>> ISTM that you miss the projection on the segment if dx=0 or dy=0.
>
> I don't need to find projection itself, I need only distance. When dx = 0
> then nearest point is on horizontal line of box, so distance to it is dy.
> Same when dy = 0. When both of them are 0 then point is in the box.
Indeed. I thought that the box sides where not parallel to the axis, but
they are. So I do not see why it should be more complex. Maybe they is a
general algorithm which works for polygons, and they use it for boxes?
--
Fabien.
