∫BD = DC，∠FDC=∠BDG，DG=DF

∴△BDG≌Rt△CDF

∴BG=FC=AC-AF=2, easy to get BG//AC.

∴∠A+∠GBA= 180 degree

* ED vertical direction finding, DG = direction finding.

∴EF=EG

∫ ef 2 = af 2+AE 2-2ae * afcos ∠ a =10-6cos ∠ a comes from cosine theorem.

eg^2=be^2+bg^2-2be*bgcos∠gba=5+4cos∠a

∴ 10-6cos∠A=5+4cos∠A

∴cos∠A= 1/2, that is ∠A=60 degrees.