Q. 344.4( 18 Votes )

Prove that opposi

Answer :

Let 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.



Then we have to prove that AOB + COD = 180° and AOD + BOC = 180°


Now, join OP, OQ, OR and OS.


Since the two tangents drawn from an external point to a circle subtend equal angles at the centre,


1 = 2, 3 = 4, 5 = 6 and 7 = 8 … (1)


Now,


1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 360° … (2)


[Since sum of all the angles subtended at a point is 360°]


From (1) and (2),


2 (2 + 3 + 6 + 7) = 360°


(2 + 3) + (6 + 7) = 180°


AOB + COD = 180°


Similarly,


2 (1 + 8 + 4 + 5) = 360°


(1 + 8) + (4 + 5) = 180°


AOD + BOC = 180°


Hence proved.


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