问题描述: 如图,在三角形abc中,D是AB的中点,过点D的直线交AC于E,交BC的延长线于F.(1)求证BF/CF=AE/EC 1个回答 分类:数学 2014-10-24 问题解答: 我来补答 证明:作DG平行AC交BC于G,则DG=1/2 AC,GC=1/2 BC.由△CEF相似于△GDF,则:CE/DG=CF/GF则:CE/(1/2AC)=CF/(GC+CF)CE/(1/2AE+1/2CE)=CF/(1/2BC+1/2CF+1/2CF)将1/2销去,得:CE/(AE+CE)=CF/(BC+CF+CF)即CE/(AE+CE)=CF/(BF+CF)即(AE+CE)/CE=(BF+CF)/CFAE/CE+1=BF/CF+1所以:BF/CF=AE/EC 展开全文阅读