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Question
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find: ∠ADB.
Solution
Again, Arc AB subtends ∠AOB at the centre and
∠ACB at the remaining part of the circle.
∠ ACB = `1/2` ∠AOB
=` 1/2 xx 108°`
= 54
In cyclic quadrilateral ADBC
∠ADB + ∠ACB = 180° ...[sum of opposite angles]
⇒ ∠ADB + 54° = 180°
⇒ ∠ADB = 180° − 54°
⇒ ∠ADB = 126°
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An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
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