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Question
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
1) Prove that AC is a diameter of the circle.
2) Find ∠ACB
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Solution
1) In ΔABD, ∠DAB + ∠ABD + ∠ADB = 180°
⇒ 65° + 70° + ∠ADB = 180°
⇒ ∠ADB = 180° - 70° - 65° = 45°
Now, ∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°
⇒ ∠ADC is the angle of semi-circle
So, AC is a diameter of the circle.
2) ∠ACB = ∠ADB (angle subtended by the same segment)
∠ACB = 45°
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