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Question
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
- Prove that AC is a diameter of the circle.
- Find ∠ACB.
Solution
i. In ΔABD,
∠DAB + ∠ABD + ∠ADB = 180°
`=>` 65° + 70° + ∠ADB = 180°
`=>` 135° + ∠ADB = 180°
`=>` ∠ADB = 180° – 135° = 45°
Now, ∠ADC = ∠ADB + ∠BDC
= 45° + 45°
= 90°
Since ∠ADC is the angle of semicircle, so AC is a diameter of the circle.
ii. ∠ACB = ∠ADB ...(Angles in the same segment of a circle)
`=>` ∠ACB = 45°
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