8/5 or 16/7 can be divided into two situations: point f is on the reverse extension line and extension line of BC; First, prove △ACF∽△BAF, then prove DM∨BA. From the proportion theorem of parallel lines, we can get the length of AM.

Connect DM, let AM=x,

∵EF bisects AD vertically, ∴AE=DE, DM=AM=x,

∴DF=AF=6,BD=2，

∴BF=6-2=4

∴∠DAF=∠ADF，

∫AD splitting∠ ∠BAC,

∴∠BAD=∠CAD，

∴∠BAF=∠C，

∴△ACF∽△BAF，

∴AF/BF=FC/AF

∴AF2=BF？ CF，

I.e. 36=4CF,

The solution is CF=9,

∴CD=3，

∠∠DAM =∠ADM，∠BAD=∠CAD，

∴∠BAD=∠ADM，

∴DM∥BA，

∴AM/AC=BD/BC

That is, x/4=2/5.

The solution is x=8/5.

∴AM=8/5

Connect DM, let AM=x,

∵EF bisects AD vertically, ∴AE=DE, DM=AM=x,

∵AF=6,∴DF=AF=6，

∴∠DAF=∠ADF，

∫AD splitting∠ ∠BAC,

∴∠BAD=∠CAD，

∴∠B=∠CAF，

∴△ACF∽△BAF，

∴AF/BF=FC/AF

∴AF2=BF？ CF，

That is 36=(2+6)? CF，

The solution is CF=9 /2.

∠∠DAM =∠ADM，∠BAD=∠CAD，

∴∠BAD=∠ADM，

∴DM∥BA，

∴am/ac=bd/ BC

The solution is x= 16/7.

∴AM= 16/7