Question

如图,已知点D在三角形ABC的边BC上,且与B、D不重合,AC平行DE,AB平行DF,BC=5,.

Answer 1

（1）∵DE∥AC,DF∥AB,

∴△BDE∽△BCA∽△DCF,

记S△BDE=S1,

S△DCF=S2,

∵SAEFD= 2／5S,

∴S1+S2=S- 2／5S= 3／5S．①

√S1／√S= BD／BC,√S2／√S= CD／BC,

于是 √S1／S+ √S2／S= ﹙BD+CD﹚／BC=1,即 √S1+√ S2= √S,

两边平方得S=S1+S2+2 √﹙S1S2﹚,

故2 √﹙S1S2﹚=SAEFD= 2／5S,即S1S2= 1／25S2．②

由①、②解得S1= ﹙3±√5﹚／10S,即 S1／S=﹙ 3±√5﹚／10．

而 S1／S= (BD／BC)²,即 ﹙3±√5﹚／10= (BD／5)2,解得BD= ﹙5±√5﹚／2．

（2）由G是△ABC的重心,DF过点G,且DF∥AB,可得 CD／CB= 2／3,则DF= 2／3AB．

由DE∥AC,CD／CB= 23,得DE= 1／3AC,

∵AC= ∨2AB,∴ AC／AB=√ 2,DF／ED= 2AB／√﹙2AB﹚=√ 2,

得 DF／DE= AC／AB,即 DF／AC= DE／AB,

又∠EDF=∠A,故△DEF∽△ABC,

得 EF／BC= DE／AB,所以EF= 5√2／3．