（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