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

AOP AOS

[By Side - Side - Side Criterion]

AOP = AOS

[Corresponding parts of congruent triangles are congruent]

1 = 2 …

Similarly, We can Prove

3 = 4 ….

5 = 6 ….

7 = 8 ….

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] 

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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