Cyclic Quadrilateral and Intersecting / Non-intersecting Circles

We all have played (and many still play) games where participants (children and adults alike) were made to sit in the formation of a circle and different games like Cat and Mouse, Passing the Parcel, etc. were played. Well, the Mathematical games involving geometrical figures like a quadrilateral, are no less fun. Let us know more!

Suggested Videos

previous arrow
next arrow
previous arrownext arrow


What is a Cyclic Quadrilateral?

A Cyclic Quadrilateral is a four-sided figure whose all four vertices lie on the circle. In other words, it is a quadrilateral whose all four vertices touch the circle.


[Source: BBC]


  • The sum of opposite angles of a cyclic quadrilateral is 180º.
  • The sum of opposite angles of a cyclic quadrilateral is equal to the sum of other two opposite angles.
  • These observations hold TRUE for any shape of a cyclic quadrilateral.


In the figure given above, ABCD is a cyclic quadrilateral. O is any point inside this quadrilateral joined by line segments to vertices B and D. We make the following observations: –

  1. \( \angle \) DCB = (\( \angle \) DOB) / 2
  2. \( \angle \) DAB = (\( \angle \) BOD) / 2


The angle subtended by a chord at the centre of the circle is twice the angle subtended by the same arc at any other point. Adding equations (a) and (b), we get: –

\( \angle \) DCB +\( \angle \) DAB = (\( \angle \) DOB +\( \angle \) BOD) / 2 — (c)

Looking at the figure, it is clear that the sum of angles about point O {forming the R.H.S. of equation (c)} is 360º. Therefore,\( \angle \) DCB +\( \angle \) DAB = 360 / 2 = 180 º. And similarly,\( \angle \) ABC +\( \angle \) ADC = 180º.

Theorem 1.0

Converse Theorem: “If the sum of either pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.”


[Source: ekShiksha]

To prove: \( \angle\)AEB = 60º

Proof: AB is the diameter of a circle and CD is a chord, whose length is equal to the radius of the circle. So,

CD = OA = OB = OC = OD = AB / 2 — (d)

AC and BD have been extended to meet at point E. By equation (d) we can deduce that –\( \angle \) COD = 60º (Δ COD is an equilateral Δ). Therefore, CD subtends an angle of 30º on any other point on the circle.

\( \angle\) CBE = 30º

Since AB is the diameter, so it subtends an angle of 90º at point C on the circle; giving:\( \angle \) ACB = 90º =\( \angle \) BCE (ACE is a straight line segment). In Δ ECB,\( \angle \) BEC = 60º (Sum of all the angles of a Δ is 180º). Hence proved.

What are Intersecting and Non-intersecting Circles?


  • A circle is a 2-D geometrical shape where every point when measured from its centre is at an equal distance, called the radius.
  • A chord is a straight line joining any two points situated on the perimeter of a circle.
  • The longest chord of the circle is the diameter; twice the length of the radius.
  • A continuous piece of a circle is called an arc.

Intersecting Circles

Now, two circles of equal or unequal radii are said to be intersecting, if any one of the following two conditions are met:

  • Either of the two circles meets at two distinct points and intersects the other circle at those points. Here, the two points of either circle intersect the perimeter of the other circle at blue dots. The circles “just touch each – other” at a single point of contact.


  • Here, the single point of contact of either circle intersects the other circle at blue dot.


Non – Intersecting Circles

  • Two circles are said to be non – intersecting if they do not intersect each other at any point in the given plane.


  • Concentric circles are also non-intersecting circles because although they have a single centre point, no point of either touches the rest of the circles.


Theorem 2.0

Theorem: Prove that if 2 circles intersect at 2 points, then the line segment through their respective centres is perpendicular to the common chord.


Given: 2 circles with centres O and O’ with points of intersection A and B.
To prove: OO’ is perpendicular bisector of AB.
Construction: Draw line segments OA, OB, O’A and O’B.

Proof: In ΔOAO’ and ΔOBO’,

  • OA = OB = r (radius)
  • O’A = O’B = s (radius)
  • And, OO’ = OO’

Therefore, by SSS criterion of congruency, ΔOAO’ and ΔOBO’ are congruent. So, Angle AOO’ = Angle BOO’ and Angle AOM = Angle BOM. Marking M as the point of intersection of OO’ and AB –

In ΔAOM and ΔBOM, OA = OB = r (radius)
\( \angle \) AOM =\( \angle \) BOM
And, OM = OM (common side)
Therefore, by SAS criterion of congruency, ΔAOM and ΔBOM are congruent. So, AM = BM &\( \angle \) AMO =\( \angle \) BOM

Now,\( \angle \) AOM +\( \angle \) BMO = 180º
2(\( \angle \) AOM) = 180º
Therefore,\( \angle \) AOM = 90º
Thus, AM = BM and\( \angle \) AOM =\( \angle \) BMO = 90º

Hence, OO’ is the perpendicular bisector of AB.

Solved Example for You

Q. Given that, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals.
If\( \angle \) DBC = 55º and\( \angle \) BAC = 45º, find\( \angle \) BCD.


[Source: LearnCBSE]

Sol: Since\( \angle \) DAC and\( \angle \) DBC are subtended by the same chord DC,
\( \angle \) DAC =\( \angle \) DBC = 55º
By the property of cyclic quadrilateral,
\( \angle \) BAD +\( \angle \) BCD = 180º [\( \angle \) BAD = 45º + 55º = 100º]
So,\( \angle \) BCD = 180º – 100º
Therefore,\( \angle \) BCD = 80º

Question- What is a quadrilateral and its properties?

Answer- If we look at Euclidean geometry, we see that a quadrilateral is a four-sided 2D figure where the sum of its internal angles is 360°. All in all, there are two properties of quadrilaterals, one is that a quadrilateral must be closed shape having four sides. Next, all its internal angles must sum up to 360°.

Question- Who invented the quadrilateral?

Answer- The Ancient Greeks invented the quadrilaterals. We also see that according to a popular saying, Pythagoras was the first to draw it. Back in those days which we are referring to here, quadrilaterals comprised of three sides. Moreover, their properties were also just faintly understood to people.

Question- What is a Cyclic Quadrilateral?

Answer- When we talk about a cyclic quadrilateral, we see that it is a figure with four sides and where the vertices lie on the circle. In other words, it is a quadrilateral where its four vertices are touching the circle.

Question- Are all rectangles Quadrilaterals?

Answer- As we have discussed, a quadrilateral is any polygon having four sides. Now, when we look at the definition of a rectangle, we see that it is a polygon with four sides and four right angles. Thus, it is clear that every rectangle is a quadrilateral.

Share with friends

Customize your course in 30 seconds

Which class are you in?
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Dr. Nazma Shaik
Gaurav Tiwari
Get Started

Leave a Reply

Your email address will not be published. Required fields are marked *

Download the App

Watch lectures, practise questions and take tests on the go.

Customize your course in 30 seconds

No thanks.