Cyclic Quadrilateral

Cyclic Quadrilateral

 

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

********************

10 Math Problems officially announces the release of Quick Math Solver and 10 Math Problems, Apps on Google Play Store for students around the world.

********************

********************

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.

 

Was this article helpful? LIKE and SHARE with your friends…