Prove that the angle bisector of any angle of a triangle and the perpendicular bisector of the opposite side,if they intersect,will intersect on the circumcircle of the triangle.

(N/A) Given: $A$ $\Delta ABC$ and a line $l$ which is the perpendicular bisector of $BC$.
To prove: The angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$.
Proof: Let the angle bisector of $\angle A$ intersect the circumcircle of $\Delta ABC$ at point $P$. Join $BP$ and $CP$.
Since angles in the same segment are equal,we have:
$\angle BAP = \angle BCP$
Since $AP$ is the bisector of $\angle A$,we have:
$\angle BAP = \angle PAC = \frac{1}{2} \angle A$
Therefore,$\angle BCP = \frac{1}{2} \angle A$ $...(1)$
Similarly,in the same segment,$\angle PAC = \angle PBC$.
Since $\angle PAC = \frac{1}{2} \angle A$,we have:
$\angle PBC = \frac{1}{2} \angle A$ $...(2)$
From equations $(1)$ and $(2)$,we get:
$\angle BCP = \angle PBC$
In $\Delta PBC$,since the base angles are equal,the sides opposite to them are equal:
$BP = CP$
Since $P$ is equidistant from $B$ and $C$,$P$ must lie on the perpendicular bisector of $BC$.
Hence,the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect on the circumcircle of $\Delta ABC$.