A Cyclic Quadrilateral is a four-sided polygon encircled by a circle. The word “cyclic” is derived from the Greek word “kuklos”, which means “circle” or “wheel”, and the word “quadrilateral” is derived from the ancient Latin word “Quadri”, which means “four-side” or “latus”. It is a particular type of quadrilateral whose four vertices lie on the circumference of a circle. In a cyclic quadrilateral, the four sides of the quadrilateral are the chords of the circle.

The cyclic quadrilateral is also known as an inscribed quadrilateral. The circumcircle or circumscribed circle is a circle that contains all of the vertices of any polygon on its circumference. This article will discuss in detail the cyclic quadrilateral, its definition, theorems, properties, angles, and cyclic quadrilateral solved examples.

Cyclic Quadrilateral Definition

A quadrilateral \(ABCD\) is a cyclic quadrilateral if all the four vertices of it lie on a circle. In other words, a quadrilateral is a cyclic quadrilateral if all its four vertices lie on the circumference of the circle.

For example, in the figure given below, the quadrilateral \(ABCD\) is cyclic since all the four vertices \(A, B, C\) and \(D\) lies on the circumference of the circle.

Cyclic Quadrilateral Angles

Consider the quadrilateral \(ABCD\) whose vertices lie on a circle. The opposite angles of a cyclic quadrilateral are supplementary. Let us prove that.

Connect the centre \(O\) of the circle with each vertex. We now see four radii \(OA, OB, OC\) and \(OD,\) giving rise to four isosceles triangles \(\Delta OAB,\Delta OBC,\Delta OCD\) and \(\Delta ODA.\) The sum of the angles around the centre of the circle is \({360^ \circ }.\) The angle sum of each isosceles triangle is \({180^ \circ }.\)

Therefore, we get from the figure,

\(2(\angle 1 + \angle 2 + \angle 3 + \angle 4) + \) Angle at centre \(O = 4 \times {180^ \circ }\)

\(2(\angle 1 + \angle 2 + \angle 3 + \angle 4) + {360^ \circ } = {720^ \circ }\)

\((\angle 1 + \angle 2 + \angle 3 + \angle 4) = {180^ \circ }\)

We now interpret this as

1. \((\angle 1 + \angle 2) + (\angle 3 + \angle 4) = {180^ \circ }\) (Sum of opposite angles \(B\) and \(D\))

2. \((\angle 1 + \angle 4) + (\angle 2 + \angle 3) = {180^ \circ }\) (Sum of opposite angles \(A\) and \(C\))

Exterior Angle of a Cyclic Quadrilateral

An exterior angle of a quadrilateral is an angle formed by one of its sides and the extension of an adjacent side.

Let the side \(AB\) of the cyclic quadrilateral \(ABCD\) be extended to \(E.\)

Here \(\angle ABC\) and \(\angle CBE\) are linear pairs, their sum is \({180^ \circ }\) and the angles \(\angle ABC\) and \(\angle ADC\) are the opposite angles of a cyclic quadrilateral, and their sum is also \({180^ \circ }.\) 

From this, \(\angle ABC + \angle CBE = \angle ABC + \angle ADC\) and finally we get \(\angle CBE = \angle ADC.\)  Similarly, we can prove it for all angles.

Radius of Cyclic Quadrilateral

Let \(ABCD\) be a cyclic quadrilateral and let \(AB=a, BC=b, CD=c\) and \(AD=d\). And the semi-perimeter \(s = \frac{{a + b + c + d}}{2}\) then the radius of the cyclic quadrilateral is given by:

\(R = \frac{1}{4}\sqrt {\frac{{(ab + cd)(ac + bd)(ad + bc)}}{{(s – a)(s – b)(s – c)(s – d)}}}\)

Diagonals of a Cyclic Quadrilateral

Let \(ABCD\) be a cyclic quadrilateral and let \(AB=a,BC=b,CD=c, AD=d, AC=p\) and \(BD=q,\) then the diagonals of a cyclic quadrilateral are given by

\(p = \sqrt {\frac{{(ac + bd)(ad + bc)}}{{(ab + cd)}}} \)

\(q = \sqrt {\frac{{(ab + cd)(ac + bd)}}{{(ad + bc)}}} \)

Area of Cyclic Quadrilateral (Brahmagupta Formula)

Let \(ABCD\) be a cyclic quadrilateral and let \(AB=a,BC=b,CD=c\) and \(AD=d,\) then the area of the cyclic quadrilateral is given by:

\(A = \sqrt {(s – a)(s – b)(s – c)(s – d)} \)

Where the semi-perimeter \(s = \frac{{a + b + c + d}}{2}\)

Cyclic Quadrilateral Theorems

There are three important theorems related to the cyclic quadrilateral.

Theorem 1: In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary.

