如图,点A和点B分别在x轴y轴的正半轴上,AD平分∠BAO交y轴于D,OC⊥AB于C,交AD于E,过E作EF平行AB,交

解:(1)∵∠AOE+∠DOE=90°; ∠DBA+∠DOE=90°.∴∠AOE=∠DBA.(同角的余角相等)又∠OAE=∠BAD.(已知)∴∠AOE+∠OAE=∠DBA+∠BAD.即∠OED=∠ODE,得OD=OE=3;作EH垂直OA于H,HE的延长线交AB于P.∵∠APE=∠AOE(均为∠CAH的余角);AE=AE;∠PAE=∠OAE.∴⊿PAE≌⊿OAE(AAS),PE=OE=3.∵EF∥PB;PE∥BF.∴四边形BFEP为平行四边形,BF=PE=3.故OB=OD+DF+BF=3+2+3=8,即点B为(0,8);(2)∵EF∥AB;OC⊥AB.∴OE⊥EF,则EF=√(OF²-OE²)=4.作EM垂直OF于M,由面积关系可知:EF*OE=OF*EM,4*3=5*EM, EM=12/5.则:CE=HE=OM=√(OE²-EM²)=9/5.所以,S⊿ACE/S⊿OAE=CE/EO=(9/5)/3=3/5.(同高三角形的面积比等底边之比)