In the figure (i) given below, chords AB, BC and CD of a circle with center O are equal. If ∠BCD = 120°, find

(i) ∠BDC

(ii) ∠BEC

(iii) ∠AEB

(iv) ∠AOB.

Hence, prove that △OAB is equilateral.

(i) In △BCD, BC = CD

∴ ∠CBD = ∠BDC (As angles opposite to equal sides are equal.)

Since sum of angles in a triangle = 180°.

In △BCD,

∠BCD + ∠CBD + ∠CDB = 180°
⇒ 120° + ∠CBD + ∠CBD = 180°
⇒ 120° + 2∠CBD = 180°
⇒ 2∠CBD = 180° - 120°
⇒ ∠CBD = $\dfrac{60°}{2}$ = 30°.

As ∠CBD = ∠BDC,

∴ ∠BDC = 30°.

Hence, the value of ∠BDC = 30°.

(ii) From figure,

∠BDC = ∠BEC (Angles in same segment are equal.)

∴ ∠BEC = 30°

Hence, the value of ∠BEC = 30°.

(iii) Given AB = CB

∴ ∠BEC = ∠AEB (Equal chords subtend equal angles.)

∴ ∠AEB = 30°

Hence, the value of ∠AEB = 30°.

(iv) Arc AB subtends ∠AOB at the centre and ∠AEB at the remaining part of the circle.

⇒ ∠AOB = 2∠AEB (As angle subtended on centre is twice the angle subtended on the remaining part of the circle).

⇒ ∠AOB = 2 × 30°

⇒ ∠AOB = 60°.

Hence, the value of ∠AOB = 60°.

Since, OB = OA (Radius of same circle)

∴ ∠OBA = ∠OAB = x (let) (As angles opposite to equal sides are equal)

In △OAB,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ x + x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\dfrac{120°}{2}$ = 60°.

∠OBA = ∠OAB = 60°.

Since,

∠AOB = ∠OBA = ∠OAB = 60°

As, each angle of equilateral triangle = 60°.

Hence, proved that OAB is an equilateral triangle.