In any triangle ABC, if the angle bisector of `/_A`and perpendicular bisector of BCintersect, prove that they intersect on the circumcircle of the triangle `ABC`

Explanation

Given: Angle bisector of angle `A` and perpendicular bisector of `BC` intersect.
Let the bisector of A meet the circumcircle at `E`.
In `/_\ABE` and ` /_\AEC`,
`/_BAE=/_CAE` {`AE` Angle bisector of `A`}
`:.BE=EC .....(1)` {converse of the angle subtended by the chord of equal length at the same point are equal}
Let `D` be the midpoint of `BC` and now connect `ED`.
In `/_\BDE` and `/_\CDE,`
`BE=CE` ....from eq. `(1)`

`BD=CD` ...{`D` is mid point of `BC`}

`DE=DE` ...common side
`:./_\BDE ~= /_\CDE` ...`SSS` test of congruence
`=>/_BDE=/_CDE` .....c.p.c.t.
Also,
`/_BDE+/_CDE=180^@`
...angles in linear pair

`:./_BDE=/_CDE=90^@`
Therefore, `DE` is the right angled bisector of `BC`.

Hence, they intersect on the circumcircle of triangle `ABC`