Question

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

Answer 1

Given: Let 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. 

R.T.P: ∠AOB + ∠COD = 180° ∠AOD + ∠BOC = 180° 

Construction: Join OP, OQ, OR and OS. 

Proof: Since the two tangents drawn from an external point of a circle subtend equal angles. 

At the centre, 

∴ ∠1 = ∠2 

∠3 = ∠4 (from figure) 

∠5 = ∠6 

∠7 = ∠8 

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

[∵ Sum of all the angles around a point is 360°] 

So, 2 (∠2 + ∠3 + ∠6 + ∠7) = 360° 

and 2 (∠1 + ∠8 + ∠4 + ∠5) = 360° 

(∠2 + ∠3) + (∠6 + ∠7) = 360/2 = 180° 

Also, (∠1 + ∠8) + (∠4 + ∠5) = 360/2 = 180° 

So, ∠AOB + ∠COD = 180° 

[∵ ∠2 + ∠3 = ∠AOB; 

∠6 + ∠7 = ∠COD 

∠1 + ∠8 = ∠AOD 

and ∠4 + ∠5 = ∠BOC [from fig.]] 

and ∠AOD + ∠BOC = 180°