Points D and E are symmetrical about AB, then BE = BD = BC/2 = AC/2; ∠EBF=∠DBF=45？ .

∴∠EBD+∠ACB= 180？ ，BE∑AC，BF/AF=BE/AC= 1/2，BF =( 1/2)AF =( 1/3)AB =√2。

Let FH be vertical AB and intersect BC at H, then ∠FHB=∠FBH=45? .

∴bh=√2bf=2； FH = FB∠FHM=∠FBN= 135？ ;

∠ MFN treatment =∠HFB=90? , then ∠MFH=∠NFB.

∴⊿MFH≌⊿NFB(ASA),HM=BN.

So BM-BN=BM-HM=BH=2.