# 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.

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.

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.

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, $\angle \text{A} + \angle \text{C} = \angle \text{B} + \angle \text{D} = 180^{\circ}$

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

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 $\text{O}$.

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

## 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.

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.

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

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 opposite angles in a cyclic quadrilateral are complementary
• Pair of opposite angles in a cyclic quadrilateral are supplementary
• square
• parallelogram
• trapezium
• rhombus
• rectangle
• kite

## FAQs 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.