Q. 185.0( 3 Votes )

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

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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