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.