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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 ⇒ ∠DAO = ∠BAO [Because, AB and AD are tangents in the circe] So , we take this angls as 1 , that is ,

⇒ ∠DAO = ∠BAO = 1

Also in quad. ABCD , we get,

⇒ ∠ABO = ∠CBO { Because , BA and BC are tangents }

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

⇒ 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.