Question

want a proper explanation answer of this question plss answer this properly ​

Answer 1

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.

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Answer 2

Step-by-step explanation:

Given : Let ABCD be the quadrilateral circumscribing the circle with centre O.

ABCD touches the circle with points P,Q,R and S.

To prove : Opposite sides subtends suplementary angles at the centre.

i.e., ∠AOB + ∠COD =  180

& ∠AOD + ∠BOC = 180

Construction: Join OP, OQ, OR & OS.

Proof: Rename the angles as shown in the figure.

In ΔAOP and ΔAOS

AP = AS (Legths of tangent drawn from external point to a circle are equal)

AO=AO (Common)

OP = OS (Radius)

∴ Δ AOP ≅ Δ AOS by SSS Congruency rule.

∠AOP = ∠AOS (CPCT)

i.e., ∠1=∠8   ...(1)

Similarly, we can prove that

∠2 = ∠3 ....(2)

∠4 = ∠5 ....(3)

∠6 = ∠7 ... (4)

Now

∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8 = 360 (Complete Angle)

∠1+∠2+∠2+∠5+∠5+∠6+∠6+∠1 = 360 (From 1,2,3,4)

2(∠1+∠2+∠5+∠6) = 360

∠1+∠2+∠5+∠6 = 360/2

(∠1+∠2)+(∠5+∠6) = 180

∠AOB + ∠COD =  180

Hence both angles are supplementary.

Similarly, we can prove that

∠AOD + ∠BOC = 180

Hence, proved.

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NOTE : Figures are in the attachement.