Question

solve this question​

Answer 1

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.