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

(N/A) Let $\Delta ABC$ be inscribed in a circle with center $O$.
Let the internal bisector of $\angle A$ intersect the circumcircle at point $E$. We need to show that $E$ lies on the perpendicular bisector of $BC$.
Since $AE$ is the bisector of $\angle BAC$,we have $\angle BAE = \angle CAE$.
Equal angles subtend equal arcs at the circumference,so $\text{arc } BE = \text{arc } EC$.
Consequently,the chords corresponding to these arcs are equal,i.e.,chord $BE = \text{chord } CE$.
Let $D$ be the midpoint of $BC$. In $\Delta BDE$ and $\Delta CDE$:
$BE = CE$ (Proved above)
$BD = CD$ ($D$ is the midpoint of $BC$)
$DE = DE$ (Common side)
By the $SSS$ congruence criterion,$\Delta BDE \cong \Delta CDE$.
Therefore,$\angle BDE = \angle CDE$ (Corresponding parts of congruent triangles).
Since $BC$ is a straight line,$\angle BDE + \angle CDE = 180^{\circ}$.
Thus,$\angle BDE = \angle CDE = 90^{\circ}$,which means $DE \perp BC$.
Since $DE$ passes through the midpoint $D$ of $BC$ and is perpendicular to $BC$,$DE$ is the perpendicular bisector of $BC$. Thus,the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect at point $E$ on the circumcircle.