#### Question

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

∠OBD + ∠ODB

=∠CBD +∠CDB

#### Solution

∠ 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

Is there an error in this question or solution?

Solution Abcd is a Quadrilateral Inscribed in a Circle, Having ∠ = 60°; O is the Center of the Circle. Show That: ∠Obd + ∠Odb =∠Cbd +∠Cdb Concept: Arc and Chord Properties - the Angle that an Arc of a Circle Subtends at the Center is Double that Which It Subtends at Any Point on the Remaining Part of the Circle.