(2010? Jiading District three models) In the figure, in the square ABCD, point E is a point on the side BC (and point B, C does not overlap), connecting AE intersects the diagonal BD at point F.

Solution: (1) ∵ in the square ABCD, diagonal BD,

∴∠BDA=∠BDC,

in △ADF and △CDF,

AD = CD
∠BDA = ∠BDC
DF = DF
,

∴△ADF??CDF,

∴∠DAF = ∠DCF,

∵∠DAF = ∠BEF,

∴∠BEF = ∠ DCF;

(2) ∵AB∥CD,

∴∠BAE=∠CGE,

∵∠BEF=∠DCF;

∵∠BAE=90°-∠DAF,∠FCE=90°-∠DCF,

∴∠BAE=∠FCE,

∴∠CGE=∠FCE,< /p>

∴△FCE∽△FGC,

FC
EF
=
FG
FC
,

∵ FC=AF,

∴AF2=FE?FG.