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What is Cyclic Quadrilateral – (Definition, Properties, Theorems & Proofs)

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The word ‘cyclic’ comes from the Latin word ‘cyclicus’ or the Greek word ‘kyklikos’, meaning ‘moving in a circle’.  In mathematics, a cyclic quadrilateral is a quadrilateral that has all its four vertices lying on a circle. It is also called an inscribed quadrilateral. 

Let’s understand what is cyclic quadrilateral, its properties, and theorems related to cyclic quadrilaterals along with their proofs.

What is Cyclic Quadrilateral?

A cyclic quadrilateral is a quadrilateral that is inscribed in a circle. It means that all four vertices of a quadrilateral lie on a circle or a circle passes through all four vertices of a quadrilateral. The vertices of such a quadrilateral are said to be concyclic. The centre of that circle is known as the circumcenter and its radius the circumradius.

what is cyclic quadrilateral

In the above figure, the quadrilateral $\text{ABCD}$ is a cyclic quadrilateral, and thus we can say that the vertices $\text{A}$, $\text{B}$, $\text{C}$, and $\text{D}$ are concyclic.

The point $\text{O}$ (centre of the circle) is called circumcenter and $\text{OC}$ is called circumradius. (The line segments joining the points $\text{O}$, $\text{A}$ and $\text{O}$, $\text{B}$, and $\text{O}$, $\text{D}$ are also called circumradius.

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Properties of Cyclic Quadrilateral

The following are the important properties of a cyclic quadrilateral.

  • All four vertices of a cyclic quadrilateral lie on the circumference of a circle
  • All four sides of a cyclic quadrilateral lie inside the circle and we say either of the following two 
  • the quadrilateral is inscribed in the circle 
  • the circle circumscribes the quadrilateral
  • The measure of an exterior angle at a vertex is equal to the opposite interior angle
  • In a cyclic quadrilateral, the sum of the product of the length of opposite sides is equal to the product of the length of the diagonals
  • The centre of the circle lies at the perpendicular bisectors of the sides of a cyclic quadrilateral
  • The opposite angles of a cyclic quadrilateral are supplementary. In the above figure, $\angle \text{A} + \angle \text{C} = \angle \text{B} + \angle \text{D} = 180^{\circ}$

Cyclic Quadrilateral Angle Properties

In a cyclic quadrilateral, the opposite angles are supplementary, i.e., the sum of opposite angles in a cyclic quadrilateral in $180^{\circ}$.

what is cyclic quadrilateral

In the above figure, $\text{ABCD}$ is a cyclic quadrilateral, therefore, $\angle \text{A} + \angle \text{C} = \angle \text{B} + \angle \text{D} = 180^{\circ}$.

Let’s prove the above statement.

To prove the statement, join the vertices $\text{A}$, $\text{B}$, $\text{C}$, and $\text{D}$ with the centre of the circle $\text{O}$.

what is cyclic quadrilateral

In $\triangle \text{AOB}$, $\angle \text{OAB} = \angle \text{OBA}$ ———————— (1)

In $\triangle \text{BOC}$, $\angle \text{OBC} = \angle \text{OCB}$ ———————— (2)

In $\triangle \text{COD}$, $\angle \text{OCD} = \angle \text{ODC}$ ———————— (3)

In $\triangle \text{DOA}$, $\angle \text{ODA} = \angle \text{OAD}$ ———————— (4)

According to the Angle Sum Property of a triangle,

In $\triangle \text{AOB}$, $\angle \text{OAB} + \angle \text{OBA} + \angle \text{AOB} = 180^{\circ}$ 

In $\triangle \text{BOC}$, $\angle \text{OBC} + \angle \text{OCB} + \angle \text{BOC} = 180^{\circ}$

In $\triangle \text{COD}$, $\angle \text{OCD} + \angle \text{ODC} + \angle \text{COD} = 180^{\circ}$

