In the adjoining figure, AB = AC = CD and ∠ADC = 35°. Calculate :

(i) ∠ABC

(ii) ∠BEC

(i) Given,

AC = CD

In triangle ACD,

∠DAC = ∠ADC = 35° [Angles opposite to equal sides in a triangle are equal]

In △ACD,

⇒ ∠DAC + ∠ADC + ∠ACD = 180°

⇒ 35° + 35° + ∠ACD = 180°

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 180° - 70° = 110°.

From figure,

⇒ ∠ACB + ∠ACD = 180° [Linear pair]

⇒ ∠ACB + 110° = 180°

⇒ ∠ACB = 180° - 110° = 70°.

Given,

AB = AC

∴ ∠ABC = ∠ACB = 70°. [As angles opposite to equal sides are equal]

Hence, ∠ABC = 70°.

(ii) In △ABC,

⇒ ∠BAC + ∠ACB + ∠ABC = 180° [Angle sum property of triangle]

⇒ ∠BAC + 70° + 70° = 180°

⇒ ∠BAC + 140° = 180°

⇒ ∠BAC = 180° - 140° = 40°.

We know that,

Angles in same segment are equal.

⇒ ∠BEC = ∠BAC = 40°.

Hence, ∠BEC = 40°.