In the adjoining figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: (i) ∠BCD (ii) ∠ADB Hence, show that AC is a diameter.

(i) We know that,

Sum of opposite angles in a cyclic quadrilateral = 180°.

In cyclic quadrilateral ABCD,

∴ ∠BCD + ∠BAD = 180°

⇒ ∠BCD + 65° = 180°

⇒ ∠BCD = 180° - 65° = 115°.

Hence, ∠BCD = 115°.

(ii) In △ABD,

⇒ ∠ADB + ∠BAD + ∠DBA = 180° [Angle sum property of triangle]

⇒ ∠ADB + 65° + 70° = 180°

⇒ ∠ADB + 135° = 180°

⇒ ∠ADB = 180° - 135° = 45°.

We know that,

Angles in same segment are equal.

∴ ∠ACB = ∠ADB = 45°.

From figure,

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.

Since, angle in a semi-circle is a right angle. Thus, AC is the diameter.

Hence, ∠ADB = 45° and proved that AC is a diameter.