Proof: let the intersection of CF and BE be be o.

BF = BC，∠ 1=∠2

Divided vertically in half.

∴BF=BC,EF=EC

∫EF∨AD

∴∠EFC=∠BCF=∠BFC

△ BEF is an isosceles triangle.

∴BF=FE

∴BF=FE=CE=BC

∴ quadrilateral BCEF is a diamond.

(2)

The quadrilateral BCEF is a diamond.

∴BC=EF

BC = AB，EF∨AB

∴ quadrilateral ABEF is a parallelogram.

∴AF=BE

Similarly, CF=CE can be obtained.

AC = BD

∴△ACF≌△BDE