ABCD is quadrilateral inscribed in circle, having ∠A = 60°, O is the centre of the circle, show that ∠OBD + ∠ODB = ∠CBD + ∠CDB
Answers:
Here, ∠BOD = 2 × ∠BAD
= 2 × 60 ̊
= 120 ̊
Now in ΔBOD,
∠OBD + ODB = 180 ̊ – 120 ̊
= 60 ……..(1)
Also, ∠DAB + ∠DCB = 180 ̊ (ABCD is a cyclic quadrilateral )
⇒ ∠DCB = 180 ̊ – 60 ̊ = 120 ̊
∴ In ΔBCD,
∠CBD + ∠CDB = 180 ̊ – ∠DCB
= 180 ̊ – 120 ̊
= 60 ̊ ……..(2)
From (1) and (2) we get the required result.
Hence proved.