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

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