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Investigation and Test on Basic Subjects of Junior Two in Suzhou (Mathematics 20 14. 1)
1、AD=AB= 1

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.