic quadrilateral.

A cyclic quadrilateral means a quadrilateral that is inscribed in a circle. That means there is a circle that passes through all four vertices of the quadrilateral.

Cyclic quadrilaterals are important in solving various types of geometry problems, where angle chasing is required. ## Lesson PLan

 1 What Are Cyclic Quadrilaterals? 2 Challenging Questions 3 Solved Examples 4 Interactive Questions on Cyclic Quadrilaterals 5 Important Notes

### Definition

A cyclic quadrilateral means a quadrilateral that is inscribed in a circle. That means there is a circle that passes through all four vertices of the quadrilateral. The vertices are said to be concyclic.
The center of the circle is known as the circumcenter and the radius of the circle is known as the circumradius.

The word "cyclic" is from the Greek word "kuklos", which means "circle" or "wheel".

The word "quadrilateral" is derived from the ancient Latin word "Quadri", which means "four-side" or "latus".

In the figure given below, ABCD is a cyclic quadrilateral. ### The Sum of a Pair of Opposite Angles is $$180^{\circ}$$

In a cyclic quadrilateral, the sum of a pair of opposite angles is $$180^{\circ }$$(supplementary).

Let $$\angle A$$, $$\angle B$$, $$\angle C$$, and $$\angle D$$ are the four angles of an inscribed quadrilateral. Then,

$$\angle A + \angle C = 180^{\circ }$$

$$\angle B + \angle D = 180^{\circ }$$

 ∠A + ∠C =∠B + ∠D =\begin{align}\pi \ radians\end{align}

Therefore, the sum of all the angles an inscribed quadrilateral is equals to $$360^{\circ}$$. Hence,

∠A + ∠C +∠B + ∠D =\begin{align}2\pi \ radians\end{align}

### The Area of a Cyclic Quadrilateral

The area of a cyclic quadrilateral is $$K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}$$ where $$a$$, $$b$$, $$c$$, and $$d$$ are the four sides of the quadrilateral. Heron's formula for a triangle is also derived from this equation

 $$K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}$$

where s, the semi perimeter, is defined as:

$$s=(\frac{1}{2})\times (a+b+c+d)$$

All vertices of a cyclic quadrilateral always lie on the circumference of the circle.
The exterior angle formed if any one side of the cyclic quadrilateral extended and is equal to the sum of the interior angle opposite to it.
In a cyclic quadrilateral, $$d1 / d2 = \text{sum of product of opposite sides}$$, which shares the diagonals endpoints.
In a cyclic quadrilateral, the perpendicular bisectors always concurrent.
In a cyclic quadrilateral, the perpendicular bisectors of the four sides of the cyclic quadrilateral meet at the center O.

Coming back to Max's problem. Sam solved it by using the cyclic quadrilateral's property "the sum of a pair of opposite angles is $$180^{\circ }$$(supplementary)". i.e.  $$\angle a+ \angle c=180^{\circ}$$

And we also know that the sum of all angles formed on the same side of a line at a given point on the line is  $$180^{\circ }$$.  i.e.  $$\angle c+ \angle e=180^{\circ}$$.

From the above two statements, Sam And Max infer that  $$\angle a=\angle e=50^{\circ}$$.

### Ptolemy Theorem

According to Ptolemy Therorem, in a cyclic quadrilateral with successive vertices $$A,\; B,\; C,\; D$$ , sides $$a = AB$$, $$b = BC$$, $$c = CD$$, $$d = DA$$, and  diagonals  $$p = AC$$ , $$q = BD$$.

we can express diagonals in terms of the sides as:

 $p\times q=(a\times c)+ (b\times d)$

### Brahmgupta Theorem

In geometry, Brahmagupta's formula is used to find the area of any quadrilateral given the sides.

Brahmagupta's formula:  the area$$K$$ of a cyclic quadrilateral whose sides have lengths $$a$$, $$b$$, $$c$$, $$d$$ as:

 $$K={\sqrt {(s-a)(s-b)(s-c)(s-d)}}$$

Where $$s$$, the semi perimeter of the cyclic quadrilateral , is :

 $$s=(\frac{1}{2})\times (a+b+c+d)$$ Challenging Questions
1. How do you prove a quadrilateral is cyclic?
2. What is a non-cyclic quadrilateral?

## Solved Examples

 Example 1

In the given problem, PQRS is a cyclic quadrilateral, and $$\triangle$$ PQS is an equilateral triangle, then find the measure of $$\angle QRS$$. Solution

Given,$$\triangle$$ PQS is an equilateral triangle.

$$\therefore$$ $$\angle QPS=60^{\circ}$$

Now,we know that opposite angle are supplementary

$$\therefore \; \angle QRS + \angle QPS=180^{\circ}$$

From above two equations we have

$$\angle QRS +60^{\circ} =180^{\circ}$$

$$\angle QRS=180^{\circ}-60^{\circ}$$

$$\angle QRS=120^{\circ}$$

 $$\therefore$$ The measure of $$\angle QRS=120^{\circ}$$
 Example 2

In the given figure, ABCD is a cyclic quadrilateral such that$$\angle ADB=50^{\circ}$$ and $$\angle DCA=70^{\circ}$$, then find the measure of $$\angle DAB$$. Solution

Given, $$\angle ADB=50^{\circ}$$ and  $$\angle DCA=70^{\circ}$$

As we know that, angles in the same segment of the circle are always equal,

$$\therefore \angle BCA=\angle ADB=50^{\circ}$$

Now, from above two statements we have:

$$\angle BCD =\angle BCA + \angle DCA$$

$$\angle BCD =70^{\circ}+50^{\circ}$$

$$\angle BCD =120^{\circ}$$

Opposite angles are Supplimentary

$$\therefore \angle DAB +\angle BCD=180^{\circ}$$

Now, $$\angle DAB+120^{\circ}=180^{\circ}$$

$$\angle DAB=180^{\circ}-120^{\circ}$$

$$\angle DAB=60^{\circ}$$

 $$\therefore$$ The measure of $$\angle DAB=60^{\circ}$$

## Interactive Questions

Here are a few questions for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

Circles
grade 9 | Questions Set 2
Circles
grade 9 | Questions Set 1
Circles