Question

a circle touches the sides of quadrilateral ABCD at P Q R S respectively show that the angles subtended at the centre by a pair of opposite sides are supplementary

Answer 1

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

Join OA, OB, OC and OD.

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

∴ ∠1 = ∠8

Similarly, ∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ........ (1)

Now,

Sum of the angles at the centre = 360°

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

⇒ 2(∠1 + ∠2 + ∠5 + ∠6) = 360° [From equation (1)]

and 2(∠7 + ∠8 + ∠3 + ∠4) = 360°

⇒ ∠1 + ∠2 + ∠5 + ∠6 = 180°

and ∠3 + ∠4 + ∠7 + ∠8) = 180°

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

Answer 2

Thank you for asking this question. Here is your answer:

∠AOD + ∠BOC = 180°

∠AOB + ∠COD = 180°

Now the area OSDR = ΔODS x ΔODR

∡1 + ∡4 + ∡5 + ∡8 = 180°

∡2 + ∡3 + ∡6 + ∡7 = 180°

∠AOD + ∠BOC = 180° (∡1 + ∡4 + ∡5 + ∡8 = 180°)

∠AOB + ∠COD = 180° (∡2 + ∡3 + ∡6 + ∡7 = 180°)

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