A circle touches the sides of a quadrilateralABCD at PQRS respectively

1.	A circle touches the sides of a quadrilateralABCD at PQRS respectively. Show thatthe angles subtended at the centre by a pairof opposite sides are supplementary
Answer» Step-by-step explanation: 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. 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°)

A circle touches the sides of a quadrilateralABCD at PQRS respectively. Show thatthe angles subtended at the centre by a pairof opposite sides are supplementary

Answer»

Step-by-step explanation:

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.

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°)