Proof: Given: \(ABCD\) is a cyclic quadrilateral of a circle with a centre at \(O\).
To prove: \(\angle BAD + \angle BCD = {180^ \circ }\)
\(\angle ABC + \angle ADC = {180^ \circ }\)
\(AB\) is the chord of the circle
\(\angle 5 = \angle 8\,\,\,\,\,…\left( 1 \right)\) (Angles in the same segment are equal)
\(BC\) is the chord of the circle
\(\angle 1 = \angle 6\,\,\,\,\,…\left( 2 \right)\) (Angles in the same segment are equal)
\(CD\) is the chord of the circle
\(\angle 2 = \angle 4\,\,\,\,…\left( 3 \right)\) (Angles in the same segment are equal)
\(AD\) is the chord of the circle
\(\angle 7 = \angle 3\,\,\,\,…\left( 4 \right)\) (Angles in the same segment are equal)

By angle sum property of a quadrilateral
\(\angle A + \angle B + \angle C + \angle D = {360^ \circ }\)
\(\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 7 + \angle 8 + \angle 5 + \angle 6 = {360^ \circ }\)
\(( – 1 + \angle 2 + \angle 7 + \angle 8) + ( – 3 + \angle 4 + \angle 5 + \angle 6) = {360^ \circ }\)
\(( – 1 + \angle 2 + – 7 + \angle 8) + ( – 7 + \angle 2 + \angle 8 + \angle 1) = {360^ \circ }\)

From equation \(\left( 1 \right),{\rm{ }}\left( 2 \right),{\rm{ }}\left( 3 \right),\) and \(\left( 4 \right)\)
\(2(\angle 1 + \angle 2 + \angle 7 + \angle 8) = {360^ \circ }\)
\((\angle 1 + \angle 2 + \angle 7 + \angle 8) = {180^ \circ }\)
\((\angle 1 + \angle 2) + (\angle 7 + \angle 8) = {180^ \circ }\)
\(\angle BAD + \angle BCD = {180^ \circ }\)

Similarly, \(\angle ABC + \angle ADC = {180^ \circ }\)
Hence proved.

The converse of this theorem is also true, which states that if opposite angles of a quadrilateral are supplementary, then the quadrilateral is a cyclic quadrilateral.

Draw a quadrilateral \(PQRS\). Since in quadrilateral \(PQRS,\angle P + \angle R = {180^ \circ }\) and \(\angle S + \angle Q = {180^ \circ }\) 

Therefore, draw a circle passing through the points \(P, Q\) and \(R\) and observe that it also passes through the point \(S.\) Hence, we conclude that quadrilateral \(PQRS\) is a cyclic quadrilateral.

Theorem 2 – Ptolemy’s theorem: If there is a quadrilateral inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.

If \(ABCD\) is a cyclic quadrilateral, \(AB\) and \(CD\), and \(AD\) and \(BC\) are opposite sides. \(AC\) and \(BD\) are the diagonals.

\((AB \times CD) + (AD \times BC) = AC \times BD\)

Theorem 3: If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

To Prove: \(\angle CBE = \angle ADC\)

Proof: Let the side \(AB\) of the cyclic quadrilateral \(ABCD\) is extended to \(E\). 

Here \(\angle ABC\) and \(\angle CBE\) are linear pair, their sum is \({180^ \circ }\) and the angles \(\angle ABC\) and \(\angle ADC\) are the opposite angles of a cyclic quadrilateral, and their sum is also \({180^ \circ }\). 

From this, \(\angle ABC + \angle CBE = \angle ABC + \angle ADC\) and finally we get \(\angle CBE = \angle ADC\). Similarly, we can prove it for all angles.

Properties

1. In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary.

2. If opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

3. If there is a quadrilateral inscribed in a circle, then the product of the diagonals is equal to the sum of the product of its two pairs of opposite sides.

4. If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle

5. The quadrilateral formed by the angle bisector of a cyclic quadrilateral is also cyclic

6. A cyclic trapezium is isosceles, and its diagonals are equal.

Solved Examples

Q.1. If \(PQRS\) is a cyclic quadrilateral in which \(\angle PSR = {70^ \circ }\) and \(\angle QPR = {40^ \circ },\) then find the measure of \(\angle PRQ.\)

Ans: It is given that \(PQRS\) is a cyclic quadrilateral.
\(\angle PSR = {70^ \circ }\) and \(\angle QPR = {40^ \circ }\)
\(\angle PSR + \angle PQR = {180^ \circ }\) (In a cyclic quadrilateral, the sum of either pair of opposite angles is supplementary) 
\( \Rightarrow {70^ \circ } + \angle PQR = {180^ \circ }\)
\( \Rightarrow \angle PQR = {180^ \circ } – {70^ \circ }\)
\( \Rightarrow \angle PQR = {110^ \circ }\)
In \(\Delta PQR\), we have
\(\angle PQR + \angle PRQ + \angle QPR = {180^ \circ }\) (By the angle sum property of a triangle)
\( \Rightarrow {110^ \circ } + \angle PRQ + {40^ \circ } = {180^ \circ }\)
\( \Rightarrow \angle PRQ = {180^ \circ } – {150^ \circ }\)
\( \Rightarrow \angle PRQ = {30^ \circ }\)

