2. Since DF is parallel to BE and DE is parallel to BF, the quadrilateral DEBF is a parallelogram, so BE=DF, so AE=CF, so all triangles ADE are equal to triangle CBF(SAS), so CF=AE.

3. It is proved that triangle ABE is equal to triangle CDF(SAS), so angle AEB= angle CFD and angle AEF= angle CFE, so AE is parallel to CF.