Question

In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°

1. Prove that AC is a diameter of the circle.
2. Find ∠ACB.

Answer 1

i. In ΔABD,

∠DAB + ∠ABD + ∠ADB = 180°

⇒ 65° + 70° + ∠ADB = 180°

⇒ 135° + ∠ADB  = 180°

⇒ ∠ADB  = 180° – 135° = 45°

Now, ∠ADC = ∠ADB + ∠BDC

=  45° + 45°

= 90°

Since ∠ADC is the angle of a semicircle, AC is a diameter of the circle.

ii. ∠ACB = ∠ADB  ...(Angles in the same segment of a circle)

⇒ ∠ACB = 45°