Question

1.如图6,点E是正方形ABCD边上BC的一点（不与B、C重合）,点F在CD边的延长线上,且满足DF=BE.联结EF,点

Answer 1

正方形中∠B=∠ADF=90°,AB=AD

∵BE=DF∴⊿ABE≌⊿ADF∴AE=AF ∠BAE=∠DAF

∵∠BAE+∠EAD=90°∴∠EAF=∠EAD+∠DAF=90°

∴∠AFE=∠AEF=45°

(2)证明：在正方形中,∠ECA=∠ACF=45°∴∠ACF=∠AFE

∵∠CME=∠ACF+∠CFE,∠CFA=∠AFE+∠CFE

∴∠CME=∠CFA∴⊿CME∽⊿CFA

∴：.CE/CM=AC/FC