已知在三角形ABC中,点D、E、F分别在边BC、AC、AB上,AD、BE、CF交于G,FH平行于AD交BC于H,FH延长后与ED的延长线交于K,求证：FH=HK.

Question

已知在三角形ABC中,点D、E、F分别在边BC、AC、AB上,AD、BE、CF交于G,FH平行于AD交BC于H,FH延长后与ED的延长线交于K,求证：FH=HK.

Answer 1

证明:
过E作EN‖AD交BC于N
FH/AD=BF/AB,EN/AD=CE/AC
==>FH/EN=(BF/CE)*(AC/AB)
DH/BD=AF/AB,DN/CD=AE/AC
==>DH/DN=(AF*BD)/(AE*CD)*(AC/AB)
==>FH/EN*(AF*BD)/(AE*CD)=DH/DN*(BF/CE)
AD,BE,CF交于G==>(AF/BF)*(BD/CD)*(CE/AE)=1
==>(AF*BD)/(AE*CD)=BF/CE
==>FH/EN=DH/DN=HK/EN
==>FH=HK