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Question
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution
∠ADB = ∠ACB ...(i) (Angles in same segement)
Similarly,
∠ABD = ∠ACD ...(ii)
But, ∠ACB = ∠ACD ...(AC is bisector of ∠BCD)
∴ ∠ADB = ∠ABD ...(From (i) and (ii))
TAS is a tangent and AB is a chord
∴ ∠BAS = ∠ADB ...(Angles in alternate segment)
But, ∠ADB = ∠ABD
∴ ∠BAS = ∠ABD
But these are alternate angles
Therefore, TS || BD
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