In the given 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) We know that,

Angles in same segment are equal.

∴ ∠BDC = ∠BAC = 25°.

AB || DC and BD is transversal.

So, ∠ABD = ∠BDC = 25°. [Alternate angles are equal]

From figure,

∠ABC = ∠ABD + ∠CBD = 25° + 55° = 80°.

In cyclic quadrilateral ABCD,

⇒ ∠ABC + ∠ADC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° - 80° = 100°.

Hence, ∠ADC = 100°.