Solution: According to the meaning of the question, BC=AD=AF= 10, DE = EF.

At △ABF, ∠ ABF = 90.

∴BF=6，

∴FC= 10-6=4，

Let EC=x, then ef = de = 8-X.

∫∠C = 90，

∴EC2+FC2=EF2，

∴x2+42=(8-x)2，

Solution: x=3,

∴EC=3(cm).