Question

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

Answer 1

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)/CF

AE/CE+1=BF/CF+1

所以：BF/CF=AE/EC