GIVEN ;-

⇒ ABCD is a quadrilateral and it hascircumscribing a circle Which has centreO.

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 is360°}

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 centreO

⇒Hence it is proved that -opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.