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Question
In the given figure, PAT is tangent to the circle with centre O at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
- ∠BAP = ∠ADQ
- ∠AOB = 2∠ADQ
- ∠ADQ = ∠ADB
Solution
i. Since PAT || BC
∴ ∠PAB = ∠ABC (Alternate angles) ...(i)
In cyclic quadrilateral ABCD,
Ext ∠ADQ = ∠ABC ...(ii)
From (i) and (ii)
∠PAB = ∠ADQ
ii. Arc AB subtends ∠AOB at the centre and ∠ADB at the remaining part of the circle.
∴ ∠AOB = 2∠ADB
`=>` ∠AOB = 2∠PAB ...(Angles in alternate segments)
`=>` ∠AOB = 2∠ADQ ...(Proved in (i) part)
iii. ∴ ∠BAP = ∠ADB ...(Angles in alternate segments)
But ∠BAP = ∠ADQ ...(Proved in (i) part)
∴ ∠ADQ = ∠ADB
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