Df⊥ag be⊥ag, that is ∠ AFD = ∠ BEA = 90.

∠ DAF+∠ BAE = 90, ∠ DAF+∠ ADF = 90, that is ∠ADF=∠BAE.

∴△ABE≌△DAF(AAS)

2、∴AF=BE

∵DF？ +AF？ =AD？

1/2DF×AF= 1/8, that is 2DF×AF= 1/2.

∴DF？ +AF？ = 1

DF？ +AF？ -2DF= 1- 1/2

(DF-AF)？ = 1/2

|DF-AF|=√2/2

AF = BE

∴|DF-BE|=√2/2

That is |BE-DF|=√2/2.