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Prove that opposite side of a …

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

Posted by Pihu ✨ 4 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 4 years, 5 months ago

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

= 2(1+2+3+4)=360^o
=1+2+3+4=180^o
In ΔAOB, ∠BOA=180−(1+2)
In ΔCOD, ∠COD=180−(3+4)
∠BOA+∠COD=360−(1+2+3+4)

=360^o–180^o
=180^o

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.

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