Question

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

Answer 1

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

Join the vertices of the quadrilateral ABCD to the center of the circle as shown in the above figure.

Now, in ΔOAP and ΔOAS,

AP = AS [Tangents from the same point]

OP = OS [Radii of the circle]

OA = OA [Common side]

By SSS congruence condition,

ΔOAP ≅ ΔOAS

So, ∠POA = ∠AOS

=> ∠1 = ∠8

Similarly we get, ∠2 = ∠3, ∠4 = ∠5, ∠6 = ∠7

Adding all these angles, we get

∠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 proved