∫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.