#### Question

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

Find:

(i) ∠CAD (ii) ∠CBD (iii) ∠ADC

#### Solution

In the given figure,

ABCD is a cyclic quad in which AB ∥ DC

∴ ABCD is an isosceles trapezium

∴ AD = BC

Ext. ∠BCE = ∠BAD [Exterior angle of a cyclic qud is equal to interior opposite angle]

∴ ∠BAD = 80° [ ∵ ∠BCE = 80°]

But ∠BAC = 25°

∴ ∠CAD = ∠BAD - ∠BAC

= 80° - 25°

= 55°

(ii) ∠CBD = ∠CAD [Angle of the same segment]

= 55°

(iii) ∠ADC = ∠BCD [Angles of the isosceles trapezium]

= 180° - ∠BCE

= 180° - 80°

= 100°

Is there an error in this question or solution?

Solution In the Given Figure, Ab is Parallel to Dc ∠Bce = 80° and ∠Bac = 25°.Find: (I) ∠Cad (Ii) ∠Cbd (Iii) ∠Adc Concept: Cyclic Properties.