Answer :

Consider a quadrilateral, ABCD circumscribing a circle with center O and AB, BC, CD and AD touch the circles at point P, Q, R and S respectively.

Joined OP, OQ, OR and OS and renamed the angles (as in diagram)

To Prove : Opposite sides subtends supplementary angles at center i.e.

∠AOB + ∠COD = 180° and ∠BOC + ∠AOD = 180°

Proof :

In △AOP and △AOS

AP = AS

[Tangents drawn from an external point are equal]

AO = AO

[Common]

OP = OS

[Radii of same circle]

△AOP ≅ △AOS

[By Side - Side - Side Criterion]

∠AOP = ∠AOS

[Corresponding parts of congruent triangles are congruent]

∠1 = ∠2 …[1]

Similarly, We can Prove

∠3 = ∠4 ….[2]

∠5 = ∠6 ….[3]

∠7 = ∠8 ….[4]

Now,

As the angle around a point is 360°

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

∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6 + ∠7 + ∠7 = 360° [From 1, 2, 3 and 4]

2(∠2 + ∠3 + ∠6 + ∠7) = 360°

∠AOB + ∠COD = 180°

[As, ∠2 + ∠3 = ∠AOB and ∠5 + ∠6 = ∠COD] [5]

Also,

∠AOB + ∠BOC + ∠COD + ∠AOD = 360°

[Angle around a point is 360°]

∠AOB + ∠COD + ∠BOC + ∠AOD = 360°

180° + ∠BOC + ∠AOD = 360° [From 5]

∠BOC + ∠AOD = 180°

Hence Proved

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