C,∵DE⊥BC,∴OD⊥DE,∴DE为⊙O的切线;Р(3)作OH⊥BC于H,则四边形ODEH为矩形,∴OD=EH,∵CE=1,AC=4,∴OC=OD=2,Р∴CH=HE﹣CE=2﹣1=1,在Rt△OHC中,∠HOC=30°,∴∠COD=60°,∴阴影部分的面积=S扇形OCD﹣S△OCD==.Р Р(1)证明:∵EF∥AB,∴∠E=∠CAB,∠EFA=∠FAB,Р∵∠E=∠EFA,∴∠FAB=∠CAB,在△ABC和△ABF中,,∴△ABC≌△ABF;Р(2)当∠CAB=60°时,四边形ADFE为菱形.Р证明:∵∠CAB=60°,∴∠FAB=∠CAB=∠CAB=60°,∴EF=AD=AE,Р∴四边形ADFE是菱形.故答案为60.Р(3)解:∵四边形AEFD是菱形,设边长为a,∠AEF=∠CAB=60°,Р∴△AEF、△AFD都是等边三角形,由题意:2×a2=6,∴a2=12,Р∵a>0,∴a=2,∴AC=AE=2,Р在RT△ACB中,∠ACB=90°,AC=2,∠CAB=60°,∴∠ABC=30°,Р∴AB=2AC=4,BC==6.故答案为6.Р解:(1)r=2.5;(2)AB:AD=.Р解:(1)∵AC与⊙O相切,∴∠OAC=90°.Р∵∠OCA=60°,∴∠AOC=30°.∵OC⊥OB,∴∠AOB=∠AOC+∠BOC=120°.Р∵OA=OB,∴∠OAB=∠OBA=30°,∴OD=AD,∠DAC=60°∴AD=CD=AC.Р∵OA=1,∴OD=AC=OA•tan∠AOC=.Р(2)∵OC⊥OB,∴∠OBE=∠OEB=45°.∵BE∥OA,∴∠AOC=45°,∠ABE=∠OAB,Р∴OA=AC,∠OAB=∠OBA=22.5°,∴∠ADC=∠AOC+∠OAB=67.5°.Р∵∠DAC=90°﹣∠OAB=67.5°=∠ADC,∴AC=CD.∵OC==,∴OD=OC﹣CD=﹣1.
全文预览
