Solution: these two paths are equal in length, that is, AF=BE, and AF is perpendicular to BE. The reasons are as follows: Let AF and BE intersect at point P, because the quadrilateral ABCD is a square, so AB=AD=DC, angle BAD= angle ADF=90 degrees. And because DE=CF and AE=DF. So the triangle ABE is equal to the triangle DAF(SAS). So BE=AF, triangle DAF= triangle ABE, triangle DFA= triangle AEB. Triangle DAF+ triangle AEB= triangle DAF+ triangle DFA=90 degrees. So the triangle APE=90 degrees, that is, AF is perpendicular to BE.