13.) Prove that opposite sides of a quadrilateral
circumscribing a circle subtend supplementary
angles at the centre of the circle.
Share
13.) Prove that opposite sides of a quadrilateral
circumscribing a circle subtend supplementary
angles at the centre of the circle.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Step-by-step explanation:
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.