Q. 13 4.2( 335 Votes )

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

Answer :

To Prove :  AOB + COD = 180°, ∠ BOC + ∠ DOA = 180°
Given: ABCD is circumscribing the circle.

Proof: 

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


join the vertices of the quadrilateral ABCD to the centre of the circle.


Consider ΔOAP and ΔOAS,


AP = AS (Tangents from the same point)


OP = OS (Radii of the same circle)


OA = OA (Common side)


ΔOAP ΔOAS (SSS congruence criterion)


And thus, POA = AOS


1 = 8 Similarly,


2 = 3


4 = 5


6 = 7


1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360°


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


2 1 + 2 2 + 2 5 + 2 6 = 360°


2( 1 + 2) + 2( 5 + 6) = 360°


( 1 + 2) + ( 5 + 6) = 180°


AOB + COD = 180°


Similarly, we can prove that BOC + DOA = 180°


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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.