Question

prove that an opposite side of a quadrilateral circymscribing the circle substended a supplementry angle at the centre of circle

Answer 1

Let ABCD be a quadrilateral circumscribing a circle with centre O. 

Now join AO, BO, CO, DO.
From the figure, ∠DAO = ∠BAO [Since, AB and AD are tangents] 
Let ∠DAO = ∠BAO = 1 
Also ∠ABO = ∠CBO [Since, BA and BC are tangents] 
Let ∠ABO = ∠CBO = 2 
Similarly we take the same way for vertices C and D
Sum of the angles at the centre is 360° 
Recall that sum of the angles in quad. ABCD = 360° 
⇒ 2(1 + 2 + 3 + 4) = 360° 
⇒ 1 + 2 + 3 + 4 = 180° 
In ΔAOB, ∠BOA = 180 – (a + b)
In ΔCOD, ∠COD = 180 – (c + d)
Angle BOA + angle COD = 360 – (a + b + c + d) 
= 360°  – 180° 
= 180° 
Hence AB and CD subtend supplementary angles at O
Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.