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Question
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: ∠ ACB.
Hence, show that AC is a diameter.
Solution
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)
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