∴∠boa=45∠coa = 90

∵bo=do,∴∠bdo=∠dbo= 1/2∠boa=22.5

∴∠OED=∠BEC=90 -22.5 =67.5

(2) It is easy to know △BCE∽△DOE, and it is known that OD=OE= radical number 2, BC= 1, ∴CE:EO= 1: radical number 2, and CO= 1.

∴ Found OE=2- radical number 2, ∴E(0,2- radical number 2)

(3) Let the intersection of CP and BD be Q,∵CQ⊥BE, ∴ CQE = ∠ CQB = 90, and Yi Zhi ∠CEQ=∠QCB.

∴△CEQ∽△CQB,CE:CO=CE:CB=QE:CQ

∵△CQE∽△COF,∴QE:CQ=FO:CO=FO:BC

∫△CPB∽△FPO

∴OP:BP=FO:BC=QE:CQ=CE:CO

∴OP:BP=CE:CO