Question

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.

Answer 1

Given: A circle with 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°.

Construction: Join OP, OQ, OR and OS.

Proof: Since the two tangents drawn from an external point to a circle subtends equal angles at centre.
∴ ∠1 = ∠2, ∠3 = ∠4, ∠5 = ∠6 and ∠7 = ∠8     ...(i)
Now,
∠1 +∠2 +∠3 + ∠4 +∠5 +∠6 +∠7 +∠8 = 360°  ...(Sum of all the angles subtended at a point is 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° 
⇒ ∠AOB + ∠COD = 180°                                   ...(∵∠2 + ∠3 =∠AOB, ∠6 + ∠7 = ∠COD, ∠1 +∠8 = ∠AOD and ∠4 +∠5 = ∠BOC)
and ∠AOD + ∠BOC = 180°