In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.

Join OB.

Given,

BC = OD = radius of circle

⇒ BC = OB.

As, angles opposite to equal sides are equal.

⇒ ∠BOC = ∠BCO = 20°.

An exterior angle is equal to the sum of two opposite interior angles.

∴ ∠ABO = ∠BCO + ∠BOC = 20° + 20° = 40° …………(1)

Now in ∆OAB,

OA = OB [Radii of the same circle]

As, angles opposite to equal sides are equal.

∠OAB = ∠ABO = 40° [from (1)]

By angle sum property of triangle,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°

As DOC is a straight line,

⇒ ∠AOD + ∠AOB + ∠BOC = 180°

⇒ ∠AOD + 100° + 20° = 180°

⇒ ∠AOD = 180° - 120° = 60°.

Hence, ∠AOD = 60°.