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Question

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

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Solution

Let ABCD be a quadrilateral circumscribing a circle with centre O.

Now join AO, BO, CO, DO.

From the figure, DAO=BAO [Since, AB and AD are tangents]

Let DAO=BAO=1

Also ABO=CBO [Since, BA and BC are tangents]

Let ABO=CBO=2

Similarly we take the same way for vertices C and D

Sum of the angles at the centre is 360o

Recall that sum of the angles in quadrilateral, ABCD = 360o

=2(1+2+3+4)=360o

=1+2+3+4=180o

In ΔAOB,BOA=180(1+2)

In ΔCOD,COD=180(3+4)

BOA+COD=360(1+2+3+4)

=360o180o

=180o

Since AB and CD subtend supplementary angles at O.

Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.


553959_494207_ans_79e310cf3ae04dd68b6a3cbd8106eca6.png

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