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Cyclic Quadrilateral

Circles III - Concepts

Angle at the Centre is Double The Angle at Circumference

Angles in the Same Segment are Equal

Angle in a Semicircle

Cyclic Quadrilateral

Class - 9th CBSE Subjects

Concept Explanation

Cyclic Quadrilateral

CYCLIC QUADRILATERAL

DEFINITION A quadrilateral is called a cyclic quadrilateral if its all vertices lie on a circle.

A cyclic quadrilateral has some special properties which other quadrilaterals, in general, need not have . We shall state and prove these properties as theorems.

Theorem 1 The sum of either pair of opposite angles of a cyclic quadrilaterals is $large 180^{circ}$

The opposite angles of a cyclic quadrilateral are supplementary.

Given A cyclic quadrilateral ABCD. TO PROVE $large angle A + angle C = 180^{circ} and angle B + angle D = 180^{circ}$ CONSTRUCTION Join AC and BD. PROOF Consider side AB of quadrilateral ABCD as the chord of the circle. Clearly, $large angle ACB ;; and;; angle ADB$ are angles in the same segment determined by chord AB of the circle. $large therefore angle ACB = angle ADB$ [ since angles in the same segment are equal] ....(i) Now, consider the side BC of quadrilateral ABCD as the chord of the circle. We find that $large angle BAC ;; and ;;angle BDC$ are angles in the same segment $large therefore angle BAC = angle BDC$ [ $large because$ Angles in the same segment are equal ] ....(ii) Adding equations (i) and (ii) , we get $large Rightarrow ;;;angle ACB + angle BAC = angle ADB + angle BDC$ $large Rightarrow ;;; angle ACB + angle BAC = angle ADC$ $large Rightarrow ;;; angle ABC + angle ACB + angle BAC = angle ABC + angle ADC$ $large Rightarrow ;;; 180^{circ}= angle ABC + angle ADC$ [ $large because$ The sum of the angle of a triangle is $large 180^{circ}$ ] $large Rightarrow ;;; angle ABC + angle ADC = 180^{circ}$ $large Rightarrow ;;; angle B + angle D = 180^{circ}$ $large But, ;;; angle A + angle B + angle C + angle D = 360^{circ}$ $large therefore ;;;angle A + angle C = 360^{circ}-(angle B +angle D)$ $large Rightarrow ;;;angle A + angle C = 360^{circ} -180^{circ}=180^{circ}$ Hence, $large angle A + angle C = 180^{circ};;and;;angle B + angle D = 180^{circ}$ The converse of this theorem is also true as given below.
Converse Theorem: If the sum of any pair of opposite angles of a quadriletral is $180^0$ , then the quadrilateral is cyclic.
Given: A quadrilateral ABCD in which $angle B + angle D = 180^0$ Tp Prove: ABCD is a cyclic quadrilateral Construction: Draw a circle passing through A, B and C. Proof: Suppose the circle meets CD or CD produced at E. Join AE Now ABCE is a cyclic Quadrilateral. As Opposite angles of a cyclic quadrilateral are supplementary $Rightarrow angle B + angle E = 180^0$ .........(1) But, $angle B + angle D = 180^0$ ........(2) [Given] From Equation (1) and (2) we get $angle D = angle E$ But this is not possible $angle E$ is an exterior angle to the $Delta ADE$ in acse (1) or $angle D$ is an exterior angle to the $Delta ADE$ in case (2) And we know that the exterior angle of a triangle is equal to sum of interior opposite angles That is exterior angle is always greater than interior opposite angle Thus our assumption is wrong Hence ABCD is a cyclic quadrilateral.	Case (1) Case (2)
Illustration: In the figure ACDF is a cyclic quadrilateral. A circle passing through A and F meets AC and DF in the points B and E respectively. Prove that BE \|\| CD.
Solution: ACDF is a cyclic quadrilateral and In a cyclic quadrilateral opposite angles are supplementary $angle ACD + angle AFD = 180^0$ .....................(1) Also ABEF is a cyclic quadrilateral and In a cyclic quadrilateral opposite angles are supplementary $angle ABE + angle AFE = 180^0$ ..................(2) $angle AFE ;is; same; as ; angle AFD$ $Rightarrow;; angle ABE + angle AFD = 180^0$ ..............(3) From Equation 1 and 3 we get $Rightarrow;; angle ABE = angle ACD$ But they are corresponding angles when BE and CD are two straight lines and BC is the transversal As they are equal, So BE \|\| CD

Sample Questions

(More Questions for each concept available in Login)

Question : 1

If $angle x=60^o;and ; angle y=120^o,;then; angle w-angle z$ is equal to

$large 0^o$

$30^o$

$120^o$

$large 60^o$

Right Option : D

View Explanation

Question : 2

In the figure given, $angle ABC$ is equal to

$frac {x}{2}$

$90^o-x$

$180^o-x$

Right Option : D

View Explanation

Question : 3

In the figure given below, ABCD is a cyclic quadrilateral.Which of following relations is correct ?

$x^o+w^o=180^o$

$x^o=y^o$

$z^o+w^o=180^o$

$x^o+y^o+z^o=180^o$

Right Option : A

View Explanation

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Chapters

Polynomials II

Linear Equations in One Variable

Introduction to Euclids Geometry

Herons Formula Complete

Polynomial I

Surface Area and Volume II

Surface Area and Volume I

Coordinate Geometry

Areas Of Parallelograms And Triangles II

Areas Of Parallelograms And Triangles I

Linear Equations in Two Variables

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Cyclic Quadrilateral

Cyclic Quadrilateral

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Shivam Rana

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Cheshta

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Yatharthi Sharma

Om Umang

Eshan Arora

Archit Segal

Shreya Shrivastava

Prisha Gupta