Solution: Let the bisector OC of ∠MON connect AB, and the perpendicular bisector and OC of the line segment intersect at point P, then point P is the capture point.

As shown in the figure,

Reason: the point on the bisector of the angle is equal to the distance between the two sides of the angle (that is, the criminal is on the bisector of ∠MON and the point P is also on it).

The point on the vertical line of the line segment is equal to the distance between the two ends of the line segment (so the point P is on the vertical line of the line segment AB).

The intersection of two straight lines, that is, point P, meets the requirements.