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 center,

∠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 °]

So , 2(∠2+∠3+∠6+∠7) = 360°

and 2(∠1+∠8+∠4+∠5) = 360°

[∠2+∠3=∠AOB;

∠6+∠7=∠COD

∠1+∠8=∠AOD

and ∠4+∠5 = ∠BOC]

and ∠AOD +∠BOC = °