In the given figure, ABCDE is a pentagon inscribed in a circle. If AB = BC = CD, ∠BCD = 110° and ∠BAE = 120°, find :

(i) ∠ABC

(ii) ∠CDE

(iii) ∠AED

(iv) ∠EAD

Join AD, AC and BD.

We know that,

In the same circle, equal chords cut off equal arcs.

Since, chord AB = chord CD

Thus, arc AB = arc CD.

We know that,

Equal chords subtend equal angles at the circumference of the same circle.

Thus,

∠ACB = ∠DAC

Since, these are alternate angles as well thus, AD must be parallel to BC.

Since, in quadrilateral ABCD, AD // BC and non parallel sides AB = DC.

Thus, ABCD is an isosceles trapezium.

In an isosceles trapezium,

Base angles are equal.

Thus, ∠ABC = ∠BCD = 110°.

ABCD is a cyclic quadrilateral.

⇒ ∠BCD + ∠BAD = 180° (Sum of opposite angles = 180°)

⇒ 110° + ∠BAD = 180°

⇒ ∠BAD = 180° - 110°

⇒ ∠BAD = 70°.

Given, ∠BAE = 120°

From figure,

⇒ ∠EAD = ∠BAE - ∠BAD = 120° - 70° = 50°.

In triangle ABC,

AB = BC

⇒ ∠BAC = ∠BCA = x (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ABC + ∠BAC + ∠BCA = 180°

⇒ 110° + x + x = 180°

⇒ 2x = 180° - 110°

⇒ 2x = 70°

⇒ x = 35°.

Thus, ∠BAC = 35°.

Since, equal chords subtends equal angles at the circumference of the same circle.

Since, chord BC = chord CD. Thus,

∠CBD = ∠BAC = 35°.

Also,

∠CBD = ∠ADB = 35° (Alternate angles are equal)

In triangle ABD,

⇒ ∠ABD + ∠ADB + ∠BAD = 180°

⇒ ∠ABD + 35° + 70° = 180°

⇒ ∠ABD + 105° = 180°

⇒ ∠ABD = 180° - 105° = 75°.

In cyclic quadrilateral ABDE,

⇒ ∠ABD + ∠AED = 180° (Sum of opposite angles = 180°)

⇒ 75° + ∠AED = 180°

⇒ ∠AED = 180° - 75° = 105°.

In triangle CBD,

BC = CD

⇒ ∠CDB = ∠CBD = 35° (Angles opposite to equal sides are equal)

In cyclic quadrilateral ABDE,

⇒ ∠BAE + ∠BDE = 180° (Sum of opposite angles = 180°)

⇒ 120° + ∠BDE = 180°

⇒ ∠BDE = 180° - 120° = 60°.

From figure,

∠CDE = ∠CDB + ∠BDE = 35° + 60° = 95°.

(i) Hence, ∠ABC = 110°.

(ii) Hence, ∠CDE = 95°.

(iii) Hence, ∠AED = 105°.

(iv) Hence, ∠EAD = 50°.