In a cyclic quadrilateral $A B C D$ the diagonal $A C$ bisects the angle $B C D$ . Prove that the diagonal $B D$ is parallel to the tangent to the circle at point $A$ .

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Solution

$∠ A D B = ∠ A C B . . . . (1)$ [Angles in same segment]
Similarly,
$∠ A B D = ∠ A C D . . . . (2)$
But
$∠ A C B = ∠ A C D$ [AC is bisector of $∠ B C D$ ]
$∴ ∠ A D B = ∠ A B D$ [From (1) and (2)]
TAS is a tangent and AB is a chord
$∴ ∠ B A S = ∠ A D B$ [Angles in alternate segment]
But,
$∠ A D B = ∠ A B D$
$∴ ∠ B A S = ∠ A B D$
But these are alternate angles
Therefore, $T S | | B D$

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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.

In a cyclic quadrilateral $A B C D$ the diagonal $A C$ bisects the angle $B C D$ . Prove that the diagonal $B D$ is parallel to the tangent to the circle at point $A$ .