(1) extends the intersection of BM and EF at point n. It is not difficult to prove that if △ABM is equal to △FNM, then AB=FN, and then EB=EN, △EBN is also an isosceles Rt△, so BM is parallel to CF (in fact, it can be thought that B and E are at both ends of C respectively, but the process should be similar. )

(2) Continuing the operation of (1), it has been proved that △EBN is isosceles Rt△, and the waist length is a ... At the same time, since △ABM is all equal to △FNM, it can be found that BM=MN and m is the midpoint of the hypotenuse, so BM=ME=a* radical 2/2. (same as above)

(3) By extending the intersection of BM and CF at point n, it is not difficult to prove that △ABM is equal to △FNM and BM is equal to Mn. At the same time, it can also be proved that △EBC is all equal to △ENF, so BE=EN. At the same time, we can find that ∠BEN is a right angle, so EM is the center line of Rt△BEN, so BM=ME.