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.