Answer: (1) Proof: ∫E is the midpoint of AC,

∴EC=

1 /2

Communication,

DB =

1 /2

Communication,

∴DB=EC，

∫DB∑AC，

∴ Quadrilateral DBCA is a parallelogram,

∴bc=de；

②△ABC plus BA=BC,

Prove that the quadrangle DBEA is a parallelogram,

ba = BCBC=DE，

∴AB=DE，

∴ Quadrilateral DBEA is a rectangle;

(3) Solution: ∵ Quadrilateral DBEA is a square,

∴BE=AE∠BEC=90，

∴△BEC is a right triangle,

And ∵E is the midpoint of AC,

∴AE=EC，

∴BE=EC，

∵△BEC is a right triangle,

∴△BEC is an isosceles right triangle,

∴∠C=45。