Current location - Training Enrollment Network - Mathematics courses - Dfq mathematics
Dfq mathematics
Square ABCD and get: AC=BD=√2AB=3√2, OA=OC=√2/2AC=3√2/2.

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