Question

In cyclic quadrilateral ABCD; AD = BC, ∠BAC = 30° and ∠CBD = 70°; find:

1. ∠BCD
2. ∠BCA
3. ∠ABC
4. ∠ADC

Answer 1

ABCD is a cyclic quadrilateral and AD = BC

∠BAC = 30°, ∠CBD = 70°

∠DAC = ∠CBD   ...[Angles in the same segment]

`=>` ∠DAC = 70°   ...[∵ CBD = 70°]

`=>` ∠BAD = ∠BAC + ∠DAC = 30° + 70° = 100°

Since the sum of opposite angles of cyclic quadrilateral is supplementary

∠BAD + ∠BCD =180°

`=>` 100°+ ∠BCD = 180°   ...[From (1)]

`=>` ∠BCD = 180° – 100° = 80°

Since, AD = BC, ∠ACD = ∠BDC  ...[Equal chords subtends equal angles]

But ∠ACB = ∠ADB   ...[Angles in the same segment]

∴  ∠ACD + ∠ACB = ∠BDC + ∠ADB

`=>` ∠BCD = ∠ADC = 80°

But in ∆BCD,

∠CBD + ∠BCD + ∠BDC = 180°   ...[Angles af a triangle]

`=>` 70° + 80° + ∠BDC = 180°

`=>` 150° + ∠BDC = 180°

∴  ∠BDC = 180° – 150° = 30°

`=>` ∠ACD = 30°   ...[∵ ∠ACD = ∠BDC]

∴ ∠BCA = ∠BCD – ∠ACD = 80° – 30° = 50°

Since the sum of opposite angles of cyclic quadrilateral is supplementary

∠ADC + ∠ABC = 180°

`=>` 80° + ∠ABC = 180°

`=>` ∠ABC = 180° – 80° = 100°