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

Let` AE` be the angle bisector of `∠A`.
We need to prove that `ED`is the perpendicular bisector of `BC`.
`∠BAE = ∠CAE `....... (1) [Since, `AE` is the angle bisector of `∠A`]
Now, `∠EBC = ∠CAE `....... (2) [Angles subtended by the same arc `EC`]
Also, `∠ECB = ∠BAE `......... (3) [Angles subtended by the same arc `BE`
But we know that, `∠BAE = ∠CAE` [From equation (1)]
Hence, `∠EBC = ∠ECB` [From equations (2) and (3)]
Therefore, `BE = EC` [Sides opposite to equal angles are equal]
Thus, point `E`is equidistant from the points `B` and `C`. This is only possible when `E` lies on the perpendicular bisector of `BC`.
Thus, `ED`is the perpendicular bisector of `BC`.
Therefore, the perpendicular bisector of side `BC` and the angle bisector of `∠A` meet on the circumcircle of triangle `ABC` at point `E`.