Question

ABCD is a quadrilateral inscribed in a circle, having ∠ = 60°; O is the center of the circle.
Show that: ∠OBD + ∠ODB =∠CBD +∠CDB.

Answer 1

∠ BOD =  2 ∠BAD =  2 × 60°  = 120°
And  ∠ BCD  = `1/2` Refelx  (∠BOD) ( 360° - 120°) =  120°
(Angle at the centre is double the angle at the circumference subtended by the same chord
∴  ∠CBD + ∠CDB =180° - 120° = 60°
(By angle sum property of triangle CBD)
Again, ∠OBD+ ∠ODB=180° - 120° = 60°
(By angle sum property of triangle OBD)
∴ ∠OBD + ∠ODB = ∠CBD+ ∠CDB

Answer 2

Here, ∠ BOD =  2 ∠BAD =  2 × 60°  = 120°.

Now in Δ BOD,
∠OBD + ∠ODB = 180° - 120° = 60°   ....(i)

Also, ∠DAB + ∠DCB = 180°      ...(ABCD is cyclic quadrilateral)
∠DCB = 180° - 60° = 120°                   ...(ii)

∴ In Δ BCD,
∠ CBD + ∠CDB = 180° - ∠DCB = 180° - 120° = 60°

from (i) and (ii), we get the required result.
∴ ∠OBD + ∠ODB = ∠CBD+ ∠CDB.
Hence proved.