已知:E为正方形ABCD的BC边延长线上一点.AE交CD于F,FN平行AD交DE于N,求证:CF=NF

Question

已知:E为正方形ABCD的BC边延长线上一点.AE交CD于F,FN平行AD交DE于N,求证:CF=
NF

Answer 1

∵ABCD是正方形,
∴DA＝AB、FC∥AB.
∵NF∥DA,
∴△ENF∽△EDA,
∴NF/DA＝EF/EA.
∵FC∥AB,
∴△EFC∽△EAB,
∴CF/AB＝EF/EA,
∴CF/AB＝NF/DA,又AB＝DA,
∴CF＝NF.