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

$\angle ADB=\angle ACB....\left(1\right)$ [Angles in same segment]

Similarly,

$\angle ABD=\angle ACD....\left(2\right)$

But

$\angle ACB=\angle ACD$ [AC is bisector of $\angle BCD$]

$\therefore \angle ADB=\angle ABD$ [From (1) and (2)]

TAS is a tangent and AB is a chord

$\therefore \angle BAS=\angle ADB$ [Angles in alternate segment]

But,

$\angle ADB=\angle ABD$

$\therefore \angle BAS=\angle ABD$

But these are alternate angles

Therefore, $TS||BD$