Cyclic Quadrilateral

Cyclic Quadrilateral

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.

Cyclic Quadrilateral ABCD

 

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.

A, B, C, D are concyclic points

 

Circle Theorems on Cyclic Quadrilateral

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

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

 

Proofs:

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

PQRS is a cyclic quadrilateral

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 the 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.

                                                                              Proved.

 

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

ABCD is a cyclic quadrilateral with exterior angle ∠DCE

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.

                                                                              Proved.

 

Look at the following worked out examples on circle and cyclic quadrilateral:

 

Workout Examples

Example 1: Find COD in the given figure.

Example 1: Circle ABCD

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°

 

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

Example 2: Circle ABCD

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°

 

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

Example 3: Circle ABCD

Solution: From the figure,

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

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

           

                  x = BAD - CAD

                     = 96° - 56°

                     = 40°

 

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.

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

Solution:

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.

                                                                                      Proved.

 

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

PQRS is a cyclic quadrilateral. ∠RPS = ∠SPT

Solution:

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.

                                                                                      Proved.

 

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