Question

In the figure prove that angle aob+ angle boc+ angle dao+ angle cod = 360°

Answer 1

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°
Recall that sum of the angles in quad. ABCD = 360°
⇒ 2(1 + 2 + 3 + 4) = 360°
⇒ 1 + 2 + 3 + 4 = 180°
In ΔAOB, ∠BOA = 180 – (a + b)
In ΔCOD, ∠COD = 180 – (c + d)
Angle BOA + angle COD = 360 – (a + b + c + d)
= 360° – 180°
= 180°
Hence 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.