Question

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

Find:

1. ∠CAD
2. ∠CBD
3. ∠ADC

Answer 1

In the given figure,

ABCD is a cyclic quad in which AB || DC

∴ ABCD is an isosceles trapezium

∴ AD = BC

i. Join BD

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°