Question

In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find:

1. ∠BCD 
2. ∠ACB

Hence, show that AC is a diameter.

Answer 1

i. In cyclic quadrilateral ABCD,

∠BCD = 180° – ∠BAD

= 180° – 65°

= 115°

(Pair of opposite angles in a cyclic quadrilateral are supplementary)

ii. By angle sum property of ∆ABD,

∠ADB = 180° – 65° – 70° = 45°

Again, ∠ACB = ∠ADB = 45°

(Angle in the same segment)

∴ ∠ADC = ∠ADB + ∠BDC

= 45° + 45°

= 90°

Hence, AC is a semicircle.

(Since angle in a semicircle is a right angle)