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.

Answer 1

∠ADB = ∠ACB      ...[i] (Angles in the same segment)

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