In $\triangle \text{DOA}$, $\angle \text{ODA} + \angle \text{OAD} + \angle \text{DOA} = 180^{\circ}$

$=> \angle \text{OAB} + \angle \text{OBA} + \angle \text{AOB} + \angle \text{OBC} + \angle \text{OCB} + \angle \text{BOC}$

$ + \angle \text{OCD} + \angle \text{ODC} + \angle \text{COD} + \angle \text{ODA} + \angle \text{OAD} + \angle \text{DOA} = 4 \times 180^{\circ}$

$=> \angle \text{OAB} + \angle \text{OBA} + \angle \text{OBC} + \angle \text{OCB} + \angle \text{OCD} + \angle \text{ODC}$

$ + \angle \text{ODA} + \angle \text{OAD} + \left(\angle \text{AOB} + \angle \text{BOC} + \angle \text{COD} + \angle \text{DOA} \right) = 720^{\circ}$

$=> \angle \text{OAB} + \angle \text{OBA} + \angle \text{OBC} + \angle \text{OCB} + \angle \text{OCD} + \angle \text{ODC} + \angle \text{ODA} + \angle \text{OAD} + 360^{\circ} = 720^{\circ}$

$=> \angle \text{OAB} + \angle \text{OBA} + \angle \text{OBC} + \angle \text{OCB} + \angle \text{OCD} + \angle \text{ODC} + \angle \text{ODA} + \angle \text{OAD} = 720^{\circ} – 360^{\circ} $

$=> \angle \text{OAB} + \angle \text{OBA} + \angle \text{OBC} + \angle \text{OCB} + \angle \text{OCD} + \angle \text{ODC} + \angle \text{ODA} + \angle \text{OAD} = 360^{\circ} $

From (1), (2), (3), and (4)

$=> 2 \left(\angle \text{OAD} + \angle \text{OAB} \right) + 2 \left(\angle \text{OCB} + \angle \text{OCD} \right) = 360^{\circ}$

$=> 2 \left(\angle \text{OAD} + \angle \text{OAB} + \angle \text{OCB} + \angle \text{OCD} \right) = 360^{\circ}$

$=> \angle \text{OAD} + \angle \text{OAB} + \angle \text{OCB} + \angle \text{OCD} = 180^{\circ}$

$=> \left( \angle \text{OAD} + \angle \text{OAB} \right) + \left( \angle \text{OCB} + \angle \text{OCD} \right) = 180^{\circ}$

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

Similarly, we can prove that $=> \angle \text{B} + \angle \text{D} = 180^{\circ}$.

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Exterior Angle Theorem of Cyclic Quadrilateral

Another important theorem of a cyclic quadrilateral is an exterior angle theorem. It states that If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.

what is cyclic quadrilateral

In the above figure, $\text{ABCD}$ is a cyclic quadrilateral. $\angle \text{B}$, $\angle \text{D}$ are opposite angles and $\angle \text{CBE}$ is $\text{Ext} \angle \text{B}$, then $\angle \text{D} = \angle \text{CBE}$, i.e., $\angle \text{D} = \text{Ext} \angle \text{B}$.

Let’s prove the statement.

In the above figure, $\angle \text{ABC} + \angle \text{CDA} = 180^{\circ}$ (Opposite angles of a cyclic quadrilateral are supplementary) ————————— (1)

Also, $\angle \text{ABC} + \angle \text{CBE} = 180^{\circ}$ ——————————– (2)

From (1) and (2), we get $\angle \text{CDA} = \angle \text{ABC}$, i.e., $\angle \text{D} = \angle \text{B}$.

Ptolemy’s Theorem of Cyclic Quadrilateral

Ptolemy’s theorem of cyclic quadrilateral states that 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.

In the above figure, $\text{ABCD}$ is a cyclic quadrilateral, $\text{AB}$, $\text{CD}$, and $\text{AD}$, $\text{BC}$ are opposite sides and $\text{AC}$, $\text{BD}$ are the diagonals.

