A and B are two points lying on the same side of a straight line ℓ. A point X is to be chosen on ℓ so that the total length AX + XB is as small as possible. Which construction locates that point X?

Reflection in ℓ keeps distances to points of ℓ unchanged, so XB = XB′ for every X on ℓ. The quantity to minimise therefore becomes AX + XB′, the length of a path from A to B′ through a point of ℓ, and the shortest such path is the straight segment AB′. So X is where AB′ meets ℓ. The perpendicular-bisector point makes AX = XB, which balances the two pieces instead of shortening their sum; the nearest point to A ignores B completely; and the foot of the perpendicular from the midpoint minimises nothing relevant.