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__GIVEN ;-__⇒ ABCD is a quadrilateral and it has circumscribing a circle Which has centre O.

__CONSTRUCTION ;-__

⇒ Join - AO, BO, CO, DO.

__TO PROVE :-__

⇒

**Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.**

__PROOF ;-__

⇒ In the given figure , we can see that

⇒

**Because, AB and AD are tangents in the circe]**

__∠DAO = ∠BAO [__So , we take this angls as 1 , that is ,

⇒

**= 1**

__∠DAO = ∠BAO__Also in quad. ABCD , we get,

⇒

**{ Because , BA and BC are tangents }**

__∠ABO = ∠CBO__⇒Also , let us take this angles as 2. that is ,

⇒

**= 2**

__∠ABO = ∠CBO__⇒ As same as , we can take for vertices C and as well as D.

⇒

**Sum. of angles of quadrilateral ABCD = 360° { Sum of angles of quad is 360°}**

Therfore ,

⇒ 2 (1 + 2 + 3 + 4 ) = 360° {

**Sum. of angles of quad is - 360°**}

⇒ 1 + 2 + 3 + 4 = 180°

Now , in Triangle AOB,

⇒

**∠BOA = 180 – ( a + b )**

⇒ { Equation 1 }

Also , In triangle COD,

⇒

**∠COD = 180 – ( c + d )**

⇒ { Equation 2 }

⇒From Eq. 1 and 2 we get ,

⇒ Angle BOA + Angle COD

= 360 – ( a + b + c + d )

= 360° – 180°

= 180°

__⇒So , we conclude that the line AB and CD subtend supplementary angles at the centre O__

⇒Hence it is proved that -

__opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.__