In the given figure, O is the centre of the circle and AB is a diameter. If AC = BD and ∠AOC = 72°, find:

(i) ∠ABC

(ii) ∠BAD

(iii) ∠ABD

(i) We know that,

Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

∠ABC = $\dfrac{1}{2}$ ∠AOC

∠ABC = $\dfrac{72^{\circ}}{2}$ = 36°

Hence, ∠ABC = 36°.

(ii) Given,

AC = BD

∠BAC = ∠ABC = 36° [Angles opposite to equal sides are equal]

Hence, ∠BAC = 36°.

(iii) In ∆ABD,

∠ABD + ∠BAD + ∠ADB = 180° [Angle sum property of triangle]

∠ABD + 36° + 90° = 180°

∠ABD + 126° = 180°

∠ABD = 180° - 126°

∠ABD = 54°

Hence, ∠ABD = 54°.