Answers

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

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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.

0