Question

如图,在ABC中,S三角形COE=S三角形DOF,且AF/AD=1/2(根号5+1).

Answer 1

⑴连结CD

由S△COE＝S△DOF可得：S△CDE＝S△CDF

∴EF∥CD

∴AF/AD＝EF/CD＝OF/CO

这个结论楼主写错了

⑵设AF/AD＝OF/CO＝a

则AE/AC=AF/AD＝OF/CO＝a

∴S△AEF/S△ACF＝AE/AC＝a

∴S△AEF/(S△ACF－S△AEF)＝a/(1－a) (比例的性质)

∴S△AEF/S△CEF＝a/(1－a)

∴S△AEF＝a/(1－a)S△CEF

∵S△EOF/S△COE＝OF/CO＝a

∴S△EOF/(S△EOF＋S△COE)＝a/(a＋1)

∴S△EOF/S△CEF＝a/(a＋1)

∴S△EOF＝a/(a＋1)·S△CEF

∴S△AEF∶S△EOF＝a/(1－a)∶a/(a＋1)＝(a＋1)∶(1－a)＝√5＋2