112.

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

Given: Bisector AP of angle A of ∆ABC and the perpendicular bisector PQ of its opposite side BC intersect at P.

In ∆BMP and ∆CMP.
BM = CM    | Given
BP = CP    | By (1)
MP = MP    | Common
∴ ∆BMP ≅ ∆CMP    | SSS
∴ ∠BMP = ∠CMP    | C.P.C.T.
But ∠BMP + ∠CMP = 180°
| Linear Pair Axiom
∴ ∠BMP = ∠CMP = 90°
⇒ PM is the right bisector of BC.
Aliter:
Assume that C does not lie on the circle through A, B and P. Let this circle intersect the side AC at C'. (Say)
∠APB = ∠ACB    | Given
∠APB = ∠AC'B
| Angles in the same segment
∴ ∠ACB = ∠AC'B
⇒ C and C' coincide
⇒ The assumption that the point C does not lie on the circle is false.
∴ A, B, P and C are concyclic.