Question

9. A circle touches the sides of a quadrilateral ABCD at P, Q , R, S respectively. Show that the

angles subtended at the centre by a pair of opposite sides are supplementary.





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Answer 1

A circle the centre O touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P,Q,R and S respectively.

To prove: ∠AOB + ∠COD = 180°

and ∠AOD + ∠BOC = 180°

Proof:

Since the two tangents drawn from an external point to a circle subtend equal angles at the centre.

∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8

Now, ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8 = 360°

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

(∠2+∠3+) + (∠6+∠7) = 180° and (∠1+∠8) + (∠4+∠5) = 180°

[∵∠2+∠3=∠AOB, ∠6+∠7=∠COD, ∠1+∠8=∠AOD \ and \ ∠4+∠5=∠BOC]

⇒ ∠AOB + ∠COD = 180°

⇒ ∠AOD + ∠BOC = 180°