In any triangle $ABC$, if the angle bisector $\angle A$ and perpendicular bisector of $BC$ intersect, prove that they intersect on the circumcircle of the triangle $ABC$.

$\angle BAE=\angle CAE$     ....Angles inscribed in the same arc

$\therefore arcBE=arcCE$

$\therefore$ chord $BE\cong$ chord $CE$   ...(1)

In $△BDE$  and $△CDE$,  
$BE=CE$      ....from (1)
$BD=CD$     ...$ED$ is perpendicular bisector of $BC$

$DE=DE$     ...common side

$\therefore △BDE\cong △CDE$      ...SSS test of congruence

$\Rightarrow \angle BDE=\angle CDE$     .....c.a.c.t.

Also, $\angle BDE+\angle CDE=180^{\circ }$      ...angles in linear pair

$\therefore \angle BDE=\angle CDE=90^{\circ }$     

Therefore, $DE$ is the right bisector of $BC$.