# Prove that opposite sides of a quadrilateral circumscribing a circle

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by dhavalRamroop887

2015-05-10T12:12:00+05:30

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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.