**Cyclic Quadrilateral**

A quadrilateral having its all four vertices on the
circumference of a circle is called a **cyclic
quadrilateral**.

ABCD, in the given figure, is a cyclic quadrilateral.

**Concyclic Points**

The points lying on the circumference of the circle are called **concyclic points**.

In the given circle, A, B, C, D are concyclic points.

**Theorems on Cyclic
Quadrilateral and Their Proofs:**

**THEOREM 1: **

The sum of opposite angles of a cyclic quadrilateral is always
180°.

PROOFS:

__Given__: PQRS is a cyclic
quadrilateral.

__To Prove__: (i) ∠QPS + ∠QRS = 180°

(ii) ∠PQR + ∠PSR = 180°

__Proof__:

__Statements__ __Reasons__

1.
∠QPS = ½ Arc QRS -----> Inscribed
angle is equal to the half of its opposite arc.

2.
∠QRS = ½ Arc QPS ----->
Inscribed angle is equal to half of its opposite arc.

3.
∠QPS + ∠QRS = ½ Arc QRS + ½ Arc QPS -----> Adding statements 1 and 2.

or, ∠QPS + ∠QRS = ½ Arc (QRS + QPS)

or, ∠QPS + ∠QRS = ½ Circle PQRS

or, ∠QPS + ∠QRS = ½ 360°

or, ∠QPS + ∠QRS = 180°

4.
∠PQR + ∠PSR = 180° -----> Same as above.

Hence proved.

**THEOREM 2:**

In a cyclic quadrilateral, the exterior angle is equal to its
opposite interior angle.

PROOFS:

__Given__: ABCD is a cyclic
quadrilateral with exterior angle ∠DCE.

__To Prove__: ∠DCE = ∠BAD

__Proof__:

__Statements__ __Reasons__

1.
∠DCE + ∠BCD = 180° -----> Opposite angles of a cyclic quadrilateral
are supplementary.

2.
∠BAD + ∠BCD = 180° -----> Linear pair of angles.

3.
∠DCE + ∠BCD = ∠BAD + ∠BCD -----> From statements 1 and 2.

4.
∠DCE = ∠BAD -----> Cancelling common angle ∠BCD.

Hence proved.

**Problems on Cyclic Quadrilaterals**

**Worked Out Examples**

**Example 1:** Find ∠COD in the given figure.

**Solution:** From the figure,

110° + ∠ODC = 180° [Opposite
angles of a cyclic quadrilateral]

or, ∠ODC = 180° – 110°

or, ∠ODC = 70°

∠ODC = ∠OCD = 70° [Being OC = OD,
radii of same circle]

∠COD + ∠ODC + ∠OCD = 180° [Sum of angles of DCOD]

or, ∠COD + 70° + 70° = 180°

or, ∠COD + 140° = 180°

or, ∠COD = 180° – 140°

or, ∠COD = 40° Ans.

**Example 2:** Calculate the value of x and y in the given figure.

**Solution:** From the figure,

x = ½ ∠BOD [Inscribed
angle is half of central angle]

or, x = ½ 160°

or, x = 80°

y + x = 180° [Opposite
angles of a cyclic quadrilateral]

or, y + 80° = 180°

or, y = 180° – 80°

or, y = 100°

Hence, x = 80° and y = 100°
Ans.

**Example 3:** Find the angle x in the given figure.

**Solution:** From the figure,

∠CAD = 56° [Inscribed angle on same arc CD]

∠BAD = 96° [Exterior angle of a cyclic quadrilateral]

Now,

x = ∠BAD – ∠CAD

= 96° – 56°

= 40°
Ans.

**Example 4:** In the adjoining figure, ABCD is a cyclic quadrilateral, side CD is
produced to the point E, where BC = DE. If CA bisects ∠BCD prove that DACE is an
isosceles triangle.

**Solution:** Here,

__Given__: ABCD is a cyclic
quadrilateral. BC = DE and ∠BCA = ∠ACD.

__To Prove__: DACE is an isosceles triangle.

__Proof__:

__Statements__ __Reasons__

1. ∠BCA = ∠ACD -----> Given.

2. Arc AB = arc AD -----> Arcs subtended by equal inscribed angles.

3. AB =AD ------> Equal chords cut off equal arcs in a circle.

4. In DABC
and DADE

i. AB = AD (S) -----> From statement 3

ii. ∠ABC = ∠ADE (A) -----> Exterior
angle of a cyclic quadrilateral

iii. BC = DE (S) -----> Given.

5. DABC ≅ DADE
-----> By SAS axiom

6. AC = AE -----> Corresponding sides of ≅ triangles.

7. ACE is an isosceles triangle -----> Being AC = AE.

Hence proved.

**Example 5:** In the given figure, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.

**Solution:** Here,

__Given__: PQRS is a cyclic
quadrilateral. ∠RPS = ∠SPT.

__To Prove__: SQ = SR.

__Proof__:

__Statements__ __Reasons__

1.
∠SPT = ∠RPS -----> From given.

2.
∠SPT = ∠QRS -----> Exterior angle of cyclic quadrilateral PQRS.

3.
∠SQR = ∠RPS -----> Inscribed angle on same arc RS.

4.
∠SQR = ∠QRS -----> From statements 1, 2 and 3.

5.
SQ = SR -----> Being
base angles of DSQR
equal, statement 4.

Hence proved.

If you have any question or problems regarding the Cyclic
Quadrilaterals, you can ask here, in the comment
section below.

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