prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answers
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.
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Answer:
ANSWER
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.
Step-by-step explanation:
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