OB=OD=3√2/2
∴RT△AOF: Yes? =AF？ -OA？ =(√5)? -(3√2/2)? =2/4，OF=√2/2
Then: DF=OD+OF=3√2/2+√2/2=2√3.
BF=BD-DF=3√2-2√2=√2
Do PM⊥AB and hand in BD extension cable to M.
Then it is easy to get that △BPM is an isosceles right triangle, then PB=PM=DQ.
Easy to obtain: ∠ MPP = ∠ DAP = 90
∴AD∥PM
Then ∠QDF=∠M, ∠DGF=∠MPF.
∴△PFM≌△QFD(ASA)
∴DF=MF=3√2, then BM = MF-BF = 3 √ 2-2 = 2 √ 2.
∴PM=PB=2, then DQ=2.
∵ AD ∨ BC, and then △DFQ∽△BNF.
∴DQ/BN=DF/BF
That is 2/BN=3√2/√2.
BN=2/3
∴CN=BC-BN=3-2/3=7/3
AQ = Ade -DQ=3-2= 1
Then AQ/CN= 1/(7/3)=3/7.
∫ AD ∨ BC
∴△AEQ∽△CEN
∴AE/CE=AQ/CN=3/7
Then AE/AC=3/ 10.
AE = 3/ 10AC =(3/ 10)×3√2 = 9√2/ 10
∴oe=oa-ae=3√2/2-9√2/ 10=6√2/ 10=3√2/5
Then in RT△OEF:
EF? =OE？ +OF？ =(3√2/5)? +(√2/2)? = 122/ 100
EF= 1 1/ 10