Question

prove that opposite sides of a quadrilateral circumscribing a circle substended suplementary angles at center of the circle​

Answer 1

GIVEN ;-

ABCD is a quadrilateral and it has circumscribing a circle Which has centre  O.

CONSTRUCTION ;-

Join -  AO, BO, CO, DO.

PROOF ;-

⇒ In the given figure , we can see that

 

∠DAO = ∠BAO [Because, AB and AD are tangents in the circle] 

So , we take this angles 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°

Therefore,

⇒ 2 (1  + 2 + 3 + 4 )  =  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° 

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.

Amannnscharlie

Answer 2

Hey Mate :D

Your Answer :---

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S.

Let us join the vertices of the quadrilateral ABCD to the center of the circle.

[ plz see attached file also :) ]

Consider ∆OAP and ∆OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

∆OAP ≅ ∆OAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus,

∠POA = ∠AOS ∠1 = ∠8 Similarly, ∠2 = ∠3 , ∠4 = ∠5, ∠6 = ∠7

now,

∠1 + ∠2 + ∠ 3 + ∠4 + ∠5 + ∠ 6 + ∠7 + ∠ 8 = 360º

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

2 ∠1 + 2 ∠2 + 2∠ 5 + 2∠ 6 = 360º

2( ∠1 + ∠ 2) + 2( ∠5 + ∠ 6) = 360º

( ∠1 + ∠2 ) + ( ∠5 + ∠ 6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that

∠BOC + ∠DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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