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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.