Question

如图,点D在△ABC的边BC上,过点D作AC的平行线DE交AB于E,作AB的平行线AE交AC于F

Answer 1

（1）∵DE∥AC,DF∥AB,

∴△BDE∽△BCA∽△DCF,

记S△BDE=S1,S△DCF=S2,

∵SAEFD=25S,

∴S1+S2=S-25S=35S．①

S1S=BDBC,S2S=CDBC,

于是S1S+S2S=BD+CDBC=1,即S1+S2=S,

两边平方得S=S1+S2+2S1S2,

故2S1S2=SAEFD=25S,即S1S2=125S2．②

由①、②解得S1=3±

510S,即S1S=3±

510．

而S1S=(

BDBC)2,即3±

510=(

BD5)2,解得BD=30±10

52=(5±

5)22=5±

52．

（2）由G是△ABC的重心,DF过点G,且DF∥AB,可得CDCB=23,则DF=23AB．

由DE∥AC,CDCB=23,得DE=13AC,

∵AC=2AB,∴ACAB=2,DFED=2AB2AB=2,

得DFDE=ACAB,即DFAC=DEAB,

又∠EDF=∠A,故△DEF∽△ABC,

得EFBC=DEAB,所以EF=5

23．