∴△ADF？ △CDF

∴AF=CF

∠DAF=∠DCF

∴∠BAF=∠BCF (complementary angles of equal angles are equal)

Because ∠ABG =∠AFG = RT∞.

∴∠ABG+∠AFG= 180

∴∠FAB+∠FGB= 180

∴∠ FGC =∠ Fab (two complementary angles of ∞∞∞∠ FGB)

∴∠FGC=∠FCG

∴AF=FG

Note: It is very convenient to use a four-point certificate.

2. Even AG, △AFG are isosceles right triangles,

∴∠FAG=45

∴∠DAE+∠BAG=45

Turn the position of △ADE to △ABH clockwise around point A,

Then AH=AE AG=AG DE=BH.

∠HAG=∠DAE

∴∠HAB+∠BAG=∠DAE+∠BAG=45 =∠EAG

△ bag? △EAG

∴EG=HG=HB+BG=DE+BG

So EG=3+2=5.