Question

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Answer 1

Answer:

First draw a quadrilateral ABCD which will circumscribe a circle with its centre O in a way that it touches the circle at point P, Q, R, and S. Now, after joining the vertices of ABCD we get the following figure:

Now, consider the triangles OAP and OAS,

AP = AS (They are the tangents from the same point A)

OA = OA (It is the common side)

OP = OS (They are the radii of the circle)

So, by SSS congruency △OAP ≅ △OAS

So, ∠POA = ∠AOS

Which implies that∠1 = ∠8

Similarly, other angles will be,

∠4 = ∠5

∠2 = ∠3

∠6 = ∠7

Now by adding these angles we get,

∠1+∠2+∠3 +∠4 +∠5+∠6+∠7+∠8 = 360°

Now by rearranging,

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

2∠1+2∠2+2∠5+2∠6 = 360°

Taking 2 as common and solving we get,

(∠1+∠2)+(∠5+∠6) = 180°

Thus, ∠AOB+∠COD = 180°

Similarly, it can be proved that ∠BOC+∠DOA = 180°

Therefore, the opposite sides of any quadrilateral which is circumscribing a given circle will subtend supplementary angles at the center of the circle.

Answer 2

Answer:

• 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°}

⇒ 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.