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Question
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find:
- ∠BCD
- ∠ACB
Hence, show that AC is a diameter.
Solution
i. In cyclic quadrilateral ABCD,
∠BCD = 180° – ∠BAD
= 180° – 65°
= 115°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
ii. 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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