Q.2. In the figure given below, find the value of \({x^ \circ }\) and \({y^ \circ }\)

Ans: By the exterior angle property of a cyclic quadrilateral, if one side of a cyclic quadrilateral is produced, then the exterior angle equals the interior opposite angle.
So, we get, \({y^ \circ } = {100^ \circ }\)
and \({x^ \circ } + {30^ \circ } = {60^ \circ }\)
\( \Rightarrow {x^ \circ } = {60^ \circ } – {30^ \circ } = {30^ \circ }\)
So, \({x^ \circ } = {30^ \circ }\) and \({y^ \circ } = {100^ \circ }\)

Q.3. If in cyclic quadrilateral \(ABCD,AD\parallel BC\),then prove that \(\angle A = \angle D\)
Ans: Given that \(ABCD\) is a cyclic quadrilateral on which \(AD\parallel BC\).

We have to prove that \(\angle A = \angle D\)
We know that sum of opposite angles of a cyclic quadrilateral is \({180^ \circ }.\)
So, \(\angle A + \angle C = 180^\circ \,\,\,\,\,….\left( 1 \right)\)
And \(\angle B + \angle D = 180^\circ \,\,\,\,\,\,\,\,…\left( 2 \right)\)
but \(\angle A+\angle B=180^{\circ}\) (co-interior corresponding angles made when the transversal, side \(A B\) is intersecting the parallel line \(AD\) & \(BC\,\,\,\,\,…\left( 2 \right)\)
From equation \(\left( 2 \right)\) and \(\left( 3 \right)\)
\(\angle A + \angle B = \angle B + \angle D\)
\( \Rightarrow \angle A = \angle D\)
Hence, proved.

Q.4. In the figure given below, \(ABCD\) is a cyclic quadrilateral in which \(AC\) and \(BD\) are its diagonals. If \(\angle DBC = {55^ \circ }\) and \(\angle BAC = {45^ \circ },\) find the measure of \(\angle BCD.\)

Ans: Given that \(ABCD\) is a cyclic quadrilateral in which \(AC\) and \(BD\) are its diagonals. 
Also, \(\angle DBC = {55^ \circ }\) and \(\angle BAC = {45^ \circ },\) 
We need to find the measure of \(BCD\).
In a circle, angles in the same segment are equal.
\(\angle CAD = \angle DBC = {55^ \circ }\)
Therefore, \(\angle DAB = \angle CAD + \angle BAC = {55^ \circ } + {45^ \circ } = {100^ \circ }\)
But \(\angle DAB + \angle BCD = {180^ \circ }\) (Opposite angles of a cyclic quadrilateral are supplementary) 
So, \(\angle BCD = {180^ \circ } – {100^ \circ } = {80^ \circ }\)

Q.5 In cyclic quadrilateral \(ABCD\), if the ratio \(\angle A:\angle C = 5:4,\) then find the measure of \(\angle A\) and \(\angle C.\) 

Ans: In cyclic quadrilaterals \(ABCD\), given that, \(\angle A:\angle C = 5:4\)
Let the common multiple be \(x\), and we get,
\(\angle A = 5x,\angle C = 4x\)
Then, we know that,
\(\angle A + \angle C = {180^ \circ }\)
\(5x + 4x = {180^ \circ }\)
\(9x = {180^ \circ }\)
\(x = {20^ \circ }\)
Now, putting the value of \(x\), we get
\(\angle A = 5x\)
\(\angle A = 5 \times {20^ \circ }\)
\(\angle A = {100^ \circ }\)
And the value of \(\angle C = 4x = 4 \times {20^ \circ } = {80^ \circ }.\)

Summary

This article discusses the cyclic quadrilateral, its definition, interior angles, exterior angles, the formula for finding its radius, the formula for finding its diagonals, the formula for finding the area, and important theorems related to the cyclic quadrilateral. Also, we have learnt the properties of a cyclic quadrilateral along with the solved examples.

FAQs

Q.1. How do you prove a shape is a cyclic quadrilateral?
Ans: To prove a shape is a cyclic quadrilateral, it must satisfy the property that the sum of either pair of opposite angles is supplementary.

Q.2. What makes a cyclic quadrilateral? 
Ans: When all the four vertices of quadrilateral lie on the circumference of a circle, then we get a cyclic quadrilateral.

Q.3. Can a cyclic quadrilateral be in a semicircle?
Ans: Yes, if we take the diameter as the one side of the quadrilateral and the other sides lie on the circumference, then a cyclic quadrilateral can be in a semicircle.  

Q.4. Are opposite angles of a cyclic quadrilateral equal?
Ans: Yes. The opposite angles of a cyclic quadrilateral are equal if the quadrilateral is a parallelogram, a square or a rectangle.

Q.5. Is a parallelogram a cyclic quadrilateral?
Ans: No, every parallelogram is not a cyclic quadrilateral, the reason being it is not always possible that the sum of the opposite angles of a parallelogram is supplementary.