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。