I obviously have to modify my assertion. That the angles of a polygon have rational tangents is necessary for it to be similar to a polygon with vertices at lattice points, because any angle of such a polygon is the difference of two angles of rational tangent and the formula tan(x - y) = (tan x - tan y)/(1 + tan x tan y) preserves rationality. I think the converse is true for completely triangulated polygons. Thus if you put in the diagonals of the 1 by e square you get an angle with an irrational tangent, and the regular octagon is shown to be non-lattice by the same device. Anything that can be built from the (0,0)-(1,0) line segment by gluing triangles with rational tangents onto edges will have rational vertices, again because of the addition formula for the tangent. Any finite figure with rational vertices can be expanded to a figure with integer vertices. The remaining gap is on whether every completely triangulated polygon can be built up in this way.