如图,△ABC中,AC=AB,∠BAC=90°,BE平分∠ABC交BC于E,CD⊥BE于D,DM⊥AB交BA的延长线于M.求BM／AB=BC的值

延长CD交BM的延长线于F.
∠FBD=∠CBD,BD=BD,∠BDF=∠BDC=90°,则⊿BDF≌⊿BDC,BF=BC;DF=DC.
DM与CA都垂直于BF,则:DM∥CA,FM/MA=DF/DC=1,FM=MA.
∴BM/(AB+BC)=BM/[(BM-AM)+(BM+FM)]=BM/(2BM)=1/2;
AM/(BC-AB)=AM/(BF-AB)=AM/(2AM)=1/2.