Then, according to Ptolemy’s theorem of cyclic quadrilateral $\left( \text{AB} \times \text{CD} \right) + \left( \text{AD} \times \text{BC} \right) = \text{AC} \times \text{BD}$.

To prove the theorem, let’s draw an $\angle \text{ABQ} = \angle \text{CBD}$ to intersect $\text{AC}$ at $\text{Q}$ as shown in the figure below.

what is cyclic quadrilateral

In $\triangle \text{ABQ}$ and $\triangle \text{CBD}$, we have

$\angle \text{ABQ} = \angle \text{CBD}$ (By Construction)

$\angle \text{BAQ} = \angle \text{CDB}$ (Angles in the same arc)

Therefore, $\triangle \text{ABQ} \sim \triangle \text{CBD}$ (AA Similarity Criterion)

Thus, $\frac{\text{AB}}{\text{BD}} = \frac{\text{AQ}}{\text{CD}}$

$=> \text{AB} \times \text{CD} = \text{BD} \times \text{AQ}$ —————- (1)

Similarly, as $\angle \text{ABQ} = \angle \text{CBD}$, we get

$\angle \text{ABQ} + \angle \text{QBP} = \angle \text{CBD} + \angle \text{QBP}$

And in $\triangle \text{ABD}$ and $\triangle \text{QBC}$, 

$\angle \text{ABD} = \angle \text{QBC}$ (Angles in the same arc)

$\angle \text{ADB} = \angle \text{QCB}$ (Angles in the same arc)

Therefore, $\triangle \text{ABD} \sim \triangle \text{QBC}$ (AA Similarity Criterion) 

Thus, $\frac{\text{AD}}{\text{QC}} = \frac{\text{BD}}{\text{BC}}$

$=> \text{AD} \times \text{BC} = \text{BD} \times \text{QC}$ —————- (2)

Adding (1) and (2), we get

$\text{AB} \times \text{CD} + \text{BC} \times \text{AD} = \left( \text{AQ} + \text{QC} \right) \times \text{BD}$  

$=> \text{AB} \times \text{CD} + \text{BC} \times \text{AD} = \text{AC} \times \text{BD}$

Practice Problems

  1. What is Cyclic Quadrilateral?
  2. State Ptolemy’s theorem of cyclic quadrilateral.
  3. Write True or False
    • Pair of opposite angles in a cyclic quadrilateral is equal
    • Pair of adjacent angles in a cyclic quadrilateral is equal
    • Pair of opposite angles in a cyclic quadrilateral are complementary
    • Pair of adjacent angles in a cyclic quadrilateral are complementary
    • Pair of opposite angles in a cyclic quadrilateral are supplementary
    • Pair of adjacent angles in a cyclic quadrilateral are supplementary
  4. Which of the following quadrilaterals are always cyclic quadrilaterals?
    • square
    • parallelogram
    • trapezium
    • rhombus
    • rectangle
    • kite

FAQs

What is cyclic quadrilateral?

what is cyclic quadrilateral

A quadrilateral whose vertices lie on the circumference of a circle is a cyclic quadrilateral. It is also called an inscribed quadrilateral.

What is the property of cyclic quadrilateral?

An important property of cyclic quadrilateral is that the opposite pairs of angles are supplementary to each other.

What is the sum of the opposite angles of the cyclic quadrilateral?

The sum of the opposite angles of a cyclic quadrilateral equals $180^{\circ}$.

Is square a cyclic quadrilateral?

Yes, a square is a cyclic quadrilateral. Since, each interior angle of square measures $90^{\circ}$, therefore, the sum of any pair of opposite angles of a square is $180^{\circ}$ and hence is a cyclic quadrilateral.

Conclusion

A quadrilateral whose vertices lie on the circumference of a circle is a cyclic quadrilateral. It is also called an inscribed quadrilateral. In the case of any cyclic quadrilateral, the opposite angles are supplementary.

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