Question

求这道题第三问解法步骤

Answer 1

2) ∵DE//AF
∴OD/OF=DE/AF=1/2
∴OF=2OD=2R
∵OF^2=AF^2+OA^2
(2R)^2=2^2+R^2
∴R=2√3/3
3)作HN⊥AC , 由(2)得OA=R=2√3/3
∵tan∠AFO=OA/AF=(2√3/3)/2=√3/3
∴∠AFOF=30
∴∠CAB=∠AFO=30,
∴BC=1/2AB=1/2*4√3/3=2√3/3 ,OM=1/2OA=√3/3
∴OF=2OA=4√3/3
∵点D是弧AC中点
∴OD⊥AC
∵BC⊥AC
∴OD//BC
∴HC/HO=BC/OF
∴HC/HO=(2√3/3)/(4√3/3)=1/2
∴HC/OC=1/3
又HN/OM=HC/OC
∴HN/OM=1/3
∵OM=1/3*OM=1/3*√3/3=√3/9
∵OM=√3/9

Answer 2