Question

prove that the oppsite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre

Answer 1

Let ABCD be a quadrilateral with center o.
join AO,BO,CO,DO.
The angle BAo,DAO is equal as AB and AD are tanglents 
let each angle equal = a;
The angles at B are similarly equal to each other. Let each of them equal b. 
Similarly for vertices C and D. 

The sum of the angles at the centre is 360 deg. 
The sum of the angles of ABCD is 360 deg. 

Therefore: 
2(a + b + c + d) = 360 
a + b + c + d = 180. 

From triangle AOB, angle BOA = 180 - (a + b). 
From triangle COD, angle COD = 180 - (c + d). 

Angle BOA + angle COD = 360 - (a + b + c + d) 
= 360 - 180 
= 180 deg. 

Thus AB and CD subtend supplementary angles at O.

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

∠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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