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

(i) Prove that AC is a diameter of the circle.

(ii) Find ∠ACB

(i) In ∆ABD,

⇒ ∠DAB + ∠ABD + ∠ADB = 180° [Angle sum property of a triangle]

⇒ 65° + 70° + ∠ADB = 180°

⇒ 135° + ∠ADB = 180°

⇒ ∠ADB = 180° - 135° = 45°

From figure,

⇒ ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

As the angle in a semicircle is a right angle,

∴ Arc ADC is semi-circle and AC is the diameter.

Hence, proved that AC is the diameter.

(ii) We know that,

Angles in the same segment of a circle are equal.

∴ ∠ACB = ∠ADB = 45°.

Hence, ∠ACB = 45°.