Abcd Is Quadrilateral Inscribed In Circle Having В€ A 60в° O Is Centre Of Circle

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.