解:(1)∵AD⊥BC, ∴∠DAC+∠C =90° ∵∠BAC =90° ∴ ∠BAF=∠C. ∵OE⊥OB,∠BOA+∠COE =90°, ∴∠BOA+ ∠ABF= 90° ∴∠ABF= ∠COE ∴△ABF∽△COE. (2)作OG⊥AC,交AD的延长线于G, ∵AC= 2AB,O是AC边的中点, ∴AB= OC= OA. 由(1)有△ABF∽△COE, ∴△ABF≌△COE. ∴BF= OE, ∠BAD+∠DAC =90°, ∠DAB+ ∠ABD =90°, ∴∠DAC =∠ABD. 又∠BAC= ∠AOG= 90°,AB= OA,△ABC≌△OAG. ∴OG =AC= 2AB, ∵OG⊥OA, ∴△ABC≌△OAG. ∴OC =AC= 2AB, ∵OG⊥OA ∴AB∥OG ∴△ABF∽△GOF, ∴ (3). |