Question

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

Hence, show that AC is a diameter.

Answer 1

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)