The attached drawings are inconsistent with the original title, and the title is wrong. It seems that it should be △DEC∽△OBF, and point B should be BF⊥AB.

Prove:

1, link AD, OD,

∫AB is the diameter,

∴〈ADB=90，

CD = BD，

AD=AD，

∴RT△ADC≌RT△ADB，

∴〈CAD=〈DAB，

OA = OD = R，

∴〈OAD=〈ODA，

∴〈ODA=〈CAD，

∴OD//AC, (internal dislocation angles are equal)

∵DE⊥AC, (known),

∴OD⊥EF，

OD is the radius of circle o,

∴EF is the tangent of circle O.

2.∵FB⊥OB, (known)

∴FB is the tangent of circle O,

∴FD=FB, (a point outside the circle makes two tangents equal to the circle),

∴△FDB is an isosceles△,

∴〈FDB=〈FBD，

∫< FDB = < CDE, (equal to vertex angle),

∴〈CDE=〈FBD，

∵DE⊥CE, (known)

∴〈FBO=〈CED=90，

Let BD and OF meet m,

OD = OB，

ODF = OBF = 90 degrees,

∴△OBF≌△ODF，

∴〈OFD=〈OFB，

∴FM is the height and center line of the isosceles triangle △FDB,

∴〈MOB+〈MBO=90，

∵〈OBM+〉FBM=90，

∴〈BOM=〈FBM，

∴〈FOB=〈COE，

∴△DEC∽△OBF。