In the figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find :

(i) ∠CAD

(ii) ∠CBD

(iii) ∠ADC

(i) We know that,

Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.

∠BAD = Exterior ∠BCE = 80°.

From figure,

∠CAD = ∠BAD - ∠BAC = 80° - 25° = 55°.

Hence, ∠CAD = 55°.

(ii) We know that,

Angles in same segment are equal.

∴ ∠CBD = ∠CAD = 55°.

Hence, ∠CBD = 55°.

(iii) Since, AB ∥ DC,

∠ACD = ∠BAC = 25° [Alternate angles are equal]

In triangle ADC,

By angle sum property of triangle,

⇒ ∠ADC + ∠CAD + ∠ACD = 180°

⇒ ∠ADC + 55° + 25° = 180°

⇒ ∠ADC + 80° = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°.

Hence, ∠ADC = 100°.